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\begin{center}
\vskip 1cm{\LARGE\bf Mahler's Expansion and Boolean Functions
}
\vskip 1cm
\large
Jean-Francis Michon\footnote{All correspondence should be directed to this author.}\\
LITIS - Universit\'e de Rouen\\
Avenue de l'Universit\'e - BP 8\\
76801 Saint-\'Etienne-du-Rouvray Cedex\\
France\\
\href{mailto:jean-francis.michon@univ-rouen.fr}{\tt jean-francis.michon@univ-rouen.fr}\\
\ \\
Pierre Valarcher\\
LACL - Universit\'e Paris 12 Val de Marne\\
Facult\'e des Sciences\\
61, avenue du G\'en\'eral de Gaulle\\
94010 Cr\'eteil Cedex\\
France\\
\href{mailto:valarcher@univ-paris12.fr}{\tt valarcher@univ-paris12.fr}\\
\ \\
Jean-Baptiste Yun\`es\\
LIAFA - Universit\'e Denis Diderot Paris 7\\
175, rue du chevaleret\\
75013 Paris\\
France\\
\href{mailto:Jean-Baptiste.Yunes@liafa.jussieu.fr}{\tt Jean-Baptiste.Yunes@liafa.jussieu.fr}\\
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\vskip .2 in

\begin{abstract} 
The substitution of $X$ by $X^2$ in binomial polynomials generates
sequences of integers by Mahler's expansion. 
We give some properties of these integers and a combinatorial  
interpretation with covers by projection.
We also give applications to the classification of
boolean functions.
This sequence arose from our previous research on classification and
complexity of Binary Decision Diagrams (BDD) associated with boolean
functions.  \end{abstract}



\section{Mahler's expansion}
We recall some standard facts about binomial polynomials and Mahler's expansion (see \cite{MA58,MA81,RO}. 

A binomial polynomial $B_j(X) = \binom{X}{j} \in \mathbb Q[X]$, for any integer $j\geq 1$, is defined by:
\[ B_j(X) = \frac{X(X-1)\cdots(X-j+1)}{j!} \]
and $ B_0(X)=\binom{X}{0} =1$ by convention. For example:
$\binom{X}{1} =X $, $\binom{X}{2} = \frac{X(X-1)}{2}$.

The degree of $B_j$ is $j$, so they form a \textbf{basis} of $\mathbb Q[X]$. The expansion 
of a polynomial in this basis is called its \textbf{Mahler expansion},
also known as the Newton interpolation formula.

From the definition, the $j$ roots  of $B_j$ are $0,\ldots,j-1$. This
can be interpreted as an extension of the definition of binomial
coefficients: for $n,j\in \mathbb N$, $\binom{n}{j}=0 \mbox{ if }
n&lt;j$.

The Pascal triangle equality is
\[ \binom{X+1}{j} = \binom{X}{j} + \binom{X}{j-1}  \]
for $j&gt;0$. This equality says that, in this basis, the endomorphism $P(X)\rightarrow P(X+1)$ of $\mathbb{Q}[X]$ has a Jordan form. 

Let $f$ any function from $\mathbb Q_p\rightarrow \mathbb Q_p$, where $\mathbb Q_p$ is the field of $p$-adic numbers, using the difference operators:
\begin{align*}
\Delta f &amp;= f(X+1)-f(X)\\
\Delta^2 f &amp;=f(X+2)-2f(X+1)+f(X)\\
&amp;\vdots\\
\Delta^j f &amp;=\sum_{r=0}^j (-1)^r\binom{j}{r}f(X+j-r).
\end{align*}
Then the Mahler expansion of $f$ is
\begin{equation}\label{diff} \sum_{j=0}^\infty (\Delta^j f)(0) \binom{X}{j} 
\end{equation}
Mahler's theorem says that, for any prime $p$, this expansion converges
uniformly towards $f$ if $f$ is any continuous mapping $f:\mathbb
Z_p\rightarrow \mathbb Q_p$.




\section{Squaring variable operator}

Consider the $\mathbb Q$-linear endomorphism of $\mathbb{Q}[X]$ defined by:
\begin{equation}
f(X)\mapsto f(X^2)\label{endo}
\end{equation}
This endomorphism is clearly injective because $f(X^2)=g(X^2)$ implies that $f-g$ has infinitely many roots: the squares of $\mathbb Q$.
It is also an algebra endomorphism because $(fg)(X^2)=f(X^2)g(X^2)$ and constant $1$ is invariant.

We study the effect of this operator on the basis $B_j$. 
\begin{defi}We define $a_{k,m}\in\mathbb Q$ for all $k,m\in \mathbb N$ as the coefficients of the Mahler expansion of $\binom{X^2}{k}$:
\begin{equation} \label{basic} B_k(X^2)= \binom{X^2}{k} = \sum_{m=0}^{\infty} a_{k,m} B_m 
\end{equation}
\end{defi}
This double sequence is the sequence \seqnum{A100344} in Sloane's Online Encyclopedia \cite{S2005}.


\subsection{General properties of the $a_{k,m}$}
For fixed $k$, all the $a_{k,m}$ are $0$, except a finite number of them.

From this definition we compute the first values of $a_{k,m}$: 
\[a_{0,0} =1, \ \ \  a_{0,r} = a_{r,0} = 0 \mbox{ for } r&gt;0\]
We give in the last section the table for the first values of $a_{k,m}$. The binomial polynomials $B_j(X)$ are a $\mathbb Z$-basis for the $\mathbb Z$-module of integer polynomials, therefore trivially $a_{k,m}\in\mathbb Z$, but we shall now prove that $a_{k,m}\in \mathbb N$ .

From (\ref{basic}), substituting $X$ with all integral values in $\mathbb N$, we get an infinite linear system. The study of this system gives many important properties of the $a_{k,m}$.

\begin{prop} \label{bound} $a_{k,m}=0$ if $m&gt;2k$ or $m &lt; \sqrt{k}$.
\end{prop}
\begin{proof}To establish the first inequality consider that the degree of the left-hand side of (\ref{basic}) is $2k$.

The second inequality is obviously true for $k=0$. Suppose $k&gt;0$, 
if $n\in \mathbb N$ and $n&lt;\sqrt k$ then $n^2&lt;k$ and $n$ is a root of $\binom{X^2}{k}$.
Replacing $X$ by $n$  in (\ref{basic}) with $n=0,\ldots,m$ we get 
\begin{align*}
0 &amp; = a_{k,0} \\
0 &amp; = a_{k,0}+ a_{k,1}\\
0 &amp; = a_{k,0}+ 2a_{k,1} + a_{k,2}\\
\vdots\\
0 &amp; = a_{k,0} + \binom{m}{1}a_{k,1} + \binom{m}{2}a_{k,2} +\cdots + a_{k,m}
\end{align*}
and so $a_{k,0} = a_{k,1} =\cdots=a_{k,m}=0$.
\end{proof}

\begin{prop}[First recursive formula] For all $k,n\in \mathbb N$ 
\begin{equation}\label{basic3}a_{k,n}  = \binom{n^2}{k} - \sum_{m=0}^{n-1} a_{k,m}  \binom{n}{m} 
\end{equation}
\end{prop}
\begin{proof} We suppose first $0\leq n \leq 2k$.
We use (\ref{basic}) and write 
$\binom{X^2}{k} = \sum_{m=0}^{2k} a_{k,m}B_m $.
Make $X=n$ and use the property that $B_m(n)=0$ if $m&gt;n$, and $B_n(n)=1$. 

If $n&gt; 2k$ we must show that the right-hand side of (\ref{basic3}) is $0$. But
$\sum_{m=0}^{n-1} a_{k,m}B_m =\sum_{m=0}^{2k} a_{k,m}B_m $ by Proposition $1$ and this 
sum is $0$ by definition of the $a_{k,m}$.
\end{proof}
A consequence of this Proposition is that $a_{k,m}\in\mathbb Z$.
\begin{prop} For all integers $k,m$ we have 
\begin{equation}
a_{k,m} = \sum_{i=0}^m (-1)^{m-i}\binom{m}{i} \binom{i^2}{k}
\end{equation}
\end{prop}
\begin{proof}This is a just a translation of Mahler's coefficient computation (\ref{diff}).
\end{proof}

\begin{prop} \label{properties}
\begin{itemize}
\item[(a)] $a_{k,2k}=\frac{(2k)!}{k!}$
\item[(b)] $a_{k,m} = \binom{m^2}{k} $ if $k&gt;(m-1)^2$
\end{itemize}
\end{prop}
\begin{proof} 
The first identity (a)
is easily obtained by comparing the leading coefficients of the 
polynomials of left-hand side and right-hand side of (\ref{basic}) which are
$\frac{1}{k!}$ and $a_{k,2k} \frac{1}{(2k)!}$ respectively.

To prove the second equality (b), use (\ref{basic3}) and 
Proposition \ref{bound},
which shows that all the terms in the sum are $0$.
\end{proof}

It is useful for computing to formally generalize the definition when $m$ is a negative integer. 
We shall set $a_{k,m}=0$ if $k\in \mathbb N$ and $m &lt;0$.

We now prove a more difficult identity:

\begin{theo}[Second recursive formula]\label{triangle}For $k\geq 1$
\[a_{k,m} =\frac{1}{k} [(m^2-k+1) a_{k-1,m} + m(2m-1)a_{k-1,m-1} + m(m-1)a_{k-1,m-2}] . \] 
\end{theo}
\begin{proof}
Consider the endomorphism of $\mathbb Q[X]$ defined by $f(X)\rightarrow Xf(X)$
(multiplication by $X$). We study its effect on the $B_m$ basis. Clearly, for all $m\geq 0$
\[X B_m = (X-m +m)B_m =(m+1)B_{m+1}+mB_{m} \] 
We consider now the endomorphism $f(X)\rightarrow X^2f(X)$
(multiplication by $X^2$). Its effect on the binomial basis is, by iteration of the preceding formula
\begin{align*}
X^2 B_m &amp;= (m+1)[(m+2)B_{m+2}+ (m+1)B_{m+1}]+m(m+1)B_{m+1}+ m^2B_m \\
&amp;= (m+1)(m+2)B_{m+2}+ (m+1)(2m+1)B_{m+1}+ m^2B_m
\end{align*}
We start from ($k\geq 1$): 
\[\binom{X^2}{k}= \binom{X^2}{k-1} \frac{X^2-k+1}{k} \]
and expand the right-hand side 
\[ \frac{X^2-k+1}{k} \sum_{m=0}^{2k-2} a_{k-1,m} B_m\]
\[=\frac{1}{k}\sum_{m=0}^{2k-2} a_{k-1,m} [(m+1)(m+2)B_{m+2}+ (m+1)(2m+1)B_{m+1}+ (m^2-k+1)B_m ]\]
Grouping together the coefficients of $B_m$ we get the formula.
\end{proof}

We applied this formula to construct the table in Section~\ref{annexe}. We started from the first column et derived all others.
\begin{coro} $0\leq a_{k,m}\leq \binom{m^2}{k}$
\end{coro} 
\begin{proof} From Theorem~\ref{triangle} if $m^2-k+1&lt;0$ or if $m=0$ then $a_{k,m} =0$ or $1$, in all other cases 
the coefficient used in Theorem~\ref{triangle} are $\geq 0$. The higher bound is an immediate consequence 
of the positivity and of the recurrence formula. 
\end{proof}
\begin{coro} $a_{k,2k-1} = a_{k,2k}.\frac{2k-1}{2} =\frac{(2k)!}{k!} \frac{2k-1}{2} $
\end{coro}
\begin{proof} Easy consequence.
\end{proof}
\begin{coro} Fix $m$, then the sequence $a_{k,m}$ is increasing with $k$ for $0\leq k \leq \frac{m^2}{2}$.
\end{coro} 
\begin{proof} By Theorem~\ref{triangle}, for $k&gt;0$: 
\[a_{k,m} \geq \frac{m^2-k+1}{k}a_{k-1,m}\]
and $\frac{m^2-k+1}{k}=\frac{m^2}{k}-1+\frac{1}{k}\geq 1 + \frac{1}{k} &gt; 1$.
\end{proof}

\textit{Questions}: Fix $m$ or $k$, prove the $a_{k,m}$ are increasing then decreasing and find good bound for them. Are there other simple expressions for the $a_{k,m}$ ?




 
\section{Covering of a finite set by projection}

Let $M=[1..m]$ the set of integers from $1$ to $m$, and a fixed integer
$k$. We look for families $F$ of $k$ distinct pairs
$F=\{(a_1,b_1),\ldots,(a_k,b_k)\}\subset M^2$. If
$\bigcup_{i=1}^k\{a_i,b_i\}=M$ we say that $F$ is a \textbf{covering of
$M$ by projection}.

This definition easily generalizes to any exponent $r$ of $M$; in this
way, we get families of $k$ distinct $r$-uples covering of $M^r$ by
projection. All the results of this article could have been written in this
perspective.

This allows a straightforward combinatorial interpretation of the $a_{k,m}$.
\begin{theo}\label{combi} The number of parts $F\subset M^2$ of $k$ distinct pairs covering $M$ by projection is $a_{k,m}$.  
\end{theo}
\begin{proof} Let $\mathcal X$ any finite set with $X$ elements.  The number of subsets of $\mathcal X^2$ having $k$ elements is $\binom{X^2}{k}$. Each of these subsets is a covering of some subset $M\subset \mathcal X$ with $m$ elements by projection and $m$ may take values between $0$ and $X^2$. This enumeration gives each term of the sum in the right-hand side of (\ref{basic}). The coefficients $a_{k,m}$ are uniquely determined by (\ref{basic}) because the binomial polynomials form a basis of the polynomial ring $\mathbb Q[X]$.
\end{proof}



\section{Profiles of boolean functions in $n$ variables }

The set \boon of boolean functions in $n$ variables is the set of all 
functions
\[f:\{0,1\}^n\rightarrow \{0,1\} .\] 
It is in bijective correspondence with the set of parts of $ \{0,1\}^n$.
For $n=0$, $\mathcal B_0$ is the set of the two constant boolean functions $0$ and $1$.
The number of elements of \boon is $2^{2^n}$ for all $n\in \mathbb N$.
\begin{defi}Let $\mathcal F=\{f_1,\ldots,f_r\}\subset\boon$ any finite and non-empty family of distinct boolean functions in $n$ variables. We associate with $\mathcal F$ a sequence of $n+1$ positive integers 
\[p(\mathcal F)= (p_0(\mathcal F),\ldots,p_n(\mathcal F)) ,\] 
where $p_i(\mathcal F)$ is the number of distinct boolean functions in $n-i$ variables obtained from $\mathcal F$ by substituting all possible boolean values to the \textbf{first} $i$ boolean variables $x_1,\ldots,x_i$. For $i=0$ we set $p_0(\mathcal F)=r$. 

We call $p(\mathcal F)$ the \textbf{profile} of the family $\mathcal F$ or the profile of $f$ if $\mathcal F$ is reduced to one boolean function $f$. 
\end{defi}

\begin{quote}Example with $r=1$, $\mathcal F=\{f(x_1,x_2,x_3)= x_2\}$. 

We have $f(0,x_2,x_3)=f(1,x_2,x_3)=x_2$, so $p_1(f)=1$. If we give all boolean values to $x_1$ and $x_2$ ($4$ possible pairs of values), in all cases we get the $0$ (resp. $1$) constant function if $x_2=0$ (resp. $x_2=1$), so we have $p_2(f)=2$. When we give any boolean values to the three variables we get the constants $0$ or $1$. Finally $p(f)=(1,1,2,2)$. 
\end{quote}
The profile is a very interesting ``classifier'' which is connected to complexity questions. It is related to the Binary Decision Diagram theory (a BDD is a boolean graph canonically associated with any boolean function). A way to define complexity of $f\in\boon$ is to consider its profile $p(f)=(1,p_1,\ldots,p_n)$ and to define its complexity as
\[c(f)=p_0+\cdots+p_n\]
This complexity measures the number of different ``subfunctions'' inside $f$ generated by our sequential affectations of values to the variables. In 
BDD theory it is the number of vertices of the canonical boolean graph associated with $f$. 
We refer the reader to our paper \cite{MVY} for the details and related results. 

The important thing about the profile and the complexity, is that they are not invariant by permutations of variables in general. This can be easily verified on our example: if $f(x_1,x_2,x_3)=x1$ then $p(f)=(1,2,2,2)$ and if $f(x_1,x_2,x_3)=x_3$ then $p(f)=(1,1,1,2)$.

Now we can state our main result: 

\begin{theo}\label{enum} The number of families of boolean functions in $n\geq 1$ variables whose profile is $(p_0,\ldots,p_n)$ is the product
\[a_{p_0,p_1}a_{p_1,p_2}\cdots a_{p_{n-1},p_n}\]
with $p_n=1$ or $2$. For $n=0$ the number is $a_{p_0,1}$. 
\end{theo}
\begin{proof} The last profile value $p_n$ of any boolean function in $n$ variables is always $1$ or $2$ because there are only $2$ boolean constant functions namely $0$ and $1$. 

We proceed by recurrence. For $n=0$ the Theorem is true by simple inspection. 

The number of families of $p_1$ distinct boolean functions in the variables $x_2,\ldots,x_n$ whose profile is $(p_1,\ldots,p_n)$ is $a_{p_1,p_2}\cdots a_{p_{n-1},p_n}$ by the recurrence hypothesis. Let $\mathcal F'=\{f_1',\ldots,f_{p_1}'\}$ such a family. 
Then we can construct an unique boolean function $f\in \boon$ from each pair $(u,v)\in \mathcal F'^2$ by using the well known Boole identity 
 \[f(x_1,\ldots,x_n)=(1-x_1)u\oplus x_1v\]
and we must choose $p_0$ distinct pairs $(u,v)$ in $\mathcal F'^2$. In this manner we can construct a family of $p_0$ distinct boolean functions in $n$ variables whose profile is $(p_0,\ldots,p_n)$. The number of such families constructed from $\mathcal F'$ coincides with the combinatorial definition of the $a_{p_0,p_1}$ as a covering of $\mathcal F'$ by projection of $p_0$ elements.  

We conclude that for each $\mathcal F'$ we can construct $a_{p_0,p_1}$ families with profile $(p_0,...,p_n)$,and the Theorem is proved.
\end{proof}

We can specialize the last formula with $p_0=1$. We get immediately
\begin{remark}The set of $f\in\boon$ with profile $(1,p_1,\ldots,p_{n-1},2)$ has 
\[a_{1,p_1}\cdots a_{p_{n-1},2}\]
elements.
\end{remark}

We list all possible profiles for $n\leq 4$ in lexicographical order and give the number of boolean functions with each profile.
\[\begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline
\multicolumn{2}{|c|}{n=0}&amp;\multicolumn{2}{c|}{n=1}&amp;\multicolumn{2}{c|}{n=2}&amp;\multicolumn{2}{c|}{n=3}&amp;\multicolumn{2}{c|}{n=4}\\
\hline
1&amp;2 &amp;1,1&amp;2 &amp;1,1,1&amp;2  &amp;1,1,1,1 &amp;2   &amp; 1,1,1,1,1&amp;2 \\
 &amp;  &amp;1,2&amp;2 &amp;1,1,2&amp;2  &amp;1,1,1,2 &amp;2   &amp; 1,1,1,1,2&amp;2 \\
 &amp;  &amp;   &amp;  &amp;1,2,2&amp;12 &amp;1,1,2,2 &amp;12  &amp; 1,1,1,2,2&amp;12 \\
 &amp;	&amp;   &amp;  &amp;    &amp;    &amp;1,2,2,2 &amp;72  &amp; 1,1,2,2,2&amp;72 \\
 &amp;  &amp;   &amp;  &amp;    &amp;    &amp;1,2,3,2 &amp;144 &amp; 1,1,2,3,2&amp;144\\
 &amp;  &amp;   &amp;  &amp;    &amp;    &amp;1,2,4,2 &amp;24  &amp; 1,1,2,4,2&amp;24\\
 &amp;  &amp;   &amp;  &amp;	&amp;    &amp;        &amp;    &amp; 1,2,2,2,2&amp;432\\
 &amp;  &amp;   &amp;  &amp;	&amp;    &amp;        &amp;    &amp; 1,2,2,3,2&amp;864\\
 &amp;  &amp;   &amp;  &amp;    &amp;    &amp;        &amp;    &amp; 1,2,2,4,2&amp;144\\
 &amp;  &amp;   &amp;  &amp;    &amp;    &amp;        &amp;    &amp; 1,2,3,3,2&amp;864\\
 &amp;  &amp;   &amp;  &amp;    &amp;    &amp;        &amp;    &amp; 1,2,3,4,2&amp;10368\\
 &amp;  &amp;   &amp;  &amp;	&amp;    &amp;        &amp;    &amp; 1,2,4,2,2&amp;8928\\
 &amp;  &amp;   &amp;  &amp;	&amp;    &amp;        &amp;    &amp; 1,2,4,2,2&amp;144\\
 &amp;  &amp;   &amp;  &amp;	&amp;    &amp;        &amp;    &amp; 1,2,4,3,2&amp;11808\\
 &amp;  &amp;   &amp;  &amp;	&amp;    &amp;        &amp;    &amp; 1,2,4,4,2&amp;31728\\
\hline
\end{array}\]
For a given $n$, we ignore which profile gives the largest number of boolean functions. This question is connected to interesting works on ``Shannon effect'' (for short: random functions have almost surely a maximal complexity) \cite{G99}. Our computations can be seen as a enumerative and effective approach to this problem. We recall that Shannon's theorem is relative to ``circuit'' complexity, and for this complexity ``almost'' nothing effective is known about functions achieving maximal complexity. 


\section{Conclusion}
The $a_{k,m}$ numbers where first introduced in a combinatorial way in
our article \cite{MVY}. Theorem~\ref{triangle} was also proved in a
combinatorial way. We were unsuccessful in the search of a generating
series for these numbers and realized after a while that the Mahler
expansion of the $B_k(X^2)$ is an answer. This permits a whole
algebraic reinterpretation of the formulas of our article \cite{MVY}
and the enlargement of the scope to all boolean functions, giving the
Theorem~\ref{enum}.  Moreover, all the results of this paper can be
generalized using other exponent than $2$, and give other
interpretations with  non boolean functions and $m$-ary trees instead
of binary trees. We choose to keep our scope restricted to the squaring
and to the boolean functions formulas for simplicity. We think that the
interested reader will have no great difficulties in constructing more
general formulas if needed.


\section{Table of the $a_{k,m}$ }
\label{annexe}
{
\scriptsize
\begin{tabular}{r|rrrrrrrrrrr}
 &amp; k\\
 &amp; 0 &amp; 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6 &amp; 7 &amp; 8 &amp; 9 &amp; 10\\
\hline
m 0 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
  1 &amp; 0 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\
2 &amp; 0 &amp; 2 &amp; 6 &amp; 4 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\
3 &amp; 0 &amp;0 &amp; 18 &amp; 72 &amp; 123 &amp; 126 &amp; 84 &amp; 36 &amp; 9 &amp; 1 &amp; 0  \\
4 &amp; 0 &amp;0 &amp; 12 &amp; 248 &amp; 1322 &amp; 3864 &amp; 7672 &amp; 11296 &amp; 12834 &amp; 11436 &amp; 8008  \\
5 &amp; 0 &amp;0 &amp; 0 &amp; 300 &amp; 4800 &amp; 32550 &amp; 137900 &amp; 423860 &amp; 1017315 &amp; 1985785 &amp; 3228720  \\
6 &amp; 0 &amp;0 &amp; 0 &amp; 120 &amp; 7800 &amp; 121212 &amp; 1003632 &amp; 5634360 &amp; 23963760 &amp; 82057010 &amp; 234694416  \\
7 &amp; 0 &amp;0 &amp; 0 &amp; 0 &amp; 5880 &amp; 235200 &amp; 3791032 &amp; 37162384 &amp; 261418626 &amp; 1437954784 &amp; 6506878224  \\
8 &amp; 0 &amp;0 &amp; 0 &amp; 0 &amp; 1680 &amp; 248640 &amp; 8280272 &amp; 141626144 &amp; 1605962556 &amp; 13627345424 &amp; 92665376496 \\
9 &amp; 0 &amp;0 &amp; 0 &amp; 0 &amp; 0 &amp; 136080 &amp; 10886400 &amp; 336616560 &amp; 6156764640 &amp; 79330914540 &amp; 790034244120  \\
10 &amp; 0 &amp;0 &amp; 0 &amp; 0 &amp; 0 &amp; 30240 &amp; 8517600 &amp; 516327840 &amp; 15590248560 &amp; 305402753240 &amp; 4409098539560  \\
11 &amp; 0 &amp;0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 3659040 &amp; 512265600 &amp; 26837228880 &amp; 812355376800 &amp; 17025823879944 \\
12 &amp; 0 &amp;0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 665280 &amp; 318003840 &amp; 31638388320 &amp; 1529756532480 &amp; 47104037930928\\
13 &amp; 0 &amp;0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 112432320 &amp; 25184839680 &amp; 2058204788640 &amp; 95321107801920\\
14 &amp; 0 &amp;0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 17297280 &amp; 12955662720 &amp; 1968191184960 &amp; 142446885060480\\
15 &amp; 0 &amp;0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 3891888000 &amp; 1307674368000 &amp; 157084383456000\\
16 &amp; 0 &amp;0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 518918400 &amp; 574269696000 &amp; 126281698583040\\
17 &amp; 0 &amp;0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 149967417600 &amp; 71984360448000\\
18 &amp; 0 &amp;0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 17643225600 &amp; 27576361612800\\
19 &amp; 0 &amp;0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 6369204441600\\
20 &amp; 0 &amp;0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 670442572800\\
\end{tabular}}




\begin{thebibliography}{100}
\bibitem{G99} C. Gr\"opl, Binary decision diagrams for random boolean functions, PhD thesis, Humboldt-Universit\"at zu Berlin, 1999.

\bibitem{MA58} K. Mahler, An interpolation series for continuous
functions of a $p$-adic variable, {\it J. Reine Angew. Math.}
\textbf{199} (1958), 23--34. Correction, \textbf{208} (1961),
pp.~70--72.

\bibitem{MA81} K. Mahler, \textit{$p$-adic Numbers and Their Functions}, Cambridge Tracts in Mathematics, Cambridge University Press (second edition), 1981. 

\bibitem{MVY} J.~F. Michon, P. Valarcher, J.-B. Yun\`es, On maximal QROBDD's of boolean functions, {\it RAIRO Theor. Infor. Appl.}, {\bf 39} (2005),
677-686.

\bibitem{RO} A. Robert, \textit{A Course in $p$-adic Analysis}, 
GTM \textbf{198}, Springer, 2000. 

\bibitem{S2005} N.~J.~A.~Sloane, \href{http://www.research.att.com/~njas/sequences/}{The On-Line Encyclopedia of Integer Sequences}.
\end{thebibliography}

\bigskip
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\noindent 2000 {\it Mathematics Subject Classification}:
Primary 05A10; Secondary 94C10. 

\noindent \emph{Keywords: } Mahler's expansion, binomial polynomials,
boolean functions, BDD,  enumeration, complexity.


\bigskip
\hrule
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\noindent (Concerned with sequence
\seqnum{A100344}.)

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\vspace*{+.1in}
\noindent
Received July 18 2005;
revised version received March 28 2007.
Published in {\it Journal of Integer Sequences}, March 28 2007.

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\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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