number of arrangements of $n$ zeroes and $n$ ones so that each zero has a neighbor zero and each one has a neighbor one this question is similar to a question in whitworth's choice and chance.
he gives the answer as  $1 + (C(n-2,0) + C(n-3,1))^2 + (C(n-3,1) + C(n-4, 2))^2 + \ldots$
i found a combinatorial "proof" by distributing the  zeroes and ones in boxes and arranged them in alternating sequence. i am not too happy with that arrangement.
i am wondering if there is a better combinatorial proof?
 A: To get the generating function for the number of ways to arrange $m$ $0$s and $n$ $1$s, I generated all blocks of at least $2$ $x$s and at least $2$ $y$s, starting with an optional block of at least $2$ $y$s and ending with an optional block of at least $2$ $x$s
$$
\begin{align}
&\left(1+\frac{y^2}{1-y}\right)\left(1+\left(\frac{x^2}{1-x}\frac{y^2}{1-y}\right)+\left(\frac{x^2}{1-x}\frac{y^2}{1-y}\right)^2+\dots\right)\left(1+\frac{x^2}{1-x}\right)\\[6pt]
&=\frac{\left(1+\frac{y^2}{1-y}\right)\left(1+\frac{x^2}{1-x}\right)}{1-\frac{x^2}{1-x}\frac{y^2}{1-y}}\\[6pt]
&=\frac{(1-x+x^2)(1-y+y^2)}{1-x-y+xy-x^2y^2}\\[18pt]
&=\color{#C00000}{1}+x^2+\color{#C00000}{0xy}+y^2+x^3+y^3+x^4+\color{#C00000}{2x^2y^2}+y^4+x^5+2x^3y^2+2x^2y^3+y^5\\
&+x^6+3x^4y^2+\color{#C00000}{2x^3y^3}+3x^2y^4+y^6+\dots
\end{align}
$$
Picking out the coefficients of the $x^ny^n$ terms, I get, starting with $n=1$,
$$
0,2,2,6,14,34,84,208,518,1296,\dots
$$

The coefficient of $x^ny^m$ is
$$
\begin{align}
&\sum_{k\ge0}\overbrace{\binom{n-k-1}{n-2k}}^{\text{$k$ blocks of $x$s}}\overbrace{\binom{m-k-1}{m-2k}}^{\text{$k$ blocks of $y$s}}\overbrace{\left(2-\binom{k-1}{k}\right)}^{\text{$x$ or $y$ first if $k\ne0$}}\\
&+\sum_{k\ge0}\overbrace{\binom{n-k-2}{n-2k-2}}^{\text{$k+1$ blocks of $x$s}}\overbrace{\binom{m-k-1}{m-2k}}^{\text{$k$ blocks of $y$s}}\\
&+\sum_{k\ge0}\overbrace{\binom{n-k-1}{n-2k}}^{\text{$k$ blocks of $x$s}}\overbrace{\binom{m-k-2}{m-2k-2}}^{\text{$k+1$ blocks of $y$s}}
\end{align}
$$
Note that $\binom{n}{0}=1$ even if $n\lt0$ and $\binom{n}{k}=0$ if $k\lt0$.
Thus, the coefficient of $x^ny^n$ is
$$
\sum_{k\ge0}\binom{n-k-1}{n-2k}^2\left(2-\binom{k-1}{k}\right)+2\sum_{k\ge0}\binom{n-k-2}{n-2k-2}\binom{n-k-1}{n-2k}
$$
which, if we assume $n\gt0$, can be simplified to
$$
2\sum_{k\ge1}\binom{n-k-1}{n-2k}^2+2\sum_{k\ge1}\binom{n-k-2}{n-2k-2}\binom{n-k-1}{n-2k}
$$
A: I'm not sure how to interpret the summation formula you gave. The combinatorial method you mentioned in your question works fine. Using that method I get the formula
$$2\sum_{k=1}^{\lfloor\frac n2\rfloor-1}\binom{n-1-k}{k-1}\binom{n-2-k}k+2\sum_{k=1}^{\lfloor\frac n2\rfloor}\binom{n-1-k}{k-1}^2$$
which is valid for $n\gt 1$. Here $\binom{n-1-k}{k-1}$ is of course the number of solutions of $x_1+\cdots+x_k=n$ in integers $x_i\ge2$ ("stars and bars"), and $\binom{n-2-k}k=\binom{n-1-(k+1)}{(k+1)-1}$, and the factor $2$ is there because the arrangement can start with a zero or a one. This answer can be rewritten in the form
$$1+\binom{n-1-\lfloor\frac n2\rfloor}{\lfloor\frac n2\rfloor-1}^2+\sum_{k=1}^{\lfloor\frac n2\rfloor-1}\left(\binom{n-1-k}{k-1}+\binom{n-2-k}k\right)^2$$
which is vaguely reminiscent of the formula in your question.
