text:
\section{Introduction}\label{sec:intro}
Twin smooth integers are pairs of consecutive smooth integers and have been studied for various reasons at least since the 18th century. They have been mentioned by Gauss in the context of computing logarithms of integers, and they can be used to find particular solutions to certain diophantine equations~\cite{lehmer}.

\begin{definition}[Smooth and twin smooth integers]
Let $B>1$ and $m\in \ensuremath{\mathbb{Z}}$. The integer $m$ is called \emph{$B$-smooth} if no prime factor of $m$  is larger than $B$. The pair $(m, m+1)$ is called a pair of \emph{twin $B$-smooth integers} if $m\cdot(m+1)$ is $B$-smooth.
\end{definition}
St{\o}rmer~\cite{stormer} proved that for a fixed smoothness bound $B>1$, there is a finite number of twin $B$-smooth integers and that all such pairs can be described via solutions of Pell equations of the form $x^2 - D y^2 = 1$.
Lehmer~\cite{lehmer} improved upon St{\o}rmer's work in utilizing the structure of the set of solutions to the Pell equation and strongly reduced the number of Pell equations that must be solved in order to find all twin $B$-smooth pairs. Lehmer computed the full set for $B=41$. Unfortunately, the number of Pell equations that must be solved is still exponential in $B$, and therefore finding all $B$-smooth pairs is computationally prohibitive for $B$ larger than 200.

In this paper, we focus on the problem of finding some B-smooth pairs that lie in a certain interval of possibly very large integers, rather than finding all pairs; for example, given a smoothness bound $B$ as well as a bitlength $b$, finding twin $B$-smooth integers that lie in the interval $[2^{b-1}, 2^b)$, i.e. finding such pairs $(m, m+1)$, for which $m$ can be written with $b$ bits. A slight variant of the problem only assumes the bitlength $b$ is fixed and then asks to find twin $B$-smooth integers with $b$ bits and the lowest possible smoothness bound $B$.

\subsubsection{Cryptographic applications.}
The latter variant of the problem is motivated by applications in cryptography. It has occurred recently in the context of selecting parameters for instantiating public key cryptography that uses isogenies between elliptic curves. For example, the digital signature scheme SQISign~\cite{sqisign} requires a large prime number $p$ such that $p-1$ and $p+1$ have large smooth factors. For security reasons, the prime $p$ (and therefore the integer $m$) needs to have a specific minimal size, usually given as a minimal number of bits, to ensure that the schemes cannot be easily broken.

If one aims to find $p$ such that both $p-1$ and $p+1$ are fully smooth, the problem is to find twin smooth integers $m = (p-1)/2$ and $m+1 = (p+1)/2$ with the additional condition that they sum to a prime number $p = m + (m+1)$. While this was an interesting scenario for the (now broken) key exchange scheme B-SIDH~\cite{b-sidh}, it is not strictly necessary. Allowing a small non-smooth cofactor in $p-1$ and $p+1$ often leads to a lower smoothness bound and a more efficient implementation of the cryptosystem. Still, algorithms for finding (fully) $B$-smooth twins, typically with much shorter bitsize, can be used to construct cryptographically sized primes.

This motivates to look for twin smooths $(m, m+1)$ in a prescribed interval, or more specifically with a given bitlength $b$, i.e. $\lfloor\log_2(m)\rfloor + 1 =b$ Furthermore, the smoothness bound $B$ for which a suitable twin $B$-smooth pair $(m, m+1)$ of the given size needs to be found, has a direct impact on the efficiency of cryptographic primitives. The smaller $B$ is, the more efficient the protocol becomes.

\subsubsection{Algorithms for finding twin smooth integers.}
As mentioned above, Lehmer's paper~\cite{lehmer} suggests to enumerate all twin $B$-smooth pairs for fixed $B$ by solving the Pell equation $x^2 - 2\Delta y^2 = 1$ for an exponential number of values for $\Delta$. This becomes infeasible for even moderately large smoothness bound $B$. Luca and Najman~\cite{luca-najman,luca-najman-err} succeeded in computing all twin 100-smooth pairs with a modified version of Lehmer's approach. This computation was very costly because of the large number of Pell equations to be solved and the solutions' potentially huge sizes. Costello \cite[\S 5.3]{b-sidh} used this approach to search for B-SIDH parameters and reports to have solved all $2^{30}$ Pell equations for $B=113$ (the 30-th prime) in a large computational effort, discovering the largest one to be a 76-bit $113$-smooth pair.
An interesting algorithm that approximates this problem and appears to be much more efficient is presented in~\cite{Conrey}.

\subsubsection{The Conrey-Holmstrom-McLaughlin approach.}
Conrey, Holmstrom, and McLaughlin propose an algorithm in \cite{Conrey} that quickly generates a large set of moderately-sized twin smooth integers for a given smoothness bound $B$. Experiments for relatively small $B$ demonstrate that their method succeeds in generating almost all $B$-smooth pairs, only missing a few that are often picked up when running the algorithm for a slightly larger $B$. The above example of finding 100-smooth twins was easily run, but missed 37 pairs of the Luca-Najman results. A significantly more costly search for 200-smooth twins recovered all but one of those.

The algorithm works by observing that two twin smooth integer pairs, $(m, m+1)$ and $(M, M+1)$ with $m < M$, generate a new pair of twin smooths $(\mu, \mu+1)$ if $\mu$ is an integer and $\frac{\mu}{\mu+1} = \frac{m}{m+1}\frac{M+1}{M}$. This can be used to iteratively expand any set of twin smooths. Begin with the set $S$ of integers $m < B$, though $S$ can be any set of integers $m$ such that $(m, m+1)$ forms a smooth pair. At each iteration, take all pairs $m < M$ in $S$ and collect all new $\mu$ of the above form in a set $S'$; set $S = S \cup S'$; if $S' = \phi$, the algorithm stops. The largest member of $S'$ is bounded by the square of the largest member of $S$, so the best case scenario is doubling the number of bits at each iteration.

While much more efficient in constructing most twin smooth pairs, this algorithm seems to quickly become infeasible when aiming for large integers that require a larger smoothness bound. The main reason is that the cardinality of $S$ balloons rapidly, which makes successive iterations infeasible before reaching a large enough bitlength. Nonetheless, this algorithm is a powerful tool for quickly generating large sets for experimentation, and an interesting avenue for future research, as demonstrated recently in \cite{CHM4SQISign}, where it was used as a subroutine for successfully constructing parameters for SQISign. The paper also reports that the authors ran the full algorithm for $B=547$. The largest $547$-smooth twin smooth pair found in these experiments has 122 bits.

Moreover, the following algorithms have been used in the context of finding cryptographic parameters. They do not attempt to compute the complete finite set of twin $B$-smooth integers, but instead have the goal of generating just enough ``good'' examples of a specific bit size. Ultimately the goal is to find a pair with prime sum, but the probability of this happening is often large enough within a large enough set of pairs. Specifically, the Prime Number Theorem gives an estimate that random numbers in the cryptographic range (around 256 bits) that we are looking for have a chance on the order of one in a hundred of being prime. In comparison, the chances of random consecutive integers in that range being B-smooth for B at most $2^{19}$ are tiny. Thus assuming there is no interaction between smoothness of $(m, m+1)$ and primality of $2m+1$, an algorithm that efficiently finds smooth pairs and then checks if their sum is prime is an efficient way to find cryptographic parameters, whereas an algorithm that checks whether $(\frac{p-1}{2}, \frac{p+1}{2})$ is a smooth pair for large primes $p$ is not. Therefore, the main task is to find the twin smooth pairs.

\subsubsection{Sieving.}
A simple algorithm to search a given interval is to use a variant of the sieve of Eratosthenes to list all $B$-smooth integers $m$ in the interval and check for each whether $m+1$ is also smooth. This is inefficient because for large $m$, the probability that $m+1$ is also smooth is very low and the chance to encounter such pairs is so low that searchable intervals are too small and likely do not even contain a solution or the search space is too large to be covered. Nevertheless, plain sieving has been successfully used as a subroutine in other methods. For example, some search algorithms increase the probability of hitting twin smooth integers by using polynomial parameterizations.

\subsubsection{Sieving with $x^n - 1$.}\label{sec:Poly-sieve}
There are various polynomial parameterizations leading to twin smooth integers that can be exploited by sieving for patterns of smaller smooth integers. A simple polynomial that was found to work well in~\cite{pte}, \cite{b-sidh}, and \cite{CHM4SQISign} is $f(x) = x^n - 1$ for small values of $n$. For example, if $n = 2$, then $(m, m+1) = (f(x), f(x) + 1)$ is a $B$-smooth pair when $(x-1)$, $x$, and $(x+1)$ are all smooth by the factorization $x^2 - 1 = (x - 1)(x + 1)$. One can then use the sieve to find three consecutive smooth integers $x-1, x, x+1$ and obtain the twin pair $(x^2-1, x^2)$ of twice as many bits. Because each of these factors are about the square root of the resulting smooth pair, it is easier to find the smooth triple of small integers. This phenomenon extends similarly to larger values of $n$ such as $4$ or $6$. We give further explanation of the reasons that certain polynomials work well in \S\ref{sec:pell} (see in particular the polynomials $p_n(x)$).

\subsubsection{Sieving with Solutions to the Prouhet-Tarry-Escott Problem.}
The Prouhet-Tarry-Escott (PTE) method is a generalization of the above idea that finding smooth pairs parameterized by polynomials can be easier because the polynomial factors of $(m, m+1)$ required to be smooth are much smaller than the $m$ and $m+1$ themselves. Solutions to the PTE problem correspond to polynomials $f(x)$ such that both $f(x)$ and $f(x) + 1$ factor into linear terms over $\mathbb{Z}$. Given such polynomials, we can search for pairs $(m = f(x), m + 1 = f(x) + 1)$ by checking for smoothness of all of the linear factors of $f$ and $f + 1$ for the same value of $x$. This is again done using a sieve. Because smaller integers are much more likely to be smooth, it is often more efficient to search for many small integers that are all smooth than for a single large pair directly. The PTE method has found many of the best-known pairs of cryptographic size; for details see \cite{pte}. However, this method is restricted in that only a finite number of solutions to the PTE problem are known, and all have degree at most $12$. Furthermore, it only finds pairs of this special form, and it is unclear what type of pairs might be missed.

\subsubsection{Extended Euclidean Algorithm.}
 Another improvement, discussed for example in ~\cite{b-sidh}, uses the extended Euclidean algorithm. It chooses two coprime, $B$-smooth integers $\alpha$, $\beta$ roughly the square root of the size of the desired pair and runs the Euclidean algorithm. This produces an equality of the form $\alpha  s + \beta t = 1$, and we take $| \alpha s |,  | \beta t |$ as our candidate smooth pair. The benefit is that the portions $\alpha$ and  $\beta$ of these numbers are already known to be smooth, increasing the probability that $|\alpha \beta s t|$ will be smooth overall. For more details and a detailed analysis of the probabilities for various sizes of $B$ and $m$, see~\cite{pte} and~\cite{b-sidh}.

\subsubsection{Contributions.}
In this paper, we describe how Pell equations can be used to search for large twin-smooth pairs. Pell equations have been believed to be mainly a theoretical tool useful for ensuring that all pairs have been found; we show that they can also provide an efficient way of searching. We give an algorithm generalizing and explaining the polynomial sieving described above and another based on heuristically choosing which Pell equations to solve. We give experimental evidence for our algorithms through the pairs that they found in our search of the interval $[2^{240}, 2^{256}]$, including pairs that are the largest known for their smoothness bound.

\subsubsection{Outline.}
In Section \ref{sec:pell}, we explore the structure of the solutions to the Pell equation, deducing some results on explicit polynomial parameterizations from the work of Lehmer. Then we specialize to the case of searching for pairs in a specified large interval and describe how solving Pell equations can be improved tremendously with this assumption. In Section \ref{sec:twins}, we apply this assumption and our results on the structure of solutions to choosing which Pell equations and which of their solutions to consider.

\section{Solutions of the Pell equation}\label{sec:pell}
For a fixed, non-square positive integer $D$, the Pell equation is a Diophantine equation of the form
\begin{equation}\label{eqn:pell}
    x^2 - Dy^2 = 1
\end{equation}
and its solutions are pairs of integers $(x, y)$ that satisfy~\eqref{eqn:pell}. The Pell equation was studied in the seventh century by Brahmagupta, and since at least the seventeenth century by mathematicians including Euler, Fermat, and Lagrange. We give a brief overview of the results that interest us, generally omitting proofs as these are readily available in resources such as \cite{lehmer}, \cite{jacobson-williams} and \cite{lenstra02}.

Equation~\eqref{eqn:pell} always has the trivial solutions $(\pm 1, 0)$, but our main interest lies in describing all non-trivial solutions $(x, y)$ and without loss of generality, we focus on those with $x, y > 0$. The following well-known result describes the structure of all such solutions (see, e.g.~\cite[Chapter 1]{jacobson-williams}, \cite[\S1]{lenstra02} and \cite[\S2]{lehmer}).

\begin{theorem}\label{thm:Pell_Solutions}
Let $D>0$ be an integer that is not a perfect square. Then, the Pell equation $x^2 - Dy^2 = 1$ has an infinite number of integer solutions $(x,y)$ with $x,y > 0$. The solution with the smallest value of $x+y\sqrt{D}$ among them is called the \emph{fundamental solution} and is denoted by $(x_1, y_1)$. All nontrivial solutions as above can be written as $(x_n, y_n)$ for some integer $n \geq 1$, where
\begin{equation}\label{eqn:solutions}
    x_n + y_n \sqrt{D} = (x_1 + y_1 \sqrt{D})^n.
\end{equation}
The solution $(x_n, y_n)$ is called the \emph{$n$-th solution}.
\end{theorem}

\subsection{Expressing solutions in terms of the fundamental solution}
Lehmer~\cite[\S2]{lehmer} describes all solutions of \eqref{eqn:pell} in terms of the Lucas function $U_n$. Let $\alpha = x_1 + y_1 \sqrt{D}$ and $\beta = x_1 - y_1\sqrt{D}$. Then, $\alpha\beta=1$ and for all $n \ge 1$, we get $\alpha^n+\beta^n = 2x_n$, $\alpha^n-\beta^n = 2y_n\sqrt{D}$. The Lucas function $U_n$ is defined as
\begin{equation}\label{eqn:Un}
    U_n = (\alpha^n-\beta^n)/(\alpha-\beta) = y_n/y_1.
\end{equation}
Hence $U_1 = 1$ and the above identities yield recursive formulas to compute $U_n$ for all $n$, since $x_{2n} = 2x_n^2 -1$, $U_{2n} = 2x_n U_n$, $x_{n+1} = x_n x_1 + Dy_n y_1 = x_n x_1 + U_n(x_1^2 - 1)$ and $U_{n+1} = x_n + x_1 U_n$. Therefore, $U_n$ assumes integer values and $y_1 \mid y_n$ for all $n \ge 1$. Also, writing out \eqref{eqn:solutions} gives the explicit formulas
\begin{equation}\label{eqn:xnUn}
    x_n = \sum_{i=0}^{\lfloor\frac{n}{2}\rfloor} \binom{n}{2i} (Dy_1^2)^ix_1^{n-2i},\quad
    U_n = \sum_{i=0}^{\lceil\frac{n}{2}\rceil-1} \binom{n}{2i+1} (Dy_1^2)^ix_1^{n-2i-1}.
\end{equation}
Note that we can replace $Dy_1^2 = x_1^2 - 1$ via the Pell equation to obtain univariate polynomials in $x_1$. We write these as polynomials $p_n(x)\in \ensuremath{\mathbb{Z}}[x]$ such that $x_n = p_n(x_1)$, i.e. we have $p_n= \sum_{i=0}^{\lfloor\frac{n}{2}\rfloor} \binom{n}{2i} (x^2-1)^ix^{n-2i}$ and $\deg(p_n) = n$. Expanding $(x^2-1)^i$ and reordering the terms gives
\begin{equation}\label{eqn:pn}
    p_n(x) = \sum_{i=0}^{\lfloor\frac{n}{2}\rfloor} \sum_{j=0}^i \binom{n}{2i} \binom{i}{j} (-1)^j x^{n-2j}
    = \sum_{j=0}^{\lfloor\frac{n}{2}\rfloor} \left((-1)^j \sum_{i=j}^{\lfloor\frac{n}{2}\rfloor} \binom{n}{2i} \binom{i}{j}\right) x^{n-2j}.
\end{equation}
The first few $p_n$, $n \in \{1,2,\dots,12\}$ are given as follows:
\begin{align*}
    p_1(x) & = x,\\
    p_2(x) & = 2x^2 - 1,\\
    p_3(x) & = x (4x^2 - 3),\\
    p_4(x) & = 8x^4 - 8x^2 + 1,\\
    p_5(x) & = x (16x^4 - 20x^2 + 5),\\
    p_6(x) & = (2x^2 - 1) (16x^4 - 16x^2 + 1),\\
    p_7(x) & = x (64x^6 - 112x^4 + 56x^2 - 7),\\
    p_8(x) & = 128x^8 - 256x^6 + 160x^4 - 32x^2 + 1,\\
    p_9(x) & = x(4x^2 - 3)(64x^6 - 96x^4 + 36x^2 - 3),\\
    p_{10}(x) & = (2x^2 - 1) (256x^8 - 512x^6 + 304x^4 - 48x^2 + 1),\\
    p_{11}(x) & = x (1024x^{10} - 2816x^8 + 2816x^6 - 1232x^4 + 220x^2 - 11),\\
    p_{12}(x) & = (8x^4 - 8x^2 + 1) (256x^8 - 512x^6 + 320x^4 - 64x^2 + 1).
\end{align*}
Similarly, Equation~\eqref{eqn:xnUn} gives an explicit formula for polynomials $u_n(x)\in \ensuremath{\mathbb{Z}}[x]$ of degree $n-1$ such that $U_n = u_n(x_1)$, namely $u_n(x) = \sum_{i=0}^{\lceil\frac{n}{2}\rceil-1} \binom{n}{2i+1} (x^2-1)^ix^{n-2i-1}$. As above, by expanding $(x^2-1)^i$ and reordering, $u_n(x)$ can be rewritten as
\begin{equation}\label{eqn:un}
    u_n(x) = \sum_{j=0}^{\lceil\frac{n}{2}\rceil - 1} \left((-1)^j \sum_{i=j}^{\lceil\frac{n}{2}\rceil - 1} \binom{n}{2i+1} \binom{i}{j}\right) x^{n-2j-1}.
\end{equation}
Here are the first $12$ polynomials $u_n(x)$:
\begin{align*}
    u_1(x) & = 1,\\
    u_2(x) & = 2x,\\
    u_3(x) & = (2x - 1)(2x + 1),\\
    u_4(x) & = 4x(2x^2 - 1),\\
    u_5(x) & = (4x^2 - 2x - 1)(4x^2 + 2x - 1),\\
    u_6(x) & = 2x(2x - 1)(2x + 1)(4x^2 - 3),\\
    u_7(x) & = (8x^3 - 4x^2 - 4x + 1)(8x^3 + 4x^2 - 4x - 1),\\
    u_8(x) & = 8x(2x^2 - 1)(8x^4 - 8x^2 + 1),\\
    u_9(x) & = (2x - 1)(2x + 1)(8x^3 - 6x - 1)(8x^3 - 6x + 1),\\
    u_{10}(x) & = 2x(4x^2 - 2x - 1)(4x^2 + 2x - 1)(16x^4 - 20x^2 + 5),\\
    u_{11}(x) & = (32x^5 - 16x^4 - 32x^3 + 12x^2 + 6x - 1)(32x^5 + 16x^4 - 32x^3 - 12x^2 + 6x + 1),\\
    u_{12}(x) & = 4x(2x - 1)(2x + 1)(2x^2 - 1)(4x^2 - 3)(16x^4 - 16x^2 + 1).
\end{align*}
Using the above polynomials, one can easily write down the $n$-th solution to the Pell equation~\eqref{eqn:pell} in terms of the fundamental solution $(x_1, y_1)$ as
\begin{equation}
    (x_n, y_n) = (p_n(x_1), y_1u_n(x_1)).
\end{equation}

Because the $p_n$ and $u_n$ are directly derived from \eqref{eqn:solutions} and \eqref{eqn:pell}, we obtain the following lemma stating some divisibility properties.

\begin{lemma}\label{lem:pnun}
Let $p_n(x), u_n(x)\in \ensuremath{\mathbb{Z}}[x]$ be the above polynomials such that $x_n = p_n(x_1)$, $U_n = u_n(x_1)$ for $0 < n \in \ensuremath{\mathbb{Z}}$, $\deg(p_n) = n$ and $\deg(u_n) = n-1$.
\begin{enumerate}
    \item[(a)] If $n$ is odd, then $x \mid p_n(x)$.
    \item[(b)] If $n = km$ for $0 < k, m \in \ensuremath{\mathbb{Z}}$, then $p_n(x) = p_m(p_k(x))$ and therefore, if $m$ is odd, $p_k(x) \mid p_n(x)$.
    \item[(c)] Furthermore, $u_n(x) = u_k(x) u_m(p_k(x))$. In particular, if $k\mid n$, then $u_k \mid u_n$.
    \item[(d)] The leading coefficients of $p_n(x)$ and $u_n(x)$ are both equal to $2^{n-1}$.
    \item[(e)] If $n=2^k$, $k\geq 1$, then $p_n(x)$ is irreducible in $\ensuremath{\mathbb{Z}}[x]$.
\end{enumerate}
\end{lemma}
\begin{proof}
Statement (a) follows from the explicit formula for $x_n$ in \eqref{eqn:xnUn}. Note that if $n$ is odd, the exponent $n-2i$ of $x_1$ is always at least $1$.

If $n$ factors as $n=km$, then the exponentiation in Equation~\eqref{eqn:solutions} can be split up into two parts, $x_n + y_n \sqrt{D} = (x_1 + y_1 \sqrt{D})^n = ((x_1 + y_1 \sqrt{D})^k)^m = (x_k + y_k \sqrt{D})^m$. Writing out the last exponentiation by $m$ in terms of $x_k$ and $y_k$ explicitly gives $x_n + y_n\sqrt{D} = p_m(x_k) + u_m(x_k) y_k \sqrt{D} = p_m(x_k) + u_m(x_k) y_1 u_k(x_1) \sqrt{D}$. Now writing $x_k$ in terms of $x_1$ gives $p_m(p_k(x_1)) + u_m(p_k(x_1)) y_1 u_k(x_1) \sqrt{D}$. The exact same calculations can be carried out as polynomials in $x$ instead of expressions in $x_1$, which gives $x_n(x) + y_n(x)\sqrt{D} = p_m(p_k(x)) + y_1 u_k(x) u_m(p_k(x))\sqrt{D}$, which proves (b) and (c).

Expression~\eqref{eqn:pn} shows that the leading coefficient of $p_n(x)$ is $\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{2i} = \binom{n}{0} + \binom{n}{2} + \dots + \binom{n}{2\lfloor\frac{n}{2}\rfloor} = 2^{n-1}$. From~\eqref{eqn:un}, one obtains the leading coefficient of $u_n(x)$ as $\sum_{i=0}^{\lceil\frac{n}{2}\rceil - 1}\binom{n}{2i+1} = \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{2\lceil\frac{n}{2}\rceil - 1} = 2^{n-1}$, and thus (d) holds.

To prove (e), we first take a look at the coefficients of $p_{2^k}(x)$ when we write it as $p_{2^k}(x) = \sum_{j=0}^{2^{k-1}} a_j x^{2^k - 2j} = a_0x^{2^k} + a_1x^{2^k-2} + \dots + a_{2^{k-1}-1}x^2 + a_{2^{k-1}}$ with
$$a_j = (-1)^j \sum_{i=j}^{2^{k-1}} \binom{2^k}{2i} \binom{i}{j}$$
as in \eqref{eqn:pn}. Note that $\lfloor n/2\rfloor = 2^{k-1}$. The constant coefficient is $a_{2^{k-1}} = 1$ for $k>1$ and $a_{2^{k-1}} = a_1 = -1$ for $k=1$. As we have seen in (d), the leading coefficient is $a_0 = \sum_{i=0}^{2^{k-1}}\binom{2^k}{2i} = 2^{2^k-1}$.

We now show by induction on $k$ that, for all coefficients $a_j$ with $1\leq j \leq 2^{k-1}$, $2^{2^k-2j}$ divides $a_j$. For the polynomial $p_2(x) = 2x^2 -1$ ($k=1$), this is true. For $k = 2$, we have $p_4(x) = 8x^4 - 8x^2 + 1$ and $a_1 = -8$ is divisible by $4 = 2^{2^2-2}$. For the induction step, note that $p_{2^{k+1}}(x) = 2(p_{2^k}(x))^2 - 1$ by the recursive formula for $x_{2n}$ discussed right after Equation~\eqref{eqn:Un}. If we write $(p_{2^{k}}(x))^2 = \sum_{j=0}^{2^{k}} b_j x^{2^{k+1} - 2j} = b_0x^{2^{k+1}} + b_1x^{2^{k+1}-2} + \dots + b_{2^{k}-1}x^2 + b_{2^{k}}$, then
$$
b_j = \sum_{\substack{0\leq l,m \leq 2^{k-1}\\ l+m = j}}a_la_m.
$$
If $lm \neq 0$, then $2^{2^k-2m} \mid a_m$ and $2^{2^k-2l}\mid a_l$, so $2^{2^{k+1}- 2(l+m)}\mid a_la_m$, i.e. if $l+m=j$, then $2^{2^{k+1}- 2j}\mid a_la_m$. If one of $l$ or $m$ is $0$, the term involves $a_0=2^{2^k-1}$. In all $b_j$ with $j>0$, $a_0$ only occurs in products with $a_j$, which appear twice. Therefore, in these sums, the two terms can be combined to $2a_0a_j$ and it follows again that $2^{2^{k+1}-2j} \mid 2a_0a_j$. Overall, this shows that $2^{2^{k+1}-2j} \mid b_j$. Since $p_{2^{k+1}} = 2b_0x^{2^{k+1}} + 2b_1x^{2^{k+1}-2} + \dots + 2b_{2^{k}-1}x^2 + (2b_{2^{k}}-1)$, the divisibility conditions still hold, which concludes our induction proof.

Now consider the polynomial $q(x) = 2p_{2^k}(x/2)$. The leading coefficient cancels exactly, making $q(x)$ monic. The constant term in $p_{2^k}(x)$ is equal to $1$, which means that the constant term in $q(x)$ is equal to $2$. The property we just proved ensures that all other coefficients of $p_{2^k}(x/2)$ are integers. Thus $q(x) = x^{2^k} + c_1x^{2^k-2} + \dots + c_{2^{k-1}-1}x^2 + 2$ for even integers $c_1, \dots, c_{2^{k-1}-1}$. The Eisenstein irreducibility criterion with the prime $p=2$ shows that $q(x)$ is irreducible in $\ensuremath{\mathbb{Z}}[x]$. Then, $p_{2^k}(x/2) = q(x)/2$ is irreducible in $\ensuremath{\mathbb{Q}}[x]$, and so is $p_{2^k}(x)$. Because $p_{2^k}(x)\in \ensuremath{\mathbb{Z}}[x]$ is primitive, it is also irreducible in $\ensuremath{\mathbb{Z}}[x]$.
\qed
\end{proof}
This lemma shows that $p_n(x)$ is irreducible if and only if $n = 2^k$ for some $k\geq 1$.

\subsection{Finding the fundamental solution via continued fractions}\label{sec:cf}
In light of the above, solving the Pell equation is reduced to finding its fundamental solution $(x_1, y_1)$. This can be done via continued fraction expansions as we explain next.

We focus on simple continued fractions (CF) for square roots, which are expressions of the form
$$\sqrt{D} = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{a_4 + \dots}}}}$$
for $D$ a squarefree integer and a sequence of integers $a_0, a_1, \dots$, called the quotients, where $a_i > 0$ for all $i$.

Every rational number can be expressed in two different ways via CF, and every irrational number has a unique CF representation that is an infinite sequence (\cite{unger}, 2.2); $\sqrt{D}$ falls in this second case.
We form the sequence of convergents $\frac{A_i}{B_i}$ to $\sqrt{D}$ by truncating the CF expansion after a finite number of terms. Specifically, the $i$th convergent is given by
$$a_0 + \cfrac{1}{a_1 + \cfrac{1}{\ddots + \cfrac{1}{a_i}}}.$$
These convergents can also be computed by the recurrence relations
\begin{eqnarray*}
A_i & = a_i A_{i-1} + A_{i-2},\\
B_i & = a_i B_{i-1} + B_{i-2},
\end{eqnarray*}
beginning with the values $A_0 = a_0$, $A_1 = a_1 a_0 + 1$, $B_0 = 1$, and $B_1 = a_1$ (\cite{unger}, 2.4). An important observation is that the numerators and denominators $A_i$ and $B_i$ of these convergents are monotonically increasing; further because the $a_i$ are positive integers, these sequences are bounded below by the Fibonacci sequence.

The quotients $a_i$ can be found almost directly from the definition by recursively reciprocating and rounding down, however, the following recurrence relations (\cite[Theorem 10.19]{Rosen}) give a faster implementation.
\begin{align*}
\alpha_k & = \left\lfloor\frac{P_k+\sqrt{D}}{Q_k}\right\rfloor,\\
P_{k+1} & = a_k Q_{k}-P_{k},\\
Q_{k+1} & = \frac{d - P_{k+1}^2}{Q_k},
\end{align*}
where $Q_0 = 1, P_0 = 0$.

The connection between continued fractions and the Pell equation is given by the following theorem (\cite[Theorem 3.3]{jacobson-williams}).

\begin{theorem}\label{thm:Continued_Fractions}
Let $D>0$ be an integer that is not a perfect square. Then if $(x, y)$ is a solution to the Pell equation $x^2 - Dy^2 = 1$ then $\frac{x}{y}$ is a convergent in the continued fraction expansion for $\sqrt{D}$.
\end{theorem}

This theorem suggests that to find the fundamental solution, we generate the quotients $a_i$ with our recurrence relations, using these to generate $A_i$ and $B_i$ for each $i$, and check whether $A_i$ and $B_i$ satisfy the Pell equation as $x$ and $y$. Because a fundamental solution exists, this algorithm always terminates. We will only be interested in solutions to the Pell equation that lie in a given range; this additional restriction and our bounds on the monotonically increasing sequence $A_i$ will ensure that this algorithm is very fast. Specifically, if we give an upper bound $\Tilde{M}$ on the size of $x$ in the solution to the Pell equation that we want to find, then we can cut off our algorithm after the first convergent $A_i$ is generated that is larger than $\Tilde{M}$. Because the $A_i$ are bounded below by the Fibonnaci sequence, this means that our algorithm will find any solution with $x \leq \Tilde{M}$ in time proportional to log($\Tilde{M}$). In our search for cryptographically large twin smooth pairs by solving Pell equations, $\Tilde{M}$ is $2^{258}$, and a single core running our C-code implementing this algorithm can find any solutions with $x \leq \Tilde{M}$ for tens of thousands of Pell equations a second.

\section{Twin smooth integers as solutions of Pell equations}\label{sec:twins}
Lehmer states the correspondence between $B$-smooth twins and certain solutions to Pell equations in Theorem 1 of \cite{lehmer}. The theorem is more general because it allows a specific finite set of primes as the prime factors of the twin smooth integers in question. But we are mainly interested in capturing $B$-smoothness, so we state it here specialized to consider all primes not exceeding $B$.
\begin{theorem}[Lehmer~\cite{lehmer}, Thm.~1]\label{thm:lehmer}
Let $B>2$ and $0< m\in \ensuremath{\mathbb{Z}}$. Then $(m, m+1)$ is a pair of twin $B$-smooth integers if and only if there exists a $B$-smooth integer $y$ and a square-free $B$-smooth integer $\Delta\neq 2$ such that $(x_n, y_n)$ with $x_n = 2m+1$, $y_n = y$ is the $n$-th solution to the Pell equation
\begin{equation}\label{eqn:twin_pell}
    x^2 - 2\Delta y^2 = 1
\end{equation}
with $1 \leq n \leq \max\{3, (q+1)/2\}$, where $q$ is the largest prime not exceeding $B$.
\end{theorem}

\begin{remark}\label{rem:xoddyeven}
If $\Delta$ in Equation~\eqref{eqn:twin_pell} is odd then the equation is a Pell equation in the sense of \S\ref{sec:pell} with a square-free, positive integer $D=2\Delta$. But Theorem~\ref{thm:lehmer} allows $\Delta$ to be even, i.e. in which case $\Delta = 2D$ for an odd, square-free, positive $D$. In this case, a solution $(x_n',y_n')$ to Equation~\eqref{eqn:twin_pell} corresponds to the solution $(x_n=x_n', y_n=2y_n')$ of the Pell equation $x^2 - Dy^2 = 1$, i.e. here $D = \Delta/2$.

Furthermore, Equation~\eqref{eqn:twin_pell} clearly implies that $x_n$ must be odd for any solution $(x_n,y_n)$. So, if $x_n = 2m+1$, then $\Delta y_n^2 = (x_n^2-1)/2 = 2m(m+1)$ is divisible by $4$. Since $\Delta$ is square-free, $y_n$ must be even.
\end{remark}

If $2 = q_1, q_2, \dots, q_t \in \ensuremath{\mathbb{Z}}$ are all the prime numbers that are not larger than $B$, listed consecutively in ascending order, then Lehmer's theorem shows that in order to find all twin $B$-smooth pairs, it suffices to solve $2^t-1$ Pell equations of the form~\eqref{eqn:twin_pell}, because that is the number of possible values $\Delta$ can take in Theorem~\ref{thm:lehmer}. Indeed, following Lehmer's notation, let $Q$ denote the set of positive $B$-smooth integers, i.e.
\[Q = \{q_1^{\alpha_1}\cdot q_2^{\alpha_2}\cdot \ldots \cdot q_t^{\alpha_t} \mid 0 \le \alpha_i \in \ensuremath{\mathbb{Z}},\ 1 \le i \le t\},\]
and let $Q'$ be the subset of square-free elements of $Q$ except $2$, i.e.
\[Q' = \{q_1^{\epsilon_1}\cdot q_2^{\epsilon_2}\cdot \ldots \cdot q_t^{\epsilon_t} \mid \epsilon_i \in \{0,1\},\ 1 \le i \le t\} \setminus \{2\}.\]
Then $Q'$ is the set of possible values of $\Delta$ that can occur in Theorem~\ref{thm:lehmer} and has $2^t-1$ elements.

Theorem~\ref{thm:lehmer} gives an explicit correspondence between twin smooth pairs $(m, m+1)$ and Pell equation coefficients $\Delta\in Q'$ together with solutions $(x,y) \in \ensuremath{\mathbb{Z}}_{>0}\times Q$.
Define the set
\[P = \{(\Delta, x, y) \in Q'\times \ensuremath{\mathbb{Z}}_{>0}\times Q \mid x^2 - 2\Delta y^2 = 1\}\]
 of triples $(\Delta, x, y)$ of such coefficient-solution triples and the set
\[T = \{m \in Q \mid m(m+1) \in Q\}\subset Q\]
of elements $m$ that describe all $B$-smooth twins.
The correspondence is given in the following lemma, which is simply a reformulation of the proof of Theorem~\ref{thm:lehmer} as given in \cite{lehmer}.
\begin{lemma}\label{lem:twin_pell_corr}
Let the sets $P$ and $T$ be defined as above. The map
$\delta: P \rightarrow T,\ (\Delta, x, y) \mapsto m = (x-1)/2$
is a bijection.
\end{lemma}
\begin{proof}
Let $(\Delta, x, y) \in P$. As noted in Remark~\ref{rem:xoddyeven}, $x$ must be odd and $y$ must be even. Hence both $m = (x-1)/2$ and $y/2$ are integers, as is $m+1 = (x+1)/2$. This means that $m\cdot(m+1) = (x^2-1)/4 = 2\Delta (y/2)^2 \in Q$ because $\Delta, y \in Q$, showing that $\delta(\Delta, x, y)$ lies in $T$.

Conversely, given $m\in T$ representing the twin $B$-smooth pair $(m, m+1)$, we have $m(m+1) \in Q$ and we can uniquely write
$m(m+1) = 2q_1^{\alpha_1}\cdot q_2^{\alpha_2}\cdot \ldots \cdot q_t^{\alpha_t}$,
where $\alpha_i = \epsilon_i + 2\beta_i$ for $\epsilon_i\in \{0,1\}$ and integers $\alpha_i,\beta_i \ge 0$. Setting $x=2m+1$, $y = 2q_1^{\beta_1}\cdot q_2^{\beta_2}\cdot \ldots \cdot q_t^{\beta_t}$ and $\Delta = q_1^{\epsilon_1}\cdot q_2^{\epsilon_2}\cdot \ldots \cdot q_t^{\epsilon_t}$, we see that $x^2-1 = 4m(m+1) = 2\Delta y^2$. Therefore, given $m$, we obtain $x=2m+1$ and unique $\Delta$ and $y$ by the square-free and square parts of $4m(m+1)$ via $x^2 - 1 = 2\Delta y^2$, so we can define $\delta^{-1}$ by $\delta^{-1}(m) = (\Delta, x, y)\in P$. We have $\delta\circ\delta^{-1}(m) = m$ for all $m\in T$ and $\delta^{-1}\circ\delta(\Delta, x, y) = (\Delta, x, y)$ for all $(\Delta, x, y) \in P$, which implies the lemma.\qed
\end{proof}
Any solution $(x, y)$ to Equation~\eqref{eqn:twin_pell} is the $n$-th solution for some positive integer $n$ as defined in Theorem~\ref{thm:Pell_Solutions}. Lehmer~\cite[Thm.~1]{lehmer} showed that $n \leq M = \max\{3, (q_t+1)/2\}$. Therefore, the set $P$ is finite.
An algorithm to compute all twin $B$-smooth pairs is thus to enumerate all coefficients $\Delta\in Q'$, compute the fundamental solution of the corresponding Pell equation $x^2 - 2\Delta y^2 = 1$ and then to compute all $n$-th solutions from that for $n\leq M$. To check whether any of those solutions yields a pair of twin $B$-smooth integers, it is sufficient to check whether the $y$-component of the solution is $B$-smooth. Clearly, this requires solving $2^t - 1$ Pell equations and quickly becomes infeasible for even moderate values of $B$.

\subsection{Twin smooth integers from $n$-th solutions for $n>1$}\label{sec:successive}
Next we consider for which $n$ the $n$-th solution to a Pell equation leads to a twin smooth pair via the above correspondence. We start by showing that if the $n$-th solution leads to a twin smooth pair then so does the fundamental solution. In other words, an $n$-th solution for $n>1$ can only correspond to a twin smooth pair if the corresponding fundamental solution corresponds to a twin smooth pair.
\begin{lemma}\label{lem:nth-to-1st-pair}
Let $\Delta\in Q'$, $(x_1, y_1)$ be the fundamental solution and $(x_n, y_n)$ the $n$-th solution to $x^2 - 2\Delta y^2 = 1$. Let $m_1 = (x_1-1)/2$ and $m_n = (x_n - 1)/2$. If $m_n \in T$, then so is $m_1$.
\end{lemma}
\begin{proof}
Since $m_n \in T$, we have $\delta^{-1}(m_n) = (\Delta, x_n, y_n) \in P$ and therefore, $y_n \in Q$. We have seen in \S\ref{sec:pell} that $y_1 \mid y_n$, so $y_1 \in Q$, which means that $(\Delta, x_1, y_1) \in P$. Then $ m_1 = \delta(\Delta, x_1, y_1) \in T$. \qed
\end{proof}
This lemma shows that when solving Pell equations to find twin smooth pairs, an equation can be discarded if the fundamental solution does not yield a twin smooth pair and there is no need to check any of the successive solutions $(x_n, y_n)$, $n>1$ for that particular value of $\Delta$. From now on, let $m_n = (x_n-1)/2$ for $n\geq 1$, where $(x_n, y_n)$ is the $n$-th solution to the Pell equation~\eqref{eqn:twin_pell}.

Whenever the fundamental solution $(x_1, y_1)$ yields a twin smooth pair $(m_1, m_1+1)$, successive solutions might yield one as well. To decide whether for a fixed $(\Delta, x_1, y_1) \in P$ we also have $(\Delta, x_n, y_n)\in P$, it is sufficient to check whether $U_n \in Q$ because  then $y_n = U_ny_1 \in Q$. For that purpose, we can simply write $U_n=v_n(m_1)$ as a polynomial in $m_1$, where $v_n(x) = u_n(2x+1)$. Then $U_n = u_n(x_1) = u_n(2m_1+1) = v_n(m_1)$. These expressions for $n\leq 12$ are shown here:
\begin{align*}
v_1(m) & = 1,\\
v_2(m) & = 2(2m + 1),\\
v_3(m) & = (4m + 1)(4m + 3),\\
v_4(m) & = 4(2m + 1)(8m^2 + 8m + 1),\\
v_5(m) & = (16m^2 + 12m + 1)(16m^2 + 20m + 5),\\
v_6(m) & = 2(2m + 1)(4m + 1)(4m + 3)(16m^2 + 16m + 1),\\
v_7(m) & = (64m^3 + 80m^2 + 24m + 1)(64m^3 + 112m^2 + 56m + 7),\\
v_8(m) & = 8(2m + 1)(8m^2 + 8m + 1)(128m^4 + 256m^3 + 160m^2 + 32m + 1),\\
v_9(m) & = (4m + 1)(4m + 3)(64m^3 + 96m^2 + 36m + 1)(64m^3 + 96m^2 + 36m + 3),\\
v_{10}(m) & = 2(2m + 1)(16m^2 + 12m + 1)(16m^2 + 20m + 5)(256m^4 + 512m^3 + 304m^2 + 48m + 1),\\
v_{11}(m) & = (1024m^5 + 2304m^4 + 1792m^3 + 560m^2 + 60m + 1)\\
& \quad\cdot (1024m^5 + 2816m^4 + 2816m^3 + 1232m^2 + 220m + 11),\\
v_{12}(m) & = 4(2m + 1)(4m + 1)(4m + 3)(8m^2 + 8m + 1)(16m^2 + 16m + 1)\\
& \quad\cdot (256m^4 + 512m^3 + 320m^2 + 64m + 1).
\end{align*}
For the $n$-th solution to give a twin $B$-smooth pair, $U_n = v_n(m_1)$ must be smooth, where $(m_1, m_1+1)$ is the twin smooth pair given by the fundamental solution. This means that each of the irreducible terms in the polynomial representation of $v_n$ needs to evaluate to a $B$-smooth value at $m_1$. In the same way as above, we may write polynomials for the potential twin-smooth pair $(m_n, m_n+1)$ corresponding to the $n$-th solution, namely, we have that
\begin{equation}\label{eqn:mn}
m_n = (p_n(2m_1+1)-1)/2.
\end{equation}
In the proof of Lemma~\ref{lem:twin_pell_corr} we have seen that $m_1(m_1+1) = (x_1^2-1)/4 = 2\Delta (y_1/2)^2$ and $m_n(m_n+1) = (x_n^2-1)/2 = 2\Delta (y_n/2)^2$. Since $y_n = y_1U_n$, this means that
\begin{equation}\label{eqn:mnmnp1}
    m_n (m_n+1) = m_1 (m_1+1) U_n^2 = m_1 (m_1+1) v_n(m_1)^2.
\end{equation}
The following lemma gives a more specific description about how $m_1$ and $m_1+1$ divide $m_n$ and $m_n+1$.
\begin{lemma}
Let $\Delta\in Q'$, $(x_1, y_1)$ be the fundamental solution and $(x_n, y_n)$ the $n$-th solution to $x^2 - 2\Delta y^2 = 1$. Let $m_1 = (x_1-1)/2$ and $m_n = (x_n - 1)/2$. Then $m_1 \mid m_n$.
If $n$ is even, then $(m_1+1) \mid m_n$, if it is odd, then $(m_1+1) \mid (m_n+1)$.
\end{lemma}
\begin{proof}
Note that $x_1^2-1 = 4m_1(m_1+1)$ and therefore, the explicit formula for $p_n$ in \S\ref{sec:pell} shows that $p_n(2m_1+1) \equiv 1 \mod (2m_1)$. It follows that $2m_1 \mid p_n(2m_1+1) - 1$ and thus $m_1 \mid (p_n(2m_1+1)-1)/2 = (x_n-1)/2 = m_n$.

We also see that $p_n(2m_1+1) \equiv (-1)^n \mod (2(m_1+1))$ such that $2(m_1+1) \mid p_n(2m_1+1)-1$ if $n$ is even and $2(m_1+1) \mid p_n(2m_1+1)+1$ if $n$ is odd.
\qed
\end{proof}

Below are the expressions for $(m_n, m_n+1)$ with $1\leq n \leq 12$.
\begin{align*}
(m_2, m_2+1) & = (4 m (m + 1),\ (2m + 1)^2),\\
(m_3, m_3+1) & = (m (4m + 3)^2,\ (m + 1) (4m + 1)^2),\\
(m_4, m_4+1) & = (16 m (m + 1) (2m + 1)^2,\ (8m^2 + 8m + 1)^2),\\
(m_5, m_5+1) & = (m (16m^2 + 20m + 5)^2,\ (m + 1) (16m^2 + 12m + 1)^2),\\
(m_6, m_6+1) & = (4 m (m + 1) (4m + 1)^2 (4m + 3)^2,\ (2m + 1)^2 (16m^2 + 16m + 1)^2),\\
(m_7, m_7+1) & = (m (64m^3 + 112m^2 + 56m + 7)^2,\ (m + 1) (64m^3 + 80m^2 + 24m + 1)^2),\\
(m_8, m_8+1) & = (64 m (m + 1) (2m + 1)^2 (8m^2 + 8m + 1)^2,\\
 & \qquad (128m^4 + 256m^3 + 160m^2 + 32m + 1)^2),\\
(m_9, m_9+1) & = (m (4m + 3)^2 (64m^3 + 96m^2 + 36m + 3)^2,\\
 & \qquad (m + 1) (4m + 1)^2 (64m^3 + 96m^2 + 36m + 1)^2),\\
(m_{10}, m_{10}+1) & = (4 m (m + 1) (16m^2 + 12m + 1)^2 (16m^2 + 20m + 5)^2,\\
 & \qquad (2m + 1)^2 (256m^4 + 512m^3 + 304m^2 + 48m + 1)^2),\\
(m_{11}, m_{11}+1) & = (m (1024m^5 + 2816m^4 + 2816m^3 + 1232m^2 + 220m + 11)^2,\\
 & \qquad (m + 1) (1024m^5 + 2304m^4 + 1792m^3 + 560m^2 + 60m + 1)^2),\\
(m_{12}, m_{12}+1) & = (16 m (m + 1) (2m + 1)^2 (4m + 1)^2 (4m + 3)^2 (16m^2 + 16m + 1)^2,\\
 & \qquad (8m^2 + 8m + 1)^2 (256m^4 + 512m^3 + 320m^2 + 64m + 1)^2).
\end{align*}

\begin{remark}
Lemma~\ref{lem:pnun}(b) shows that as soon as $n$ has an odd factor, the polynomial $p_n(x)$ used to represent $x_n = p_n(x_1)$ is reducible. This also means that $2m_n + 1 = p_n(2m_1+1)$ as a polynomial in $m_1$ is reducible and cannot assume a prime value for any $m_1$. Only $p_{2^k}(x)$ is irreducible for any $k\geq 1$ according to Lemma~\ref{lem:pnun}(e). Therefore, when we are interested in finding twin smooth pairs $(m_n, m_n+1)$, where $p = m_n + (m_n+1)$ is prime, only those solutions need to be checked where $n = 2^k$ is a power of $2$.
\end{remark}

Since pairs $(m_n, m_n+1)$ are polynomial expressions in the corresponding $m_1$, one can easily obtain bounds on the size of $m_n$ in terms of the size of $m_1$. For example, equations \eqref{eqn:mn} and \eqref{eqn:mnmnp1} allow to deduce a simple lower bound for $m_n$ in terms of the leading coefficient of $p_n$ or $v_n$. We obtain the following lemma.
\begin{lemma} \label{lemma:nth_solution_size}
Let $\Delta\in Q'$, $(x_1, y_1)$ be the fundamental solution and $(x_n, y_n)$ the $n$-th solution to $x^2 - 2\Delta y^2 = 1$, and let $m_1 = (x_1-1)/2$, $m_n = (x_n - 1)/2$, $U_n = y_n/y_1$ and $v_n(x) \in \ensuremath{\mathbb{Z}}[x]$ the above polynomial such that $U_n = v_n(m_1)$. Then
\[
m_1 v_n(m_1) \leq m_n < (m_1+1) v_n(m_1).
\]
It follows that $m_n > 4^{n-1}m_1^n$.
\end{lemma}
\begin{proof}
We use Equation~\eqref{eqn:mnmnp1} to see that $m_n^2 < m_n (m_n+1) = m_1 (m_1+1) v_n(m_1)^2 < (m_1+1)^2 v_n(m_1)^2$, which yields $m_n < (m_1+1) v_n(m_1)$. Similarly, $(m_n+1)^2 > m_n (m_n+1) = m_1 (m_1+1) v_n(m_1)^2 > m_1^2 v_n(m_1)^2$ and therefore $m_n \geq m_1 v_n(m_1)$. Because $v_n(x) = u_n(2x+1)$, $\deg(u_n) = n-1$ and the leading coefficient of $u_n$ is $2^{n-1}$ according to Lemma~\ref{lem:pnun}, the leading coefficient of $v_n$ is $(2^{n-1})^2 = 4^{n-1}$. The explicit formulas for $u_n$ in Section~\ref{sec:pell} show that all coefficients of $v_n(x)$ are positive and therefore, $v_n(m_1) > 4^{n-1}m_1^{n-1}$, which concludes the proof.\qed
\end{proof}

The last inequality in the lemma provides an upper bound on $m_1$ if we are given an upper bound on $m_n$. Assume, we are interested in twin smooth integers given by $m_n$ corresponding to an $n$-th solution to the Pell equation with $n>1$ with an upper bound $m_n < R$. Such pairs must then come from a fundamental solution pair given by $m_1$ such that $4^{n-1}m_1^n < R$, or, equivalently, $m_1 < \frac{1}{4}(4R)^{1/n}$.
For example, to find twin smooth integers where $m_n$ has at most $b$ bits, we have $R = 2^b$, i.e. $m_1 < R_b :=  2^{(b-2n+2)/n}$. Table~\ref{tbl:m1bounds} shows concrete values for the maximal possible bitsize of $m_1$ when $b=256$ and $n\in \{1,2,\dots,12\}$.

\begin{table}[]
    \centering
    \renewcommand{\tabcolsep}{0.2cm}
    \renewcommand{\arraystretch}{1.5}
    \begin{tabular}{c|cccccccccccc}
        \toprule
        $n$ &  $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ & $11$ & $12$\\
        $\lceil 258/n \rceil - 2$ & $256$ & $127$ & $84$ & $63$ & $50$ & $41$ & $35$ & $31$ & $27$ & $24$ & $22$ & $20$\\
        \bottomrule
    \end{tabular}
    \vspace{0.3cm}
    \caption{Maximal bitsizes $\lceil\log(R_{256})\rceil = \lceil 258/n \rceil - 2$ of $m_1$ such that the corresponding $m_n$ has at most $256$ bits for $n = 1, 2, \dots, 12$.}
    \label{tbl:m1bounds}
\end{table}

Further we have the following absolute upper bound on $n$ for which we can have twin smooth pairs corresponding to the $n$-th solution to a Pell equation (see \cite{lehmer}, Theorem 1).

\begin{theorem}
If $(m, m+1)$ is a pair of twin $B$-smooth integers corresponding to the $n$-th solution to a Pell equation, then $n \leq \frac{B+1}{2}$.
\end{theorem}

This upper bound is in many ways much weaker than the results of Lemma~\ref{lemma:nth_solution_size} because Lemma~\ref{lemma:nth_solution_size} shows that for large $n$, the $n$-th solutions to Pell equations under our bound all correspond to fundamental solutions that are exponentially smaller. Together, this suggests an algorithm for finding all smooth pairs in an interval corresponding to $n$-th solutions to Pell equations for large $n$.
The algorithm works as shown in Algorithm~\ref{algorithm}.

\begin{algorithm}
\begin{algorithmic}[1]
\State Fix a lower bound $s$ on $n$.
\For {$n \in \{s, s+1, \dots, \frac{B+1}{2}$\}}
    \State Compute $T = \lceil M/n \rceil - 2$ as in Table~\ref{tbl:m1bounds}.
    \For {$w \leq T$}
        \If {$(w, w+1)$ is a twin $B$-smooth pair}
            \If {$v_n(w)$ is $B$-smooth}
                \State Record $(m_n, m_n+1)$ as a twin $B$-smooth pair.
            \EndIf
        \EndIf
    \EndFor
\EndFor
\end{algorithmic}
\caption{Algorithm to search for twin $B$-smooth pairs of size at most $M$ bits.}\label{algorithm}
\end{algorithm}

The key benefit of this algorithm is that it finds smooth pairs of size roughly $2^M$ while only searching through integers up to $2^T$, an exponentially smaller range. In our running example of searching for pairs in the range $[2^{240}, 2^{256}]$, we ran the above algorithm with a lower bound of $s = 6$. Then the most costly iteration above with $n = 6$ only required checking approximately $2^{41}$ possibilities for $w$. Notice further that as $n$ gets larger, the possibilities for $w$ grow extremely restricted, and the algorithm terminates quickly. The overall runtime is dominated by the iteration with $n = s$.

For example, our run with $s = 6$ and a maximum of $M = 2^{275}$ found the following smooth pairs. A $16759$-smooth, $245$-bit pair $(m, m+1)$ with
\begin{align*}
m & = 44746808406030847930450201970587971020922341276429366152081686798603000000\\
  & = 2^6 \cdot 3^4 \cdot 5^6 \cdot 7^5 \cdot 11^2 \cdot 13^2 \cdot 41 \cdot 43^2 \cdot 53 \cdot 97 \cdot 241^2 \cdot 337 \cdot 509 \cdot 673 \cdot 4703^2\\
  & \quad \cdot 5981 \cdot 9413^2 \cdot 13669^2 \cdot 16759^2,\\
m+1 & = 31^2 \cdot 157^4 \cdot 181^2 \cdot 251^2 \cdot 349^2 \cdot 359^2 \cdot 457^2 \cdot 1427^2 \cdot 2617^2 \cdot 9109^2 \cdot 9649^2 \cdot 10253^2\\
\end{align*}
and a $24551$-smooth, $260$-bit pair with
{\small
\begin{align*}
m & = 1248045507865502270977250845951694434798578493856490782548653674169732908101560\\
 & = 2^3 \cdot 3^5 \cdot 5 \cdot 7^2 \cdot 17 \cdot 19 \cdot 29 \cdot 31 \cdot 43^2 \cdot 53^2 \cdot 149^2 \cdot 211^2 \cdot 227^2 \cdot 233 \cdot 827 \cdot 919^2 \cdot 2659 \cdot 4723^2\\
 & \quad \cdot 6907^2 \cdot 10831 \cdot 16691^2 \cdot 24551,\\
m+1 & = 11^2 \cdot 13^2 \cdot 23^2 \cdot 71^2 \cdot 107^2 \cdot 263^2 \cdot 587^2 \cdot 1021^2 \cdot 4297^2 \cdot 6491^2 \cdot 7309^2 \cdot 8089^2 \cdot 9049^2 \cdot 19009^2.
\end{align*}}
This second pair is larger than any previously known for its smoothness bound.

\subsection{Lower order solutions}

The above algorithm finds all twin-smooth pairs corresponding to $n$-th solutions for $n$ large. In order to complete our approach, we now consider algorithms for finding pairs corresponding to Pell equation solutions with $n$ small. In this case, the bounds from Lemma~\ref{lemma:nth_solution_size} are too weak, and the algorithm from the previous section is intractable. Therefore, we search for such smooth pairs directly by solving Pell equations via continued fractions to obtain the fundamental solution, and then checking the successive solutions for small $n$ with our polynomials $v_n$.

The obvious difficulty with the approach is that there are exponentially many Pell equations which could have corresponding smooth pairs, and we cannot solve all of them. While we believe (see discussion) that this difficulty may be intrinsic, and that finding all pairs may not be tractable, we can still cleverly choose which Pell equations to solve in order to find some large twin-smooth pairs.

Recall that each of the $2^t - 1$ Pell equations $x^2 - Dy^2 = 1$ with potential corresponding smooth pairs has a unique coefficient $D = 2 \Delta$ where $\Delta$ is a smooth, square-free integer and $\Delta \neq 2$. Given any particular $\Delta$, solving the corresponding Pell equation can be done extremely quickly (see \ref{sec:cf}). Choosing which Pell equations to solve means choosing the values of $\Delta$. Generally, the difficulty is in finding $\Delta$s for which the size of the solution to the Pell equation is in the desired range. Fundamental solutions to Pell equations can be extremely large (exponential in $\Delta$), and such large solutions $(x,y)$ are very unlikely to have smooth $y$, and are thus not helpful to our search. It is therefore beneficial to choose coefficients $\Delta$ for which we expect the resulting pairs to generally be small.

\subsection{The smallest coefficient heuristic}\label{sec:smallCoeff}
A very simple heuristic for choosing $\Delta$ is to choose the smallest values possible. This is beneficial for a simple combinatorial reason. Note that if $(x,y)$ is a solution to the Pell equation
$$x^2 - 1= 2 \Delta y^2$$
then $y = \sqrt{\frac{x^2 - 1}{2 \Delta}}$ is an integer, and as we have seen in Section~\ref{sec:twins}, it must be smooth if $(x,y)$ corresponds to a smooth pair. Now, if $x$ lies in a certain range, such as $[2^{240}, 2^{256}]$ in our running example, then a small $\Delta$ means that there is a larger range for possible values of $y$. Thus we expect that having smaller $\Delta$ increases the probability that some $y$ solving the Pell equation will meet our criteria. Assuming that this is the only property of $\Delta$ affecting the chance of having a solution in the desired range, the probability of having a solution scales like $\frac{1}{\sqrt{\Delta}}$. Since $\Delta$ can range from $3$ to the square of the maximum bound on the size of smooth pairs we are interested in ($2^{512}$ in our example), the effect of this combinatorial heuristic is very strong.

This reasoning agrees with experiments. In our search for pairs in the range $[2^{240}, 2^{256}]$, we solved all Pell equations with $\Delta \leq 2^{44}$. This computation ran in parallel on a 64-core computer for 3 days and found a $19949$-smooth, $215$-bit pair $(m, m+1)$ corresponding to a fundamental solution with
\begin{align*}
m & = 51963397732665557125190357543988479960188331933248699616266017360\\
& = 2^4 \cdot 3^5 \cdot 5 \cdot 7 \cdot 11^2 \cdot 647 \cdot 911 \cdot 919^2 \cdot 1103^2 \cdot 2099^2 \cdot 2203^2 \cdot 2423^2 \cdot 5279^2 \cdot 8641^2 \cdot 19949,\\
m+1 & = 19^2 \cdot 23^2 \cdot 47^2 \cdot 277^2 \cdot 359^2 \cdot 419^2 \cdot 541^2 \cdot 887^2 \cdot 1993^2 \cdot 4549^2 \cdot 4813^2 \cdot 12721^2.
\end{align*}
While this pair is comparatively small to many others found, it is different in that it corresponds directly to a fundamental solution. Certain cryptographic applications require smooth pairs $(m, m + 1)$ for which the sum $m + (m + 1) = 2m + 1$ is prime. This property is not achievable by pairs corresponding to $n$-th solutions unless $n = 2^k$ for some integer $k$ by Lemma \ref{lem:pnun}.

Checking the $n$-th solutions of the pairs found using the previously discussed methods for $n \in \{1, 2, \dots, 12\}$ also found a pair
{\small
\begin{align*}
m & = 13751398930221343029252446890555634360426232035770955214339102967231667741706727080\\
 & =  2^3 \cdot 3^5 \cdot 5 \cdot 59 \cdot 103^2 \cdot 113^2 \cdot 1697 \cdot 2381 \cdot 2383^2 \cdot 2399^2 \cdot 3623^2 \cdot 4733 \cdot 9151^2 \cdot 10607^2\\
 & \quad \cdot 16267 \cdot 18059^2 \cdot 18289 \cdot 23603,\\
m+1 & = 11^4 \cdot 17^2 \cdot 53^2 \cdot 83^2 \cdot 109^2 \cdot 227^2 \cdot 263^2 \cdot 347^2 \cdot 373^2 \cdot 599^2 \cdot 2341^2 \cdot 7883^2 \cdot 10223^2 \cdot 10883^2 \cdot 12511^2,
\end{align*}}

which has $273$ bits and is $23603$-smooth. It corresponds to the sixth solution to a Pell equation. This pair, though larger than the target cryptographic size of 256 bits, is strong evidence for the efficiency of this algorithm. It is both larger and more smooth than the second pair presented in \ref{sec:successive}, and is far larger than any previously known for its smoothness bound. This pair is also a nice example of how the methods in \ref{sec:successive} can be combined well with other algorithms to yield $B$-smooth pairs larger than either algorithm could individually.

\subsection{The small primes heuristic}\label{sec:smallPrime}

While the above heuristic of choosing small values of $\Delta$ works well, a limitation of it is that there are relatively few small $\Delta$, and the probability of each Pell equation resulting in smooth pairs in the desired range decreases strongly as $\Delta$ increases. Furthermore, once we relax the restrictions on the maximum size of $\Delta$, exhausting all of its possible values becomes computationally infeasible. Another heuristic that narrows the search space is to choose values of $\Delta$ with smaller prime factors.

Among coefficients $\Delta$ of roughly the same size, those having more (and hence on average smaller) prime factors correspond to Pell equations that are more likely to result in smooth pairs in the desired range. We again give a combinatorial argument. If $p \mid \Delta$, then from the Pell equation, we have
$$x^2 - 1 \equiv 0 \pmod p \iff (x - 1) (x + 1) \equiv 0 \pmod p \iff x \equiv \pm 1 \pmod p$$
because the integers modulo $p$ form a field. Suppose that $\Delta$ is divisible by $n$ primes, say $\Delta = p_1 \cdot p_2 \cdot \ldots \cdot p_n$. Any integer $x$ that solves the Pell equation must satisfy the above condition, so we can associate to it a sequence $s \in \{-1, +1\}^n$, where the $i$th element of the sequence is $x \text{ mod } p_i$.

Now suppose we fix such a sequence $s$, and consider all $x$ in our desired range that are associated to $s$. By the Chinese remainder theorem, there is exactly one such $x$ in every interval of length $\Delta$. This gives further evidence for the above heuristic of choosing $\Delta$ small, because there are more $x$s in our desired range with the potential of solving the Pell equation.

Further, notice that $x$ in a solution to the Pell equation with coefficient $\Delta$ can correspond to any of the length $n$ sequences. Since there are $2^n$ such sequences and there are (up to $\pm 1$) the same number of $x$'s in the range corresponding to each, the implicit search space in which we look for solutions when solving Pell equations with coefficients $\Delta$ of a specified size is much larger when $\Delta$ has more prime factors.

Small searches with this method over coefficients larger than $2^{42}$ yielded a $201$-bit, $29881$-smooth pair
\begin{align*}
m & = 2317395102010090979961970844394697249269453177012017537196644\\
 & = 2^2 \cdot 13 \cdot 53^2 \cdot 113 \cdot 139^2 \cdot 269^2 \cdot 347^2 \cdot 509^4 \cdot 569^2 \cdot 743^2 \cdot 12823^2 \cdot 14149 \cdot 29881,\\
m+1 & = 3^9 \cdot 5 \cdot 23^2 \cdot 43^2 \cdot 167^2 \cdot 179^2 \cdot 227 \cdot 349^2 \cdot 613^2 \cdot 661^2 \cdot 2927^2 \cdot 4099^2 \cdot 6421^2.
\end{align*}
It corresponds to $\Delta = 13 \cdot 113 \cdot 14149 \cdot 29881$.
This pair is smaller than those given previously, which is expected given the increase in the sizes of $\Delta$ searched over. However, it is still close to cryptographic size, indicating that more precise heuristics could extend our methods to search effectively over large values of $\Delta$.

\section{Discussion}

We have seen how solutions to Pell equations give both a theoretical and practical approach to finding twin smooth integers, including several new algorithms that have produced new pairs. We conclude by discussing two observations for exploration in further research.

Based on experiments, our algorithms are more likely to find smooth pairs $(m, m+1)$ with $2m + 1$ composite, which is undesirable for some applications. Based on Lemma \ref{lem:pnun}, the only pairs that can sum to a prime are those corresponding to $n$-th solutions with $ n = 2 ^ k $ for integer $k$. Generally such pairs are easy to exhaust with the algorithms given in \ref{sec:successive} for $k$ even moderately large; for our example we found that no pairs in $[2^{240}, 2^{256}]$ exist with $k \geq 3$. Therefore, it is interesting to consider whether we can improve our search through fundamental solutions to directly find large pairs more often.

One approach is to consider the structure of the continued fractions expansions. Because it was not useful to our algorithms, we have omitted discussion of many remarkable structures in the continued fractions of square roots, such as the periodicity of the quotients. Because the size of fundamental solutions is determined by the length of the period and size of the quotients of $\sqrt{\Delta}$, one idea is to choose coefficients $\Delta$ by fixing the length of quotients in the corresponding continued fraction expansion.

Unfortunately, while this approach guarantees that we have pairs $(x,y)$ satisfying the Pell equation $x^2 - 1 = 2\Delta y^2$ with $x$ in a given range, however, it does not guarantee smoothness of $y$ or $D$. It is interesting to consider whether a clever way of choosing the period lengths and continued fractions quotients could result in a high probability of $y$ and $D$ being smooth, thus resulting in an efficient algorithm for finding twin smooth pairs. This is a promising approach because of the strong and well-studied structure of continued fractions. An overview of the theory of continued fractions can be found in \cite{jacobson-williams}, chapters 3 and 6.

The tension described above between searching for solutions to Pell equations (or equivalently smooth pairs) with a fixed smoothness bound and searching for solutions (or smooth pairs) in a given size range is common across many algorithms for finding twin smooth integers. This difficulty seems intrinsic, as the problem of predicting the size of the solution to a Pell equation is related to questions about computing class numbers of quadratic fields, which remain highly mysterious.

Solving the Pell equation is equivalent to finding the fundamental unit in the associated real quadratic field $\mathbb{Q}(\sqrt{D})$. Because of the structure of units in its ring of integers, given the fundamental unit $\epsilon_D = t + u\sqrt{D}$, then either $x = t, y = u$ or $x = t^2 + u^2, y = 2tu$ solve the Pell equation $x^2 - D y^2 = 1$. The two cases arise from the fact that the fundamental unit can have norm $\pm 1$. For simplicity we focus on the first case; the analysis is analogous for the second. Since $x \approx \sqrt{D}y$, $\epsilon_D \approx 2x$.

Dirichlet's class number formula for quadratic fields relates the size of the fundamental unit to that of the fields' class number $h(D)$,

$$h(D) = \frac{\mathcal{L}(\chi, 1) \sqrt{D}}{2 \text{log}(\epsilon_D)}.$$

Based on both experimental evidence and theoretical bounds, $\mathcal{L}(\chi, 1)$ grows extremely slowly with $D$, see for example the following plot.

\includegraphics[scale = 0.8]{L_functions.png}

Therefore if we want to find coefficients $D$ with corresponding $x \approx M$, we want

$$h(D) \approx \frac{\sqrt{D}}{\text{log}(M)}.$$

Since $M$ is fixed, choosing smooth, squarefree coefficients $D$ with Pell equation solutions in the right range is a problem equivalent to choosing $D$ with class number roughly $\sqrt{D}$. This problem is likely very difficult, since little is know about the sizes of class numbers of real quadratic fields. For example, Gauss conjectured in 1801 that there are infinitely many real quadratic fields with class number 1; this conjecture remains open. In the absence of new breakthroughs in predicting the sizes of class numbers, heuristics such as those presented in \ref{sec:smallCoeff} and \ref{sec:smallPrime} appear to be the best way to search for large twin smooth integers corresponding to fundamental solutions.

redpajama_set_name: RedPajamaArXiv


JACKSONVILLE, FL. --- Drone Aviation Holding Corp., a developer of specialized lighter-than-air aerostats and tethered drones, today announced that its wholly owned subsidiary, Drone Aviation Corp., has received a follow-on order to support activities at Eglin Air Force Base in Florida.
The additional ground and aerial-based equipment will work together with a previously supplied mobile aerostat launcher system delivered in the third quarter of 2013.
The Drone equipment and aerostat system will be used to support Department of Defense research, engineering, test and evaluation efforts. The follow-on order substantially increases the capabilities of the persistent range instrumentation required by the customer for furthering weapons and platform engineering and design efforts. These programs include the F-35 Joint Strike Fighter and related munitions testing and integration.
Felicia Hess, CEO of Drone Aviation Holding Corp., stated, "This follow-on order demonstrates the Air Force's ongoing commitment to our aerostat platform in support of their operations. The Eglin order also demonstrates our ability to customize our technologies to meet a customer's unique needs, ensuring that our aerostat system remains a highly capable, adaptable and cost effective tactical aerial platform. Moving forward, we are committed to expanding our existing governmental relationships and developing new customer relationships by leveraging our current customers as well as other branches of the military."
Drone Aviation Holding Corp. provides critical aerial and land based surveillance and communications solutions to government and commercial customers. Utilizing a unique tethering capability, drone systems operate in the National Airspace within FAA guidelines for safe operations.


text:
Follow Statista
Topics›
Internet usage worldwide›
Worldwide Internet Usage
Half of the World Will be on the Worldwide Web
Internet usage worldwide
More than half of the world's population will be online, according to new estimates released by the International Telecommunication Union. The U.N. estimates that about 3.9 billion users are set to go online as 2018 comes to a close.
Accelerating growth across the many nations in Africa are responsible for this continued uptick. In 2005, only 2 percent of people in Africa used the internet. As of 2018, about a quarter of people in Africa used the internet. Between 2005 and 2018, about 4 times more people across the world are on the internet.
This chart shows the number of individuals using the internet between 2004 and 2018 (in millions).
Download Chart
URL to be used as reference link:
https://cdn1.statista.com/chart/16467/internet-usage-worldwide/
HTML code to embed chart
You will find more infographics at Statista
Average daily internet usage worldwide 2019, by age and device
Number of worldwide internet users 2009-2019, by region
Number of fixed broadband internet subscriptions worldwide 2005-2019
Internet usage share of tablet operating systems in Germany 2012-2019
Daily mobile internet usage per capita worldwide 2011-2021
Global number of internet users 2005-2019
Infographic Newsletter
Statista offers daily infographics about trending topics covering Media and Society.
Related Infographics
Wikipedia Day
The Most Read Wikipedia Articles of 2019
The Rise of the Internet Kill Switch in India
Cyber Attacks Often Target Microsoft Office
Tariff Hike Would Hit U.S. Wine Lovers Where It Hurts
Cashing in on Free-to-Play Games
New Year's Peak Illustrates Airbnb's Growing Stature
Americans Most Commonly Use Facebook and YouTube
End of the Decade
Tech Booms of the New Decade?
U.S. Energy Consumption
Why U.S. Homeowners Want Solar Power
Star Wars: The Rise of the Blockbuster
Disney Dominates at the Box Office
In 2100, Five of the Ten Biggest Countries in the World Will Be in Africa
Who may use the "Chart of the Day"?
The Statista "Chart of the Day", made available under the Creative Commons License CC BY-ND 3.0, may be used and displayed without charge by all commercial and non-commercial websites. Use is, however, only permitted with proper attribution to Statista. When publishing one of these graphics, please include a backlink to the respective infographic URL. More Information
Which topics are covered by the "Chart of the Day"?
The Statista "Chart of the Day" currently focuses on two sectors: "Media and Technology", updated daily and featuring the latest statistics from the media, internet, telecommunications and consumer electronics industries; and "Economy and Society", which current data from the United States and around the world relating to economic and political issues as well as sports and entertainment.
Does Statista also create infographics in a customized design?
For individual content and infographics in your Corporate Design, please visit our agency website www.statista-content-design.de
Do you still have questions?
Feel free to contact us anytime using our contact form or visit our FAQ page.
Your contact to the Infografik Newsroom
Felix Richter
Data Journalist
felix.richter@statista.com +49 (40) 284 841 557
Statista Content & Information Design
Research, storytelling, infographics & presentation design on any topic in your corporate design.
The Statista Infographic Newsletter
Receive a new up-to-date issue every day for free
Our infographics team prepares current information in a clear and understandable format
Relevant facts covering media, economy, e-commerce, and FMCG topics
Use our newsletter overview to manage the topics that you have subscribed to

redpajama_set_name: RedPajamaCommonCrawl


Sacramento Kings 2nd overall pick Marvin Bagley III will miss the rest of the NBA Summer League after an MRI revealed a pelvic bone bruise, according to Yahoo Sports' Chris Mannix.
Per the report, Bagley will miss the rest of summer league action while resting for approximately 1-2 weeks when he is expected to resume full basketball activity.
Bagley was injured during Saturday's 71-63 to the Phoenix Suns and #1 overall draft pick Deandre Ayton.
The Sacramento Kings wasted no time in finding a replacement for fired head coach Dave Joerger, as they have officially hired Luke Walton as their next head coach, according to an official team press release.
Walton was fired by the Los Angeles Lakers after a 98-148 record in three seasons as the helm, and was immediately snatched up by the Kings on a deal that will run through 2022/2023, according to The Athletic's Sam Amick.
It can be argued that Walton got an unfair shake with the Lakers, but the same can also be said for Joerger, who led the team to marked improvement this season under his watch.
When it comes to the handling of the Sacramento Kings, former Kings' superstar Chris Webber has faith in his former teammate, and current head of basketball operations, Vlade Divac's handling of the team.
"Vlade has believed in these guys, especially [De'Aaron] Fox and Buddy [Hield], telling me a long time ago they were going to change the trajectory of the team," Webber said, according to Jason Anderson of The Sacramento Bee.
"The great part about it as a Sacramento fan is, I know Vlade, he doesn't just work for the organization, but he is a fan of the city. He wants it to be great, not only on the court but in the community as well.
The Kings have made marked strides under Divac's watch, after a chaotic start, and growth is visible.


text:
Netanyahu's corruption trial resumes amid COVID-19 protests, economic fallout
Sunday, 19 July 2020 4:11 PM [ Last Update: Sunday, 19 July 2020 4:33 PM ]
Israeli prime minister Benjamin Netanyahu wears a face mask as he opens the weekly cabinet meeting at the foreign ministry, in Jerusalem al-Quds, on July 5, 2020. (Photo by AP)
An Israeli court has resumed the corruption trial of prime minister Benjamin Netanyahu after a two-month delay amid mounting protests over his corruption and poor handling of the coronavirus crisis.
On Sunday, the Jerusalem District Court decided that the evidentiary phase in Netanyahu's corruption trial would begin in January next year, with witnesses being heard three times a week.
The decision was taken after the judges accepted a request by Netanyahu's lawyers to give them additional time due to the coronavirus crisis and having more time to study the case's investigative materials.
Netanyahu did not attend the Sunday session as a spokesman for the prosecution said the session would be focused on a technical discussion
Netanyahu, the first serving Israeli prime minister to go on trial, was indicted last November in at least three cases of alleged bribery, fraud and breach of trust. He is accused of receiving extravagant gifts from his billionaire friends, and granting media tycoons some regulatory favors in return for more agreeable coverage of himself and his family.
Israel's top court allows scandal-tainted Netanyahu to form new administration
Israel's top court has ruled that Prime Minister Netanyahu can form a coalition administration despite facing criminal charges.
In the first hearing in May, the premier denied the charges. At the time, the court agreed to give an additional two-month break to his lawyers due to the pandemic.
The second hearing was focused on the additional amount of time his lawyers requested and the timeline of the trial. Furthermore, in Sunday's hearing, which was described as being technical, only the defense attorneys were present, not the accused.
Netanyahu, 70, clinched in April a power-sharing deal with centrist and bitter rival Benny Gantz, after the pair failed to secure parliament seats sufficient to form a majority administration following three inconclusive elections since April 2019.
Netanyahu's corruption charges, annexation plan spell the beginning of the end for Israel
Could Israel's current Prime Minister be overstepping, falling, and dragging down the entire Israeli regime slowly down behind him?
If found guilty, Netanyahu will face a sentence of up to 10 years in prison and/or a fine for his bribery charges, and a jail term of up to three years for the allegations of fraud and breach of trust.
Netanyahu's second hearing on Sunday opened as the Tel Aviv regime is contending with mounting public discontent over his poor handling of the health crisis following a sharp rise in COVID-19 cases and the re-imposition of coronavirus curbs in recent weeks.
Rising unemployment, part of which caused by the pandemic's economic fallout, has also led to the public outrage over Netanyahu's handling of the crisis.
Hundreds of protesters swarm Netanyahu's home again
Hundreds hold fresh demonstrations in front of Benjamin Netanyahu's residence again, urging the embattled and scandal-hit Israeli prime minister to resign.
On Saturday, Israeli police resorted to water cannons to disperse protesters around Netanyahu's residence in Jerusalem al-Quds. In Tel Aviv, thousands of demonstrators convened to demand better aid from the regime to help their businesses badly hurt in the health crisis.
Israel Palestine Benjamin Netanyahu corruption protests Coronavirus bribery trial
Protesters, police scuffle at anti-Netanyahu demo


A guest master class by Philippe Skow, violin.
Mr. Skow has most recently performed for the President of France, Emmanuel Macron and wife, the Danish Royal Family, and the Danish Prime Minister.
TV2 Good Morning Denmark and Danish newspapers Berlingske and Børsen have named him "a leading violinist," "born soloist," and "deserved winner".


text:
Nesslau é uma comuna da Suíça, situada no distrito de See-Gaster, no cantão de São Galo. Tem  de área e sua população em 2018 foi estimada em 3.575 habitantes.

Foi criada em 1 de janeiro de 2013, a partir da fusão das antigas comunas de Nesslau-Krummenau e Stein.

Comunas de São Galo (cantão)

redpajama_set_name: RedPajamaWikipedia


Doctors are recognizing the low amount of magnesium in American diets is causing a lot of pain. Without proper levels of magnesium, the muscles and tendons in the body cannot function effectively, and the result is often a painful condition called tendonitis. For many people, the good news is that taking a magnesium supplement of 350 mg can put them on the road to recovery in no time.
It has been said that 95% to 98% of the American population is not receiving the proper amount of magnesium in the diet for several reasons. One is because our soils are depleted of many nutrients. One of those nutrients being depleted is magnesium. Obviously, because the plants don't have the magnesium, it is not being delivered in the food to consumers. New studies show that another reason is from food processing. From 30%-85% of the magnesium that is present in foods today, gets stripped away through various food processes, like refining. As one can see, it is very difficult, without taking a supplement, to achieve 100% of RDA's total from the food we eat.
The muscles and tendons function together during body movement. The muscle requires a certain level of magnesium within the cells to fully relax and to allow the muscle fibers to fully lengthen. How the body contracts the muscle will be explained in a different article. Without the magnesium, the fibers remain partially contracted and shortened and begin to put pressure on the tendons. Over time, the pressure on the tendons causes inflammation, and the tendons do not glide easily back and forth when stretched. This is where pain comes in the picture. The pain from tendonitis begins as a dull pain that can be tolerated and over time it becomes unbearable.
Magnesium is available at any store that carries vitamins and minerals. There are several different forms of magnesium; the absorption rate depends on the form. Some are absorbed faster than others. Magnesium Citrate seems to be the most soluble form thus absorbing quickest. The bottom line is that any magnesium supplement can correct the deficiency. Some combinations are more expensive and claim they are the best but, in the long run, they all work the same. The goal should be to take the supplement each day in order to maintain a daily 350-400 mg amount.
If tendonitis is a problem, magnesium can be a simple solution. There is nothing better than finding a simple inexpensive solution that makes long term pain disappear. And more times than not, the pain disappears very quickly. Living pain-free is being free.
Thank you Lori for this article. I have struggled with tendonitis on and off for years. I went trhough a period where I took a magnesium supplement for muscle pain, but had never made the connection between my improved tendonitis and taking the supplement. I think it's time to get back on the magnesium supplement.


Experience and photograph the spectacular fall color of the Aspen trees and the snow-capped San Juan Mountains of Colorado. We'll spend four days photographing the area around the Dallas Divide, Silverton, Kebler Pass, historic ghost towns and local ranch culture, all while surrounded by the beautiful fall colors of the aspen forests that blanket the San Juan Mountains.
During our trip, participants will learn how to photograph these scenes under a number of different lighting situations including sunrise and sunset, backlight, and nighttime photography. Photo critiques and post-processing techniques will also be conducted.
Hotels, meals and transportation are not included in the ticket price. Our trip will be based out of Ridgway, CO so your best bet is to find a hotel in town. Below is a link to the Ridgway Lodge & Suites. We've stayed there before and have had good service there.
Transportation must be done on your own or through carpooling. You must have a 4×4 or AWD SUV and be willing to drive on unpaved back roads. Most of our shooting locations are within 50 miles of Ridgway but the participants must be willing to drive within 100 miles of Ridgway at least once or twice. Because of the nature of this trip, the participants will need to be able to stay awake at different times of the day and night if they want to participate in both the day and night shoots.
Our latest go/no go date is August 15th. If by August 15th, there are not enough tickets sold, the charter will be canceled. All tickets are non refundable after our go/no go date is set. If a patron can't make the event after our go/ no go date is set, the patron must find someone else to purchase their ticket or the ticket charge gets credited to a future event. Lerro Productions does reserve the right to cancel a photo charter at their discretion. If the charter is canceled, Lerro Productions will refund the photo charter ticket price. All refunds will be issued by check only. The Credit Card Convenience fee in non-refundable regardless of patron or charter cancellation. Lerro Productions is not responsible for loss of additional expenses.
Lerro Productions reserves the right to make changes to the schedule, dates and or lighthouses if need be. We are not responsible for bad weather, delays or cancelations. We are also not responsible for lost or stolen items, injuries, sickness or damages to personal equipment and lost of personal travel expense.


text:
line leadership
(Field Marshal) Ayub Khan 's regimes. He served as Ambassador of Pakistan to Netherlands in 1962 and later as Secretary of Information and Education. He resigned after clash with the new regime of Yahya Khan and opted for a self-imposed exile at UK. Shahab was elected a member of the executive board of UNESCO in 1968. Legacy From the early days of Pakistan (History of Pakistan), Shahab worked with front-line leadership of the country till the regime of Yahya Khan. Shahab unveiled in Shahab Nama, and Mufti (Mumtaz Mufti) conferred it in Alakh Nagri, that the idea of giving a constitutional name to Pakistan as Islamic Republic was actually proposed by him to Ayub Khan (Ayub Khan (Field Marshal)). Shahab did argumentation in the parliament in the favor of this idea, which was unanimously accepted by the then leaders. ****'''J: (Haplogroup J (mtDNA))''' The highest frequency is in the Near East (12%), 21% in Saudi Arabia. J declines towards Europe at 11%, Caucasus 8%, North Africa 6% and becomes practically missing in East Asia. Serk, Piia. 2004. Human Mitochondrial DNA Haplogroup J in Europe and Near East. Thesis University of Tartu, Estonia ****'''T: (Haplogroup T (mtDNA))''' The highest frequency is in the Caspian region (Caucasus, Northern Iran, Turkmenistan). It is important in Europe (almost 10%), oxfordancestors.com Maternal Ancestry Middle East, Central Asia, Pakistan and North Africa. Small frequency in the Horn of Africa and India. **'''R3''': Found in Armenia. Mannis van Oven's PhyloTree.org – mtDNA subtree R The '''Greater Middle East''' is a political (Political geography) term coined by the administration (George W. Bush administration) Commons:Category:Pakistan WikiPedia:Pakistan Dmoz:Regional Asia Pakistan


If your answer to this query is yes, then I will hopefully be able to support you with the suggestions I give on this hub. This technology is very pricey at this point with the LCS glasses alone costing about $100. Flash memory is based on memory cells, and the most widespread (and most reasonably priced) variety is the MLC (Multi-Level Cell) – a technology that enables writing much more data to a single cell of the drive. The very same will be correct as pc technology evolves during the rest of the 21st century. In the contemporary age, a number of factors have made the use of assistive technology more critical. The use of fundamental technology is also a function of other animal species apart from humans. The technology turns the dilemma of nuclear waste into a supply of secure nuclear energy.
There are diverse approaches to develop this illusion and that is exactly where the different sorts of 3D technology comes in. Let's assessment 3 different kinds of 3D technology. I am content to share that I will be attending the UIST 15 , the 28th ACM User Interface Software and Technology Symposium, from November eight-11. It is constructed with massive, tightly interlocking stones that are so large and heavy we can not even move them right now with our present technology. The details availability is largely due to the advancement of technology know as computers.
Integrating technology and education is a challenging activity – that is if the inventor truly wants to scale-up education practices by way of correct use of technology then he has to come out of the typical and feel out of box. The theme is closely linked to 1 of the hosting university's (UEF's) existing research orientations on human sciences and technology research. Let International Details Technology upgrade your staff with the most current certifications and expertise established to keep your entire company on the cutting edge of technology whilst preserving productivity.
The notion of acceptable technology was created in the 20th century by thinkers such as E. F. Schumacher and Jacques Ellul to describe conditions exactly where it was not desirable to use quite new technologies or these that necessary access to some centralized infrastructure or components or capabilities imported from elsewhere. As a cultural activity, technology predates both science and engineering , every single of which formalize some elements of technological endeavor. Several phenomenal teachers, Technology Tailgate Coaches and bloggers are joining forces for an amazing classroom technology blog hop! We wrap issues up with our Technology Picks of the Week this time featuring Virtual Reality. For her element, Patel has been approached about expanding the technology to help a lot more men and women.
This technology has been typically utilised because of its low expense, even so, the image high quality is not that excellent. Studies have indicated that persons who use a fantastic deal of technology on a everyday basis do not consider that sending a message, or uploading a video off the net should take longer than a matter of minutes, or even seconds, so it can trigger aggravation to wait.


text:
Q: Mixed targets in VSTS build We have an aspnet-core application which consists of several .net standard libraries, two aspnet-core mvc applications and, exceptionally, a .net framework console application.
These all build successfully, both in solution in Visual Studio, and also individually using msbuild/dotnet build commands.
We are now trying to implement continuous integration/build with VSTS.
We have found that the aspnet-core web applications build successfully, but the .net framework console application fails with multiple error messages like the following:
2017-12-24T07:20:26.4090447Z C:\Program
Files\dotnet\sdk\2.0.3\Microsoft.Common.CurrentVersion.targets(1988,5):
warning MSB3245: Could not resolve this reference. Could not locate the
assembly "Xxx.Core, Version=1.1.0.0, Culture=neutral,
processorArchitecture=MSIL". Check to make sure the assembly exists on disk.
If this reference is required by your code, you may get compilation errors.
[d:\a\1\s\WinApps\ListBuilder\ListBuilder.csproj]

C:\Program Files\dotnet\sdk\2.0.3\Microsoft.Common.CurrentVersion.targets(1988,5): warning MSB3245: Could not resolve this reference. Could not locate the assembly "Microsoft.EntityFrameworkCore, Version=2.0.1.0, Culture=neutral, PublicKeyToken=adb9793829ddae60, processorArchitecture=MSIL". Check to make sure the assembly exists on disk. If this reference is required by your code, you may get compilation errors. [d:\a\1\s\WinApps\TelemListBuilder\TelemListBuilder.csproj]

The console application follows the same folder structure as the web applications but seems unable to find any of the referenced packages.
Our Agent queue is Hosted VS2017.
We have two VSTS tasks: Restore and Build from the asp.net core build template.
If necessary we would be happy if we could exclude the console app from the build.

A: Since there are general .net framework project and .net core project and you want to build them together, you need to use Visual Studio Build task.
So, remove .Net Core Restore and .Net Core build tasks and add NuGet Tool Installer (4.3.0), Nuget restore and Visual Studio Build tasks (Visual Studio Version: Latest or Visual Studio 2017)
If you want to generate the deployment package during the build, you can specify MSBuild arguments like this:
/p:DeployOnBuild=true /p:WebPublishMethod=Package /p:PackageAsSingleFile=true /p:SkipInvalidConfigurations=true /p:PackageLocation="$(build.artifactstagingdirectory)"

redpajama_set_name: RedPajamaStackExchange


The clean, straightforward design of the 2019 Chevrolet Equinox Lt Interior is let down only by the turbo model's excessive add-ons.
After a complete redesign for 2018, the Equinox will probably not receive any significant updates for the 2019 model year.
The Equinox is positioned above the smaller Trax and below the larger Traverse, Tahoe, and Suburban in Chevrolet's crossover and SUV lineup.
The 2018 Equinox trims should carry over into 2019 and include the base L, LS, LT, and Premier. Standard features on the base trim for 2018 included 17-inch wheels, a 7.0-inch multimedia touchscreen, Apple CarPlay and Android Auto, OnStar Guidance, and heated and power adjustable side view mirrors. The top Premier trim comes standard with 19-inch wheels, LED headlights and taillights, heated front seats, eight-way power adjustable driver's seat, blind-spot monitoring with rear-cross traffic alert, a hands-free liftgate, and an 8.0-inch touchscreen.
The 2018 powertrains should carry over and include a 170-hp 1.5-liter turbo-four, a 252-hp 2.0-liter turbo-four, and a 1.6-liter turbodiesel I-4 that produces 137 hp and 240 lb-ft of torque. A six-speed automatic backs the 1.5-liter turbo-four while the 1.6-liter turbodiesel engine and the 2.0-liter turbo-four uses a nine-speed automatic. All powertrains come standard with front-wheel drive and are available with all-wheel drive. In 2018, EPA-rated fuel economy numbers ranged from 22/28 mpg city/highway with the AWD 2.0-liter turbo-four to 28/39 mpg with the front-drive turbodiesel.
Driver-assist safety tech will likely continue into 2019 unchanged. Chevrolet's safety tech on the Equinox includes forward collision alert, low-speed automatic emergency braking, lane keep assist with lane departure warning, blind-spot monitoring with lane change alert, and rear-cross traffic alert. We hope the automatic emergency braking system is improved to a wider range of speeds.
A surround vision camera system and a rear seat reminder feature are also available on the Equinox. OnStar also offers several safety services including automatic crash response, emergency services, roadside assistance, and stolen vehicle assistance.
The Equinox's tech will likely carry over into 2019 with a standard 7.0-inch multimedia touchscreen with Apple CarPlay and Android Auto that can be upgraded to an 8.0-inch touchscreen that adds SiriusXM satellite radio and an optional navigation system. The upgraded audio system is an optional seven-speaker Bose system. A 4G LTE Wi-Fi hotspot is available and the myChevrolet mobile app allows the owner to start/stop the engine, lock/unlock the doors, honk the horn and flash the lights. The Teen Driver feature allows owners to monitor the driving characteristics of a new driver.
"The competition has already set a benchmark far ahead of the Equinox," we concluded.
The best value in 2019 Chevrolet Equinox Lt Interior is a base model with manual gearbox, but it's still decently priced after adding fancy audio, leather, and a moonroof.


text:
ABOUT ANTHROPOLOGY
DEPTDIRECTORIES
Faculty » Scott L. Fedick
Ph.D. 1988 Arizona State University
E-mail: scott.fedick@ucr.edu
Professor Fedick's primary research interests are in prehistoric settlement patterns, agriculture, and the development of social complexity, with an emphasis on Mesoamerica (particularly the Maya area) and the American Southwest. He also has an active interest in historical archaeology of the western United States. Professor Fedick is also involved in archaeological applications of Geographic Information Systems (GIS). Professor Fedick's field research has been conducted in the Maya Lowlands of Belize, Mexico, and Guatemala, the Mexican highlands of Oaxaca, and the western and southwestern United States. He is currently serving as director of the Yalahau Regional Human Ecology Project, a multidisciplinary, multinational research effort focusing on ancient Maya settlement, land use, and political organization in an extensive wetland area of northern Quintana Roo, Mexico.
SELECTED PUBLICATIONS: The Lowland Maya Area: Three Millennia at the Human-Wildland Interface (2003, edited with Arturo Gomez-Pompa, Michael Allen, and Juan Jiménez-Osornio); Wetland Manipulation in the Yalahau Region of the Northern Maya Lowlands (2000, with 4 co-authors); The Managed Mosaic: Ancient Maya Agriculture and Resource Use (1996, editor and author of 3 chapters); Indigenous Agriculture in the Americas, Journal of Archaeological Research (1995); Land Evaluation and Ancient Maya Land Use in the Upper Belize River Area, Belize, Central America, Latin American Antiquity (1995); The View from Yalahau: 1993 Archaeological Investigations in Northern Quintana Roo, Mexico (1995, editor with K. Taube, and author of four chapters); Ancient Maya Agricultural Terracing in the Upper Belize River Area: Computer-Aided Modeling and the Results of Initial Field Investigations, Ancient Mesoamerica (1994); Prehistoric Maya Settlement Patterns in the Upper Belize River Area: Initial Results of the Belize River Archaeological Settlement Survey, Journal of Field Archaeology (1992, with A. Ford); Shell Working in Ejutla, Oaxaca (Mexico): Findings from an Exploratory Field Season, Mexicon (1991, with G. M. Feinman and L. M. Nicholas); Chert Tool Production and Consumption among Classic Period Maya Households, in Maya Stone Tools: Selected Papers from the Second Maya Lithics Conference (1991, edited by T. R. Hester and H. J. Shafer); The Prehistoric Agricultural Landscape of the Central Maya Lowlands: An Examination of Local Variability in a Regional Context, World Archaeology (1990, with A. Ford); and The Economics of Agricultural Land Use and Settlement in the Upper Belize Valley, in Prehistoric Maya Economies of Belize (1989, edited by P. A. McAnany and B. L. Isaac).
© Regents of the University of California. All rights reserved.
Page created and maintained by CHASS College Computing.


Geof Harries (@geofharries of subvert.ca) pointed me at Pulse (pulseapp.com), which encapsulates much of my process in a nice web tool. I haven't had the chance to try it out, but it looks like it captures the key steps of estimating, scheduling, repeating, and editing. One benefit of using our Excel spreadsheet was that we could do 30-second "what-if" experiments. Move a customer payment to a later date and see if you still make payroll. Increase your monthly billing by 10% and see if you can hire another developer. This flexibility and control is key to this being something you are willing to use.
Fred Wilson's (@fredwilson) excellent MBA Mondays series has included a few posts that introduce the basic accounting reports of balance sheet, profit & loss, and cash flow. My post focused on the cash flow, as it's the most important real-time report for a bootstrapped / low-capital startup. He also has his own post on Budgeting in a Small, Early-Stage Company. I'll explain why I disagree with his recommendation (at least for the very early stages) but you should go read everything he's written.
I still think that annual budgeting, as he recommends, is not useful in the early stages, when your business plan, revenue model and cost structure are changing frequently. Focus on your market fit, your repeatable revenue model, and watch your cash. Once you've accomplished those (reaching the Optimize and Scale levels of the Startup Pyramid – @seanellis) then your company will be static enough to spend the time on annual or semi-annual budgets.
Moving forward, I hope to crank out a few more posts on lean processes that I've used in the past so we can get away from old BDUF time wasters throughout the company.
Budgets are a waste of time.
In startups and small companies, the business or strategy can change every few months and the revenue and expenses can swing wildly monthly. In an environment of constant change, it's a complete waste of time to put the effort into an annual, quarterly or even rolling budget. What you can do instead is implement a few monitoring and control systems, so that you always know exactly what the financial picture is but you spend the minimal time on the processes. I'd like to put together a few standard tools that startups can use to manage their finances in an agile way – continuous improvement, low process overhead and immediate feedback. Let's work on Agile Budgeting.
UPDATE: I should note here that, as we discuss in the comments, this post and this process really apply mostly to bootstrapped operations with small amounts of capital. Doesn't have to be early-stage, I used this process in years 8-10 of my company. Use it when you don't have huge amounts of padding.
The most critical piece is your cash flow management. You need a system that will give you an accurate picture of your bank account balance right now, as well as at various points in the future. Will you make payroll on the 15th? Will you make rent on the first? When in the next two months should you buy those Aerons? I'm going to describe a process that should give you this transparency without spending more than a few minutes a day on it.
Before starting: Get a bookkeeper as early as possible. They're inexpensive compared to software engineers or biz dev folks, and they should keep your finances up to date with few mistakes. Both are critical to your company's success. No amount of light process will save you from doing hours of data entry without a bookeeper. This model is something that you should (we did, at least) update every day, so when there's a decision to be made you can pull up the spreadsheet and know that the numbers were accurate this morning.
Before working on your cash flow, you need a clear view of your situation today. A worksheet like this can be quickly updated from your online banking by your bookeeper every morning, and as cheques are written and payments received. The available balance calculated at the end here is the starting balance for your cash flow, so it must be accurate. That's why we include outstanding cheques and undeposited payments. We made a policy of keeping a minimum balance in our main account, shown here as $5,000 to guard against bounced cheques or variances in timing.
We always preferred to use a separate account for tax liabilities, so here you see the payroll witholdings going on when payroll is made, and coming back out before the payment is due the next month. Things are pretty red here! But that's ok, we haven't entered any income yet.
Next step is to get a clear picture of your expense patterns. Summarize ALL of your expenses for the past few months and identify the things that happen regularly. This spreadsheet lists the expenses, recent monthly sums (doesn't have to happen in every month, but often enough to be a normal part of your operation), and what dates you usually pay it on. Things without set payment dates I've split evenly between the first and second halfs of the month (eg. Lawyer). The only things being left out of this list are the ones we've already entered on the cash flow.
I split the expenses into month-halfs because generally it's not important what date it happens on, but whether it happens before or after your big fixed-date expenses: rent and payroll. Also you'll see in a minute that we can fine-tune this much more.
In this list you have captured all of your outflows in the past three months. This spreadsheet goes into your cashflow Excel file and gets updated probably twice a year.
Now, depending on the level of detail you want, you pull out the regularly recurring and fairly fixed stuff and enter it as new row items on the cashflow sheet. I would enter items such as Dell Lease, Health Insurance, Hosting, etc… because they happen monthly, I know what date they happen on, and I have a good idea of the amount.
These numbers are based on the averages from the previous three months, but when the expense rows are being entered for a new month, use the expected value. For instance, LocalHost should be entered at the recent rate of $1,800 not the average of $1,467.
Now comes the process part. You've constructed a great starting point, now you make it repeatable, and you ingrain it as part of your routines.
Update balances – Usually performed every morning, the bookkeeper will check the online banking and update the balances sheet of this workbook. The updated Total Available carries through to the Cash Flow sheet, perhaps significantly departing from what was expected for this point in the month. Preferably he would also scan the Balance column of the Cash Flow sheet for any red numbers and bring them up for discussion. In our case he also updated the current exchange rate.
Pre-fill months – Done monthly by the bookkeeper, the template rows (the set of standard expenses and revenues) are filled in to keep the timeline extending about two months out. The owners should review after this to see how realistic the numbers are – the goal here is that the template set should be a pretty good picture of reality, and there will frequently be tweaks needed.
Enter specific numbers when known – The bookkeeper will update the template rows when they learn of the real amount for any of the transactions. For example, you may receive the invoice for your hosting and update the amount for the payment row that occurs a few weeks later. The bookkeeper should also enter any additional transactions that they learn about, and flag them for discussion or review (surprise payments are no fun).
Move the horizon – As you go through the month, the bookkeeper should be removing (or hiding – sometimes you need to look back) rows that have actually happened, leaving any that haven't, and updating the "Misc Expenses" rows with new amounts as some of those component transactions occur.
Use it – If these are happening, you will be able to pull up the spreadsheet at any time, see where cash crunhes occur in the next month, and shuffle things around. Most cash flow issues are related to timing, so opening up the sheet and seeing a red balance in a few weeks allows you to get on the phone with a client and speed up their payment. You're reacting in real time to accurate information about the future. When we used this at my company, my partners and I would pull it up a few times a week, adjust the entries with what we knew, and check for cash crunches as far out as two months away. This is invaluable.
Some of the expenses are more for padding than anyting, such as the Lawyer row. It doesn't happen every month, but if you've budgeted a normal amount you'll never get caught without the money to pay. You'll never be annoyed by having some extra money in the bank. In fact, we would usually enter specific padding transactions in our monthly template for expenses that only happened annually, such as an amount for the annual tax preparation. These can be entered as a monthly transaction for a portion of the amount, which is then put into a different account, or added as a running "earmark" item on the Bank Accounts spreadsheet (like the "Minimum $5,000 balance" item in my example at top).
If you can make this group of worksheets a part of your (well, your bookkeeper's) daily routine, the ongoing effort is very small and the accuracy is very high. Depending on your needs, I'm sure it would also work fine being updated weekly, but I find it's best to have it accurate continuously as the situation changes.
PLEASE let me know if you apply this, I'd love to hear what modifications you make to the framework in your situation. Let's keep improving on this model and passing it around.
UPDATE: This process and this workbook are both released under CC – Attribution Share Alike.