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Prove the following identity:

$$\sum\limits_{k=-\infty}^{\infty}{h \choose \left(\frac{h+1+3k}{2}\right)}=\frac{2^h-(-1)^h}{3}$$

Here, ${n \choose k}$ are the Binomial coefficients, defined as $0$ when $k$ is not an integer, $k<0$ or $k>n$. The right expression is the number of paths with $h$ hops in a $n$ vertex fully connected graph when that graph has three nodes and the left is the number of such paths in an $n$ sided polygon when $n=3$. Of course, a three node graph is simultaneously both, hence the expression. See here: Number of $h$ hop paths between two vertices with shortest path $s$ on an $n$ sided polygon. and here: Closed form solution to recurrence: $g(n)=(k-2)g(n-1)+(k-1)g(n-2)$. The accepted answer in the first link alludes to the "roots of unity" filter as a potential method for proving this, but I'm unable so far to use it.

Can this then be extended to the general summation:

$$S = \sum\limits_{k=-\infty}^{\infty}{h \choose \left(\frac{h+s+nk}{2}\right)}$$

Rohit Pandey
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1 Answers1

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Let $\omega=\exp(2\pi i/3)$ be the primitive cube root of unity. Because $$\frac{1+\omega^k+\omega^{2k}}{3}=\begin{cases}1&\text{if $3 \mid k$}\\0&\text{otherwise}\end{cases}$$ we have $$\sum_k a_{3k} = \sum_k a_k \frac{1+\omega^k+\omega^{2k}}{3}.$$ Now take $$a_k=\binom{h}{\frac{h+1+k}{2}}$$ and apply the binomial theorem to each of the resulting three sums.

RobPratt
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