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\begin{center}
\vskip 1cm{\LARGE\bf Asymptotic Estimate for the  \\
\vskip .1in
Multinomial Coefficients}
\vskip 1cm
\Large
Jiyou Li\\
School of Mathematical Sciences\\
Shanghai Jiao Tong University \\
Shanghai, 200240\\
China \\
\href{mailto:lijiyou@sjtu.edu.cn}{\tt lijiyou@sjtu.edu.cn} \\
 \end{center}

\vskip .2in

\begin{abstract}
The multinomial coefficient $\bi {n,q}{k}$ is defined to be the
coefficient of $x^{k}$ in $(1+x+x^2+\cdots
+x^{q-1})^n$.  It is conjectured that for given $n>2$,
  $T(n, q):=\bi {n,q}{cn}-\bi {n,q-1}{cn}$ is unimodal and the maximum  occurs at $q=\lfloor\log_{1+\frac 1{c}}{n}\rfloor$ or
$q=\lfloor\log_{1+\frac 1{c}}{n}\rfloor+1$. As an attempt to prove  this conjecture,  we give an asymptotic estimate for $\bi {n,q}{cn}$ as $n$ tends to infinity, where $c$ is a positive integer.
\end{abstract}


\section{Introduction}


 The multinomial coefficient  $\bi {n,q}k$ is defined by
$$\sum_{k=0}^\infty \bi {n,q}k x^k=(1+x+x^2+\cdots +x^{q-1})^n.$$
 Clearly $\bi {n,q}k$ is a natural generalization of the well-known binomial and trinomial coefficients and thus  belongs to a large class of fundamental combinatorial numbers. It was studied extensively
by many mathematicians since Euler. For details related to this number the readers are referred to  \cite{An, A, A2, DS, GJ, OR}.  Some applications in coding theory and
communication theory can be found in \cite{Gi,GO}.



Multinomial coefficients count the numbers of certain compositions.
For a positive integer $k$, a {\it composition} (also called an ordered
partition) is a finite sequence of positive integers $x_1,
x_2,\ldots, x_r$ such that $x_1+x_2+\dots+x_r = k$. The $x_i$'s are
called {\it parts} of the composition.
A composition with $n$ parts is called a {\it $n$-part composition}.

Let $b(k,n,q)$ be the number of $n$-part compositions of $k$ such
that each part is bounded by $q$. Obviously $b(k,n,q)$ equals $\bi {n,q}{k-n}$, which is the coefficient of $x^k$ in the expansion of  $(x+x^2+\cdots +x^q)^n=x^n(1+x+\cdots +x^{q-1} )^n$. It also equals the  number of different ways putting $k$ identical balls into $n$ distinct boxes with each one nonempty and containing at most $q-1$ balls, or equivalently, the number of $k$-multisets in $\{1, 2, \ldots, n\}$ such that each number appears and is repeated at most $q-1$ times.
 In this note we will mainly focus on the study of $\bi {n,q}k$.

From the multinomial theorem
 one has $$\bi {n,q}k=\sum_{\substack{i_1+i_2+\cdots+i_q=n,\\ i_2+2i_3+\cdots+(q-1)i_{q}=k}}\binom{n}{i_1, i_2, \ldots, i_q}.$$
However,  it was proved in \cite{PWZ} that when $q>2$, $\bi {n,q}k$ has no closed form, that is, it cannot be written as a sum of finite  hypergeometric terms.
A natural question is thus to ask if there are any
nice asymptotic estimates for $\bi {n,q}k$ for suitable parameters $n, q ,k$.

We are also interested in the unimodality of the multinomial coefficients.
\begin{definition}
A sequence $a_0,a_1,\ldots,a_n$ of real numbers is \textit{unimodal} if for some $0 \leq k\leq n$ one has
$$a_0\leq a_1\leq \cdots \leq a_{k-1}\leq a_k \geq a_{k+1} \geq \cdots\geq a_n.$$
  \end{definition}
For instance, the sequence ${n \choose i}, 0\leq i\leq n$ is unimodal.   Unimodality plays an important role in combinatorics, number theory and representation theory. Many interesting and important examples are surveyed by Stanley \cite{Stanley-1, Stanley-2}.

It is well known that for given $n, q$, $\binom{n, q}{k}$ is unimodal 
(see, for example, \cite{An}).
\begin{proposition}
 For given positive integers $n, q$, $\binom{n, q}{k}$ is unimodal on $k$ and reaches its maximum at $k=\lfloor \frac{qn} 2 \rfloor$.
\end{proposition}




Recall that $b(k,n,q)$ is the number of $n$-part compositions of $k$ such
that each part is bounded by $q$.
Let $a(k,n,q)$ be the number of compositions of $k$ with $n$ parts
such that the largest part is $q$. Then
$a(k,n,q)=b(k,n,q)-b(k,n,q-1)$ and in particular  $a(2n,n,q)=\bi {n,q}n-\bi {n,q-1}n$.


Let $b(k,q)=\sum_{n=1}^k b(k,n,q)$
(respectively, $a(k,q)=\sum_{n=1}^k a(k,n,q)$) be the number of compositions
of a positive integer $k$ with parts bounded by $q$ (respectively, the largest part is $q$). It is well known that \cite{Mac}
$$\sum_{k=0}^{\infty}b(k,q)x^k=\frac {1-x}{1-2x+x^{q+1}}.$$


  Based on this formula and analytical tools, Odlyzko and Richmond \cite{OR} proved the next statement.
\begin{lemma}[Odlyzko and Richmond]
Let $a(k,q)$  be defined as above. Then $a(k,q)$ is unimodal for any $k$ and the maximum value occurs for $q=\lfloor \log_2{k}\rfloor$
infinitely often and $q=\lfloor \log_2{k}\rfloor+1$ infinitely often
and always at one of these two values and no other.
\end{lemma}

Based on numerical results, an improved conjectured is proposed.
 \begin{conjecture}\label{Conjecture 1.4}
Let $a(k,q)$  be defined as above. Let $c$ be a positive integer.
Then for any $n$, $a((c+1)n,n,q)=\bi {n,q}{cn}-\bi {n,q-1}{cn}$ is unimodular on $q$ and the maximum value occurs for
$q=\lfloor\log_{1+\frac 1c}{n}\rfloor+1$ or $q=\lfloor\log_{1+\frac
1c}{n}\rfloor+1$.

In particular, $a(2n,n,q)=\bi {n,q}n-\bi {n,q-1}n$ is  unimodular
on $q$ and the maximum value occurs for
$q=\lfloor\log_2{n}\rfloor$ or $q=\lfloor\log_2{n}\rfloor+1$.
\end{conjecture}

Our attempt to establish this conjecture starts with
an investigation to the
asymptotic behaviors of $\binom{n, q}{k}$ when $k$ is linear of
$n$. We will first review some classical results.

For the simplest case $q=2$, it is well known that the binomial coefficient
$\bi {n,2}{cn}=\bi {n}{cn}$ has asymptotic estimate
$$\bi {n,2} {cn}\sim \frac 1 {\sqrt{2 \pi (c-c^2) n}}({c^{-c}(c-1)^{-c+1}})^n,$$
where $0<c<1$ is a constant.


For the case $q=3$, it is known for $k=n$, the central  trinomial coefficient
has asymptotic estimate
$$\bi {n,3}n\sim \frac {3^{n+1/2}}{2 \sqrt{\pi n}}.$$


For large $q$  and general $k$,  based on the integral representation
$$\binom{n, q}{k}=\frac 2 \pi \int_0^{\frac \pi 2} \left(\frac {\sin q\theta } {\sin\theta }\right)^n \cos( ((q-1)n -2k)\theta) d \theta,$$
Andr\'{e} \cite{C} proved that
$$
\sup_k{\binom{n, q}{k}}\sim \frac{\sqrt{6}q^n} {\sqrt{{(q^2-1)\pi n}}}, \ \
n\rightarrow \infty.
$$

This estimate has several other proofs; see, for example,
a recent one by Eger \cite{E0, E00}, by representing it as the distribution of sums of independent discrete random variables.
An asymptotic distribution in this case was given by Neuschel \cite{N}.


Star \cite{Star} generalized the
result of Andr\'{e}. Write $k=\frac 12(n-s)(q+1)$, where $s=Kn^{\theta}, 0\leq \theta\leq 1/2$ and $K>0$ is a constant.
Star proved that
 \begin{align*}{ n,q \choose k-n} &= \frac{\sqrt{6}q^n} {\sqrt{{(q^2-1)\pi n}}}\cdot\left(
 1+\frac {\sum_{j=0}^{1} h_{1,j}(q)s^{2j}}{n^1}+\cdots+
 \frac {\sum_{j=0}^{m-1} h_{m-1,j}(q)s^{2j}}{n^{m-1}}+O(\frac
{1+s^{2m}}{n^m})\right),
 \end{align*} where $h_{i,j}(q)$ are some rational
functions in the function field $\mathbb{R}(q)$.




 The main result of this note is an asymptotic estimate for
 $\binom{n, q}{k}$ for large $q>3$ and for general $k=cn$, where $c$ is fixed positive integer.
The proof uses simple analysis based on Hayman's method.

\begin{lemma}\label{Theorem 1.2}
Suppose $q>3$ and $k=cn$, where $c<q$ is an absolute positive integer.
Then we have
 \[\binom{n, q}{cn} \sim \frac {\phi (r)}{\sqrt{2\pi n}}\left ( \frac {1-r^q} {r-r^2} \right )^n,\]
as $n\rightarrow\infty$, where
$$
\phi(r)=\left(\frac r{(1-r)^2}-\frac {q^2 r^{q}}
{(1-r^q)^2}\right)^{-1/2},\ \ \
r=\frac {1}{d} +\frac {q}{c^2 d^{q+2}}+\theta \frac {q^3}{d^{2q}},
$$
 $|\theta|\leq 1$ and $d=1+\frac 1 c$.
In particular, when $c=1$ we have
 $$
r={\displaystyle \frac {1}{2}}  + {\displaystyle \frac {q}{2^{q + 2}
}}+ \theta \frac{q^3}{2^{2q}}.
$$
\end{lemma}


  The proof of Theorem \ref{Theorem 1.2} is given in Section 2.  For simplicity of the computations, we only gives details of the proof for the special case $k=n$, i.e., $c=1$. The proof of case $c>1$ is
  essentially the same as the case $c=1$.




\section{Proof of Theorem \ref{Theorem 1.2}}

\begin{definition}

Suppose that $f(z)=\sum_{n=0}^{\infty} a_n z^n$ is a complex
analytic function for $|z|<R$, where $0<R\leq \infty$. Define
 \begin{align} \label{5}
M(r)=\max_{|z|=r}{|f(z)|}.
 \end{align}
If for large enough $r$, we have $M(r)=f(r)$, then $f(z)$ is called
an admissible function.   The references \cite{H,W} present a discussion on admissible functions.

\end{definition}
Hayman  \cite{H} showed that such good functions have nice asymptotic estimates for their coefficients.

\begin{lemma}[Hayman] \label{lem2}

 Let $f(z)=\sum_{n=0}^{\infty} a_n z^n$ be an admissible function, which is analytic in the disk
 $|z|<R$. Denote
 $$
a(r)=r\frac {f'(r)}{f(r)},\ \ \ b(r)=ra'(r),
$$
and suppose $0<r_n<R$ is a positive real zero satisfying
$$ a(r_n)=n,\
\ \ \forall n \in N.
$$
Then
$$
a_n \sim \frac {f(r_n)}{r_n^n\sqrt{2\pi b(r_n)}},\ \ \ n \rightarrow
\infty .
$$
\end{lemma}

\begin{lemma}\cite{A}\label{lem0}
The function $f(z)=(1+z+z^2+\cdots+z^{q-1})^n$ is an
admissible function analytical in the disk $|z|<1$.
\end{lemma}


\begin{lemma}\label{lem1}
 For $q \geq 3$ the equation
$$
(q-2)x^{q+1}-(q-1)x^{q}+2x-1=0, \ \  \ q \in N $$ has only two
positive real roots including 1 being one of them. The second root $r$ 
satisfies
$$\left| r-\frac {1}{2}-\frac {q}{2^{q + 2}
}\right|\leq \frac{q^3}{2^{2q}}.
$$
\end{lemma}

\begin{proof}
Since the cases $q=3, 4$ can be verified directly, we may assume
$q>4$.  Suppose $f(x)=(q-2)x^{q+1}-(q-1)x^{q}+2x-1$. Then $f''(x)=q
x^{q-2}(xq^2-2x-xq-q^2+2q-1)=0$ gives two inflection points $\frac
{(q-1)^2}{(q+1)(q-2)}, 0$. This proves that there are only two
positive real roots including 1.


 Now suppose that $r=\frac 1
2 +\frac {c}{2^{q+2}}$ is a positive real zero of $f(x)$, where $c$
is regarded as a variable depending on $q$ and will be specified.
Then
 \begin{align*} f(r)
&=(q-2)(\frac 1 2 +\frac {c}{2^{q+2}} )^{q+1}-(q-1)(\frac 1 2 +\frac {c}{2^{q+2}} )^{q}+2(\frac 1 2 +\frac {c}{2^{q+2}} )-1\\
&=(\frac {-q} {2^{q+1}}+ \frac {(q-2)c} {2^{2q+2}}  )(1 +\frac
{c}{2^{q+1}})^{q}+\frac {c}{2^{q+1}}\\
&=0.
 \end{align*}
   Assume without of generality that $0\leq c\leq q^{3/2}$. By
   Taylor's theorem one has
 \begin{displaymath}
\left| (1 +\frac {c}{2^{q+1}})^{q}-1-{\frac {cq}{2^{q+1}}} \right| \leq
\frac {c^2 q^2}{2^{2q}} \leq \frac {q^5}{2^{2q}}  .
 \end{displaymath}
And hence
 \begin{displaymath}
\left| f(r)+\frac {q-c} {2^{q+1}}+\frac {c(q^2-q+2) } {2^{2q+2}}
\right|\leq\frac {q^6}{2^{3q+1}}.
 \end{displaymath}

Since $f(r)=0$, we then have
 $$
 q-c+\frac {c(q^2-q+2) } {2^{q+1}}+ \frac {\theta q^6}{2^{2q}}=0,
 $$
 where $0\leq |\theta|\leq 1$.

Let  $c=q+c'$. Putting it into the above equality, one has
$$
c'(1-\frac {q^2-q+2}{2^{q+1}})=\frac {q^3-q^2+2q}{2^{q+1}}+\frac
{\theta q^6}{2^{2q}}.
$$
This implies
$$
\left| c-q-\frac {q^3-q^2+2q}{2^{q+1}-q^2+q-2}\right|\leq \frac{2
 q^6}{2^{2q}}.
 $$
 Replace in $r=\frac 1 2
+\frac {c}{2^{q+2}}$ to obtain, for $q>16$,
  $$\frac{q^3-q^2}{2^{2q+2}} \leq  r-\frac {1}{2}-\frac {q}{2^{q + 2}
}\leq \frac{q^3}{2^{2q+2}}.
$$
The remaining cases of $q$ can be easily verified.
 \end{proof}

\begin{lemma}\label{Theorem 3.4}
Assume $q>3$. Then we have the asymptotic estimate $$
 {n,q \choose n }\sim \frac {\phi (r)}{\sqrt{2\pi n}}\left ( \frac {1-r^q} {r-r^2} \right )^n, \ \ n\rightarrow \infty$$
where
$
\phi(r)=\left(\frac r{(1-r)^2}-\frac {q^2 r^{q}}
{(1-r^q)^2}\right)^{-1/2}
$, and  $\left| r-\frac {1}{2}-\frac {q}{2^{q + 2}
}\right|\leq \frac{q^3}{2^{2q}}.
 $
\end{lemma}

\begin{proof}
Let $f(z)=(1+z+z^2+\cdots+z^{q-1})^n=(\frac {1-z^q}{1-z})^n.
  $ By Lemma \ref{lem0},  $f(z)$ is an admissible analytical function on $\mathbb{C}-\{\infty\}$.
  Applying Hayman's theorem (Lemma \ref{lem2}), we have
 \begin{align*}
  a(x)&=x\frac {f'(x)}{f(x)}=\frac {-nx(qx^{q-1}-qx^q-1+x^q) } {(1-x^q)(1- x)},\\
b(x)& =xa'(x)=\frac {nx(1-x^{q-1}q^2-2x^q+x^{2q}+2q^2x^q-x^{q+1}q^2) }{(-1+x^q)^2(x-1)^2}\\
 &=nx\left(\frac 1 {(1-x)^2}-\frac {q^2 x^{q-1}} {(x^q-1)^2}\right).
 \end{align*}

 The equation $a(x_n) = n$ yields

 $$
 \frac {-nx_n(qx_n^{q-1}-qx_n^q-1+x_n^q) } {(1-x_n^q)(1- x_n)}=n,
 $$
and thus  $$ (q-2)x_n^{q+1}-(q-1)x_n^{q}+2x_n-1=0. $$
The result now follows from Lemma \ref{lem1}.
\end{proof}



\begin{corollary}
When $q>3$, for large $n$ we have the estimate
$$
 {n,q \choose n }\sim \frac
  {(1+\frac{q^2-6q}{2^q}+\theta_1\frac{q^2}{2^{2q}}) 2^n}{\sqrt{\pi n}}(1-\frac
1{2^{q-2}}+\theta_2\frac{q^2}{2^{2q}})^n, \  n\rightarrow \infty,$$
where $|\theta_i|\leq1$ for $i=1,2$.
\end{corollary}
\begin{proof}Since
$$\left| r-\frac {1}{2}-\frac {q}{2^{q + 2}
}\right|\leq \frac{q^3}{2^{2q}},
$$
it follows that
$$\left|\frac r {(1-r)^2}-\frac 12- \frac {3q}{2^{q-1}}\right|\leq \frac{q}{2^{2q}},$$
and
$$\left|\frac {q^2r^q}
{(1-r^q)^2}-\frac{q^2}{2^q}\right|\leq \frac{q^4}{2^{2q}}.$$ Thus
$$
\left|\phi(r)-\sqrt{2}(1+\frac{q^2-6q}{2^q})\right|\leq
\frac{q^2}{2^{2q}}.
$$
Similarly $$\left|\frac {1-r^q} {r-r^2}-2+\frac
1{2^{q-1}}\right|\leq \frac{q^2}{2^{2q}},$$ and the result follows
from Theorem \ref{Theorem 3.4}.
\end{proof}

\section {Acknowledgments}

Part of this work was done during the author's master's studies at
Beijing Normal University.  This work is supported in part by the
National Science Foundation of China (11771280) and in part by the
National Science Foundation of Shanghai Municipal (17ZR1415400).

The author wishes to thank the referees for their constructive comments.

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\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 05A16; Secondary 05A15, 05A17.


\noindent \emph{Keywords: } multinomial coefficient, Hayman's method.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A002426},
\seqnum{A005725},
\seqnum{A187925}, and
\seqnum{A305161}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received January 29 2018;
revised versions received February 4 2018;
August 23 2018; June 8 2018; June 12 2018; September 5 2018; 
September 8 2018; May 30 2019; August 25 2019.
Published in {\it Journal of Integer Sequences}, December 28 2019.

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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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