Coefficient of $1$ in the expansion of $\left(1+x+\frac{1}{x}\right)^n$

Question

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.

Answer 1

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).

Answer 2

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 $i=j$: $$ \,_2F_1\left(\frac{1}{2}-\frac{n}{2},-\frac{n}{2};1;4\right)=\sum_{i=0}^n\binom{n}{i}\binom{n-i}{i}=\sum_{i=0}^n \frac{n!}{i!^2(n-2i)!} $$