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What is the coefficient of $1$ in the expansion of $(1+x+\frac{1}{x})^n$? In other words, what is the sum of the coefficients of $x^ky^k$ in the expansion of $(1+x+y)^n$? Here, $n$ is a positive integer. To be specific, the coefficient is \begin{equation*} c_n=\sum_{k=0}^{\left[\frac{n}{2}\right]}\binom{n}{2k}\binom{2k}{k}. \end{equation*} If there is no explicit formula, an asymptotic formula with some analysis is also very fine and appreciated. Thanks.

Haoran Chen
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    Does this help? https://math.stackexchange.com/questions/1612713/determine-the-coefficient-of-xayb-in-the-expansion-of-1xyn?rq=1 – Gary Jun 14 '21 at 06:53
  • @Gary That multinomial formula only gives the formula for each coefficient of $x^ay^b$. Here the coefficient of $1$ is a summation of such terms. What I prefer is a closed form or an asymptotic formula. – Haoran Chen Jun 14 '21 at 07:06
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    You should expand your question by writing down that formula and specify that you are looking for a closed form of that sum. – Gary Jun 14 '21 at 07:14
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    It appears to be https://oeis.org/A002426, it has some formulas in it (e.g. $a_n \sim d \frac{3^n}{\sqrt {n}}$ with $d=\frac{1}{2}\sqrt{3/\pi}$) – Sil Jun 14 '21 at 07:21
  • @Sil Thanks a lot! – Haoran Chen Jun 14 '21 at 07:28
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    Long ago, I determined the asymptotics of these numbers and found $$ \frac{{3^{n + 1/2} }}{{2\sqrt {\pi n} }}\left( {1 - \frac{3}{{16n}} + \frac{1}{{512n^2 }} + \frac{{263}}{{8192n^3 }} - \cdots } \right) $$ for large $n$. I used the generating function, Darboux's method and some asymptotics of gamma function ratios. There is an ugly formula for the coefficients in terms of Nörlund polynomials. – Gary Jun 14 '21 at 07:29
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    @Gary Thanks for the asymptotics, Gary! – Haoran Chen Jun 14 '21 at 07:37

2 Answers2

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A probabilistic approach: consider the random variable $X$ taking values on $\{-1,0,1\}$ with equal probability. Then its GF is $G_X(x)=\frac13 (1 + x + x^{-1})$.

Also, $E[X]=0$ and $\sigma^2_X=E[X^2]=\frac23$

Now, let $Y= \sum_{i=1}^N X_i$ where $X_i$ are iid with same distribution as above. Then $E[Y]=0$ and $\sigma^2_Y=N \frac{2}{3}$

Also $P(Y=0)=3^{-N} g_0$ where $g_0$ is the independent term in $(1 + x + x^{-1})^N$

But for large $N$, using CLT:

$$ P(Y=0) \approx \frac{1}{\sqrt{2 \pi \sigma^2_Y}}=\sqrt{\frac{3}{4 \pi N}}$$

Hence $$g_0 \approx 3^N \sqrt{\frac{3}{4 \pi N}}=3^{N+\frac12} \frac{1}{2 \sqrt{\pi N}}$$

This first order asymptotic approximation could be refined (eg).

leonbloy
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Since the coefficient of $1$ corresponds to the coefficient of $x^0$ we have: $$ \begin{align} (1+x+1/x)^n &= \sum_{i=0}^n \binom ni (1+x)^{n-i} x^{-i} \\ &= \sum_{i=0}^n \sum_{j=0}^{n-i}\binom ni \binom{n-i}jx^{j-i} \end{align} $$ so what you are looking for is the case $i=j$: $$ \,_2F_1\left(\frac{1}{2}-\frac{n}{2},-\frac{n}{2};1;4\right)=\sum_{i=0}^n\binom{n}{i}\binom{n-i}{i}=\sum_{i=0}^n \frac{n!}{i!^2(n-2i)!} $$

GuPe
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