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

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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