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Is there a simple closed expression for the following sum? $$\sum_{i=0}^{\lfloor\frac n2\rfloor}{n\choose i}{n-i\choose i}$$

I can see that this is the constant term in $\big(\frac 1x+1+x\big)^n$. But this is no closer to a simple close expression I am seeking.

Hans
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    There is a giant OEIS page about that term... – Dirk May 23 '19 at 06:28
  • @Dirk: I am not familiar with how to search for a particular general expression in OEIS. Also, even if there is a simple expression, how does one derive that? – Hans May 23 '19 at 06:52
  • Compute the first few terms for $n=1,2,3,4,5,6,...$ and put these into OEIS. It will give you http://oeis.org/A002426. I don't claim that there is an easy, closed form, but if there is, your best chance would be to find it there or in one of the sources mentioned there. – Dirk May 23 '19 at 07:34
  • @Dirk: I see. Appreciate the explanation. By the way, did you downvote and voted to close the question? If so, is your rationale the answer could be found in OEIS? – Hans May 23 '19 at 08:09
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    I did downvote because there isn't any own effort shown, but I did not vote to close. :) – Dirk May 23 '19 at 11:31
  • @Dirk: I tried the generating function route but did not get anywhere in terms of obtaining a simple closed expression. I have now put my failed attempt in the question. – Hans May 24 '19 at 00:28
  • To people vote to close: How is this question off-topic as in not pertaining to mathematics? It is a combinatorial sum, one of THE topic of combinatorics which is a main branch of mathematics. – Hans May 24 '19 at 00:30

2 Answers2

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We use the coefficient of operator $[z^i]$ to denote the coefficient of $z^i$ of a series. This way we can write for instance \begin{align*} [z^i](1+z)^n=\binom{n}{i}\tag{1} \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{i=0}^{\left\lfloor \frac{n}{2}\right\rfloor}}&\color{blue}{\binom{n}{i}\binom{n-i}{i}}\tag{2}\\ &=\sum_{i=0}^{n}\binom{n}{i}[z^i](1+z)^{n-i}\tag{3}\\ &=[z^0](1+z)^n\sum_{i=0}^{n}\binom{n}{i}\left(\frac{1}{z(1+z)}\right)^i\tag{4}\\ &=[z^0](1+z)^n\left(1+\frac{1}{z(1+z)}\right)^{n}\tag{5}\\ &\,\,\color{blue}{=[z^{n}](1+z+z^2)^n}\tag{6} \end{align*}

Comment:

  • In (3) we apply the coefficient of operator according to (1). We also set the upper limit to $n$ without changing anything, since $\binom{n-i}{i}=0$ if $i>\left\lfloor\frac{n}{2}\right\rfloor$.

  • In (4) we use the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

  • In (5) we apply the binomial theorem.

  • In (6) we do some simplifications and use again the rule as in (3).

We observe the binomial sum (2) represents essentially central trinomial coefficients \begin{align*} [z^n](1+z+z^2)^n \end{align*} for which there is no closed form available.

Notes from the experts:

D.E. Knuth gives in Concrete Mathematics, Appendix A 7.56 the following representation of a more general expression

\begin{align*} [z^n](a+bz+cz^2)^n=[z^n]\frac{1}{\sqrt{1-2bz+(b^2-4ac)z^2}} \end{align*}

He states that according to the paper Hypergeometric Solutions of Linear Recurrences with Polynomial Coeffcients by Marko Petkovšek there exists a closed form (more precisely: a closed form solution as a finite sum of hypergeometric terms) if and only if $$\color{blue}{abc(b^2-4ac)=0}$$

In case of central trinomial coefficients we have $a=b=c=1$. Since then the expression $abc(b^2-4ac)=-3\ne 0$ there is no such closed form in particular for the central trinomial coefficients.

Markus Scheuer
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As other people recommended, I computed a few terms by computer and it gives A002426, which seems that there's no exact formula for it. The generating function is $$ \frac{1}{\sqrt{1-2x-3x^{2}}} $$ and we also have an asymptotic formula $$ a(n) \sim 3^{n}\sqrt{\frac{3}{4\pi n}} $$

Seewoo Lee
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  • Can you give a derivation of the asymptotic formula, perhaps from the generating function? – Hans May 23 '19 at 17:41