text:
\section{Introduction}
Sums over compositions are an object of study in their own right, and there is a vast literature considering sums over compositions or enumeration of restricted compositions. A \textit{composition} of an integer number $n$ is any sequence of
integers $n_{i}\ge1$, called \textit{parts} of $n$, such that
\[
n=n_{1}+\dots+n_{m}.
\]
There are $2$ compositions of $2:$
\[
2,\thinspace\thinspace1+1,
\]
$4$ compositions of $3:$
\[
3,\thinspace\thinspace2+1,\thinspace\thinspace1+2,\thinspace\thinspace1+1+1
\]
and more generally $2^{n-1}$ compositions of $n.$ These compositions
can be represented as follows:
\[
2:\thinspace\thinspace\newmoon\newmoon,\thinspace\thinspace\newmoon\vert\newmoon
\]
\[
3:\thinspace\thinspace\newmoon\newmoon\newmoon,\thinspace\thinspace\newmoon\newmoon\vert\newmoon,\thinspace\thinspace\newmoon\vert\newmoon\newmoon,\thinspace\thinspace\newmoon\vert\newmoon\vert\newmoon
\]
and so on. We also introduce the following notation:
let $\mathcal{C}\left(n\right)$ denote the set of all compositions
of $n;$  for an element $\pi=\left\{ n_{1},\dots,n_{m}\right\} \in \mathcal{C}\left(n\right),$
denote $m=\vert\pi\vert$ its length, i.e the number of its parts. For a sequence $\{g_n\},$ we also use the multi-index notation
\[
g_{\pi}=g_{n_{1}}\dots g_{n_{m}}.
\]
We can now state our main result, a generating function for sums over compositions:
\begin{thm}
Let $g(z) = \sum_{n \geq 1} g_n z^n $ and $f(z) = \sum_{n\geq 0 }f_n z^n$. We then have the generating function identity

\begin{equation}
f(g(z)) = f_0 + \sum_{n \geq 1} z^n \sum_{\pi \in \mathcal{C}_n} f_{\vert \pi \vert} g_{\pi}.
\end{equation}
\end{thm}

This general result unites many previous results spread across the literature, because the composite generating function $f(g(z))$ often has a simple closed form.
We link this formula to Woon's and Fuchs' trees, which provide graphical ways of visualizing it, as described below.

In Section \ref{sec1} we introduce Woon's and Fuchs' tree, and in Section \ref{sec2} we connect them to compositions for the first time. In Section \ref{sec3}, we introduce our main results, and use them to derive a variety of closed form expressions for sums over compositions. In Section \ref{sec4} we introduce a new notation for generalized composition sums, and explore some basic identities for them.

Throughout, we apply our results to special functions and generalized Bernoulli numbers, which allows us to derive expressions for them as sums over compositions.

\section{Background}\label{sec1}

Woon's tree, as introduced by S.G. Woon in \cite{Woon}, is the following construction
\begin{center}
\begin{forest}
[$\frac{1}{2!}$[$-\frac{1}{3!}$[$\frac{1}{4!}$[$-\frac{1}{5!}$][$\frac{1}{2!4!!}$]][$-\frac{1}{2!3!}$[$\frac{1}{3!3!}$][$-\frac{1}{2!2!3!}$]]][$\frac{1}{2!2!}$[$-\frac{1}{3!2!}$[$\frac{1}{4!2!}$][$-\frac{1}{2!3!2!}$]][$\frac{1}{2!2!2!}$[$-\frac{1}{3!2!2!}$][$\frac{1}{2!2!2!2!}$]]]]
\end{forest}
\par\end{center}

with the rule
\begin{center}
\begin{forest}
[$\pm\frac{1}{a_{1}!\dots a_{k}!}$[$\mp\frac{1}{\left(a_{1}+1\right)!\dots a_{k}!}$][$\pm\frac{1}{2!a_{1}!\dots a_{k}!}$]]
\end{forest}
\par\end{center}

\begin{flushleft}
S.G. Woon proved, using Euler MacLaurin summation formula, that the successive row sums
\begin{align*}
\frac{1}{2!} & =\frac{1}{2},\thinspace\thinspace\frac{1}{2!2!}-\frac{1}{3!}=\frac{1}{4}-\frac{1}{6}=\frac{1}{12},\thinspace\thinspace\frac{1}{4!}-\frac{1}{2!3!}-\frac{1}{3!2!}+\frac{1}{2!2!2!}=0,\\
\frac{1}{2!2!2!2!} & -\frac{3}{2!2!3!}+\frac{2}{2!4!}+\frac{1}{3!3!}-\frac{1}{5!}=-\frac{1}{720},\dots
\end{align*}
coincide with the sequence $\left\{ \left(-1\right)^{n}\frac{B_{n}}{n!}\right\} $
where the Bernoulli numbers are defined by the generating function
\[
\sum_{n\ge0}\frac{B_{n}}{n!}z^{n}=\frac{z}{e^{z}-1}.
\]
\par\end{flushleft}

Later on, Fuchs \cite{Fuchs} considered  a more general construction, the {\it general PI tree}. This associates, to the sequence of real numbers $\left\{ g_{n}\right\} ,$ which we call the {\it input sequence},
the tree
\begin{center}
\begin{forest}
[$g_{1}$[$g_{1}g_{1}$[$g_{1}g_{1}g_{1}$[$g_{1}g_{1}g_{1}g_{1}$][$g_{2}g_{1}g_{1}$]][$g_{2}g_{1}$[$g_{1}g_{2}g_{1}$][$g_{3}g_{1}$]]][$g_{2}$[$g_{1}g_{2}$[$g_{1}g_{1}g_{2}$][$g_{2}g_{2}$]][$g_{3}$[$g_{1}g_{3}$][$g_{4}$]]]]
\end{forest}
\par\end{center}

built using the two operators $P$ (for "put a 1") and $I$ (for "increase") as follows
\begin{center}
\begin{forest}
[$g_{i_{1}}g_{i_{2}}\dots g_{i_{k}}$[$g_{1}g_{i_{1}}g_{i_{2}}\dots g_{i_{k}}$,
edge label={node[midway,left,font=\scriptsize]{$P$}}
][$g_{i_{1}+1}g_{i_{2}\dots}g_{i_{k}}$,
edge label={node[midway,right,font=\scriptsize]{$I$}}]]
\end{forest}
\par\end{center}
The row sums of the tree generate what will be called the {\it output sequence} $\{x_n\}.$
Note that although the $g_{n}$ are real numbers, they should
be considered as noncommutating variables in the process of construction
of the tree. Moreover, Woon's tree corresponds to a PI tree with the particular choice
\[
g_{n}=\frac{-1}{\left(n+1\right)!}.
\]

Fuchs proved that if $x_{0}=1,$ then the sequence of row sums $\left\{ x_{n}\right\} _{n\ge1}$
of the general PI tree is related to the sequence of its entries $\left\{ g_{n}\right\} _{n\ge1}$
by the convolution
\begin{equation}
x_{n}=\sum_{j=1}^{n}g_{j}x_{n-j}=g_{n}+g_{n-1}x_{1}+\dots+g_{1}x_{n-1}.\label{eq:recursion}
\end{equation}

This result translates in terms of generating functions as follows - see \cite{Fuchs}:
if $\left\{ g_{n}\right\} $ and $\left\{ x_{n}\right\} $ are the
input and output sequences of Fuchs' tree, then their generating functions
\[
x\left(z\right)=\sum_{n\ge1}x_{n}z^{n},\thinspace\thinspace g\left(z\right)=\sum_{n\ge1}g_{n}z^{n}
\]
are related as
\begin{equation}
x\left(z\right)=\frac{g\left(z\right)}{1-g\left(z\right)},
\end{equation}
and
\begin{equation}
\label{CX}
g\left(z\right)=\frac{x\left(z\right)}{1+x\left(z\right)}.
\end{equation}

\begin{example}
\label{ex:Bernoulli}
In the case of Bernoulli numbers, the generating function of the input sequence is
\[
g\left(z\right)=\sum_{k\ge1}\frac{-1}{\left(k+1\right)!}z^{k}=-\frac{1}{z}\left(e^{z}-1-z\right)=1-\frac{e^{z}-1}{z}
\]
so that the generating function of the row sums sequence is
\[
x\left(z\right)=\frac{1-\frac{e^{z}-1}{z}}{\frac{e^{z}-1}{z}}=\frac{z}{e^{z}-1}-1=\sum_{n\ge1}\frac{B_{n}}{n!}z^{n}.
\]
The recursive identity (\ref{eq:recursion}) is the well-known identity
for Bernoulli numbers \cite[24.5.3]{NIST}
\[
B_{n}=-\sum_{k=1}^{n}\binom{n}{k}\frac{B_{n-k}}{k+1}.
\]
\end{example}
\begin{example}
The Bernoulli polynomials $B_{n}\left(x\right)$ are defined by the generating function
\[
\sum_{n\ge0}\frac{B_{n}\left(x\right)}{n!}z^{n}=\frac{ze^{zx}}{e^{z}-1}.
\]
We deduce that the tree with polynomial entries
\[
g_{n}=\frac{\left(-1\right)^{n+1}}{\left(n+1\right)!}\left[x^{n+1}-\left(x-1\right)^{n+1}\right]
\]
has row sums that coincide with the Bernoulli polynomials:
\[
x_{n}=\frac{B_{n}\left(x\right)}{n!}.
\]
This gives a decomposition of Bernoulli polynomials as a sum of elementary
polynomials; for example
\[
B_{1}\left(x\right)=\frac{1}{2}\left[x^{2}-\left(x-1\right)^{2}\right]
\]
\[
\frac{B_{2}\left(x\right)}{2!}=\frac{1}{2}\left[x^{2}-\left(x-1\right)^{2}\right]-\frac{1}{6}\left[x^{3}-\left(x-1\right)^{3}\right].
\]
The general expression for $\frac{B_{n}\left(x\right)}{n!}$ is given
in Thm. \ref{Bernoulli expansion} below.
\end{example}

\begin{example}
Higher-order Bernoulli numbers, also called N\"{o}rlund polynomials, are defined for an integer parameter
$p$ by the generating function
\begin{equation}
\label{Norlund gf}
\sum_{n\ge0}\frac{B_{n}^{\left(p\right)}}{n!}z^{n}=\left(\frac{z}{e^{z}-1}\right)^{p}.
\end{equation}

The output sequence
\[
x_{n}=\frac{B_{n}^{\left(p\right)}}{n!}
\]
corresponds to the input sequence
\[
g_{n}=-\frac{p!}{\left(p+n\right)!}\left\{ \begin{array}{c}
n+p\\
p
\end{array}\right\}
\]
where $\left\{ \begin{array}{c}
n\\
p
\end{array}\right\} $ is the Stirling number of the second kind.
\begin{example}
The hypergeometric Bernoulli numbers, as introduced by F.T. Howard \cite{Howard}, are defined, for $a>0$ and $b>0,$ by the generating function
\[
\sum_{n\ge0}\frac{B_{n}^{\left(a,b\right)}}{n!}z^{n}=\frac{1}{_{1}F_{1}\left(\begin{array}{c}
a\\
a+b
\end{array};z\right)}
\]
where $_{1}F_{1}$ is the hypergeometric function
\[
_{1}F_{1}\left(\begin{array}{c}
a\\
c
\end{array};z\right) = \sum_{n\ge0} \frac{\left(a\right)_{n}}{\left(c\right)_{n}} \frac{z^n}{n!}
\]
with the notation $\left(a\right)_{n}$ for the Pochhammer symbol
\[
\left(a\right)_{n}=\frac{\Gamma\left(a+n\right)}{\Gamma\left(a\right)}.
\]
The output sequence
\[
x_{n}=\frac{B_{n}^{\left(a,b\right)}}{n!}
\]
corresponds to the input sequence
\[
g_{n}=-\frac{1}{n!}\frac{\left(a\right)_{n}}{\left(a+b\right)_{n}}.
\]
\end{example}
\end{example}

\section{Connection to compositions}\label{sec2}

\subsection{Definitions}

In the examples studied so far, we were able to compute a few row
sums of Fuchs' tree, but we still need a general and non-recursive formula that gives
the row sum sequence $\left\{ x_{n}\right\} $ explicitly as a function
of the input sequence $\left\{ g_{n}\right\} .$ This implies
a better description of the row generating process of Woon's tree,
which requires the notion of compositions.

The representation
\begin{center}
\begin{forest}
[$\newmoon$[$\newmoon$$\vert$$\newmoon$[$\newmoon$$\vert$$\newmoon$$\vert$$\newmoon$][$\newmoon$$\newmoon$$\vert$$\newmoon$]][$\newmoon$$\newmoon$[$\newmoon$$\vert$$\newmoon$$\newmoon$][$\newmoon$$\newmoon$$\newmoon$]]]
\end{forest}
\par\end{center}
suggests a natural bijection between the generation process of the next row
in Woon's tree and the generation process of the compositions of the integer
$n+1$ in terms of those of $n.$ We restate Fuchs' result \cite{Fuchs} in terms of compositions,
and then apply this result to derive several new identities for Catalan numbers and Hermite polynomials.

\subsection{Fuchs' result}
We can now restate Fuchs' main result \eqref{eq:recursion} as follows
\begin{thm}\label{thm:5}
The sequence of row sums $\left\{ x_{n}\right\} $ in Woon's tree
can be computed from its sequence of entries $\left\{ g_{n}\right\} $
as the sum over compositions
\begin{equation}
x_{n}=\sum_{\pi\in\mathcal{C}\left(n\right)}g_{\pi}=\sum_{p=1}^n\sum_{\underset{k_{i}\ge1}{k_{1}+\dots+k_{p}=n}}g_{k_{1}}\dots g_{k_{p}}.\label{eq:compositions}
\end{equation}
This sum over compositions can also be expressed
as the weighted sum over convolutions
\begin{equation}
x_{n}=\sum_{p=1}^{n}\binom{n+1}{p+1}\sum_{\underset{k_{i}\ge 0}{k_{1}+\dots+k_{p}=n}}g_{k_{1}}\dots g_{k_{p}}.\label{eq:convolutions}
\end{equation}
Moreover, these relations can be inverted by exchanging $x_n$ with $-g_n$
: the input sequence $\{g_n\}$ in Woon's tree can be recovered from its row sum sequence $\{x_n\}$ as the sum over compositions
\begin{equation}
g_{n} =\sum_{\pi\in\mathcal{C}\left(n\right)}\left(-1\right)^{\pi+1}x_{\pi}=\sum_{p=1}^n\left(-1\right)^{p+1}\sum_{\underset{k_{i}\ge1}{k_{1}+\dots+k_{p}=n}}x_{k_{1}}\dots x_{k_{p}}
\label{eq:inverse compositions}
\end{equation}
or as the weighted sum over convolutions
\begin{equation}
x_{n}=\sum_{p=1}^{n}\left(-1\right)^{p+1}\binom{n+1}{p+1}\sum_{\underset{k_{i}\ge 0}{k_{1}+\dots+k_{p}=n}}x_{k_{1}}\dots x_{k_{p}}.
\end{equation}
\end{thm}

\begin{proof}
Denote $g_{n}=-\frac{w_{n}}{n!}$ so that
\begin{align*}
x_{n} & =\sum_{m=1}^{n} \sum_{\underset{k_{i}\ge1}{k_{1}+\dots+k_{m}=n}}g_{k_{1}}\dots g_{k_{m}}=\sum_{m=1}^{n}\left(-1\right)^{m}\sum_{\underset{k_{i}\ge1}{k_{1}+\dots+k_{m}=n}}\frac{w_{k_{1}}}{k_{1}!}\dots\frac{w_{k_{m}}}{k_{m}!}\\
 & =\sum_{m=1}^{n} \frac{\left(-1\right)^{m}}{n!} \sum_{\underset{k_{i}\ge1}{k_{1}+\dots+k_{m}=n}}\binom{n}{k_{1},\dots,k_{m}}w_{k_{1}}\dots w_{k_{m}}.
\end{align*}
Denote the incomplete sum
\[
\tilde{S}_{m,n}=\sum_{\underset{k_{i}\ge1}{k_{1}+\dots+k_{m}=n}}\binom{n}{k_{1},\dots,k_{m}}w_{k_{1}}\dots w_{k_{m}}
\]
and its complete version
\[
S_{m,n}=\sum_{\underset{k_{i}\ge0}{k_{1}+\dots+k_{m}=n}}\binom{n}{k_{1},\dots,k_{m}}w_{k_{1}}\dots w_{k_{m}}
\]
so that
\[
x_{n}=\sum_{m=1}^{n}\frac{\left(-1\right)^{m} }{n!}\tilde{S}_{m,n}
\]
and both variables $S_{m,n}$ and $\tilde{S}_{m,n}$ are related as
\[
\tilde{S}_{m,n}=\sum_{p=1}^{m}\left(-1\right)^{m-p}\binom{m}{p}S_{p,n},
\]
that can be proved for example by induction on $m.$
We deduce, following the same steps as in the Bernoulli case above,
\[
x_{n}=\sum_{m=1}^{n}\frac{\left(-1\right)^{m} }{n!}\sum_{p=1}^{m}\left(-1\right)^{m-p}\binom{m}{p}S_{p,n}=\sum_{p=1}^{n}\frac{\left(-1\right)^{p}}{n!}S_{p,n}\sum_{m=p}^{n} \binom{m}{p}.
\]
Moreover
\[
S_{p,n}=\sum_{\underset{k_{i}\ge0}{k_{1}+\dots+k_{p}=n}}\binom{n}{k_{1},\dots,k_{p}}w_{k_{1}}\dots w_{k_{p}}=\left(-1\right)^{p}n!\sum_{\underset{k_{i}\ge0}{k_{1}+\dots+k_{p}=n}}g_{k_{1}}\dots g_{k_{p}}
\]
and the proof of the first part follows.
The inversion of these relations is deduced from the identity \eqref{CX}.\end{proof}

The result of Thm. \ref{thm:5} has been rediscovered many times in relation to sums over compositions. For examples of identities derived from this result, see \cite{Sills}, \cite{Hoggatt} and  \cite{Gessel}. Selecting various $g_{k} =1$ for $k$ in some set of indices $J$, and $g_i = 0$ otherwise, also gives the generating function for the number of compositions into parts from this set $J$. For example, we can let $g_{2n}=0$ and $g_{2n+1}=1$ to find a formula for the number of compositions into odd parts.

\subsection{Invariant sequences}
In this section, we look for sequences that are invariant by Fuchs' tree.
\subsubsection{Catalan numbers}

The Catalan numbers are defined by the generating function
\[
\sum_{n\ge0}C_{n}z^{n}=\frac{1-\sqrt{1-4z}}{2z}.
\]
They are invariants of Woon's tree in the following sense.
\begin{thm}
If
\[
g_{n}=\begin{cases}
C_{n-1} & n\ge1\\
1 & n=0
\end{cases}
\]
is the input sequence of Woon's tree, then the row sums are
\[
x_{n}=C_{n},\thinspace\thinspace n\ge0.
\]
As a consequence, the Catalan numbers satisfy the sum over compositions
identities
\begin{equation}
C_{n}=\sum_{m=1}^{n}\sum_{\underset{k_{i}\ge1}{k_{1}+\dots+k_{m}=n}}C_{k_{1}-1}\dots C_{k_{m}-1}.\label{eq:Catalan sum over compositions}
\end{equation}
and
\begin{equation}
C_{n-1}=\sum_{m=1}^{n}\left(-1\right)^{m+1}
\sum_{\underset{k_{i}\ge1}{k_{1}+\dots+k_{m}=n}}C_{k_{1}}\dots C_{k_{m}}.
\label{eq:inverse Catalan sum over compositions}
\end{equation}

\end{thm}
\begin{rem}
Identity \eqref{eq:Catalan sum over compositions} can be considered as a generalization of the classic convolution identity for Catalan numbers \cite[26.5.3]{NIST}
\[
C_n = \sum_{k=0}^{n-1}C_{k}C_{n-1-k}.
\]
\end{rem}
\begin{proof}
From the generating function of the input sequence
\[
g\left(z\right)=\sum_{n\ge1}g_{n}z^{n}=\sum_{n\ge1}C_{n-1}z^{n}=z\frac{1-\sqrt{1-4z}}{2z}=\frac{1-\sqrt{1-4z}}{2},
\]
we deduce the generating function of the row sums
\[
X\left(z\right)=\frac{g\left(z\right)}{1-g\left(z\right)}=\frac{1-\sqrt{1-4z}}{1+\sqrt{1-4z}}=-1+\frac{1-\sqrt{1-4z}}{2z}=\sum_{n\ge1}C_{n}z^{n},
\]
which proves the result. Both identities for Catalan numbers \eqref{eq:Catalan sum over compositions} and \eqref{eq:inverse Catalan sum over compositions}
are a consequence of \eqref{eq:compositions} and \eqref{eq:inverse compositions}.
\end{proof}

\subsubsection{Hermite polynomials}

Another invariant sequence of Woon's tree is the sequence of Hermite polynomials $\left\{ H_{n}\left(x\right)\right\}$  defined by the generating function
\[
\sum_{n\ge0}\frac{H_{n}\left(x\right)}{n!}z^{n}=e^{2xz-z^{2}}.
\]
It can be checked that if the entries of Woon's tree are chosen as
\[g_{n}=-\imath^{n}\frac{H_{n}\left(\imath x\right)}{n!}
\]
then the sequence of row sums is equal to
\[
x_{n}=\frac{H_{n}\left(x\right)}{n!}.
\]

\section{A Nonlinear Generalization}\label{sec3}
We are able to generalize these results to sums over compositions with weights, which corresponds to a further generalization of Fuchs' tree. Because Fuchs' tree graphically represents sums over compositions, we can represent the generating function $f(g(z))$ as the convolution of two trees:
\vspace{0.5cm}

\begin{minipage}{1\textwidth}%
\noindent\begin{minipage}{0.3\textwidth}%
\begin{forest}
[$g_{1}$[$g_{1}g_{1}$[$g_{1}g_{1}g_{1}$][$g_{2}g_{1}$]][$g_{2}$[$g_{1}g_{2}$][$g_{3}$]]]
\end{forest}
\end{minipage}
$*$
\noindent\begin{minipage}{0.3\textwidth}%
\begin{forest}
[$f_{1}$[$f_{2}$[$f_{3}$][$f_{2}$]][$f_{1}$[$f_{2}$][$f_{1}$]]]\end{forest}
\end{minipage}%
$=$
\noindent\begin{minipage}{0.3\textwidth}%
\begin{forest}
[$f_{1}g_{1}$[$f_{2}g_{1}g_{1}$[$f_{3}g_{1}g_{1}g_{1}$][$f_{2}g_{2}g_{1}$]][$f_{1}g_{2}$[$f_{2}g_{1}g_{2}$][$f_{1}g_{3}$]]]\end{forest}
\end{minipage}%
\end{minipage}
\vspace{0.5cm}

The analytic description of this operation is as follows.
\begin{thm}\label{thm:main}
Let $g(z) = \sum_{n \geq 1} g_n z^n $ and $f(z) = \sum_{n\geq 0 }f_n z^n$. We then have the generating function identity
\begin{equation}
\label{eq:f(g(z))}
f(g(z)) = f_0 + \sum_{n \geq 1} z^n \sum_{\pi \in \mathcal{C}_n} f_{\vert \pi \vert} g_{\pi}.
\end{equation}

\end{thm}
\begin{proof}
We note that the case $f_0=0$ and $f_n = 1$, $n \geq 1$, corresponds to Theorem \ref{thm:5}. Then $f(z) = \frac{z}{1-z}$, leading to the observed relationship between $g(z)$ and $x(z)$. In terms of trees, the sum $\sum_{\pi \in \mathcal{C}_n} f_{\vert \pi \vert} g_\pi$ corresponds to a row sum, with additional weights depending on how many different parts are in each composition of $n$. With the notation $T_n(f;a) = \frac{f^{(n)}(a)}{n!}$, we have the following result of Vella \cite{Vella}, based off the classical Fa\'{a} di Bruno formula:
\begin{equation}
T_n(f \circ g;a) = \sum_{\pi \in C_n}  T_{\vert \pi \vert}(f;g(a)) \prod_{i=1}^n [T_i(g;a)]^{\pi_i}.
\end{equation}

We let $a=0$, so that $g(a)=0$. Then letting $T_n(f;0) =f_n$ and $T_n(g;0) = g_n$ yields the result.
\end{proof}
\begin{rem}
Eger \cite{Eger} explored the case $f_k=1$ and $f_n=0$ for $n \neq k$. He used this to define 'extended binomial coefficients', which constitute a special case of our results.
\end{rem}
\begin{rem}
One consequence of identity \eqref{eq:f(g(z))} is as follows: let $\{\mu_n\}$ denote a moment sequence and $\{\kappa_n\}$ a cumulant sequence of a random variable $Z$: the moments are the expectations $\mu_n = \mathbb{E} X^n$ and the cumulants are the Taylor coefficients of the logarithm of the moment generating function $\sum_{n \ge 0} \frac{\mu_n}{n!} z^n$. Choosing
\[
g_1\left(z\right)= \sum_{n=1}^{\infty} \mu_n \frac{z^n}{n!},\thinspace\thinspace f\left(z\right)=\log\left(1+z\right)
\]
so that
\[
f\left(g_1\left(z\right)\right)=\sum_{n=1}^{\infty}  \kappa_n \frac{z^n}{n!},
\]
gives the identity between moments and cumulants
\[
\sum_{\pi\in\mathcal{C}\left(n\right)}\frac{\left(-1\right)^{\vert \pi \vert}}{\vert \pi \vert}\frac{\mu_\pi}{\pi!}=\frac{\kappa_n}{n!}
\]
Let $g_2(z) = \sum_{n=1}^{\infty}  \kappa_n \frac{z^n}{n!}$ be the cumulant generating function and $f(z) = e^z-1$, so that $f(g_2(z))$ is the moment generating function. This gives the other identity between moments and cumulants
\[
\sum_{\pi\in\mathcal{C}\left(n\right)}\frac{1}{\vert \pi \vert!}\frac{\kappa_\pi}{\pi!}=\frac{\mu_n}{n!}.
\]
This allows us to express moments and cumulants in terms of sums over compositions of each other. Note that by definition, if $\pi = k_1 \dots k_p$ then $\left(-1\right)^{\pi}=\left(-1\right)^{k_1 +\dots+ k_p}$, $\mu_\pi = \mu_{k_1}\cdots\mu_{k_p}$, $\kappa_\pi = \kappa_{k_1}\cdots\kappa_{k_p}$, $\pi! = k_{1}! \dots k_{p}!$ and $\vert \pi \vert!=p!.$
\end{rem}

\begin{thm}
\label{non linear}
We have the following identity for sums over compositions:
\begin{equation}\label{eq:convolutions}
\sum_{\pi \in \mathcal{C}_n} f_{\vert \pi \vert} g_{\pi} = \sum_{p=1}^{n} f_p\sum_{\underset{k_{i}\ge1}{k_{1}+\dots+k_{p}=n}} g_{k_{1}}\dots g_{k_{p}} = \sum_{p=1}^{n}\left(\sum_{m=p}^{n}f_m \binom{m}{p}\right)\sum_{k_{1}+\dots+k_{p}=n}g_{k_{1}}\dots g_{k_{p}}.
\end{equation}
\end{thm}
\begin{proof}
The proof is identical to that given in Theorem \ref{thm:5}.
\end{proof}

We can also, for the first time, find a general formula for sums over all parts of all compositions of $n$. This corresponds to an 'additive' Woon tree where the creation operator "I" adds $g_1$ instead of multiplying by it.
\begin{thm}
Let $g(z) = \sum_{n \geq 1} g_n z^n $ and $f(z) = \sum_{n\geq 0 }f_n z^n$. We then have the generating function identity

\begin{equation}
f'\left(\frac{z}{1-z}\right) g(z)=  \sum_{n \geq 1} z^n \sum_{\pi \in \mathcal{C}_n} f_{\vert \pi \vert} \sum_{k_i \in \pi} g_{k_i}.
\end{equation}
\end{thm}
\begin{proof}
We begin with $g(z) = \sum_{n \geq 1} g_n z^n $ and transform it to the bivariate generating function $G(z, \lambda) =  \sum_{n \geq 1} \lambda^{g_n} z^n$. We then take a partial derivative with respect to $\lambda$ and evaluate at $\lambda=1$, so
\begin{equation}
\frac{\partial}{\partial \lambda}  f(G(z,\lambda)) \Bigr|_{\lambda=1}= \frac{\partial}{\partial \lambda}   \sum_{n \geq 1} z^n \sum_{\pi \in \mathcal{C}_n} f_{\vert \pi \vert} \lambda^{ \sum_{k_i \in \pi} g_{k_i}}  \Bigr|_{\lambda=1}.
\end{equation}
Noting that $G(z,1) =\frac{z}{1-z}$ and $\frac{\partial}{\partial \lambda}  G(z,\lambda) |_{\lambda=1} = g(z)$ yields the theorem.
\end{proof}

\begin{rem}
As a consequence of this result, choosing $f\left(z\right) = \frac{z}{1-z},$ we deduce the generating function of the sequence
\[
\sum_{\pi \in \mathcal{C}_n}\sum_{k_i \in \pi} g_{k_i}
\]
of a sequence $\{g_n\}$
as
\[
\sum_{n \geq 1} z^n \sum_{\pi \in \mathcal{C}_n} \sum_{k_i \in \pi} g_{k_i}.
= \left(\frac{1-z}{1-2z}\right)^2 g\left(z\right).
\]
Generating functions for other sequences related to compositions, such as the total number of summands in all the compositions of $n,$ can be found for example in \cite{Merlini}.
\end{rem}

\subsection{Back to the Bernoulli numbers}

Going back to Example \ref{ex:Bernoulli} and Equation \eqref{eq:convolutions}, we deduce the expression
\[
\frac{B_{n}}{n!}=\sum_{\pi\in\mathcal{C}\left(n\right)}\frac{\left(-1\right)^{\vert \pi \vert}}{\left(\pi+1\right)!}= \sum_{p=1}^{n}\binom{n+1}{p+1}\left(-1\right)^{p}\sum_{k_{1}+\dots+k_{p}=n}\frac{1}{\left(k_{1}+1\right)!\dots\left(k_{p}+1\right)!}
\]
with the notation
\[
\left(\pi+1\right)!=\prod_{i}\left(k_{i}+1\right)!.
\]
From the multinomial identity, we know that, in terms of Stirling numbers of the second kind,
\begin{equation}
\label{Stirling}
\sum_{k_{1}+\dots+k_{p}=n}\binom{n}{k_{1},\dots,k_{p}}\frac{1}{\left(k_{1}+1\right)\dots\left(k_{p}+1\right)} =\frac{n!p!}{\left(n+p\right)!}\left\{ \begin{array}{c}
n+p\\
p
\end{array}\right\} .
\end{equation}
We deduce the expression
\[
B_{n}=\sum_{p=1}^{n}\left(-1\right)^{p}\frac{\left\{ \begin{array}{c}
n+p\\
p
\end{array}\right\} }{\binom{n+p}{p}}\binom{n+1}{p+1},\,\,n \ge 1.
\]
Another expression for the Bernoulli numbers can be obtained by choosing
\[
g\left(z\right)=e^{z}-1=\sum_{n\ge1}\frac{z^{n}}{n!}
\]
and
\[
f\left(z\right)=\frac{\log\left(1+z\right)}{z}=\sum_{n\ge0}\frac{\left(-1\right)^{n}}{n+1}z^{n}
\]
giving
\begin{eqnarray}
\frac{B_{n}}{n!} &=&\sum_{\pi\in C\left(n\right)}\frac{\left(-1\right)^{\vert \pi \vert}}{\vert \pi \vert+1}\frac{1}{\pi!}\\
&=&\sum_{p=1}^{n}\frac{\left(-1\right)^p}{p+1}\sum_{\underset{k_{i}\ge1}{k_{1}+\dots+k_{p}=n}}\frac{1}{k_{1}!\dots k_{p}!} = \sum_{p=1}^{n}\frac{\left(-1\right)^p}{p+1}p!
\left\{
\begin{array}{c}
n\\p
\end{array}
\right\}
\end{eqnarray}
which can be found under a slightly different form as \cite[Entry 24.6.9]{NIST}.

We note that we have two distinct representations of the Bernoulli numbers as sums over compositions:
\[
\frac{B_{n}}{n!} =\sum_{\pi\in\mathcal{C}\left(n\right)}\frac{\left(-1\right)^{\vert \pi \vert}}{\left(\pi+1\right)!} =\sum_{\pi\in C\left(n\right)}\frac{\left(-1\right)^{\vert \pi \vert}}{\vert \pi \vert+1}\frac{1}{\pi!}.
\]
We also note that in \cite{Vella},
D.C. Vella obtains other expressions of the Bernoulli and Euler numbers
as sums over compositions.

\subsection{Back to Bernoulli polynomials}

We can now provide a general formula for the Bernoulli polynomials
as follows:
\begin{thm}
\label{Bernoulli expansion}
The Bernoulli polynomials can be expanded as
\[
B_{n}\left(x\right)=\sum_{p=1}^{n}\binom{n+1}{p+1}\left(-1\right)^{p}B_{n}^{\left(-p\right)}\left(-px\right),\thinspace\thinspace n\ge1.
\]
where $B_{n}^{\left(p\right)}\left(x\right)$ is the higher-order Bernoulli polynomial with parameter $p$ with generating function
\begin{equation}
\label{gf generalized Bernoulli}
\sum_{n\ge0}\frac{B_{n}^{\left(p\right)}\left(x\right)}{n!}z^{n}=e^{zx}\left(\frac{z}{e^{z}-1}\right)^{p}.
\end{equation}

\end{thm}
\begin{proof}
Using
\[
g_{n}=\frac{\left(-1\right)^{n+1}}{\left(n+1\right)!}\left(x^{n+1}-\left(x-1\right)^{n+1}\right)=\frac{\left(-1\right)^{n+1}}{n!}\int_{0}^{1}\left(x+u-1\right)^{n}du,
\]
we deduce
\[
x_{n}=\frac{B_{n}\left(x\right)}{n!}=\sum_{p=1}^{n}\binom{n+1}{p+1}\sum_{k_{1}+\dots+k_{p}=n}g_{k_{1}}\dots g_{k_{p}}
\]
with
\begin{align*}
\sum_{k_{1}+\dots+k_{p}=n}g_{k_{1}}\dots g_{k_{p}} & =\frac{\left(-1\right)^{n+p}}{n!}\sum_{k_{1}+\dots+k_{p}=n}\binom{n}{k_{1},\dots,k_{p}}\int_{0}^{1}\dots\int_{0}^{1}\prod_{i=1}^{p}\left(x+u_{i}-1\right)^{k_{i}}du_{1}\dots du_{p}\\
 & =\frac{\left(-1\right)^{n+p}}{n!}\int_{0}^{1}\dots\int_{0}^{1}\left(px-p+u_{1}+\dots+u_{p}\right)^{n}du_{1}\dots du_{p}.
\end{align*}
This sequence has generating function
\begin{align*}
\sum_{n\ge0}\left(\frac{\left(-1\right)^{n+p}}{n!}\int_{0}^{1}\dots\int_{0}^{1}\left(px-p+u_{1}+\dots+u_{p}\right)^{n}du_{1}\dots du_{p}\right)z^{n} & =\left(-1\right)^{p}e^{-z\left(px-p\right)}\left(\frac{1-e^{-z}}{z}\right)^{p}\\
 & =\left(-1\right)^{p}e^{-zpx}\left(\frac{e^{z}-1}{z}\right)^{p};
\end{align*}
comparing to the generating function of the higher-order Bernoulli
polynomials \eqref{gf generalized Bernoulli},
we deduce
\[
\sum_{k_{1}+\dots+k_{p}=n}g_{k_{1}}\dots g_{k_{p}}=\frac{\left(-1\right)^{p}}{n!}B_{n}^{\left(-p\right)}\left(-px\right)
\]
and
\[
B_{n}\left(x\right)=\sum_{p=1}^{n}\binom{n+1}{p+1}\left(-1\right)^{p}B_{n}^{\left(-p\right)}\left(-px\right)
\]
which is the desired result.
\end{proof}

\subsection{Back to higher-order Bernoulli polynomials}

We apply the result of Thm. \ref{thm:main} to higher-order Bernoulli numbers as follows: starting
from
\[
g_{n}=-\frac{1}{\left(n+1\right)!},\thinspace\thinspace g\left(z\right)=1-\frac{e^{z}-1}{z},
\]
the desired generating function is
\[
f\left(g\left(z\right)\right)=\left(\frac{z}{e^{z}-1}\right)^{q}-1=\left(\frac{1}{1-g\left(z\right)}\right)^{q}-1
\]
so that
\[
f\left(z\right)=\left(\frac{1}{1-z}\right)^{p}-1=\sum_{k\ge1}\binom{k+p-1}{p-1}z^{k}.
\]
Applying formula \eqref{eq:convolutions}, we deduce
\[
\frac{B_{n}^{\left(q\right)}}{n!}=\sum_{p=1}^{n}\left(\sum_{m=p}^{n}\binom{m}{p}\binom{m+q-1}{q-1}\right)\sum_{k_{1}+\dots+k_{p}=n}\frac{\left(-1\right)^{p}}{\left(k_{1}+1\right)!\dots\left(k_{p}+1\right)!}.
\]
The inner sum is easily computed as
\[
\sum_{m=p}^{n}\binom{m}{p}\binom{m+q-1}{q-1}=\frac{n-p+1}{p+q}\binom{n+1}{p}\binom{n+q}{q-1}
\]
whereas the last sum is, by \eqref{Stirling}, equal to
\[
\sum_{k_{1}+\dots+k_{p}=n}\frac{\left(-1\right)^{p}}{\left(k_{1}+1\right)!\dots\left(k_{p}+1\right)!}=\left(-1\right)^{p}\frac{p!}{\left(n+p\right)!}\left\{ \begin{array}{c}
n+p\\
p
\end{array}\right\}
\]
so that
\begin{align*}
B_{n}^{\left(q\right)} & =\binom{n+q}{q-1}\sum_{p=1}^{n}\left(-1\right)^{p}\frac{\left\{ \begin{array}{c}
n+p\\
p
\end{array}\right\} }{\binom{n+p}{p}}\left(\frac{n+1}{p+q}\right)\binom{n}{p}\\
 & =\sum_{p=1}^{n}\left(-1\right)^{p}\frac{\left\{ \begin{array}{c}
n+p\\
p
\end{array}\right\} }{\binom{n+p}{p}}\binom{n+q}{n-k}\binom{q+k-1}{k}.
\end{align*}
This identity can be found as \cite[Eq.(15)]{Srivastava}.

\subsection{Compositions with restricted summands}

When the input sequence $\{ g_{n} \}$ has a simple structure, more details can be obtained about the row-sum sequence $\{x_{n}\}.$
We start with the example of Fibonacci numbers,
by explicitly constructing Woon's tree for the Fibonacci case
\[
x_{n}=F_{n}
\]
with
\[
F_{0}=1,\thinspace\thinspace F_{1}=1,\thinspace\thinspace F_{2}=2,\thinspace\thinspace F_{3}=3,\thinspace\thinspace F_{4}=5,\thinspace\thinspace F_{5}=8\dots
\]
Since the Fibonacci numbers satisfy the recurrence
\[
F_{n}=F_{n-1}+F_{n-2},\,\,n \ge 2,
\]
we deduce from (\ref{restricted}) that the sequence of entries
of the tree satisfies
\[
g_{k}=\begin{cases}
1 & k=1,2\\
0 & \text{else}
\end{cases}.
\]
The corresponding Woon's tree starts as
\begin{center}
\begin{forest}
[$1$[$1$[$1$[$1$][$1$]][$1$[$1$][$0$]]][$1$[$1$[$1$][$1$]][$0$[$0$][$0$]]]]
\end{forest}
\par\end{center}
\begin{flushleft}
We deduce the expression for the Fibonacci numbers as
\[F_{n}=\mathcal{C}_{n}^{\left\{ 1,2\right\}}=\#\left\{ \left(k_{1},\dots,k_{m}\right)\vert k_{1}+\dots+k_{m}=n,\thinspace\thinspace k_{i}=1,2\right\},
\]
the number of compositions of $n$ with parts equal to $1$ or $2$, which can be found in \cite[p.42]{Flajolet} and \cite{Alladi}. Moreover, we recover the generating function $\frac{z+z^2}{1-z-z^2} = \sum_{n\geq 1} F_n z^n$.
\par\end{flushleft}

This result can be generalized remarking that we have a correspondence between linear recurrences and compositions into restricted parts. Fix a set $J$ of integers: we adopt the notation
\[
\mathcal{C}_{n}^{\left\{J\right\} }=\#\left\{ \left(k_{1},\dots,k_{m}\right)\vert k_{1}+\dots+k_{m}=n,\thinspace\thinspace k_{i}\in J\right\}
\]
from \cite[p.42]{Flajolet} to denote the number of compositions of $n$ with parts in the set $J$.

\begin{thm}\label{restricted}
Let $J $ be a finite set of positive indices and consider the linear recurrence $x_n  =  \sum_{j \in J} x_{n-j}$. Then we have the dual identities:
\begin{equation}\label{eq:comp}
x_n  = \mathcal{C}_{n}^{\left\{J\right\} },
\end{equation}
\begin{equation}\label{eq:compgen}
\sum_{n \geq 1} x_n z^n  =\frac{\sum_{j \in J}z^j}{1-\sum_{j \in J}z^j},
\end{equation}
Furthermore, the corresponding Woon tree has input sequence
\begin{equation}\label{eq:comptree}
g_{k}=\begin{cases}
1 & k \in J\\
0 & \text{else}
\end{cases}.
\end{equation}
\end{thm}
\begin{proof}
We begin with the recurrence \eqref{eq:recursion}, $x_n = \sum_{j=1}^n g_j x_{n-j}$, where $x_n$ corresponds to the $n-$th row sum of Woon's tree. Since $x_n  =  \sum_{j\in J} x_{n-j}$, the set $J$ consists of the indices $i$ such that $g_i =1$. We also have that $g_i=0$ everywhere else. This sequence then satisfies the conditions to be considered a generalized Woon tree, so we have the realization \eqref{eq:comptree}. We can then apply the transform $x(z) = \frac{g(z)}{1-g(z)}$, which yields \eqref{eq:compgen}, since $g(z)=\sum_{n \geq 1}g_n z^n = \sum_{j \in J}z^j$. Finally, we can apply \eqref{eq:convolutions} with $f_0 =0$ and $f_n=1$ to yield
\begin{equation*}
\frac{g(z)}{1-g(z)} =  \sum_{n \geq 1} z^n \sum_{\underset{k_{i}\ge1,\thinspace m\ge1}{k_{1}+\dots+k_{m}=n}} g_{k_{1}}\dots g_{k_{m}} =   \sum_{n \geq 1} z^n \sum_{\underset{k_{i}\ge1,\thinspace k_i \in J,\thinspace m\ge1}{k_{1}+\dots+k_{m}=n}} 1 = \sum_{n \geq 1} \mathcal{C}_{n}^{\left\{J\right\} }z^n.
\end{equation*}
Comparing coefficients yields \eqref{eq:comp}.
\end{proof}

A direction for future research is to study the asymptotics of $x_n = \sum x_{j_i}$ which would allow us to obtain asymptotics for $\mathcal{C}_{n}^{\left\{J\right\} }$. The asymptotics of $x_n = \sum g_i x_{n-i}$ are complicated, but we have a much easier problem since the coefficients are either zero or one. Finding the asymptotics of $x_n = \sum g_i x_{n-i}$ reduces to studying the roots of the characteristic polynomial $x^n - \sum {x^{j_i}}$.

\subsection{Sum of digits}
For a given $n,$ the set of $2^{n-1}$ integers
$
\left\{ \vert \pi \vert\right\} _{\pi\in\mathcal{C}\left(n\right)}
$
coincides with the set
$
\left\{ 1+s_{2}\left(k\right)\right\} _{0\le k\le2^{n-1}-1},
$
where $s_{2}\left(k\right)$ is the number of $1'$s in the binary
expansion of $k.$

For example, for $n=3,$
\[
\left\{ \vert \pi \vert\right\} _{\pi\in\mathcal{C}\left(3\right)}=\left\{ 1,2,2,3\right\}
\]
while
\[
s_{2}\left(0\right)=0,\thinspace\thinspace s_{2}\left(1\right)=1,\thinspace\thinspace s_{2}\left(2\right)=1,\thinspace\thinspace s_{2}\left(3\right)=2.
\]

This remark, together with Theorem \ref{non linear}, can be used to derive some interesting finite sums that involve the sequence $s_{2}\left(k\right).$
For example, the choice
\[
f\left(z\right)=\log\left(1+z\right),\thinspace\thinspace g\left(z\right)=\frac{z}{1-z}
\]
gives
\[
f\left(g\left(z\right)\right)=-\log\left(1-z\right)
\]
so that
\[
x_{n}=\frac{1}{n}=\sum_{k=0}^{2^{n-1}-1}\frac{\left(-1\right)^{s_{2}\left(k\right)}}{s_{2}\left(k\right)+1},\thinspace\thinspace n\ge1
\]
This extends naturally as follows:
\begin{thm}
For $f\left(z\right) = \sum_{k \ge 1} f_k z^k,$ we have
\[
\sum_{k=0}^{2^{n-1}-1} f_{s_2\left(k\right)+1} = \left[z^n\right] f\left(\frac{z}{1-z}\right) = \sum_{k=1}^{n} f_k \binom{n-1}{n-k}.
\]
\end{thm}
This formula simply expresses the fact that for $0 \le k \le 2^{n-1}-1,$ the sequence $\left\{s_2\left(k\right)\right\}$ takes on the value zero $\binom{n-1}{1}$ times, the value one $\binom{n-2}{2}$ times, and so on.

\section{A general formula for composition of functions}\label{sec4}
We conclude this study with a general formula for composition of functions.
Introduce the notation
\[
\pi\models n
\]
for $\pi\in\mathcal{C}\left(n\right).$ This mirrors the notation $\lambda \vdash  n$, which states that $\lambda$ is a partition of $n$. From Thm. \ref{thm:main}, we have the formula
\[
\left(f\circ g\right)_{n}=\sum_{\pi\models n}f_{\vert \pi \vert}g_{\pi}.
\]
Now we look at
\[
\left(f\circ\left(g\circ h\right)\right)_{n}=\sum_{\pi\models n}f_{\vert \pi \vert}\left(g\circ h\right)_{\pi}
\]
where
\begin{align*}
\left(g\circ h\right)_{\pi} & =\left(g\circ h\right)_{\pi_{1}}\dots\left(g\circ h\right)_{\pi_{p}}\\
 & =\left(\sum_{\mu_{1}\models\pi_{1}}g_{l\left(\mu_{1}\right)}h_{\mu_{1}}\right)\dots\left(\sum_{\mu_{p}\models\pi_{p}}g_{l\left(\mu_{p}\right)}h_{\mu_{p}}\right)\\
 & =\sum_{\mu_{1}\models\pi_{1}}\dots\sum_{\mu_{p}\models\pi_{p}}g_{l\left(\mu_{1}\right)}\dots g_{l\left(\mu_{p}\right)}h_{\mu_{1}}\dots h_{\mu_{p}}.
\end{align*}
We introduce the new notation for a ``composition of compositions''
\[
\sum_{\mu\models\pi}x_{\mu}=\sum_{\mu_{1}\models\pi_{1}}\dots\sum_{\mu_{p}\models\pi_{p}}x_{\mu_{1}}\dots x_{\mu_{p}}
\]
so that we can write
\[
\left(f\circ\left(g\circ h\right)\right)_{n}=\sum_{\mu\models\pi\models n}f_{\vert \pi \vert}g_{\vert \mu \vert}h_{\mu}
\]
Moreover, the associativity of function compositions translates into
another equivalent expression,
\begin{align*}
\left(\left(f\circ g\right)\circ h\right)_{n} & =\sum_{\substack{\pi \models n \\ \mu\models \vert \pi \vert}}f_{\vert \mu \vert}g_{\mu}h_{\pi}.
\end{align*}
For $n=4$ we have the 5 equivalent possibilities
\begin{eqnarray*}
f^{(1)}\circ f^{(2)}\circ f^{(3)}\circ f^{\left(4\right)} & =f^{\left(1\right)}\circ\left(f^{\left(2\right)}\circ\left(f^{\left(3\right)}\circ f^{\left(4\right)}\right)\right)=\sum_{\substack{\pi\models n\\
\mu\models\pi\\
\lambda\models\mu
}
}f_{\vert \pi \vert}^{\left(1\right)}f_{\vert \mu \vert}^{\left(2\right)}f_{\vert \lambda \vert}^{\left(3\right)}f_{\lambda}^{\left(4\right)}\\
& =\left(\left(f^{\left(1\right)}\circ f^{\left(2\right)}\right)\circ f^{\left(3\right)}\right)\circ f^{\left(4\right)}=\sum_{\substack{\pi\models n\\
\mu\models \vert \pi \vert\\
\lambda\models \vert \mu \vert
}
}f_{\vert \lambda \vert}^{\left(1\right)}f_{\lambda}^{\left(2\right)}f_{\mu}^{\left(3\right)}f_{\pi}^{\left(4\right)}\\
 & =\left(f^{\left(1\right)}\circ f^{\left(2\right)}\right)\circ\left(f^{\left(3\right)}\circ f^{\left(4\right)}\right)=\sum_{\substack{\pi\models n\\
\mu\models \vert \pi \vert\\
\lambda\models\pi
}
}f_{\vert \mu \vert}^{\left(1\right)}f_{\mu}^{\left(2\right)}f_{\vert \lambda \vert}^{\left(3\right)}f_{\lambda}^{\left(4\right)}\\
 & =f^{\left(1\right)}\circ\left(\left(f^{\left(2\right)}\circ f^{\left(3\right)}\right)\circ f^{\left(4\right)}\right)=\sum_{\substack{\pi\models n\\
\mu\models\pi\\
\lambda\models \vert \mu \vert
}
}f_{\vert \pi \vert}^{\left(1\right)}f_{\vert \lambda \vert}^{\left(2\right)}f_{\lambda}^{\left(3\right)}f_{\mu}^{\left(4\right)}\\
 & =\left(f^{\left(1\right)}\circ\left(f^{\left(2\right)}\circ f^{\left(3\right)}\right)\right)\circ f^{\left(4\right)}=\sum_{\substack{\pi\models n\\
\mu\models \vert \pi \vert\\
\lambda\models\mu
}
}f_{\vert \mu \vert}^{\left(1\right)}f_{\vert \lambda \vert}^{\left(2\right)}f_{\lambda}^{\left(3\right)}f_{\pi}^{\left(4\right)}.
\end{eqnarray*}
with the corresponding tree representations

\vspace{1cm}
\noindent\begin{minipage}[t]{1\columnwidth}%
\begin{minipage}[t]{0.4\columnwidth}%
\begin{forest}calign=fixed edge angles,
[$f^{\left(1\right)}  \circ  \left( f^{\left(2\right)} \circ \left( f^{\left(3\right)} \circ f^{\left(4\right)} \right) \right)$[$f^{\left(1\right)}$,  edge label={node[midway,left,font=\scriptsize]{$\vert \pi \vert$}}][$f^{\left(2\right)}  \circ  \left( f^{\left(3\right)} \circ f^{\left(4\right)} \right)$,  edge label={node[midway,right,font=\scriptsize]{$\pi$}}[$f^{\left(2\right)}$,  edge label={node[midway,left,font=\scriptsize]{$\vert \mu \vert$}}][$f^{\left(3\right)} \circ f^{\left(4\right)}$,  edge label={node[midway,right,font=\scriptsize]{$\mu$}}[$f^{\left(3\right)}$,  edge label={node[midway,left,font=\scriptsize]{$\vert \lambda \vert$}}][$f^{\left(4\right)}$,  edge label={node[midway,right,font=\scriptsize]{$\lambda$}}]]]]
\end{forest}
\end{minipage}
\begin{minipage}[t]{0.4\columnwidth}%
\begin{forest}calign=fixed edge angles, [$\left(\left(f^{\left(1\right)}\circ f^{\left(2\right)}\right)\circ f^{\left(3\right)}\right)\circ f^{\left(4\right)}$[$\left(f^{\left(1\right)}\circ f^{\left(2\right)}\right)\circ f^{\left(3\right)}$,  edge label={node[midway,left,font=\scriptsize]{$\vert \pi \vert$}}[$f^{\left(1\right)}\circ f^{\left(2\right)}$,  edge label={node[midway,left,font=\scriptsize]{$\vert \mu \vert$}}[$f^{\left(1\right)}$,  edge label={node[midway,left,font=\scriptsize]{$\vert \lambda \vert$}}][$f^{\left(2\right)}$,  edge label={node[midway,right,font=\scriptsize]{$\lambda$}}]][$f^{\left(3\right)}$,  edge label={node[midway,right,font=\scriptsize]{$\mu$}}]][$f^{\left(4\right)}$,  edge label={node[midway,right,font=\scriptsize]{$\pi$}}]] \end{forest}
\end{minipage}%
\begin{minipage}[t]{0.33\columnwidth}%
\begin{forest}calign=fixed edge angles,
[$\left(f^{\left(1\right)}  \circ   f^{\left(2\right)} \right) \circ \left( f^{\left(3\right)} \circ f^{\left(4\right)}  \right)$[$f^{\left(1\right)} \circ f^{\left(2\right)}$,  edge label={node[midway,left,font=\scriptsize]{$\vert \pi \vert$}}[$f^{\left(1\right)}$,  edge label={node[midway,left,font=\scriptsize]{$\vert \mu \vert$}}][$f^{\left(2\right)}$,  edge label={node[midway,right,font=\scriptsize]{$\mu$}}]][$ f^{\left(3\right)} \circ f^{\left(4\right)} $,  edge label={node[midway,right,font=\scriptsize]{$\pi$}}[$f^{\left(3\right)}$,  edge label={node[midway,left,font=\scriptsize]{$\vert \lambda \vert$}}][$f^{\left(4\right)}$,  edge label={node[midway,right,font=\scriptsize]{$\lambda$}}]]]
\end{forest}
\end{minipage}
\end{minipage}
\vspace{1cm}
\noindent\begin{minipage}[t]{1\columnwidth}%
\begin{minipage}[l]{0.33\columnwidth}%
\begin{center}
\begin{forest}calign=fixed edge angles,
[$f^{\left(1\right)}\circ \left(\left(f^{\left(2\right)}\circ f^{\left(3\right)}\right)\circ f^{\left(4\right)}\right)$
[$f^{\left(1\right)}$,  edge label={node[midway,left,font=\scriptsize]{$\vert \pi \vert$}}]
[$\left(f^{\left(2\right)}\circ f^{\left(3\right)}\right)\circ f^{\left(4\right)}$,  edge label={node[midway,right,font=\scriptsize]{$\pi$}}[$f^{\left(2\right)}\circ f^{\left(3\right)}$,  edge label={node[midway,left,font=\scriptsize]{$\vert \mu \vert$}}[$ f^{\left(2\right)}$,  edge label={node[midway,left,font=\scriptsize]{$\vert \lambda \vert$}}][$ f^{\left(3\right)}$,  edge label={node[midway,right,font=\scriptsize]{$\lambda$}}]][$ f^{\left(4\right)}$,  edge label={node[midway,right,font=\scriptsize]{$\mu$}}]]]
\end{forest} \par\end{center}%
\end{minipage}%
\begin{minipage}[r]{0.33\columnwidth}%
\begin{center} \begin{forest}calign=fixed edge angles,
[$\left(f^{\left(1\right)}\circ \left(f^{\left(2\right)}\circ f^{\left(3\right)}\right)\right)\circ f^{\left(4\right)}$
[$f^{\left(1\right)}\circ \left(f^{\left(2\right)}\circ f^{\left(3\right)}\right)$,  edge label={node[midway,left,font=\scriptsize]{$\vert \pi \vert$}}[$f^{\left(1\right)}$,  edge label={node[midway,left,font=\scriptsize]{$\vert \mu \vert$}}][$f^{\left(2\right)}\circ f^{\left(3\right)}$,  edge label={node[midway,right,font=\scriptsize]{$\mu$}}[$f^{\left(2\right)}$,edge label={node[midway,left,font=\scriptsize]{$\vert \lambda \vert$}}][$f^{\left(3\right)}$,edge label={node[midway,right,font=\scriptsize]{$\lambda$}}]]][$f^{\left(4\right)}$,  edge label={node[midway,right,font=\scriptsize]{$\pi$}}]]
\end{forest} \par\end{center}%
\end{minipage}
\end{minipage}

Consider now the general case $f^{(1)}\circ\cdots\circ f^{(n)}$.
There are $C_{n+1}$ ways to compose these functions associatively,
where $C_{n}$ is the $n-$th Catalan number
\[
C_{n}=\frac{1}{2n+1}\binom{2n}{n}.
\]
The associativity of function composition gives us then $C_{n+1}-1$ equivalent representations for generalized composition sums.

While writing down each of these $C_{n+1}$ representations is very messy, we present a simple algorithmic approach for doing so. The general formula for the composition $f^{\left(1\right)} \circ \dots f^{\left(n\right)}$ is
\[
\sum_{  \substack{  \pi_1 \models n \\ \pi_2 \models \pi_1^{\pm} \\ \dots \\ \pi_{n-1} \models \pi_{n-2}^{\pm}  }   }
f_{\pi_{i_1}^{\pm}}^{\left(1\right)}
f_{\pi_{i_2}^{\pm}}^{\left(2\right)}
\dots
f_{\pi_{i_n}^{\pm}}^{\left(n\right)},
\]
with the notation $\pi_{i}^{+}=\pi_{i}$ and $\pi_{i}^{-}=\vert \pi_{i} \vert$. We draw the tree representation for function composition and determine the summation set by the following rules:
\begin{itemize}
\item
If the parent of the level corresponding to $\pi_{i}$ is a right child, we sum over $\pi_{i-1}^+ = \pi_{i-1}$. If it is a left child we sum over $\pi_{i-1}^- = \vert \pi_{i-1} \vert$
\item
Select $f^{(i)}$ and look where it is a leaf. If it is a right child, we sum over the factor $f^{(i)}_{\pi_j}$, where $j$ is the depth of the leaf $f^{(i)}$. If $f^{(i)}$ is a leaf and left child we sum over the factor $f^{(i)}_{\vert \pi_j \vert}$.
\item
At the top level, the previous rules do not apply and we always sum over $f^{(i)}_{\pi_1}$ and $\pi_1 \models n$.
\end{itemize}
This way, the subscript is determined by the \textit{leaf's} position while the summation set is determined by the \textit{parent's} position in the corresponding tree. For instance, select the following tree:

\begin{center} \begin{forest}calign=fixed edge angles,
[$\left(f^{\left(1\right)}\circ \left(f^{\left(2\right)}\circ f^{\left(3\right)}\right)\right)\circ f^{\left(4\right)}$
[$f^{\left(1\right)}\circ \left(f^{\left(2\right)}\circ f^{\left(3\right)}\right)$,  edge label={node[midway,left,font=\scriptsize]{$\vert \pi_1 \vert$}}[$f^{\left(1\right)}$,  edge label={node[midway,left,font=\scriptsize]{$\vert \pi_2 \vert$}}][$f^{\left(2\right)}\circ f^{\left(3\right)}$,  edge label={node[midway,right,font=\scriptsize]{$\pi_2$}}[$f^{\left(2\right)}$,edge label={node[midway,left,font=\scriptsize]{$\vert \pi_3 \vert$}}][$f^{\left(3\right)}$,edge label={node[midway,right,font=\scriptsize]{$\pi_3$}}]]][$f^{\left(4\right)}$,  edge label={node[midway,right,font=\scriptsize]{$\pi_1$}}]].
\end{forest} \par\end{center}%
Consider $f^{(1)}$. It's a left child with depth $2$, so it has the subscript $\vert \pi_2 \vert$. Its parent $f^{\left(1\right)}\circ \left(f^{\left(2\right)}\circ f^{\left(3\right)}\right)$ is also a left child so the summation includes the terms $\pi_2 \models \vert \pi_1 \vert$. $f^{(3)}$ is a right child with depth $3$ so we sum over $f^{(3)}_{\pi_3}$. Its parent is a right child so the summation goes over $\pi_3 \models \pi_2$. Repeating this procedure for $f^{(2)}$ and $f^{(4)}$ allows us to recover our previous expression
\begin{equation}
\left(f^{\left(1\right)}\circ\left(f^{\left(2\right)}\circ f^{\left(3\right)}\right)\right)\circ f^{\left(4\right)}=\sum_{\substack{\pi_1\models n\\
\pi_2\models \vert \pi_1 \vert\\
\pi_3\models\pi_2
}
}f_{\vert \pi_2 \vert}^{\left(1\right)}f_{\vert \pi_3 \vert}^{\left(2\right)}f_{\pi_3}^{\left(3\right)}f_{\pi_1}^{\left(4\right)}.
\end{equation}

\section{Conclusion}
In this paper, we recognized Fuchs' generalized PI tree as a graphical method to represent sums over compositions. Then, the introduction of the Fa\'{a} di Bruno formula allowed us to introduce  further set of weights based on the number of parts in every compositions. Trees are then an easy 'bookkeeping' method to visualize the classical Fa\'{a} di Bruno formula. The introduction of this Fa\'{a} di Bruno formula also allowed us to synthesize many past works, since the nonlinear weights $f_n$ generalize most of the existing literature on compositions. Through this generating function methodology,
we were also able to find a generating function for sums over all parts of all compositions of $n$. Together, our results unite the existing literature on composition sums and provide an efficient method to graphically visualize them. For the first time, we also study iterated sums over compositions and link them to tree representations.

redpajama_set_name: RedPajamaArXiv


text:
Ontario reports 184 new COVID-19 cases, hospitalizations continue to drop
Ontario's government reported 184 new COVID-19 cases on Tuesday, while the number of people hospitalized with the illness continues on a downward trend.
The 0.6 per cent increase in cases brings the cumulative total in Ontario to 32,554, and marks the third day in a row where Ontario is reporting fewer than 200 cases.
Of those confirmed cases, just over 84 per cent are resolved. Some 218 cases were marked resolved on Monday.
The province completed 21,724 tests yesterday, with 17,920 currently under investigation. That's the fewest number of tests seen in the province in six days, though Minister of Health Christine Elliott said in a tweet Tuesday that "Ontario remains the leader in daily COVID-19 testing in Canada and our positivity rate continues to be at all-time lows."
Ontario has now completed over a million COVID-19 tests since the start of the pandemic.
There are 413 patients currently hospitalized with the illness, down slightly from 419 the day before. The number of patients in intensive care units also dropped slightly, down to 98 from 104. Elliott said on Twitter that this marks the first time ICU admissions have been below 100 in the province since March 29.
The number of people on a ventilator in Ontario went up one, to 70.
"Locally, 27 of Ontario's 34 public health units reported five or fewer cases, with 14 of them reporting no new cases at all," Elliott said. "We're continuing to monitor case numbers in Toronto, Peel and Windsor as we assess when they might be ready to enter Stage 2."
Those three health units will be the only ones in the province still in the first stage of reopening as of this Friday, with the province poised to ease restrictions elsewhere.
At a news conference Tuesday, Premier Doug Ford said that as Ontario starts to reopen, the province has to ensure adequate hospital capacity, rebuild personal protective equipment stockpiles, keep testing numbers high, and put in workplace health and safety measures.
"We have to get it right. The worst thing we could do is rush into reopening," Ford said.
Ford also announced a provincial COVID-19 safety plan guide for businesses with sector-specific workplace safety guidelines. He asked Ontarians to shop local and plan vacations at provincial destinations this summer.
"It's up to all of us to support our local businesses," Ford said.
Ontario's official COVID-19 death toll grew by 11 and is now 2,538. A CBC News count based on data from regional public health units puts the real toll at at least 2,579.
The highest number of confirmed cases is found in the 40 to 59 age bracket, at 9,974. A close second, however, is the 20 to 39 age bracket, at 9,115.
Researchers at the University of Guelph have found numbers of new cases continue to climb among people under the age of 20 in Ontario, making it the group increasing at the highest rate.
Provincial data shows 1,422 cases in the 19 and under age bracket.
The province is reporting 67 long-term care homes with active outbreaks, down two from the day before, and 240 homes with resolved outbreaks.
Ontario is also reporting 458 active cases in long-term care residents, down from 490 on Monday. Resident deaths now stand at 1,794 according to provincial data, up two from the day before.
Legal groups launch court application over sharing of COVID-19 test data with police
Ontario reports 111 new cases of COVID-19, as parts of province prepare for stage 3 of reopening

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Concatenating crosstabs of different variables I have a Pandas (0.23.4) DataFrame with several categorical columns.
df = pd.DataFrame(np.random.choice([True, False, np.nan], (6,4)), columns = ['a','b','c','d'])

     a    b    c    d
0  NaN  1.0  NaN  NaN
1  NaN  1.0  NaN  0.0
2  1.0  NaN  1.0  NaN
3  0.0  NaN  0.0  1.0
4  NaN  1.0  NaN  NaN
5  NaN  1.0  0.0  1.0

I have two sets of columns of interest:
cross_cols = ['a', 'b']
type_cols = ['c', 'd']

I would like to get a cross tab of counts of each cross_col variable with each type_col variable (a with c and d, and b with c and d), excluding NaN, all displayed side-by-side. The desired result is:
         c             d
       0.0  1.0  All 0.0 1.0 All
a 0.0  0    0    0     1   1   2
  1.0  2    1    3     1   0   1
  All  2    1    3     2   1   3
b 0.0  0    0    0     0   1   1
  1.0  2    1    3     2   0   2
  All  2    1    3     2   1   3

Notice that I am not interested in counts for different combinations of a and b or of c and d, which is what I'm getting by changing the index and columns parameters of pd.crosstab.
Currently I'm using the following code:
cross_rows = []
for col in cross_cols:
    cross_rows.append(pd.concat([pd.crosstab(df[col], df[type_var],margins=True) for type_var in type_cols],axis=1,keys = type_cols,sort=True))
results = pd.concat(cross_rows, keys = cross_cols,sort=True)

It gives the following result:
         c             d
c      0.0  1.0  All 0.0 1.0 All
a 1.0  2.0  1.0  3.0   1   0   1
  All  2.0  1.0  3.0   2   1   3
  0.0  NaN  NaN  NaN   1   1   2
b 1.0  2.0  1.0  3.0   2   0   2
  All  2.0  1.0  3.0   2   1   3
  0.0  NaN  NaN  NaN   0   1   1

The result is fine, but the code is slow and a bit ugly. I suspect that there's a faster and more Pythonic approach. Is there a single function call that would get the job done, or another faster solution?

redpajama_set_name: RedPajamaStackExchange


text:
Home/Sports/EPL: Lingard rejects new Man United contract
EPL: Lingard rejects new Man United contract
Manchester United forward, Jesse Lingard, has rejected a new contract at the club, according to ESPN.
Lingard is seeking assurances about his first team prospects this season.
The 28-year-old is into the last months of his current deal and has been offered an extension.
However, Lingard has turned down United's initial approach and is keen to ensure he will get the opportunity to play regularly before making a decision about his future.
If a new contract is not agreed with United, he will be able to leave as a free agent at the end of the season.
Lingard will also be able to negotiate with clubs outside England from January.
The England international is determined to find regular football in order to keep his place in Gareth Southgate's England squad ahead of the 2022 World Cup in Qatar.
He narrowly missed out on a place in the squad for Euro 2020.
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Let our splatter screen keep your stove clean as a whistle while cooking foods like fried chicken or bubbling pasta sauce. The 13-inch stainless steel screen fits over large cooking pots to contain spits and splatters that normally end up all over your stove, and the long handle helps keep hands safely in the clear.
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The Beckman Coulter Allegra X-30R Refrigerated Centrifuge gives you high performance in an easy to operate and space saving design. This high-speed benchtop centrifuge tops out at 18000 RPM, with a maximum g force of 29765 x g when using a fixed angle rotor. The Allegra X-30R can accommodate volumes of up to 400 mL, and has a temperature range of between -20°C to 40°C adjustable in 1°C increments. This refrigerated benchtop centrifuge is an ideal choice for labs dealing in general research, or sample preparation, and features quick runs and simple set ups. At only 18" wide, this benchtop centrifuge won't take up too much bench and with a noise output of <68 dba it shouldn't distract you either.
The Beckman Allegra X30R Refrigerated Centrifuge uses a reliable brushless induction drive, and features 10 acceleration and deceleration profiles as well as storing up to 20 user defined profiles. The automatic rotor detection system of this Beckman centrifuge helps make sure your parameters are correcting before running. This benchtop centrifuge can maintain temperatures for up to 10 minutes after the completion of a run, alerting you to the completion with an audible tone, and giving you time to collect your samples.
This Beckman Coulter Allegra X-30R Refrigerated Centrifuge has been fully inspected and reached its max speed and temperatures without issue. ALl of our lab equipment comes backed with our standard warranty. Should you have any questions, please don't hesitate to contact us.


text:
James Gunn Confirms DC TV Series After Guardians 3
Campbell Clark
Where will James Gunn go after he makes Guardians 3 and the Holiday Special? Now we know the answer as James Gunn confirms DC TV series after Guardians 3 in a recent interview with Collider. To be fair, from the outside it always looked likely that after Guardians 3, Gunn would go back to the DCEU. What we did not know was whether that would be for a movie, or another TV series like Peacemaker? Gunn has now confirmed it will be another TV show.
Previously Gunn has expressed interest in more spin-off shows from The Suicide Squad like Peacemaker, but of course for now, he's not sharing details.
'When asked if he was involved with another project for DC, Gunn tried to retain the mystery about his future productions. In Gunn's words: "You know, we're stepping forward. It keeps happening. We'll see what's happening, and I'll know what's happening very shortly. I dealt with it last night and this morning." However, once asked directly if this secret project would be for TV or film, Gunn gave a direct reply: "It's TV. There you go."'
RELATED: Peacemaker Release In Britain And India Update From James Gunn
I really enjoyed The Suicide Squad and I am looking forward to seeing Peacemaker when it eventually hits my shores. I'd be happy to see a TV series about many of the surviving, or even non-surviving players in that story. What about a King Shark and Ratcatcher II show? I'd watch the heck out of that?
However until we know more about these plans, (which we won't for a while) it's all open for speculation. So get to speculating. What are your hopes for next James Gunn led TV series? Or just leave any thoughts about the news as James Gunn confirms DC TV series after Guardians 3.
Tags: DC Extended Universe James Gunn
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StarKist Co., owned by Dongwon Industries Co. Ltd.., says it's newest product launch is Chicken Creations in pouches, which will be available nationally this summer. After 100 years in the tuna business, the company marks the launch as its first foray outside of the seafood space. The line of premium white chicken is available in four flavors in StarKist's signature ready-to-eat pouches. Flavors include Bold Buffalo Style, Ginger Soy, Zesty Lemon Pepper and Chicken Salad.
"It is our goal to provide convenient, healthy options that satisfy the palates and nutritional needs for all Americans," said Andy Mecs, vice president of marketing and innovation at StarKist Co. "Chicken Creations is a natural extension to our highly successful Tuna and Salmon Creations, as we already have the pouch expertise and consumer trust in our high quality flavored protein products. We are pleased consumers will now have even more to love, as they can tear, eat; and go."
The premium, white meat chicken is made with chicken sourced from American farms and raised cage- and hormone-free. With protein snacking on the rise, StarKist Chicken Creations pouches tie into consumer demands for taste, portability and convenience. The pouch has a tear-open, no-drain design for those on the move. Each pouch of chicken has 70-90 calories and 9 or 10g of protein.
The Chicken Creations launch is part of a fully integrated marketing campaign, with TV commercials, print media, influencer outreach, social media promotions and couponing.


The UK competition authority is likely to ask for more store disposals than the merging parties are willing to allow in order for the merger to continue.
Whether or not the competition regulator sees the Aldi and Lidl supermarkets as suitable places for a "one-stop shop" is critical to the outcome of the proposed merger between ASDA and Sainsbury's.
Meanwhile, a GlobalData survey has found that almost as many grocery shoppers think Aldi (60%) and Lidl (55%) are suitable places for doing a big food shop as the big four (Tesco, Asda, Morrisons and Sainsbury's).
If all were available to you (located nearby), which of the following supermarkets would you consider suitable for doing a 'big food shop'?
Sample: 1,593 grocery shoppers who do a 'big food shop' once a week or less frequently.
The UK's Competition and Markets Authority (CMA) is currently in phase two of its investigation and is considering the impact on competition that the deal would have in each location where ASDA and Sainsbury's stores overlap. In the past, the CMA has considered only stores greater than 1,400sqm to be suitable places for "one-stop shops", where a grocery consumer could be considered to get everything they need.
While Aldi and Lidl have grown in market share over the last 10 years – opening stores in many of the catchments that the CMA is looking at – most of their stores are smaller than the 1,400sqm threshold.
Too high a number of disposals?
If the CMA uses the same distinction, and rules that Aldi and Lidl stores typically do not offer competition to larger stores, then it is likely to decide that many more catchment areas will have competitiveness significantly eroded by the merger, and demand a far higher number of disposals. That is, the CMA may rule that for the merger to go ahead the retailers must sell some of their stores in the interest of broader competition.
If the weighted share of shops methodology is used, instead of the four to three fascia test (in past supermarket investigations, a local market was deemed problematic where the number of competing players fell from four to three or less), it gives the CMA the ability to weight competitors by other factors, such as alignment of customer demographic, and closeness to the Asda and Sainsbury's stores.
But if it excludes all Aldi and Lidl stores under 1,400sqm from the analysis of Asda and Sainsbury's larger stores, then we believe that the CMA will ask for more disposals than the merging parties are willing to allow in order for the merger to continue.


text:
Entertainment Lizzo Gets Candid About the Stigma of Pop Music: "Genre's Racist Inherently"
09:50 26 november 2022 Source: msn.com
Lizzo shows off her Chrisean Rock look, for Halloween [VIDEO]
Lizzo dresses up as Chrisean Rock for Halloween outfit Halloween is almost here, as many celebrities have already began to show off their costumes. One of them is Lizzo, as she went viral with hers, on Friday. Lizzo went on to dress up as Chrisean Rock, who is in an on and off relationship with […] The post Lizzo shows off her Chrisean Rock look, for Halloween [VIDEO] appeared first on Hip Hop Vibe.
© Provided by The Hollywood Reporter
Lizzo recently opened up about the stigma of pop music, explaining the difference between pop music and race music and how the latter was used to segregate Black artists.
"Genre's racist inherently," the Emmy-winning artist told Entertainment Weekly. "I think if people did any research they would see that there was race music and then there was pop music. And race music was their way of segregating Black artists from being mainstream because they didn't want their kids listening to music created by Black and brown people because they said it was demonic and yada, yada, yada."
10 best rhythm games on Android
See games that end up as music to your earsThis is why we've compiled a list of some of the best rhythm games you can find on Android, offering well-polished gameplay for all the finger-tapping exercises you're about to do.
Charles Koppelman, Former Music Exec and Martha Stewart Living Chairman, Dies at 82
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She went on to explain that genres were created almost like code words, for example with R&B, which eventually led to the birth of hip-hop and rap.
"I think when you think about pop, you think about MTV in the '80s talking about 'We can't play rap music,' or 'We can't put this person on our platform because we're thinking about what people in the middle of America think' — and we all know what that's code for," Lizzo continued.
Now that pop is a "well-oiled" machine, she pointed out that it's important to remember that it has a "racist origin." As music has evolved, Lizzo said she finds it "so cool" that hip-hop and rap artists have made their way into the pop music sphere. "Rap is running the game," she added.
What is a good station for Christmas music? A guide for you and your radio dial
The holiday season is upon us. Need some festive tunes to get you in the Christmas spirit? We've got you covered. Here's how to listen on the radioTo get into the spirit, it might be time to turn your radio dial to the Christmas music station. Whether you're a fan of Elvis Presley's crooning "Blue Christmas" or prefer the modern synth of Wham's "Last Christmas," the right holiday tune can almost invariably be found on the airwaves come December.
In that same conversation, the artist touched on the backlash that she's faced about people thinking she and her music aren't Black enough, explaining that she thinks people just have to get used to her.
"I think anything that's new, people are going to criticize and feel like it's not for them," Lizzo said. "But once you get used to something, it might be for you. So for people who don't like pop music or don't like Black artists that make pop music, they may eventually like me," adding, "You just gotta get used to me because I'm making good s***. You missing out."
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For more stories like this, follow us on MSN by clicking the button at the top of this page.
Man! She Feels Like a Wealthy Woman! Shania Twain's Net Worth In 2022 .
Okay, so you're one of the biggest-selling female artists of all time. That does impress us much! Shania Twain's net worth is the result of her domination over both the country and pop charts over the years. It may surprise fans to learn that this country queen isn't even from the south—she's Canadian! That hasn't stopped the Ontario-born singer from becoming one of the most prolific country artists to ever live.The five-time Grammy winner topped the charts throughout the '90s and '00s and has taken on her second Las Vegas residency in recent years.
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Oklahoma quarterback Kyler Murray is in a battle for the starting quarterback spot against Austin Kendall this preseason.
By next spring, Murray is expected to be playing pro baseball, after being drafted ninth in the baseball draft by the Oakland Athletics.
For now, though, Murray is focusing on football, and getting snaps and practice time is already paying off.
Last Friday, OU head coach Lincoln Riley kept the door open for Murray to play two more seasons, after it had been reported multiple times Murray would play just one more season of football, then start his pro baseball career.
On Monday, Murray was asked about Riley's comments.