## Apéry Numbers

This is a translation of a previous Portuguese post of mine.

Show that for all integers $n\geq 0,$ the following identity holds

$\displaystyle\sum_{i=0}^{n}\dbinom{n}{i}^{2}\sum_{k=0}^{i}\dbinom{i}{k}\dbinom{n}{k}\dbinom{2n-k}{k}=\sum_{k=0}^{n}\dbinom{n}{k}^{2}\dbinom{n+k}{k}^{2}.$

Solution

In order to have a short notation and because these numbers are the Apéry Numbers, let´s denote  them $A_{n}$,

$A_{n}=\displaystyle\sum_{i=0}^{n}\dbinom{n}{i}^{2}\sum_{k=0}^{i}\dbinom{i}{k}\dbinom{n}{k}\dbinom{2n-k}{k}$$=\displaystyle\sum_{i=0}^{n}\dbinom{n}{i}^{2}\sum_{k=0}^{n}\dbinom{i}{k}\dbinom{n}{k}\dbinom{2n-k}{k}$

Since

$\displaystyle\sum_{k=0}^{i}\dbinom{i}{k}\dbinom{n}{k}\dbinom{2n-k}{k}=\sum_{k=0}^{n}\dbinom{i}{k}\dbinom{n}{k}\dbinom{2n-k}{k},$

because for $i+1\leq k\leq n$ every term equals zero and also $\dbinom{i}{k}=0$.

Multiplying both sides of the identity proved in my post “Uma proposição da análise combinatória” (here) [see “caderno” in the PDFs page] by

$\dbinom{n}{k}\dbinom{2n-k}{k},$

we have

$\displaystyle\binom{n}{k}^{2}\binom{2n-k}{n}^{2}=\binom{n}{k}\binom{2n-k}{k}\sum_{i=k}^{n}\binom{n}{i}^{2}\binom{i}{k}.$

Now summing in $k$ , we get successively

$\displaystyle\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{2n-k}{n}^{2}=\sum_{k=0}^{n}\binom{n}{k}\binom{2n-k}{k}\sum_{i=0}^{n}\binom{n}{i}^{2}\binom{i}{k}$$\displaystyle=\sum_{k=0}^{n}\sum_{i=0}^{n}\binom{n}{k}\binom{2n-k}{k}\binom{n}{i}^{2}\binom{i}{k}$$\displaystyle=\sum_{i=0}^{n}\sum_{k=0}^{n}\binom{n}{k}\binom{2n-k}{k}\binom{n}{i}^{2}\binom{i}{k}$$\displaystyle=\sum_{i=0}^{n}\binom{n}{i}^{2}\sum_{k=0}^{n}\binom{i}{k}\binom{n}{k}\binom{2n-k}{k}$

But

$\displaystyle\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{2n-k}{n}^{2}=\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{2n-k}{n-k}^{2}$ $\displaystyle=\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{2}.$

Therefore,

$\displaystyle A_{n}=\sum_{k=0}^{n}\dbinom{n}{k}^{2}\dbinom{n+k}{k}^{2}$

which proves the proposition. $\qquad\blacktriangleleft$

Anúncios

## Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Demonstração, Matemática, Math, Problem, Proof com as etiquetas , , . ligação permanente.

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