text:
\section{Introduction}
	In this paper, we will denote
	$$\zeta(s_1,\cdots,s_k) = \sum_{s_1>\cdots>s_k\geq 1} \frac{1}{n_1^{s_1} \cdots n_k^{s_k}}\qquad L_{s_1,\cdots,s_k}(a_1,\cdots,a_k) = \sum_{s_1>\cdots>s_k\geq 1}\frac{a_1^{n_1}\cdots a_k^{n_k}}{n_1^{s_1} \cdots n_k^{s_k}}$$
	to be the \textit{multiple zeta function} and \textit{colored polylogarithm} (or multiple $L$-values) respectively. For these two functions, $k$ is called the \textit{length} and $s_1+\cdots+s_k$ is called the \textit{weight}. We will also use the following common notation for alternating MZV: for example $\zeta(3,1,\bar{2},\bar{1}) = L_{3,1,2,1}(1,1,-1,-1)$. \par

	When $a_i$ are $N$-th roots of unity, $s_i$ are positive integers and $(a_i, s_i) \neq (1,1)$, $L_{s_1,\cdots,s_k}(a_1,\cdots,a_k)$ is called a colored multiple zeta values (CMZV) of weight $s_1+\cdots+s_k$ and \textit{level} $N$. Denote the $\mathbb{Q}$-span of weight $n$ and level $N$ CMZVs by $\textsf{CMZV}^N_n$.

	There have been a lot of researches on $\mathbb{Q}$-dimensions spanned by CMZV (\cite{Arakawa}, \cite{Kawashima}, \cite{Terhune}, \cite{Terhune2}, \cite{ZhaoJQ3}, \cite{ZhaoJQ2}). Let $d(w,N)$ be the dimension of $\textsf{CMZV}^N_w$, a deep result due to Deligne and Goncharov \cite{Deligne} provides an upper bound of $d(w,N)$: \begin{theorem}
		Let $D(w,N)$ be defined by
		$$1+\sum_{w=1}^\infty D(w,N) t^w = \begin{cases}(1-t^2-t^3)^{-1} \qquad &\text{ if } N = 1 \\ (1-t-t^2)^{-1} \qquad &\text{ if } N =2 \\ (1-at+bt^2)^{-1} \qquad &\text{ if } N \geq 3 \end{cases}$$
		where $a=\varphi(N)/2 + \nu(N), b=\nu(N)-1$. Here $v(N)$ denote number of distinct prime factors of $N$ and $\varphi$ is the Euler totient function. Then $d(w,N)\leq D(w,N)$.
	\end{theorem}

	The major focus of the paper (Section 3-5) is to present an algorithm (Theorem \ref{intmaintheo}) to calculate a large class of one-dimensional integral involving ordinary polylogarithm and generalized polylogarithm. The main idea is to represent the integrand as an iterated integral, but there are some subtleties. As an immediate consequence, some highly nontrivial result for infinite series are obtained. For example,
	$$\sum\limits_{n = 2}^\infty  {\frac{{H_{n - 1}^{(2)}}}{{{n^3}}}\left[ {{2^n}{\binom{2n}{n}^{ - 1}}} \right]} = \frac{\pi ^3 C}{24}-\pi  \beta(4)-\frac{3 \pi ^2 \zeta (3)}{128}+\frac{527 \zeta (5)}{256}+\frac{1}{384} \pi ^4 \log (2)$$
	\small \begin{multline*}\sum_{n=1}^\infty \frac{1}{n^5 2^n \binom{3n}{n}} = 4 \pi  \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)+3 \pi  \beta(4)-\frac{51 \text{Li}_5\left(\frac{1}{2}\right)}{2} -15 \text{Li}_4\left(\frac{1}{2}\right) \log (2)+\frac{\pi ^2 \zeta (3)}{4}+\frac{9 \zeta (5)}{2}\\ \qquad -3 \zeta (3) \log ^2(2) -\frac{97}{240} \log ^5(2)+\frac{41}{144} \pi ^2 \log ^3(2)-\frac{61}{960} \pi ^4 \log (2) \end{multline*}
	\normalsize{here} $C = \beta(2)$ denotes the Catalan constant, $\beta$ is the Dirichlet Beta function. The last series was conjectured by \cite[p.~27-28]{Browein}. Some of the series have been proved via complicated but ingenious manipulations in \cite{Zhao1}, \cite{Zhao2}.  \cite{FL3} contains related sums. We also obtained some rapidly convergent series for $\zeta(\bar{5},1)$ and $\zeta(\bar{5},1,1)$. \par
	Another line of results (Section 4.1 - 4.3) is the following: let $a_n = 4^{-n}\binom{2n}{n}$, $H_n^{s_1,\cdots,s_k} = \sum_{n\geq n_1>\cdots>n_k\geq 1}(n_1^{s_1}\cdots n_k^{s_k})^{-1}$, set $S = s+s_1+\cdots+s_k$, then:
	$$\sum_{n=1}^\infty \frac{H_n^{s_1,\cdots,s_k}}{n^s}a_n^{\pm 1} \in \textsf{CMZV}^2_{S} \qquad \sum_{n=1}^\infty \frac{H_n^{s_1,\cdots,s_k}}{n^s}a_n^{-2} \in \textsf{CMZV}^4_{S}\qquad \sum_{n=1}^\infty \frac{H_n^{s_1,\cdots,s_k}}{n^s}a_n^2 \in \frac{1}{\pi}\textsf{CMZV}^4_{S+1}$$
	~\\[0.02in]
	As immediate consequence of results in Section 3, some neat evaluation of definite integral can be obtained:
	\small $$\int_0^1 \frac{\Li_2 (-\frac{4x}{(1-x)^2}) \Li_3 (1-x^2)}{x} dx = -4 \pi ^2 \text{Li}_4\left(\frac{1}{2}\right)+\frac{7 \zeta (3)^2}{8}-\frac{93}{2} \zeta (5) \log (2)+\frac{139 \pi ^6}{3360}-\frac{1}{6} \pi ^2 \log ^4(2)+ \frac{1}{6} \pi ^4 \log ^2(2) $$
	\small \begin{multline*}
		\int_0^1 \frac{\log^2 (1-x) \log^2 x \log^3(1+x)}{x} dx = -168 \text{Li}_5\left(\frac{1}{2}\right) \zeta (3)+96 \text{Li}_4\left(\frac{1}{2}\right){}^2-\frac{19}{15} \pi ^4 \text{Li}_4\left(\frac{1}{2}\right)+\\ 12 \pi ^2 \text{Li}_6\left(\frac{1}{2}\right)+8 \text{Li}_4\left(\frac{1}{2}\right) \log ^4(2)-2 \pi ^2 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)+12 \pi ^2 \text{Li}_5\left(\frac{1}{2}\right) \log (2)+\frac{87 \pi ^2 \zeta (3)^2}{16}+\\ \frac{447 \zeta (3) \zeta (5)}{16}+\frac{7}{5} \zeta (3) \log ^5(2)-\frac{7}{12} \pi ^2 \zeta (3) \log ^3(2)-\frac{133}{120} \pi ^4 \zeta (3) \log (2)-\frac{\pi ^8}{9600}+\frac{\log ^8(2)}{6}- \\ \frac{1}{6} \pi ^2 \log ^6(2)-\frac{1}{90} \pi ^4 \log ^4(2)+\frac{19}{360} \pi ^6 \log ^2(2)
	\end{multline*}
	\normalsize{the} last expression is remarkable because it lacks level 2 weight 8 CMZV with higher length. Note that RHS is as elegant as the result of $\zeta(\bar{3},1,\bar{3},1)$ (conjectured in \cite{Browein3} and proved in \cite{Blumlein}). However, the connection between this integral and $\zeta(\bar{3},1,\bar{3},1)$ is not clear to the author.

	In Section 2, we quickly recalls relevant notations on Hoffman-Racinet algebra which will be used in subsequent sections. The last part of the section deals with the reduction of CMZV into more elementary constants. In \cite{ZhaoJQ2} and \cite{ZhaoJQ}, although ways to obtaining relations are detailed, only linear relations satisfied by CMZVs on a particular weight and level are considered. Using the method in these two papers, we calculated a complete reduction for level $4$ weight $5$ CMZVs.

	\section{Preliminaries}
	\subsection{Iterated integral}
	We quickly assemble required facts of iterated integral (\cite{Chen}, \cite{Gil}). Let functions $f_i(t)$ defined on $[a,b]$, define inductively
	$$\int_a^b f_1(t) dt \cdots f_n(t) dt = \int_a^b f_1(u) du \cdots f_{n-1}(u) \int_a^u f_n(t) dt $$
	When $r=1$, this is the usual definite integral of $\int_a^b f_1(t) dt$. When $r=0$, define its value to be $1$. \par
	The definition can be extended to manifold. Let $\gamma: [0,1]\to M$ a path on a manifold $M$, $\omega_1,\cdots,\omega_n$ be differential $1$-forms on $M$. Then
	$$\int_\gamma \omega_1\cdots \omega_n := \int_0^1 f_1(t) dt \cdots f_r(t) dt$$
	with $\gamma^\ast \omega_i = f_i(t) dt$ being the pullback of $\omega$. Then if $f: N\to M$ is a differentible map between two manifolds $N$ and $M$,
	\begin{equation}\label{itintpullback}
		\int_{f\circ \gamma} \omega_1\cdots \omega_n = \int_\gamma f^\ast \omega_1 \cdots f^\ast\omega_n
	\end{equation}

	\begin{proposition}
		Iterated integral enjoys the following properties:
		$$\int_\gamma \omega_1\cdots \omega_n = (-1)^n \int_{\gamma^{-1}} \omega_n\cdots \omega_1$$
		where $\gamma^{-1}$ is the reverse path of $\gamma$.
		$$\int_{\gamma_1 \gamma_2} \omega_1\cdots \omega_n = \sum_{r=0}^n \int_{\gamma_1} \omega_1\cdots \omega_r \int_{\gamma_2} \omega_{r+1}\cdots \omega_n$$
		where $\gamma_2(1) = \gamma_1(0)$, here $\gamma_1\gamma_2$ means composition of two paths, first $\gamma_2$, then $\gamma_1$.
		\begin{equation}\label{itintshuffle}\int_{\gamma} \omega_1\cdots \omega_n \int_{\gamma} \omega_{n+1}\cdots \omega_{n+m} = \sum_{\substack{\sigma\in S_{n+m} \\ \sigma(1)<\cdots<\sigma(n) \\ \sigma(n+1)<\cdots<\sigma(n+m)}} \int_\gamma \omega_{\sigma^{-1}(1)} \cdots \omega_{\sigma^{-1}(n+m)}\end{equation}
		the last sum is over certain elements of symmetric group $S_{n+m}$, it can also be viewed as shuffle product between $\omega_1\cdots \omega_n$ and $\omega_{n+1}\cdots \omega_{n+m}$, as defined in next subsection.
	\end{proposition}

	For use in Section 4, we define another kind of iterated integral. Let $\omega_1,\cdots,\omega_n$ be differential forms on $[0,1]$, for $0\leq a, b\leq 1$, we will put a bar over a differential form to indicate limit of integration should go from $x$ to $1$, rather than from $0$ to $x$. For example, if $\omega_i = f_i(x) dx$,
	$$\int_a^b \omega_1 \overline{\omega_2} \omega_3 \overline{\omega_4}= \int_a^b f_1(x_1) dx_1 \int_{x_1}^1 f_2(x_2) dx_2 \int_0^{x_2} f_3(x_3) dx_3 \int_{x_3}^1 f_4(x_4) dx_4$$
	A bar will never be put on the first differential form $\omega_1$, so this notation will not cause confusion. By writing $\int_{x}^1 = \int_0^1 - \int_0^{x}$, iterated integral with bars can be converted into linear combination of those without bars (i.e. all limits of integration except first one are from $0$ to $x$).

	\subsection{Hoffman-Racinet algebra}
	Let $X$ be a set, $\mathbb{Q}\langle X\rangle$ be the free non-commutative polynomial algebra over $\mathbb{Q}$ generated over $X$. Treating $X$ as alphabet, let $X^*$ be the set of words over $X$. \par
	Define a binary operation $\shuffle$ on $\mathbb{Q}\langle X\rangle$ via:
	$$w \shuffle 1 = 1 \shuffle w = w \qquad x w \shuffle y v = x(w\shuffle yv)+y(xu\shuffle v)$$
	for $w,v\in X^\ast, x,y\in X$. Then distribute $\shuffle$ over addition and scalar multiplication. The shuffle product is commutative and associative. \par
	Using shuffle product, the last property $\ref{itintshuffle}$ of iterated integral can be written as $$\int_{\gamma} \omega_1\cdots \omega_n \int_{\gamma} \omega_{n+1}\cdots \omega_{n+m} = \int_\gamma \omega_1\cdots \omega_n \shuffle \omega_{n+1}\cdots \omega_{n+m}$$
	~\\[0.02in]
	Now we specialize to the situation of CMZV. Fix a positive integer $N$, $\mu = \exp(2\pi i /N)$, set $$a = \frac{dt}{t}, b_i = \frac{dt}{\mu^{-i}-t}, z_{k,i} = a^{k-1}b_i \qquad i=0,\cdots,N-1$$
	Let $X = \{a,b_0,\cdots,b_{N-1}\}$, set $\mathfrak{A}^N = \mathbb{Q}\langle X \rangle$, $\mathfrak{A}^N_1$ be the subalgebra of $\mathbb{Q}\langle X\rangle$ generated by $z_{k,i}$;  $\mathfrak{A}^N_0$ be the subalgebra of $\mathfrak{A}^N_1$ generated by words not beginning with $b_0$ and not ending with $a$.
	then if $(s_1, i_1)\neq (1,0)$,
	\begin{equation}\label{toitint} L_{s_1,\cdots,s_n}(\mu^{i_1},\cdots,\mu^{i_n}) = \int_0^1 z_{s_1,i_1}z_{s_2,i_1+i_2}\cdots z_{s_n, i_1+i_2+\cdots i_n}\end{equation}

	Let $$\text{\L} (z_{s_1,i_1}\cdots z_{s_n,i_n}) = L_{s_1,\cdots,s_n}(\mu^{i_1},\mu^{i_2-i_1},\cdots, \mu^{i_n-i_{n-1}})$$
	then $\text{\L}(w) = \int_0^1 w$ for $w\in \mathfrak{A}^N_0$. We extend $\text{\L}$ linearly to $\mathbb{Q}\langle X\rangle$. \par

	Next we define the \textit{stuffle product} $\ast$ on $\mathfrak{A}^N_1$. Let $j\in \mathbb{Z}$, define $$\tau_j(z_{s_1,i_1}\cdots z_{s_n,i_n}) = z_{s_1,j+i_1}\cdots z_{s_n,j+i_n}$$
	here the second subscript of $z_{s,i}$ is considered modulo $N$. Then
	\begin{equation}\label{stuffle} z_{s,j} u \ast z_{t,k}v = z_{s,j} \tau_j(\tau_{-j}(u) z_{t,k}v) + z_{t,k} \tau_k(z_{s,j}u\ast \tau_{-k}v) + z_{s+t,j+k}\tau_{j+k} (\tau_{-j}(u)\ast \tau_{-k}(v))\end{equation}
	Then distribute $\ast$ over addition and scalar multiplication. The stuffle product is commutative and associative. \par
	It can be shown that, for $u,v \in \mathfrak{A}^N_0$,
	$$\text{\L}(u\shuffle v) = \text{\L}(u)\text{\L}(v) = \text{\L}(u\ast v)$$
	The above equality called \textit{finite double shuffle relation}, there is also a regularized version, but we do not state it here. Distribution (regularized or finite) also provides new relation \cite{ZhaoJQ2}. When $N$ is not a prime power, Zhao (\cite{ZhaoJQ3}, \cite{ZhaoJQ2}) also conjectures that these methods do not exhaust all relations between CMZVs. When $N=4$, Zhao \cite{ZhaoJQ3} discovers a way to generate these exceptional relations, he used these relations to reach bound predicted by Deligne $\textsf{CMZV}^4_l$, $l=2,3,4$. The author of this paper performed additional computation, Deligne's bound is also attained for $\textsf{CMZV}^4_5$.

	\begin{example}[Harmonic number]
		For future reference, we record an example here. Let $N=1$, $M$ be a fixed positive integer, $x_0 = a, x_1 = b_0, z_{k,i} = y_k$, so $y_k = x_0^{k-1} x_1$. For $w=y_{s_1}\cdots y_{s_k}$, define
		$$H_M(w) = \sum_{M\geq n_1 >\cdots > n_k \geq 1} \frac{1}{n_1^{s_1} \cdots n_k^{s_k}}$$
		and extend $H_M$ linearly to $\mathbb{Q}\langle X \rangle$, then it can be shown that $H_M(w \ast v) = H_M(w) H_M(v)$ for $w,v\in \mathfrak{A}^1_1$. For empty word, define $H_M(1) = 1$ for $M\geq 0$; for non-empty word $w$, $H_0(w) = 0$. \par
		The above definition can be generalized to arbitrary level: let $w = z_{s_1,i_1}\cdots z_{s_n,i_n}\in \mathfrak{A}^N_1$, define
		$$H_M(w) = \sum_{M\geq n_1 >\cdots > n_k \geq 1} \frac{\mu^{i_1}\mu^{i_2-i_1}\cdots \mu^{i_n-i_{n-1}}}{n_1^{s_1} \cdots n_k^{s_k}}\qquad \mu = \exp(2\pi i /N)$$
		then $H_M(w \ast v) = H_M(w) H_M(v)$ still holds for $u,v\in \mathfrak{A}^N_1$ (see \cite{Bigotte}).
	\end{example}

	\begin{proposition}
		Let $w = z_{s_1,i_1}\cdots z_{s_n,i_n}\in \mathfrak{A}^N_1$, then there exists positive integers $s_i$, $a_i \in \mathbb{C}$ such that
		$$H_M(w) = \sum a_i (\log M +\gamma)^{s_i} + c + o(1)\qquad M\to\infty$$
		with $\gamma$ the Euler-Mascheroni constant and $c\in \textsf{CMZV}^N_{s_1+\cdots+s_n}$.
	\end{proposition}
	\begin{proof}
		The existence of asymptotic expansion of this form is proved in (\cite{Racinet}, Section 2). To show that $c$ is a CMZV of level $N$, use the fact that $(\mathfrak{A}^N_1,\ast)$ is a commutative polynomial algebra over $(\mathfrak{A}^N_0,\ast)$ generated by $z_{1,0}$, (\cite{Hoffman}, \cite{Ihara}) and $H_M(z_{1,0}) = 1+\frac{1}{2}+\cdots+\frac{1}{M} = \log M + \gamma + O(1/M)$.
	\end{proof}

	\subsection{Regularization}
	We wish to extend the domain of $\text{\L}$ to all of $\mathbb{Q}\langle X\rangle$. Extend the definition as follows: let $k,m,n\geq 0$ be integers, $\xi_i \in X, \xi_1 \neq b_0, \xi_k \neq a$, set $\xi_1 \cdots \xi_k a^n = \xi_1 \cdots \xi_q$,
	\begin{equation}\label{reg}\text{\L}(b_0^m \xi_1 \cdots \xi_k a^n) = \begin{cases} 0 \qquad &\text{ if } mn=k=0 \\
			\text{\L}(\xi_1 \cdots \xi_k)  \qquad &\text{ if } m=n=0 \\
			-\frac{1}{m}\sum_{i=1}^q \text{\L}(b_0^{m-1} \xi_1 \cdots \xi_i b_0 \xi_{i+1}\cdots \xi_q)  \qquad &\text{ if } m>0 \\
			-\frac{1}{n}\sum_{i=1}^k \text{\L}(\xi_1 \cdots \xi_{i-1} a \xi_{i+1}\cdots \xi_k a^{n-1})  \qquad &\text{ if } m=0,n>0
	\end{cases}\end{equation}
	\begin{theorem}\label{regularization}
		Let $w\in \mathbb{Q}\langle X\rangle$, then there exists positive integers $s_i, t_i$ and $a_i,b_i,c\in \mathbb{C}$ such that
		$$\int_\alpha^\beta w = \sum a_i \log^{s_i}(\alpha) + \sum b_i \log^{t_i}(1-\beta) + c + o(1) \qquad \alpha\to 0^+,\beta\to 1^-$$
		the above method of extending \textup{\L} will make $c = \text{\L}(w)$.
	\end{theorem}
	\begin{proof}
		(\ref{reg}) is equivalent to saying that certain formal sum is grouplike in the bialgebra $\mathfrak{A}^N$, it is also the unique lift from certain grouplike element in $\mathfrak{A}^N_1$, satisfying the initial condition that nullifies the divergent part of iterated integral. Core ideas of such argument can be found in (\cite[Chap.~2,~13]{ZhaoJQ}, \cite{Racinet}).
	\end{proof}

	\begin{remark}
		The $c$ in the theorem will still be denoted by $\int_0^1 w$, even though the integral might not make sense from the traditional perspective. For example, $\int_0^1 \frac{dx}{x} = \int_0^1 a = \text{\L}(a) = 0$.
	\end{remark}

	\begin{example} [Generalized polylogarithm] Let $X = \{x_0, x_1\}$, $x_0 = dx/x, d_1 = dx/(1-x)$, for any word $w$ formed from $X$, define inductively:
		$$\Li_w(x) = \begin{cases}\frac{\log^n x}{n!} \qquad &\text{ if }w=x_0^n \\
			\int_0^x x_i \Li_v(x) \qquad &\text{ if }w=x_i v \end{cases}$$
		$\Li_w$ is then extended linearly to all element in $\mathbb{Q}\langle x_0,x_1\rangle$.
		If $w = x_0^{s_1-1}x_1 \cdots x_0^{s_k-1}x_1 = x_0^{s_1-1}x_1 v\in \mathfrak{A}^1_1$, we have
		$$\Li_{s_1,\cdots,s_k}(x) = \Li_w(x) = \sum_{n_1>\cdots>n_k\geq 1} \frac{x^{n_1}}{n_1^{s_1}\cdots n_k^{s_k}} = \sum_{n=1}^\infty H_{n-1}(v) \frac{x^n}{n^{s_1}}$$
		For any $y\in \mathbb{Q}\langle x_0,x_1\rangle$, $\Li_y(1) = \text{\L}(y)$ with regularization if necessary.

	\end{example}

	\subsection{Some low level, low weight reduction}
	Recall our notation that $d(w,N) = \dim_\mathbb{Q} \textsf{CMZV}^N_w$. Let $$\widetilde{\textsf{CMZV}}^N_w = \frac{\textsf{CMZV}^N_w }{ \sum_{n<w} \textsf{CMZV}^N_n \textsf{CMZV}^N_{w-n}} $$
	Let $c(w,N)$ be defined by $$\sum_{w\geq 0} c(w,N) t^w = \log\left(1+\sum_{w=1}^\infty d(w,N) t^w \right)$$
	Assuming the algebra of CMZV of level $N$ forms a graded (with respect to weight) free algebra, then $$\tilde{d}(w,N) = \dim_\mathbb{Q} \widetilde{\textsf{CMZV}}^N_w = \sum_{k\mid w}\frac{\mu(k)}{k}c(\frac{w}{k},N)$$
	with $\mu$ the Möbius function. Moreover assume Deligne's bound is tight, then the following table gives $\tilde{d}(w,N)$ for small $(w,N)$:

	\begin{table}[h]
		\begin{tabular}{l|cccccccccccccccc}
			\cline{1-1}
			\multicolumn{1}{|l|}{$N \string\ w$} & \cellcolor[HTML]{C0C0C0}1 & \cellcolor[HTML]{C0C0C0}2 & \cellcolor[HTML]{C0C0C0}3 & \cellcolor[HTML]{C0C0C0}4 & \cellcolor[HTML]{C0C0C0}5 & \cellcolor[HTML]{C0C0C0}6 & \cellcolor[HTML]{C0C0C0}7 & \cellcolor[HTML]{C0C0C0}8 & \cellcolor[HTML]{C0C0C0}9 & \cellcolor[HTML]{C0C0C0}10 & \cellcolor[HTML]{C0C0C0}11 & \cellcolor[HTML]{C0C0C0}12 & \cellcolor[HTML]{C0C0C0}13 & \cellcolor[HTML]{C0C0C0}14 & \cellcolor[HTML]{C0C0C0}15 & \cellcolor[HTML]{C0C0C0}16 \\ \hline
			\cellcolor[HTML]{C0C0C0}1 & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{0} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{2} & \multicolumn{1}{c|}{2} & \multicolumn{1}{c|}{3} & \multicolumn{1}{c|}{3} & \multicolumn{1}{c|}{4} & \multicolumn{1}{c|}{5} \\ \cline{2-17}
			\cellcolor[HTML]{C0C0C0}2 & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{2} & \multicolumn{1}{c|}{2} & \multicolumn{1}{c|}{4} & \multicolumn{1}{c|}{5} & \multicolumn{1}{c|}{8} & \multicolumn{1}{c|}{11} & \multicolumn{1}{c|}{18} & \multicolumn{1}{c|}{25} & \multicolumn{1}{c|}{40} & \multicolumn{1}{c|}{58} & \multicolumn{1}{c|}{90} & \multicolumn{1}{c|}{135} \\ \cline{2-17}
			\cellcolor[HTML]{C0C0C0}3, 4 & \multicolumn{1}{c|}{2} & \multicolumn{1}{c|}{1} & \multicolumn{1}{c|}{2} & \multicolumn{1}{c|}{3} & \multicolumn{1}{c|}{6} & \multicolumn{1}{c|}{9} & \multicolumn{1}{c|}{18} & \multicolumn{1}{c|}{30} & \multicolumn{1}{c|}{56} & \multicolumn{1}{c|}{99} & \multicolumn{1}{c|}{186} & \multicolumn{1}{c|}{335} & \multicolumn{1}{c|}{630} & \multicolumn{1}{c|}{1161} & \multicolumn{1}{c|}{2182} & \multicolumn{1}{c|}{4080} \\ \cline{2-17}
			\cellcolor[HTML]{C0C0C0}5 & \multicolumn{1}{c|}{3} & \multicolumn{1}{c|}{3} & \multicolumn{1}{c|}{8} & \multicolumn{1}{c|}{18} & \multicolumn{1}{c|}{48} & \multicolumn{1}{c|}{116} & \multicolumn{1}{c|}{312} & \multicolumn{1}{c|}{810} & \multicolumn{1}{c|}{2184} & \multicolumn{1}{c|}{5880} & \multicolumn{6}{c|}{$>10^4$} \\ \cline{2-17}
			\cellcolor[HTML]{C0C0C0}6 & \multicolumn{1}{c|}{3} & \multicolumn{1}{c|}{2} & \multicolumn{1}{c|}{5} & \multicolumn{1}{c|}{10} & \multicolumn{1}{c|}{24} & \multicolumn{1}{c|}{50} & \multicolumn{1}{c|}{120} & \multicolumn{1}{c|}{270} & \multicolumn{1}{c|}{640} & \multicolumn{1}{c|}{1500} & \multicolumn{1}{c|}{3600} & \multicolumn{1}{c|}{8610} & \multicolumn{4}{c|}{$>10^4$} \\ \cline{2-17}
		\end{tabular}
		\caption{\small{A table of $\tilde{d}(w,N)$, which can be interpreted as number of "primitive constants" of weight $w$ and level $N$. (Assuming CMZVs form a free graded algebra and Deligne's bound is tight)}. \scriptsize{They are OEIS A113788, A006206, A001037, A027376 and A072337 respectively.}}
		\label{primitiveconstants}
	\end{table}
	In the Mathematica package \textit{MultipleZetaValues} written by the author (see Appendix A), complete reduction of the following weights and levels are stored:
	\begin{itemize}
		\item Weight $\leq 16$ at level 1
		\item Weight $\leq 8$ at level 2
		\item Weight $\leq 5$ at level 4
	\end{itemize}
	check Appendix A to see the chosen basis of $\widetilde{\textsf{CMZV}}^N_w$ that is used to store these results. Such compilation of values is important when we use them to evaluate certain definite integrals and infinite series.

	\section{One dimensional definite integral}
	Before going into the full algorithm, we first do some examples.

	\subsection{Some examples}
	\begin{example}
		We first do a baby example: evaluate $$I=\int_0^1 \frac{\Li_2(x) \log(1-x)}{x} dx$$
		using notations in Section 2.2, we have $\Li_2(x) = \int_0^x ab_0, \log(1-x) = -\int_0^x b_0$, so
		$$I = -\int_0^1 (\frac{1}{x}\int_0^x ab_0 \shuffle b_0) = -\int_0^1 a(ab_0 \shuffle b_0) = -\int_0^1 (2a^2b_0^2 + ab_0ab_0)
		$$
		the integral can be converted into colored polylogarithm via (\ref{toitint}), since the level in this case is $1$, so they can be converted into multiple zeta function. We have $I = -2\zeta(3,1)-\zeta(2,2) = -\frac{\pi^4}{72}$.
	\end{example}

	\begin{example}
		Let level $N=2$, adopt notations in Section 2.2. Let $u$ be $a=\frac{dx}{x}$ or $b_0 = \frac{dx}{1-x}$ or $b_1 = \frac{-dx}{1+x}$, consider $$I = \int_0^1 \log^n x \log^m(1-x) \log^p(1+x) u$$
		Note that $$\log(1-x)^m = (-1)^m m! \int_0^x b_0^m \qquad \log(1+x)^p = (-1)^p p! \int_0^x b_1^p \qquad \log^n x = (-1)^n n! \int_x^1 a^n$$
		therefore
		$$I = (-1)^{n+m+p}m!n!p! \int_0^1 u\int_x^1 a^n \int_0^x b_0^m \shuffle b_1^p = (-1)^{n+m+p}m!n!p! \int_0^1 a^n u (b_0^m \shuffle b_1^p)$$
		the RHS can be converted directly into CMZV of level $2$. For example,
		$$\begin{aligned}\int_0^1 \frac{\log x \log(1-x) \log(1+x)}{x} dx &= -\int_0^1 a^2 (b_0\shuffle b_1) = -\int_0^1 a^2b_0b_1 + a^2b_1b_0 \\
			&= -\int_0^1 z_{3,0}z_{1,1} + z_{3,1}z_{1,0} \\
			&= -L_{3,1}(1,-1) -L_{3,1}(-1,-1) = -\zeta(3,\bar{1}) -\zeta(\bar{3},\bar{1})\\
			&= 2 \text{Li}_4\left(\frac{1}{2}\right)+\frac{7}{4} \zeta (3) \log (2)-\frac{3 \pi ^4}{160}+\frac{\log ^4(2)}{12}-\frac{1}{12} \pi ^2 \log ^2(2)
		\end{aligned}$$
		Integrals involving on $\log x, \log(1+x), \log(1-x)$ are subjected to intensive investigation in \cite{Au}, some information for weight $\leq 20$ are recorded there.
	\end{example}

	\begin{example}
		This example illustrates the principal of regularization as indicated in Section 2.3. Consider the integral (see the Remark in Section 2.3)
		$$I = \int_0^1 \frac{\Li_3(x)}{1-x} dx$$
		Theorem \ref{regularization} enables us to do the following manipulation:
		$$I = \int_0^1 \frac{1}{1-x}\int_0^x a^2 b_0 = \int_0^1 b_0 a^2 b_0 = \text{\L}(b_0 a^2 b_0) = -\text{\L}(ab_0ab_0) - 2\text{\L}(a^2 b_0^2) = -\zeta(2,2)-2\zeta(3,1)$$
		so $I = -\frac{\pi^4}{72}$, it is the value of the following (convergent) integral:
		$$\int_0^1 \frac{\Li_3(x) - \zeta(3)}{1-x}dx$$
	\end{example}

	\subsection{Admissible rational function}
	We introduce the following definition. Let $N$ be a positive integer, we say that a rational function $R(x)\in \mathbb{C}(x)$ is \textit{$N$-admissible} if both $R(x)$ and $1-R(x)$ are of the form ($\mu = \exp(2\pi i /N)$)
	$$C x^d \prod_{i=0}^{N-1} (x-\mu^i)^{c_i} \qquad C\in \mathbb{C}\qquad  d,c_i\in \mathbb{Z}$$

	The following conjecture is very plausible, but author's limited knowledge leads no rigorous proof (For $N=1$ this is easy).

	\begin{conjecture}
		For positive integer $N$, the number of $N$-admissible rational functions is finite.
	\end{conjecture}

	A large list of $4$-admissible rational function can be found in Appendix B. The list there \textit{has not been proved to be complete} by the author. \par

	Recall the generalized polylogarithm defined in Example 2.5.

	\begin{theorem}\label{Liitint}  Let $\mathfrak{A} = \mathbb{C}\langle x_0,x_1\rangle$. For $N$-admissible $R(x)$, $w\in \{x_0,x_1\}^\ast$ of weight $n$\footnote{that is, number of $x_0$ plus number of $x_1$ equals $n$}, $c = R(0)$, we have the following iterated integral representation:
		\begin{equation}\tag{*}\Li_w(R(x)) = \sum_{k=1}^n c_k \int_0^x f_n \cdots f_k + \int_0^x f_n f_{n-1} \cdots f_1\end{equation}
		where $f_i = (R'/R) dx$ or $R'/(1-R) dx$ are differential forms, depending on whether the $i$-th letter (from the left) of $w$ is $x_0$ or $x_1$ respectively, and $c_k = \int_0^c f_k \cdots f_1$.\footnote{By principal of regularization in Section 2.3, if $c=\infty$, then invoke the asymptotic expansion of $\Li_w$ at infinity, ignore term of logarithmic growth keep just the constant term.}

		Therefore for any $N$-admissible $R(x)$, any $w\in \mathfrak{A}$, $\Li_w(R(x)) \in \int_0^x y $ for some $y\in \mathbb{C}\langle a,b_0,\cdots,b_{N-1}\rangle$.
	\end{theorem}
	\begin{proof}
		The proof is not difficult. Proceed by induction, the case $n=1$ is evident. Note that (*) is true when $x=0$, thus it suffices to prove both sides are equal after differentiation, this is true by induction hypothesis.
	\end{proof}

	\begin{example}
		We compute $$I = \int_0^1 \frac{\Li_2(-\frac{1}{x}) \Li_2(\frac{4x}{(1+x)^2})}{x} dx$$
		since both $-1/x, 4x/(1+x)^2$ are $2$-admissible, let the level $N=2$, then in notation of Section 2.2, $a=dx/x, b_0 = dx/(1-x), b_1 = -dx/(1+x)$.
		Because $$\Li_2(-1/x) = -\frac{\pi^2}{6} - \frac{1}{2}\log^2 x + o(1) \qquad x\to 0^+$$ using above theorem we have the (regularized) iterated integral (with $c_2 = -\pi^2/6, c_1 = 0, f_2 = -a, f_1 = a+b_1$):
		$$\Li_2(-1/x) = -\frac{\pi^2}{6} - \int_0^x a(a+b_1)$$
		similarly, $$\Li_2(\frac{4x}{(1+x)^2}) = \int_0^x (a+2b_1)(2b_0-2b_1)$$
		The shuffle product of $-\frac{\pi^2}{6} - a(a+b_1)$ and $(a+2b_1)(2b_0-2b_1)$ equals
		\scriptsize{ \begin{multline*}-\frac{1}{3} \pi ^2 ab_0+\frac{1}{3} \pi ^2 ab_1-6 a^3b_0+6 a^3b_1-4 a^2b_0a-4 a^2b_0b_1+4 a^2b_1a-8 a^2b_1b_0+12 a^2b_1b_1 -2 ab_0a^2\\-2 ab_0ab_1+2 ab_1a^2-6 ab_1ab_0+8 ab_1ab_1-4 ab_1b_0a-4 ab_1b_0b_1+4 ab_1b_1a-8 ab_1^2b_0+12 ab_1^3-4 b_1a^2b_0+4 b_1a^2b_1- 4 b_1ab_0a\\-4 b_1ab_0b_1+4 b_1ab_1a-4 b_1ab_1b_0+8 b_1ab_1^2-4 b_1b_0a^2-4 b_1b_0ab_1+4 b_1^2a^2+4 b_1^2ab_1-\frac{2}{3} \pi ^2 b_1b_0+\frac{2}{3} \pi ^2 b_1^2
		\end{multline*} }
		\normalsize{Concatenate this with $a = dx/x$ at the front, then convert it into $\text{\L}$, some terms requires (\ref{reg}) to convert to alternating multiple zeta function. Plug in all the level 2  CMZVs that pops up, we then obtain}
		\small{$$I = 24 \text{Li}_5\left(\frac{1}{2}\right)+16 \text{Li}_4\left(\frac{1}{2}\right) \log (2)+\frac{\pi ^2 \zeta (3)}{2}-\frac{93 \zeta (5)}{4}+\frac{7}{2} \zeta (3) \log ^2(2)+\frac{7 \log ^5(2)}{15}-\frac{1}{3} \pi ^2 \log ^3(2)-\frac{41}{360} \pi ^4 \log (2)$$}
	\end{example}
	~\\[0.02in]

	For positive integer $N$, let
	$$\mathcal{C}^N = \{R(0) | R(x)\text{ is }N\text{-admissible}\}$$
	then it is not difficult to show $\mathcal{C}^1 = \{0,1,\infty\}$. $\mathcal{C}^N$ for other $N$ is not known with certainty to the author, but very likely they are
	\begin{equation}\tag{**}\mathcal{C}^2 \overset{?}{=} \{0,1,\frac{1}{2},2,\infty\} \qquad \mathcal{C}^4 \overset{?}{=} \{0,1,\frac{1}{2},2,\pm i, 1\pm i, \frac{1\pm i}{2}, \infty\}\end{equation}

	Although the set of $N$-admissible rational function remains elusive, the author has faith in the following conjecture:
	\begin{conjecture}
		Let $w\in \{x_0,x_1\}^\ast$ be a weight $n$ word, then for any $x\in \mathcal{C}^N$, $\Li_w(x) \in \textsf{CMZV}^N_w$.
	\end{conjecture}

	The conjecture is known for any word $w$ and values of $x$ in (**). Its intractability mainly comes from the possible incompleteness of values listed in (**). Assuming this conjecture, we have the main result of this paper:

	\begin{theorem}\label{intmaintheo}
		Let $R_i(x)$ be $N$-admissible rational functions, $f(x) = 1/(x-d)$, with $d=0$ or an $N$-th root of unity, $w_i\in \{x_0,x_1\}^\ast$, then
		$$\int_0^1 f(x) \prod_i \Li_{w_i}(R_i(x)) dx \in \textsf{CMZV}^N_{1+\sum |w_i|}$$
		with $|w_i|$ the weight of $w_i$.
	\end{theorem}
	\begin{proof}
		All aspects of the algorithm has been illustrated in Example 3.6.
	\end{proof}

	\begin{remark}
		Since $\Li_w$ is multi-valued, and $R_i(x)$ might wind around branch point more than one complete cycle as $x$ varies from $0$ to $1$, the above assertion is best interpreted by \textit{thinking of} $\Li_w(R(x))$ as RHS of $(*)$. If $R_i$ all satisfy $R_i([0,1])\subset [0,1]$, then such issue do not arise.
	\end{remark}

	\section{Infinite sums that are reducible to CMZVs}
	\subsection{Apéry-like series involving harmonic numbers}
	Apart from classical Euler sums (for example, in \cite{Flajolet}, \cite{Borwein3}), which are directly reducible to CMZV, we state a few more which follows nicely from our main theorem Theorem \ref{intmaintheo}. Recall the harmonic number $H_n(w)$ defined in Example 2.2.
	\begin{theorem}
		For any word $w\in \mathfrak{A}_1$\footnote{i.e. algebra generated by words ending in $x_1$}, positive integer $s$
		$$\sum_{n=1}^\infty \frac{H_{n-1}(w)}{n^s} \left[ {{4^n}{\binom{2n}{n}^{ - 1}}} \right] \in \textsf{CMZV}^2_{s+|w|} \qquad s>1$$
		$$\sum_{n=1}^\infty \frac{H_{n-1}(w)}{n^s} \left[ {{2^n}{\binom{2n}{n}^{ - 1}}} \right] \in \textsf{CMZV}^4_{s+|w|}$$
	\end{theorem}
	\begin{proof}
		Consider $$\int_0^1 \frac{1}{x}\Li_w(R(x))dx \qquad R(x) = \frac{4x}{(1+x)^2} \text{ or } \frac{2x}{(1+x)^2}$$
		the first $R(x)$ is $2$-admissible, the second $R(x)$ is $4$-admissible. Using series expansion of $\Li_w(x)$ (see Example 2.5), integrate termwise, and use
		$$\int_0^1 \frac{x^{n-1}}{(1+x)^{2n}} dx = \frac{1}{n}\binom{2n}{n}^{-1}$$
		completes the proof.
	\end{proof}

	\begin{example}
		$$\begin{aligned}
			\sum\limits_{n = 2}^\infty  {\frac{{H_{n - 1}^{(2)}}}{{{n^3}}}\left[ {{4^n}{\binom{2n}{n}^{ - 1}}} \right]}  &= \frac{1}{{12}}{\pi ^4}\log 2 - \frac{3}{4}{\pi ^2}\zeta (3) + \frac{{31}}{8}\zeta (5) \\
			\sum\limits_{n = 2}^\infty  {\frac{{H_{n - 1}}}{{{n^4}}}\left[ {{4^n}{\binom{2n}{n}^{ - 1}}} \right]}  &= 32 \text{Li}_5\left(\frac{1}{2}\right)-\frac{\pi ^2 \zeta (3)}{2}-\frac{155 \zeta (5)}{8}-\frac{1}{15} 4 \log ^5(2)+\frac{4}{9} \pi ^2 \log ^3(2)+\frac{23}{180} \pi ^4 \log (2) \\
			\sum\limits_{n = 2}^\infty  {\frac{{H_{n - 1}^{(3)}}}{{{n^2}}}\left[ {{4^n}{\binom{2n}{n}^{ - 1}}} \right]}  &= \frac{93 \zeta (5)}{4}-\frac{7 \pi ^2 \zeta (3)}{4} \qquad \qquad
			\sum\limits_{n = 2}^\infty  {\frac{{H_{n - 1}^{(2)}}}{{{n^2}}}\left[ {{2^n}{\binom{2n}{n}^{ - 1}}} \right]} = \frac{\pi ^4}{384} \\
			\sum\limits_{n = 2}^\infty  {\frac{{H_{n - 1}^{(2)}}}{{{n^3}}}\left[ {{2^n}{\binom{2n}{n}^{ - 1}}} \right]} &= \frac{\pi ^3 C}{24}-\pi  \beta(4)-\frac{3 \pi ^2 \zeta (3)}{128}+\frac{527 \zeta (5)}{256}+\frac{1}{384} \pi ^4 \log (2) \\
			\sum\limits_{n = 2}^\infty  {\frac{{H_{n - 1}^{(3)}}}{{{n^2}}}\left[ {{2^n}{\binom{2n}{n}^{ - 1}}} \right]} &= -3 \pi  \beta(4)-\frac{35 \pi ^2 \zeta (3)}{128}+\frac{1581 \zeta (5)}{128}
		\end{aligned}$$
		here $C$ is the Catalan constant, $\beta$ is Dirichlet beta function.
	\end{example}

	In order to derive more interesting examples, we need an integral operator that gives out generalized harmonic numbers $H_n(w)$. Recall our notation of a variation of iterated integral defined in last part of Section 2.1, for positive integer $s$, set
	$$\chi_s = \begin{cases}\overline{x_1} &\qquad \text{ if }s=1 \\
		\overline{x_0}x_1 &\qquad \text{ if }s=2 \\
		\overline{x_0}x_0^{s-2}x_1 &\qquad \text{ if }s\geq 3
	\end{cases}$$
	Writing $x_0 = dx/x, x_1 = dx/(1-x)$, for positive integers $s_i$, define the linear operator $D_{s_1,\cdots,s_k}$:
	$$D_{s_1,\cdots,s_k} f = \int_0^1 x_0^{s_1-1}x_1\chi_{s_2}\cdots \chi_{s_k} \int_x^1 f(x) dx $$

	\begin{lemma}
		For positive integer $n$, write $w = x_0^{s_k-1}x_1\cdots x_0^{s_1-1}x_1$
		\begin{equation}\tag{**}D_{s_1,\cdots, s_k} (nx^{n-1}) = H_n^\star (w) \end{equation}
		where (note that the order of $s_1,\cdots,s_k$ in $w$ has been reversed), $$H_n^\star(w) = \sum_{n\geq n_k\geq \cdots\geq n_1\geq 1} \frac{1}{n_1^{s_k}\cdots n_k^{s_1}}$$
	\end{lemma}
	\begin{proof}
		Proceed by induction on $k$, the case for $k=1$ is evident:
		$$D_{s}(nx^{n-1}) = \int_0^1 \frac{1}{x_1}\int_0^{x_1} \frac{1}{x_2} \cdots \int_0^{x_{s-1}} \frac{1-x_s^n}{1-x_s} dx_s = \sum_{i=1}^n \frac{1}{i^s}$$
		Now one easily computes, via induction hypothesis:
		$$D_{s_1,\cdots, s_k} (nx^{n-1} - (n-1) x^{n-2}) = \frac{1}{n^{s_k}} \sum_{n\geq n_{k-1}\geq \cdots\geq n_1\geq 1} \frac{1}{n_1^{s_{k-1}}\cdots n_{k-1}^{s_1}} \quad n\geq 1$$
		Because $(**)$ is true when $n=0$, the above forward difference easily implies the result.
	\end{proof}

	Now consider a holomorphic function $f(z)$ on $|z|<1$ defined by the power series:
	$f(z) = \sum_{n=1}^\infty a_n z^n$. We shall assume $0<z<1$, so ($x_0 = dz/z$):
	$$\sum_{n=1}^\infty \frac{a_n}{n^{s-1}}z^n  = \int_0^z x_0^{s-2} \frac{f(z) dz}{z}$$
	taking the operator $D_{s_1,\cdots,s_k}$ on both sides (this is legitimate via dominated convergence theorem, as long as LHS converges), we have
	$$\sum_{n=1}^\infty \frac{a_n}{n^s} H_n^\star(w) = D_{s_1,\cdots,s_k} \left(\frac{1}{z}\int_0^z x_0^{s-2} \frac{f(z) dz}{z}\right)\qquad w = x_0^{s_k-1}x_1\cdots x_0^{s_1-1}x_1$$
	If $s=1$, then term inside the parenthesis should be interpreted as $\frac{f(z)}{z}$. Therefore, by switching $\int_x^1$ that occurs in $\chi_s$ back to $\int_0^1$, we see that $\sum_{n=1}^\infty \frac{a_n}{n^s} H_n(w)$ is always a $\mathbb{Z}$-linear combination of terms of form $$\left(\int_0^1 \omega_1\cdots \omega_{i_1}\right)\cdots \left(\int_0^1 \omega_{i_{k-1}+1}\cdots \omega_{i_k}\right)$$
	with $i_{k} = s+s_1+\cdots+s_k = |w|+s$. ($|w|$ is the weight of word $w$) The last differential form $\omega_{|w|+s}$ is $f(z)/z dz$, and all previous $\omega_i$ are either $dz/z$ or $dz/(1-z)$.

	\begin{theorem}
		For any word $w\in \mathfrak{A}_1$, positive integer $s$, coprime integers $p<0, q>0$, $p+q\geq 1$. We have $$\sum_{n=1}^\infty \frac{H_{n-1}(w)}{n^s} (-1)^n \binom{p/q}{n} \in \textsf{CMZV}^q_{s+|w|}$$
	\end{theorem}
	\begin{proof}
		Since $H_{n-1}(w)$ can be written as a linear combination of $n^{-m_i}H_n^\star(w_i)$, with $m_i+|w_i| = |w|$. It suffices to the assertion for $H_{n}^\star(w)$. By our observation above, the series is a $\mathbb{Z}$-linear combination of $$\left(\int_0^1 \omega_1\cdots \omega_{i_1}\right)\cdots \left(\int_0^1 \omega_{i_{k-1}+1}\cdots \omega_{i_k}\right)$$
		with $i_k = s+|w|, \omega_{|w|+s} = ((1-x)^{p/q}-1)/x dx$, and all other $\omega_i$ are either $dx/x$ or $dx/(1-x)$. All terms except the last one in the above displayed equation are level $1$ CMZVs. For the last iterated integral involving $\omega_{|w|+s}$, pull it back by $g: x\to 1-x^q$ using (\ref{itintpullback}), then the path of integration is still $[0,1]$ (direction reversed), and $$g^\ast \frac{dx}{x} = \frac{-qx^{q-1}}{1-x^q}dx \qquad g^\ast \frac{dx}{1-x} = \frac{-q}{x} dx\qquad  g^\ast \omega_{|w|+s} = \frac{qx^{q-1}-qx^{p+q-1}}{1-x^q}dx$$
		since $0\leq p+q-1< q$, all above differential forms can be converted into linear combination of $dx/(1-\exp(2\pi i k /q)), k=0,1,\cdots,q-1$, so the last iterated integral can be converted into level $q$ CMZV, the completes the proof.
	\end{proof}

	\begin{corollary}
		For any word $w\in \mathfrak{A}_1$, positive integer $s$
		$$\sum_{n=1}^\infty \frac{H_{n-1}(w)}{n^s} \left[ 4^{-n} \binom{2n}{n} \right] \in \textsf{CMZV}^2_{s+|w|}$$
	\end{corollary}
	\begin{proof}
		Apply above theorem to $p=-1, q=2$ and use $\binom{-1/2}{n}(-1)^n = 4^{-n}\binom{2n}{n}$.
	\end{proof}

	\begin{example}
		$$\begin{aligned}\sum_{n=2}^\infty \frac{H_{n-1}}{n^3} \left[ 4^{-n} \binom{2n}{n} \right] &= 8 \text{Li}_4\left(\frac{1}{2}\right)+2 \zeta (3) \log (2)-\frac{11 \pi ^4}{180}+\log ^4(2) \\
			\sum_{n=2}^\infty \frac{H_{n-1}^{(2)}}{n^2} \left[ 4^{-n} \binom{2n}{n} \right] &= \zeta (3) \log (2)+\frac{\pi ^4}{120}+\frac{2 \log ^4(2)}{3}-\frac{1}{3} \pi ^2 \log ^2(2) \\
			\sum_{n=2}^\infty \frac{H_{n-1}^{(3)}}{n} \left[ 4^{-n} \binom{2n}{n} \right] &= -8 \text{Li}_4\left(\frac{1}{2}\right)-3 \zeta (3) \log (2)+\frac{7 \pi ^4}{90}+\frac{\log ^4(2)}{3}
		\end{aligned}$$
	\end{example}

	Corollary 4.5 was already obtained by Wang and Xu in \cite{Xu}, a Maple package to calculate such series was also written. More computational approaches can be found in \cite{phys1}, \cite{phys2}, \cite{phys3} and \cite{Boyadzhiev}. First half of Theorem 4.1 and (a weaker version of) Corollary 4.5 (in terms of ordinary harmonic numbers) are recently and independently proved by Zhao \cite{Zhao2}, the idea of iterated integral is already germinating in this paper. Some special Apéry-like series was also intensively investigated via WZ-method in \cite{Borwein4}, \cite{Pilehrood}.

	\subsection{Fourier-Legendre expansion of generalized polylogarithm}
	Let $P_n(x) = \frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n$ be the classical Legendre polynomials, we will focus on the shifted version $\tilde{P}_n(x) = P_n(2x-1)$. $\{\tilde{P}_n(x)\}$ forms a complete orthogonal family on $L^2(0,1)$, with
	$$\int_0^1 {{{\widetilde P}_n}(x){{\widetilde P}_m}(x)dx}  = \frac{{{\delta _{mn}}}}{{2n + 1}}$$
	For $f\in L^2(0,1)$, we will write $f\sim \sum\limits_{n \ge 0} {{c_n}{{\widetilde P}_n}(x)} $ to represent the expansion of $f$ in terms of $\tilde{P}_n(x)$. The expansion converges to $f(x)$ in $L^2$ norm (\cite{Askey}), we don't need results about pointwise convergence.

	\begin{proposition}
		Let $f\in L^2(0,1), f(x)\sim\sum\limits_{n \ge 0} {{c_n}{{\widetilde P}_n}(x)}$. Then $$f(1 - x)\sim \sum\limits_{n \ge 0} {{{( - 1)}^n}{c_n}{{\widetilde P}_n}(x)} \qquad \int_0^x {f(x)dx} \sim \sum\limits_{n \ge 0} {\left[ {\frac{{{c_{n - 1}}}}{{2(2n - 1)}} - \frac{{{c_{n + 1}}}}{{2(2n + 3)}}} \right]{{\widetilde P}_n}(x)} $$
		If $f(x)/(1-x) \in L^2(0,1)$, then $$\frac{{f(x)}}{{1 - x}}\sim \sum\limits_{n \ge 0} {(2n + 1)\left( {\int_0^1 {\frac{{f(x)}}{{1 - x}}dx}  - 2\sum\limits_{m = 1}^n {\frac{1}{m}\sum\limits_{k = 0}^{m - 1} {{c_k}} } } \right)} {\widetilde P_n}(x)$$
		If $f(x)/x \in L^2(0,1)$, then $$\frac{{f(x)}}{x}\sim \sum\limits_{n \ge 0} {{{( - 1)}^n}(2n + 1)\left( {\int_0^1 {\frac{{f(x)}}{x}dx}  - 2\sum\limits_{m = 1}^n {\frac{1}{m}\sum\limits_{k = 0}^{m - 1} {{{( - 1)}^k}{c_k}} } } \right)} {\widetilde P_n}(x)$$
	\end{proposition}
	\begin{proof}
		The expansion about $f(1-x)$ follows from $P_n(x) = (-1)^n P_n(-x)$, that of $\int_0^x f(x) dx$ follows from $\int {{{\widetilde P}_n}(x)dx}  = \frac{{{{\widetilde P}_{n + 1}}(x) - {{\widetilde P}_{n - 1}}(x)}}{{2(2n + 1)}}$. The last assertion about $f(x)/x$ follows from that of $f(x)/(1-x)$. Therefore it remains to prove the expansion of $f(x)/(1-x)$, we include a quick proof due to lack of reference. Let $F_{n+1}(x) = (n+1)[\tilde{P}_{n+1}(x)-\tilde{P}_n(x)]$, then the three-term recurrence of $\tilde{P}_n$ implies
		$$\int_0^1 \frac{f(x)}{1-x}(F_{n+1}-F_n(x)) dx = -2c_n$$
		telescoping, after that divide by $n+1$, another telescoping gives the result.
	\end{proof}

	Recall our notation about Hoffman-Racinet algebra $z_{k,i} = a^{k-1}b_i$. For $w = z_{k_1,i_1}\cdots z_{k_2,i_2}\cdots \in \mathfrak{A}^2_1$, define an $\mathbb{C}$-linear map $\theta_i: \mathfrak{A}^2_1 \to \mathfrak{A}^2_1$ ($i=0, 1$) by $\theta_i(w) = {z_{{k_1} + 1,{i_1} + i}}{z_{{k_2},{i_2}}}\cdots$.
	\begin{proposition}
		Let $w\in \mathfrak{A}^2_1$, $f(x)\in L^2(0,1)$. If
		$f(x)\sim c + \sum\limits_{n \ge 1} {(2n + 1) H_n^\star (w)} {\widetilde P_n}(x)$,
		then $$\frac{1}{x}\int_0^x {f(x)dx} \sim \sum\limits_{n \ge 0} {(2n + 1){{( - 1)}^n}H_n^\star (C - {\theta _1}w)} {\widetilde P_n}(x) \qquad C = \int_0^1 \frac{1}{x}\int_0^x {f(t)dt}$$
		$$\frac{1}{{1 - x}}\int_x^1 {f(x)dx} \sim \sum\limits_{n \ge 0} {(2n + 1)H_n^\star (C + {\theta _0}w - 2{b_0}w){{\widetilde P}_n}(x)} \qquad C = \int_0^1 \frac{1}{1-x}\int_x^1 {f(t)dt}$$
		If
		$f(x)\sim c + \sum\limits_{n \ge 1} {(2n + 1) (-1)^n H_n^\star (w)} {\widetilde P_n}(x)$,
		then $$\frac{1}{x}\int_0^x {f(x)dx} \sim \sum\limits_{n \ge 0} {(2n + 1){{( - 1)}^n}H_n^\star ({C + {\theta _0}w - 2{b_0}w})} {\widetilde P_n}(x) \qquad C = \int_0^1 \frac{1}{x}\int_0^x {f(t)dt}$$
		$$\frac{1}{{1 - x}}\int_x^1 {f(x)dx} \sim \sum\limits_{n \ge 0} {(2n + 1)H_n^\star ({C - {\theta _1}w}){{\widetilde P}_n}(x)} \qquad C = \int_0^1 \frac{1}{1-x}\int_x^1 {f(t)dt}$$
	\end{proposition}
	\begin{proof}
		Immediately follows from the previous proposition.
	\end{proof}

	\begin{proposition}
		Let $w\in \{x_0,x_1\}^\ast$ that contains $x_1$. Then there exists an $w_0\in \mathfrak{A}^2_1$ so that
		$$\frac{\Li_w(x)}{x} \sim \sum_{n\geq 0}(2n+1) (-1)^n H_n^\ast(w_0) \tilde{P}_n(x)$$
	\end{proposition}
	\begin{proof}
		Use induction on weight of $w$. For $w=x_1$, we have
		$$\frac{-\log(1-x)}{x} dx \sim \sum_{n\geq 0}(2n+1) (-1)^n \left[\frac{\pi^2}{6}+2\sum_{k=1}^n \frac{(-1)^k}{k^2}\right] \tilde{P}_n(x)$$
		For $w$ of weight $n$, applying above proposition repeatedly shows that $\frac{1}{x}\int_0^x u v_1 \cdots v_{n-1}$, with $u = x_0 = dx/x$ or $x_1 = dx/(1-x)$, $v=x_0,x_1,\overline{x_0}$ or $\overline{x_1}$ (see last paragraph of Section 2.1) has FL-expansion of desired form. Convert $\int_x^1$ into $\int_0^1 - \int_0^x$, induction hypothesis completes the proof.
	\end{proof}

	\begin{example}
		For each $w$ below, we give\footnote{We abbreviate $\sum_{n\geq n_1>n_2>n_3\geq 1}\frac{(-1)^{n_3}}{n_1 n_2 n_3^2}$ as $H_{1,1,-2}$, and similarly for other harmonic numbers.} $c_n$ in $\Li_w(x)/x \sim \sum_{n\geq 0} (2n+1) (-1)^n c_n \tilde{P}_n(x)$:
		$$w = x_0^2 x_1\qquad c_n = \frac{\pi^4}{90} + \frac{\pi^2}{3}H_2 + 2H_{-4}+\frac{2\pi^2}{3}H_{1,1}+4H_{1,-3}+4H_{2,-2}+8H_{1,1,-2}+\zeta(3)H_1$$
		$$w = x_0^2x_1x_0\qquad c_n = \frac{\pi^4}{15}H_1 + 2H_5+4H_{1,4}+4H_{2,3}+8H_{1,1,3}-4\zeta(3)H_{2}-8\zeta(3)H_{1,1}-4\zeta(5)$$
		\begin{multline*}w=x_0^3x_1\qquad c_n =  -\frac{\pi^4}{45}H_1-\frac{\pi^2}{3}H_3-2H_{-5}-\frac{2\pi^2}{3}H_{1,2}-4H_{1,-4}-\frac{2\pi^2}{3}H_{2,1}-4H_{2,-3}-4H_{3,-2}-\frac{4\pi^2}{3}H_{1,1,1}\\ -8H_{1,1,-3}-8H_{1,2,-2}-8H_{1,1,-1}-16H_{1,1,1,-2}+2\zeta(3)H_2+4\zeta(3)H_{1,1}+\zeta(5)\end{multline*}
		Note that last entry records the Fourier-Legendre expansion of $\Li_4(x)/x$.
	\end{example}

	Our main goal of introducing Legendre polynomial is to prove the following theorem. Let $K(x) = \frac{\pi}{2} {_2F_1}(\frac{1}{2},\frac{1}{2};1;x)$ be the complete elliptic integral of first kind.
	\begin{theorem}
		Let $w\in \{x_0,x_1\}^\ast$ that contains $x_1$ of weight $|w|$, then (both integrals converge)
		$$\int_0^1 \frac{K(x) \Li_w(x)}{x} dx, \int_0^1 \frac{K(1-x)\Li_w(x)}{x}dx  \in \textsf{CMZV}^4_{|w|+2}$$
	\end{theorem}
	\begin{proof}
		The proof uses $K(x) \sim \sum_{n\geq 0}\frac{2}{2n+1} \tilde{P}_n(x)$ (see \cite{FL5}). Using expansion of $\Li_w(x)/x$ obtained above, the first integral becomes a linear combination of sums of form $\sum_{n\geq 0} (-1)^n H_n(w)/(2n+1)$, with $w\in \mathfrak{A}^2_1$ in level $2$ Hoffman-Racinet algebra. This is a convergent Euler sum, and can be converted into level $4$ CMZV. This completes the proof for $\int_0^1 \frac{K(x) \Li_w(x)}{x} dx$. For the second one, it reduces into a linear combination of $\sum_{n\geq 0} H_n(w)/(2n+1)$, which diverges. However, its regularized value ($c$ in the statement of Proposition 2.3) is still a level 4 CMZV. Although individual sums might diverge, the overall sum $\sum \frac{c_n}{2n+1}$ (see notations of Examples above) must converge, hence it is legitimate to replace divergent sum by its regularized value, completing the proof.
	\end{proof}

	\begin{example} For example,
		\small $$\begin{aligned}\int_0^1 \frac{K(x)\Li_2(x)}{x}dx &= -\frac{2 \pi ^2 C}{3}-512 L_4-128 \log (2) L_3+400 \beta(4)+\frac{4}{3} \pi  \log ^3(2)-3 \pi ^3 \log (2) \\
			\int_0^1 \frac{K(1-x)\Li_2(x)}{x}dx &= -32 \pi  L_3+32 \text{Li}_4\left(\frac{1}{2}\right)+\frac{41 \pi ^4}{90}+\frac{4 \log ^4(2)}{3}-\frac{1}{3} \pi ^2 \log ^2(2)\\
			\int_0^1 \frac{K(1-x)\log x \log(1-x)}{x} dx&= 512 L_4+128 \log (2) L_3-416 \beta(4)+7 \pi  \zeta (3)-\frac{4}{3} \pi  \log ^3(2)+3 \pi ^3 \log (2)
		\end{aligned}$$
		\normalsize with $L_n = \Im\left(\text{Li}_n\left(\frac{1}{2}+\frac{i}{2}\right)\right)$.
	\end{example}

	\begin{remark}
		Let $w$ be any word in $\{x_0,x_1\}^\ast$. Since $K(0)=\pi/2, K(x) = 2\log 2 -\log(1-x)/2 + o(1)$ as $x\to 1^-$, there exists positive integers $s_i$ and $a_i,c\in \mathbb{C}$ such that
		$$\int_\alpha^1 \frac{f(x) \Li_w(x)}{g(x)} dx = \sum a_i \log^{s_i} \alpha + c + o(1) \qquad \alpha\to 0^+$$
		with $f(x) = K(x)$ or $K(1-x)$, $g(x) = x$ or $1-x$. $c$ will be defined as the regularized value of the integral. Using machinery developed, it is not difficult to prove that such regularized value is also in $\textsf{CMZV}^4$.
	\end{remark}

	Our inspiration to work with Legendre polynomials was sparked from \cite{FL2}. However, the author still hopes to find a way to prove Theorem 4.11 via integral transformations only, this has two advantages. Firstly, we already have a well-developed regularization theory for shuffle CMZV (Theorem 2.4); but when working with Fourier-Legendre expansion, one has to take care of convergence everywhere due to lack of suitable regularization theory. Secondly, there are cases not included in Theorem 4.11 but nonetheless yield closed forms in CMZVs, see \cite{Wan}, it is hoped that by figuring out the integral transformations, we can differentiate whether a certain hypergeometric-type integral is related to CMZVs. \par
	Fourier-Legendre expansion utilized to series evaluations can also found in \cite{FL6}, \cite{Chan}.

	\subsection{Series involving binomial coefficient squared and harmonic numbers}
	We elucidate the nature of the following intractable sums:
	\begin{theorem}
		For any word $w\in \mathfrak{A}_1$, positive integer $s$
		$$\sum_{n=1}^\infty \frac{H_{n-1}(w)}{n^s} \left[ 4^{-n} \binom{2n}{n} \right]^2 \in \frac{1}{\pi}\textsf{CMZV}^4_{s+|w|+1}$$
	\end{theorem}
	\begin{proof}
		It suffices to prove the assertion for $H_n^\star(w)$. Set $a_n = \left[ 4^{-n} \binom{2n}{n} \right]^2$, by our discussion preceding Theorem 4.4, we have $\sum_{n=1}^\infty a_n x^n = \frac{2}{\pi}K(x) -1$, so
		$$\sum_{n=1}^\infty \frac{H_n^\star(w)}{n^s} a_n = \frac{1}{\pi} D_{s_1,\cdots,s_k} \left(\frac{1}{x}\int_0^x x_0^{s-2} \frac{(2K(x)-\pi) dx}{x}\right)\qquad w = x_0^{s_k-1}x_1\cdots x_0^{s_1-1}x_1$$
		Apart from the factor $1/\pi$, RHS is a $\mathbb{Z}$-linear combination of $$\left(\int_0^1 \omega_1\cdots \omega_{i_1}\right)\cdots \left(\int_0^1 \omega_{i_{k-1}+1}\cdots \omega_{i_k}\right)$$
		with $i_k = s+|w|, \omega_{|w|+s} = (2K(x)-1)/x dx$, and all other $\omega_i$ are either $dx/x$ or $dx/(1-x)$. All terms except the last one in the above displayed equation are level $1$ CMZVs. For the last iterated integral involving $\omega_{|w|+s}$, pull it back by $x\to 1-x$, gives, for some $w'\in \mathfrak{A}_1$, $$\int_0^1 \frac{(2K(1-x)-\pi) \Li_{w'}(x)}{1-x} dx = \int_0^1 \frac{(2K(x)-\pi)\Li_{w'}(1-x)}{x}dx $$
		Since $\Li_{w'}(1-x)$ can be written as $\sum c_i \Li_{w_i}(x)$ for some other words $w_i$ and $c_i\in \mathbb{R}$, weight preserved, Theorem 4.11 says the above (regularized) integral is a level 4 CMZV. Completing the proof.
	\end{proof}

	\begin{theorem}
		For any word $w\in \mathfrak{A}_1$, positive integer $s\geq 3$,\footnote{This condition is imposed to ensure convergence of the series.}
		$$\sum_{n=1}^\infty \frac{H_{n-1}(w)}{n^s} \left[ 4^{-n} \binom{2n}{n} \right]^{-2} \in \textsf{CMZV}^4_{s+|w|}$$
	\end{theorem}
	\begin{proof}
		Denote the series by $S$. Set $v=x_0^{s-3}x_1w$, then
		$$S = \iint_{(0,1)^2} \frac{\Li_v(16xy(1-x)(1-y))}{xy} dxdy = \iint_{(0,1/2)^2} \frac{\Li_v(16xy(1-x)(1-y))}{x(1-x)y(1-y)} dxdy$$
		replace $x$ by $(1-\sqrt{1-x})/2$, and $y$ by $(1-\sqrt{1-y})/2$, we have $$S = \iint_{{{(0,1)}^2}} {\frac{{{{{\mathop{\rm Li}\nolimits} }_v}(xy)}}{{xy\sqrt {1 - x} \sqrt {1 - y} }}dxdy} = \iint_{0 < x < y < 1} {\frac{{{{{\mathop{\rm Li}\nolimits} }_v}(x)}}{{x\sqrt y \sqrt {y - x} \sqrt {1 - y} }}dxdy} $$
		integrating respect to $y$ gives $S =2 \int_0^1 {\frac{{K(1 - x){{{\mathop{\rm Li}\nolimits} }_v}(x)}}{x}dx}$, which is in $\textsf{CMZV}^4_{s+|w|}$.
	\end{proof}

	\begin{example}
		\small
		$$\begin{aligned}\sum_{n=1}^\infty \frac{1}{n^4} \left[ 4^{-n} \binom{2n}{n} \right]^{-2}&= -64 \pi  L_3+64 \text{Li}_4\left(\frac{1}{2}\right)+\frac{41 \pi ^4}{45}+\frac{8 \log ^4(2)}{3}-\frac{2}{3} \pi ^2 \log ^2(2) \\
			\sum_{n=1}^\infty \frac{H_{n-1}}{n^3} \left[ 4^{-n} \binom{2n}{n} \right]^{-2} &= 32 C^2-32 \pi  C \log (2)-64 \pi  L_3+\frac{3 \pi ^4}{2}+2 \pi ^2 \log ^2(2) \\
			\sum_{n=1}^\infty \frac{H_n}{n^2} \left[ 4^{-n} \binom{2n}{n} \right]^{2}& = \frac{1024 L_4}{\pi }+\frac{256 \log (2) L_3}{\pi }-\frac{800 \beta(4)}{\pi }+\zeta (3)-\frac{8}{3} \log ^3(2)+\frac{20}{3} \pi ^2 \log (2) \\
			\sum_{n=1}^\infty \frac{H_n^{(2)}}{n} \left[ 4^{-n} \binom{2n}{n} \right]^{2}& = -\frac{4 \pi  C}{3}-\frac{32 \beta(4)}{\pi }+12 \zeta (3)
		\end{aligned}$$
		\normalsize with $L_n = \Im\left(\text{Li}_n\left(\frac{1}{2}+\frac{i}{2}\right)\right)$.
	\end{example}

	Series involving square central binomial coefficients have been intensively studied (via ingenious, elementary means) in \cite{FL1}, \cite{FL2}, \cite{FL3}, \cite{FL4}, \cite{FL7} and \cite{FL8}. A lots of closed-forms in these papers are essentially CMZVs. Virtually all of them use Fourier-Legendre techniques. One might be tempted to consider series involving cubed central binomial coefficients or even higher power, but they seem to leave the realm of CMZV and are more affiliated to critical $L$-values (\cite{Rogers}).

	\subsection{Rapidly converging series}
	For $x\in \mathbb{C}, x\notin (0,1)$, denote $\omega(c) = \frac{dx}{x-c}$.

	\begin{proposition}
		$$\int_0^1 \omega(c_1)\cdots \omega(c_n) = (-1)^n \int_0^1 \omega(1-c_n)\cdots \omega(1-c_1)$$
	\end{proposition}
	\begin{proof}
		Immediate from first rule of Proposition 2.1 and (\ref{itintpullback})
	\end{proof}

	\begin{theorem}
		For positive integer $s$,
		$$\sum_{n=1}^\infty \frac{(-1)^n}{n^s 2^n \binom{2n}{n}} \in \textsf{CMZV}^2_s$$
	\end{theorem}
	\begin{proof}
		The summation equals to $$I = \int_0^1 \frac{1}{x}\Li_{s-1}\left(-\frac{1}{2}x(1-x)\right) dx$$
		by expanding $\Li_{s-1}$ and termwise integration. Using method as in (\ref{Liitint}), we have
		$$\Li_{s-1}\left(-\frac{1}{2}x(1-x)\right) = \int_0^x (\omega(0)+\omega(1))^{s-2} (-\omega(-1) - \omega(2))$$
		so $I$ will be the iterated integral of a combination of $\omega(c_1)\cdots \omega(c_s)$, with $c_i \in \{0,1,-1,2\}$ with $-1,2$ \textit{never both occur in the same word}. If a word contains only $\omega(0),\omega(1),\omega(-1)$ then it is already a level 2 CMZV, if a word contains only $\omega(0),\omega(1),\omega(2)$, then above proposition transforms it to a level 2 CMZV, so $I \in \textsf{CMZV}^2_s$.
	\end{proof}

	\begin{example}
		$$\begin{aligned}
			\sum_{n=1}^\infty \frac{(-1)^n}{n^3 2^n \binom{2n}{n}} &= \frac{\log ^3(2)}{6}-\frac{\zeta (3)}{4} \\
			\sum_{n=1}^\infty \frac{(-1)^n}{n^4 2^n \binom{2n}{n}} &= -4 \text{Li}_4\left(\frac{1}{2}\right)-\frac{13}{4} \zeta (3) \log (2)+\frac{7 \pi ^4}{180}-\frac{1}{24} 5 \log ^4(2)+\frac{1}{6} \pi ^2 \log ^2(2) \\
			\sum_{n=1}^\infty \frac{(-1)^n}{n^5 2^n \binom{2n}{n}} &= -10 \text{Li}_5\left(\frac{1}{2}\right)-6 \text{Li}_4\left(\frac{1}{2}\right) \log (2)+\frac{19 \zeta (5)}{2}-\zeta (3) \log ^2(2)-\frac{1}{120} 19 \log ^5(2)\\ &\qquad +\frac{1}{9} \pi ^2 \log ^3(2)-\frac{7}{180} \pi ^4 \log (2)
		\end{aligned}$$
		The case of weight $6$ and $7$ respectively give
		\small \begin{multline*}\zeta(\overline{5},1) = 8 \text{Li}_6\left(\frac{1}{2}\right)+3 \text{Li}_5\left(\frac{1}{2}\right) \log (2)+\frac{S_6}{2}+\frac{\zeta (3)^2}{2}-\frac{1}{6} \zeta (3) \log ^3(2)+\frac{19}{4} \zeta (5) \log (2)-\frac{\pi ^6}{112}\\ -\frac{19 \log ^6(2)}{1440}+\frac{1}{72} \pi ^2 \log ^4(2)-\frac{7}{720} \pi ^4 \log ^2(2) \end{multline*}
		\begin{multline*}\zeta(\overline{5},1,1) = \frac{1}{2} \log (2) \zeta(\overline{5},1)+\frac{11 \text{Li}_7\left(\frac{1}{2}\right)}{2}+\frac{3}{2} \text{Li}_6\left(\frac{1}{2}\right) \log (2)+\frac{S_7}{4}+\\ \frac{\pi ^4 \zeta (3)}{90}+\frac{\pi ^2 \zeta (5)}{6}-\frac{535 \zeta (7)}{64}+\frac{1}{48} \zeta (3) \log ^4(2)-\frac{19}{16} \zeta (5) \log ^2(2)-\frac{1}{4} \zeta (3)^2 \log (2)+\frac{19 \log ^7(2)}{20160}\\ -\frac{1}{720} \pi ^2 \log ^5(2)+\frac{7 \pi ^4 \log ^3(2)}{4320}+\frac{1}{224} \pi ^6 \log (2) \end{multline*}
	\end{example}

	\normalsize{where} $$S_n = \sum_{n=1}^\infty \frac{(-1)^n}{n^s 2^n} \binom{2n}{n}^{-1}$$ Note that $S_n$ converges quite fast, on geometric rate of $8^{-n}$, so the above two series are suitable for high precision ($> 10^4$ decimal digit) calculation of these two constants, this is better than conventional method on level $2$ CMZV that converges only at rate of $2^{-n}$ \cite[Sect.~7]{Browein2}. \\
	The above series for $\zeta(\bar{5},1)$ is originally due to Zhao \cite{Zhao1}. We cannot resist to mention the beautiful closed-form of $\zeta(\bar{5},1)$ conjectured by Charlton \cite{Charlton}, discovered via motivic techniques:
	\small \begin{multline*}\zeta(\bar{5},1) = -\frac{126 \text{Li}_6\left(\frac{1}{2}\right)}{13}-\frac{162 \text{Li}_6\left(-\frac{1}{2}\right)}{13}+\frac{\text{Li}_6\left(-\frac{1}{8}\right)}{39}+\frac{3 \zeta (3)^2}{8}+\frac{31}{16} \zeta (5) \log (2)-\frac{1787 \pi ^6}{589680}-\frac{1}{208} \log ^6(2)\\ +\frac{1}{208} \pi ^2 \log ^4(2)-\frac{1}{156} \pi ^4 \log ^2(2)\end{multline*}

	\begin{theorem}
		For positive integer $s$,
		$$\sum_{n=1}^\infty \frac{1}{n^s 2^n \binom{3n}{n}} \in \textsf{CMZV}^4_s$$
	\end{theorem}
	\begin{proof}
		The summation equals to $$I = 2\int_0^1 \frac{1}{x}\Li_{s-1}\left(\frac{1}{2}x^2(1-x)\right) dx$$
		Note that
		$$\Li_{s-1}\left(\frac{1}{2}x^2(1-x)\right) = \int_0^x (2\omega(0)+\omega(1))^{s-2} (-\omega(-1) - \omega(1-i) -\omega(1+i))$$
		so $I$ will be the iterated integral of a combination of $\omega(c_1)\cdots \omega(c_s)$, with $c_i \in \{0,1,-1,1-i,1+i\}$ with $-1, 1+i, 1-i$ \textit{never occur in the same word}. If a word contains only $\omega(0),\omega(1),\omega(-1)$ then it is a level 2 CMZV, if a word contains only $\omega(0),\omega(1),\omega(1\pm i)$, then above proposition transforms it to a level 4 CMZV, so $I \in \textsf{CMZV}^4_s$.
	\end{proof}

	\begin{example}
		\small $$\begin{aligned}
			\sum_{n=1}^\infty \frac{1}{n^3 2^n \binom{3n}{n}} &= \pi  C-\frac{33 \zeta (3)}{16}+\frac{\log ^3(2)}{6}-\frac{1}{24} \pi ^2 \log (2) \\
			\sum_{n=1}^\infty \frac{1}{n^4 2^n \binom{3n}{n}} &= 2 \pi  \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{21 \text{Li}_4\left(\frac{1}{2}\right)}{2}-\frac{57}{8} \zeta (3) \log (2)+\frac{61 \pi ^4}{960} -\frac{23}{48} \log ^4(2)+\frac{19}{48} \pi ^2 \log ^2(2) \\
			\sum_{n=1}^\infty \frac{1}{n^5 2^n \binom{3n}{n}} &= 4 \pi  \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)+3 \pi  \beta(4)-\frac{51 \text{Li}_5\left(\frac{1}{2}\right)}{2} -15 \text{Li}_4\left(\frac{1}{2}\right) \log (2)+\frac{\pi ^2 \zeta (3)}{4}+\frac{9 \zeta (5)}{2}\\ &\qquad -3 \zeta (3) \log ^2(2) -\frac{97}{240} \log ^5(2)+\frac{41}{144} \pi ^2 \log ^3(2)-\frac{61}{960} \pi ^4 \log (2)
		\end{aligned}$$
	\end{example}

	The last two series were conjectured by Browein \cite[p.~27-28]{Browein}. The penultimate series is solved by Zhao \cite{Zhao1}, who relies on \textit{ad hoc} integration by parts and certain level $4$ polylogarithmic integrals.

	\newpage
	\section*{Appendix A: Mathematica package}
	The package can be download at https://www.researchgate.net/publication/357601353. A short documentation on installation and funtionalities can also be found there. All explicit one-dimensional integrals that appear before Section 4 can be calculated by the package. \par
	Currently it can only handle ordinary polylogarithm $\Li_n$ as the integrand, the case for generalized polylogarithm might be added in a future version.

	Here are lists of "new constants" that appear at each weight for level $2$ and $4$. For level $1$, consult \cite{Petitot}.
	\begin{table}[h]
		\begin{tabular}{|l|c|}
			\hline
			Weight & A basis of $\widetilde{\textsf{CMZV}}^2_w$ \\ \hline
			1 & $\log 2$ \\ \hline
			2 & $\zeta(2)$ \\ \hline
			3 & $\zeta(3)$ \\ \hline
			4 & $\Li_4(1/2)$ \\ \hline
			5 & $\Li_5(1/2), \zeta(5)$ \\ \hline
			6 & $\Li_6(1/2), \zeta(\bar{5},1)$ \\ \hline
			7 & $\Li_7(1/2), \zeta(7), \zeta(\bar{5},1,1), \zeta(5,\bar{1},1)$ \\ \hline
			8 & $\Li_8(1/2), \zeta(6,2), \zeta(\bar{7},1), \zeta(\bar{5},1,\bar{1},1), \zeta(\bar{5},\bar{1},\bar{1},\bar{1})$ \\ \hline
		\end{tabular}
	\end{table}

	\begin{table}[h]
		\begin{tabular}{|l|c|}
			\hline
			Weight & A basis of $\widetilde{\textsf{CMZV}}^4_w$ \\ \hline
			1 & $\log 2, i\pi$ \\ \hline
			2 & $iC$ \\ \hline
			3 & $\zeta(3), i\Im\Li_3((1+i)/2)$ \\ \hline
			4 & $\beta(4), \Li_4(1/2), i\Im\Li_4((1+i)/2)$ \\ \hline
			5 & $\zeta(5), \Li_5(1/2), i\Im\Li_5((1+i)/2), L_{4,1}(i,1), L_{4,1}(i,-1), L_{3,1,1}(1,1,i)$ \\ \hline
		\end{tabular}
	\end{table}

	\newpage
	\section*{Appendix B: Admissible $4$-rational functions}
	If $R(x)$ is $N$-admissible, then so are
	$$\{R, 1-R, \frac{R}{R-1}, \frac{1}{R}, \frac{R-1}{R}, \frac{1}{1-R}\}$$
	this amounts to an $S_3$-action. (with $S_3$ symmetric group on $3$ letters). \par

	When $N=4$, the automorphism group of $\hat{\mathbb{C}}$ that permutes $\{0,\infty,\pm i, \pm 1\}$ is the (orientation preserving) octahedral group $S_4$. So if $R(x)$ is $4$-admissible, then for $(g,h)\in S_3\times S_4$, $gR(h^{-1}x)$ is also $4$-admissible. This defines an $S_3\times S_4$ action on the set of $4$-admissible functions.

	\begin{table}[h]
		\begin{tabular}{|l|c|}
			\hline
			$R(x)$ & Size of orbit \\ \hline
			$x$ & 72 \\ \hline
			$x^2$ & 36 \\ \hline
			$(x^2+1)/(2x)$ & 36 \\ \hline
			$x^4$ & 18 \\ \hline
			$4x^2/(1+x^2)^2$ & 6 \\ \hline
		\end{tabular}
		\caption{\small Orbits of $4$-admissible functions known to the author, there might be more}
	\end{table}

	The following Mathematica code finds all distinct elements in an orbit:
	\begin{verbatim}Clear[S4, f, g, x, S3S4orbit]; S4 = {x, (I - x)/(I + x), (-I - x)/(-I + x), (
			I + x)/(-I + x), (-I + x)/(I + x), -((I (-1 + x))/(1 + x)), (
			I (-1 + x))/(1 + x), (I (1 + x))/(-1 + x), -((I (1 + x))/(-1 + x)),
			1/x, -x, -(1/x), I x, (1 - x)/(1 + x), (-1 - x)/(-1 + x), (
			1 + x)/(-1 + x), (-1 + x)/(1 + x), -((I (I + x))/(-I + x)), (
			I (I + x))/(-I + x), (I (-I + x))/(
			I + x), -((I (-I + x))/(I + x)), -(I/x), -I x, I/x};
		f[x_] := x/(x - 1); g[x_] := 1 - x;
		S3S4orbit[rat_] :=
		DeleteDuplicatesBy[
		Flatten[{f[#], g[#], f[g[#]], g[f[#]], f[g[f[#]]], #} &[
		rat /. x -> #] & /@ S4] // Simplify //
		Sort, # /. x -> 1/11 &];
	\end{verbatim}

	Copy the above code into Mathematica, then execute \begin{verbatim} S3S4orbit[x]\end{verbatim} gives $72$ distinct $4$-admissible functions.

	\newpage

redpajama_set_name: RedPajamaArXiv


text:
In H&H: Royal International Horse Show preview
H&H VIP [Members Only]
Mark Phillips: 'Getting BE to agree to an extra day of competition is like pulling teeth' *H&H VIP*
Mark Phillips 9 June, 2016 08:55
Four-star courses do not have a new designer very often — it's not an easy job and the margin for error is very small. Part of the skill is in knowing the ground and how the different parts of the course ride.
Therefore for Mike Etherington-Smith to be taking over from me at Luhmühlen, and Eric Winter replacing Giuseppe della Chiesa at Badminton is a major move in the four years leading up to Tokyo 2020. It will take them a little time to settle in.
I'm sad to leave Luhmühlen after 10 years. I wish Mike luck as he strives for a good balance under the critical eye of the European mainstream media.
Many were surprised at Eric Winter's appointment at Badminton, but he is relatively local and knows the track well, having been closely involved with the organisation.
He will benefit from Giuseppe's greatest legacy, the loop up through the New Pond. Giuseppe brought some great ideas to Badminton, even if he found getting the overall balance less easy.
Selection puzzler
There are now only two big opportunities for riders to shine before Olympic selection, this week at Bramham and next at Luhmühlen.
The five Brits chosen for Rio will be announced on 5 July, before Barbury, and don't be surprised if you see many of them in the CIC2* there for a confidence boost.
To me, this is the most open selection ever. The dilemma is that no one knows how strong a test Pierre Michelet has made the cross-country, therefore the relative influence of the phases. The selectors may gamble on one or two better dressage performers as recent history and the 2020 Agenda suggests that the track will be on the easier side. Having said that, Pierre has always been a law unto himself. Also, the ground will be fast and there is no real terrain. The final factor that we are not privy to is soundness and the veterinary input.
Are you any the wiser? I'm just glad I'm not a selector!
Early days for ERM
The Event Rider Masters (ERM) series at Chatsworth got off to a positive start. Yes, there were problems with the live scoring, but glitches are inevitable as the ERM pushes the boundaries of presentation.
At Gatcombe we've run the British Open cross-country in reverse order since 1985, so we know the hazards. At Chatsworth excessive speed led to near misses at the penultimate. With big money at stake, riders are tempted to fly in the latter stages, which runs an inherent risk.
At Bramham the ERM class does not have a full field of 40, partly because performance manager Yogi Breisner wanted Rio contenders to showjump last, so they run in the CIC3*, and partly because some won't want to go so fast again so soon.
At Barbury we have moved the ERM start adjacent to the finish to help with the presentation. Hopefully someone has worked out how to avoid the razzmatazz that caused Kitty King issues at Chatsworth. There has to be a balance so competitors can actually get into the start box!
British Eventing (BE) says it is "doing its best" to seek additional days of competition. My experience from talking to organisers has been the opposite — getting BE to agree to an extra day is like pulling teeth.
Fixtures are still a major problem for BE. With luck this is at the top of chief executive David Holmes' agenda.
Ref: Horse & Hound; 9 June 2016

redpajama_set_name: RedPajamaCommonCrawl


Fans of 007 get the chance to live vicariously through their favorite British spy in a new digital campaign.
Launched to coincide with the release of the upcoming James Bond film, Skyfall, the interactive campaign features Daniel Craig (in the title role) and new Bond girl Bérénice Marlohe (as Sévérine).
Called "Crack the Case," the mini film takes viewers on a train through snowy Siberia, challenging them to complete several tests while dodging Soviet-esque villains.
Die-hard fans will notice that replica James Bond film props, such as the Spectre ring from Dr. No and the Faberge egg used in From Russia with Love and Goldeneye, make cameo appearances.
The adventure is a simple one that concludes by directing participants to Heineken's Facebook page; the beer company launched the campaign on Thursday.
What do you think of this interactive campaign? Tell us in the comments below.


The startup Tech Trailblazers Awards programme has announced a new bursary for early stage non-VC backed startups.
The new Firestarter programme has been created to ensure that entry fees are not a barrier for "new lean startups" who are still securing major funding from VCs, said the organisers.
The new programme is set to attract more earlier stage startups within the accelerator and incubator communities, as well as those self-funding or with angel, seed and crowdfunding as their primary source of investment.
The recent Tech Trailblazers Tech Startup Index surveyed the funding environment for nearly 200 enterprise tech startups. The Index found that 35 percent of those surveyed indicated that raising funds took "considerably longer than anticipated", and 31 percent said that it was "considerably harder than expected".
The bursary aims to ease that pressure in the early stages for a startup. It ensures that "tech firestarters" get a shot at recognition. All eligible entrants who are two years or younger, and who are at the pre-VC funding stage, receive a bursary for the $295 entry fee to their chosen Tech Trailblazers category, as well as a chance to win the new cross category early stage "Firestarter" award.
The other startup awards include the big data, cloud, emerging markets, FinTech, IoT, mobile technology, networking, IT security, storage, sustainable IT and virtualisation categories.
All startups, up to and including those with "series C" VC-backed funding, who are younger than six years old, have until 19 September - as a final extended deadline - to complete their entries.
Rose Ross, founder of the awards, said, "It feels that the time is right to give more of the newest kids on the tech startup block more help and greater opportunity for hard earned reward and recognition. That's what the Firestarter bursary programme is all about."


BINGHAMTON — Police have arrested two people who they say spray-painted swastikas on a New York high school and the surrounding area.
The Press and Sun Bulletin of Binghamton reports a 22-year-old man and a 17-year-old boy were taken into custody Wednesday and charged with felony aggravated harassment and criminal mischief, designated as a hate crime.
Authorities say the swastikas, painted backward, were found on a window at Binghamton High School and neighboring businesses in Binghamton Tuesday morning.
The symbols were removed later that day.
Both suspects have been remanded to the Broome County Jail without bail.


text:
UL overcomes tough SLU arm in one-hitter
tbuckley2@theadvertiser.com
With the arm on the other side owning a 1.17 ERA in eight starts this season, UL starting pitcher Gunner Leger understood there was little margin for error and Ragin' Cajun hitters knew they'd have to pounce on whatever little miscues Kyle Cedotal might make.
The Cajuns took care of business on both counts.
Leger threw the first seven innings of a combined one-hitter with Eric Carter and Dylan Moore while Brenn Conrad and Joe Robbins both homered for No. 22 UL in its 4-1 win over Southeastern Louisiana on Friday night at M.L. "Tigue" Moore Field.
"That kid (Cedotal) has incredible numbers. One of the best in the country," Leger said. "So we lucky enough to get to him."
The Lions, meanwhile, couldn't quite get to UL's pitching.
The combined one-hitter was UL's first since Hunter Moody and Matt Broussard had one against Texas Tech in 2008, and the win was the 11th in 13 outings for the 23-11 Cajuns, who will play the next two games of the three-game non-conference series with SLU in Hammond on Saturday night and Sunday afternoon.
"We just were able to play a complete ballgame," UL coach Tony Robichaux said. "We needed to do that. That was our only way to beat 'em."
UL went up 1-0 in the second inning, when Dylan Poncho walked, advanced to second on Conrad's sacrifice bunt, moved to third on a fielder's choice and scored as Derek Herrington reached on shortstop Brennan Breaud's throwing error to first.
Southeastern Louisiana tied it in the top of the third, when Chris Eades doubled and moved around on back-to-back sacrifice flies by Ryan Byers and Taylor Schwaner.
But that was all there would offensively for the Southland Conference's Lions, who fell to 25-10.
Cajun outfielder Brian Mills broke the tie when he doubled, and later scored on wild pitch in the bottom of the third.
Conrad and Robbins added insurance runs on solo shots to right in the fourth and sixth innings, respectively.
"Hitters did a great job of trying to have a good approach," Robichaux said, "and then we got a couple of mistakes, and got some good swings on 'em."
Robbins' homer was his team-high ninth of the season. Conrad's was his third.
"I put a good swing on it," Conrad said. "I just didn't know if it was too high, or what. But I knew I hit the ball well, so it felt good."
Conrad also realized he had to take advantage when he had a chance against Cedotal (4-4), who is part of a staff that ranked second nationally in ERA (2.32) coming into the weekend.
"This guy — he had, like, four or five pitches that he threw on any count," Conrad said. "Coach Talbot (Cajuns assistant coach Jeremy Talbot) said before the game, 'You can't try to hit all three. You've got to pick one, and sit on it until you get two strikes.'
"That's what we did as a team. We had nine hits, and this guy had a 1.17 ERA. So I think we did a good job of just sitting on it."
Robbins' blast gave UL a three-run cushion after six innings, and Robichaux asked Leger if he wanted to go one more inning or turn it over to Cajun relievers then.
"(Robichaux) came up to me about four or five times in the dugout, to make sure," Leger said. "But I felt good."
So Leger opted to throw one more, even though bullpen arms were warm and ready.
"It's awesome," Leger said, "knowing all we need, really is a one-run leading with those two guys – especially DMo (Moore), and the way Eric (Carter) has been throwing, how hot he's been in his past few outings."
Leger put the side down in order in the seventh, ending a stretch in which he retired 10 straight to finish his appearance.
Carter and Moore took over from there as the Cajuns retired 16 of the last 17 they faced.
Carter struck out the first two batters he saw in the eighth, and Moore sat the Lions down in turn, getting two groundouts before ending things with a strikeout of Carson Crites.
"We knew coming in they were a really aggressive team," Leger said. "We knew that we could kind of expand the zone.
"I was kind of struggling early with my off-speed, so we just kind of went after them with fastball, kind of went off the plate to try to get them to swing early.
"The defense played great," he added. "I'm trying to get weak, early outs, and when the defense plays like they did (Friday night) it ends up working out."
No. 22 UL (23-11) vs. SOUTHEASTERN LOUISIANA (25-10)
WHAT: Middle game of three-game non-conference weekend series
WHERE: Pat Kenelly Diamond at Alumni Field (2,500), Hammond
WHEN: 6 p.m. Saturday
RADIO: KPEL 96.5 FM with Jay Walker
TV: Cox Sports Television (CST)
PITCHING MATCHUP: UL RHP Wyatt Marks (3-3, 4.22 ERA) vs. RHP Matt Sceroler (5-2, 2.30)


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text:
using System.Linq;
using System.Threading.Tasks;
using Microsoft.AspNetCore.Authorization;
using Microsoft.AspNetCore.Mvc;
using Microsoft.EntityFrameworkCore;
using WakeboardUK2018.Data;
using WakeboardUK2018.Models;

namespace WakeboardUK2018.Controllers
{
    public class CompetitionsController : Controller
    {
        private readonly ApplicationDbContext _context;

        public CompetitionsController(ApplicationDbContext context)
        {
            _context = context;
        }

        // GET: Competitions
        public async Task<IActionResult> Index(int? page, int? year)
        {
            var pageSize = 100;
            if (year == null) year = System.DateTime.Today.Year;
            var competitions = _context.Competitions
                .OrderBy(a => a.start_date)
                .Where(a => a.start_date.Value.Year == year)
                .Where(s => s.published);
            return View(await PaginatedList<Competition>.CreateAsync(competitions, page ?? 1, pageSize));

            //return View(await _context.Competitions.OrderBy(s => s.start_date).Where(s => s.published).Where(t => t.start_date > nowish).ToListAsync());
        }

        // GET: Competitions/Details/5
        public async Task<IActionResult> Details(int? id)
        {
            if (id == null)
                return NotFound();

            var competition = await _context.Competitions
                .SingleOrDefaultAsync(m => m.CompetitionId == id);
            if (competition == null)
                return NotFound();

            return View(competition);
        }

        // GET: Competitions/Create
        [Authorize]
        public IActionResult Create()
        {
            return View();
        }

        // POST: Competitions/Create
        // To protect from overposting attacks, please enable the specific properties you want to bind to, for
        // more details see http://go.microsoft.com/fwlink/?LinkId=317598.
        [Authorize]
        [HttpPost]
        [ValidateAntiForgeryToken]
        public async Task<IActionResult> Create(
            [Bind(
                "CompetitionId,Title,Description,start_date,end_date,short_name,location_id,created_at,updated_at,country_id,photo_file_name,photo_content_type,photo_file_size,location,entry,sponsors,schedule,notes,official_url,accommodation,published,result")]
            Competition competition)
        {
            if (ModelState.IsValid)
            {
                _context.Add(competition);
                await _context.SaveChangesAsync();
                return RedirectToAction(nameof(Index));
            }
            return View(competition);
        }

        // GET: Competitions/Edit/5
        [Authorize]
        public async Task<IActionResult> Edit(int? id)
        {
            if (id == null)
                return NotFound();

            var competition = await _context.Competitions.SingleOrDefaultAsync(m => m.CompetitionId == id);
            if (competition == null)
                return NotFound();
            return View(competition);
        }

        // POST: Competitions/Edit/5
        // To protect from overposting attacks, please enable the specific properties you want to bind to, for
        // more details see http://go.microsoft.com/fwlink/?LinkId=317598.
        [Authorize]
        [HttpPost]
        [ValidateAntiForgeryToken]
        public async Task<IActionResult> Edit(int id,
            [Bind(
                "CompetitionId,Title,Description,start_date,end_date,short_name,location_id,created_at,updated_at,country_id,photo_file_name,photo_content_type,photo_file_size,location,entry,sponsors,schedule,notes,official_url,accommodation,published,result")]
            Competition competition)
        {
            if (id != competition.CompetitionId)
                return NotFound();

            if (ModelState.IsValid)
            {
                try
                {
                    _context.Update(competition);
                    await _context.SaveChangesAsync();
                }
                catch (DbUpdateConcurrencyException)
                {
                    if (!CompetitionExists(competition.CompetitionId))
                        return NotFound();
                    throw;
                }
                return RedirectToAction(nameof(Index));
            }
            return View(competition);
        }

        // GET: Competitions/Delete/5
        [Authorize]
        public async Task<IActionResult> Delete(int? id)
        {
            if (id == null)
                return NotFound();

            var competition = await _context.Competitions
                .SingleOrDefaultAsync(m => m.CompetitionId == id);
            if (competition == null)
                return NotFound();

            return View(competition);
        }

        // POST: Competitions/Delete/5
        [Authorize]
        [HttpPost]
        [ActionName("Delete")]
        [ValidateAntiForgeryToken]
        public async Task<IActionResult> DeleteConfirmed(int id)
        {
            var competition = await _context.Competitions.SingleOrDefaultAsync(m => m.CompetitionId == id);
            _context.Competitions.Remove(competition);
            await _context.SaveChangesAsync();
            return RedirectToAction(nameof(Index));
        }

        private bool CompetitionExists(int id)
        {
            return _context.Competitions.Any(e => e.CompetitionId == id);
        }
    }
}

redpajama_set_name: RedPajamaGithub


text:
THE MASK (1961) - vintage 3D horror finally on DVD
also called EYES FROM HELL
Definitely a cult movie, early Canadian horror The Mask is finally out on DVD. In 3D!
I was going to include this in the Not On DVD movies. But after another viewing, I couldn't recommend it as a must-see. It's a cheap, dull story for the most part, only the 3D 'dream sequences' are really worth seeing.
Really very good 3D effects and wild, horrific imagery, presumably shot by a different director. The 3D benefits from long tracking shots. I still think that the best 3D effects allow the viewer to 'get their bearings', like a camera moving through a deep hallway of cobwebs. The set designs and make-up effects are uniquely nightmarish and the shocks are carefully designed for 3D. A giant Aztec skull floating over a sacrificial ritual, a phantom firing fireballs from its palms, druids with razor-gloves... The stills you see from The Mask are always from these scenes, which tell their own dream-logic story. Joined together, they would make a far more famous 3D short film.
The Mask is OK to watch if you're used to low-budget fifties sci-fi. Most of the talky action takes place in small, functional three-wall sets. Where an archaeologist starts hearing voices coming from one of his artefacts, demanding that he wears it. When he does, he has nightmares about an ancient ritual... If you want a chance to see what a great 3D horror, or even LSD horror could look like, try out the dream sequences of The Mask...
The DVD is from Cheezy Flicks. It's a black and white fullscreen movie (this lobby card was hand-coloured). It's misleadingly listed as colour, because it's presented in the red/green 3D process. From the reviews, it doesn't appear to have been digitally remastered.
"Put the mask on, now..."
Labels: 1960s, 3D movie, Canada
VILLAGE OF THE DAMNED (1960) - from the author of ...
SATURN 3 (1980) - good-looking space thriller
ALTERED STATES (1980) - biological time travel
THE MASK (1961) - vintage 3D horror finally on DVD...
SOLITUDE OF BLOOD (2002) - PSYCHO meets REQUIEM FO...
BATMAN (1943), BATMAN AND ROBIN (1949) - the early...
BLACK SUNDAY (1977) - an epic terrorist thriller
SATELLITE IN THE SKY (1956) - early British spaces...


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