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It's not difficult to show that

$$(1-z^2)^{-1/2}=\sum_{n=0}^\infty \binom{2n}{n}2^{-2n}z^{2n}$$

On the other hand, we have $(1-z^2)^{-1}=\sum z^{2n}$. Squaring the first power series and comparing terms gives us

$$\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}2^{-2n}=1$$

that is,

$$\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$$

My question: is there a more direct, combinatorial proof of this identity? I've been racking my brains trying to come up with one but I'm not having much success.

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This is identity 5.39 (p. 187, 2nd ed.) in Concrete Mathematics. The proof they give there uses Vandermonde's convolution and the identity $\binom{-1/2}{n} = (-1/4)^n \binom{2n}{n}$. It's not a combinatorial proof, but it might be interesting to take a look at anyway. –  Mike Spivey May 9 '11 at 16:15
I gave an elementary combinatoric proof here. –  barto Feb 20 at 20:53

2 Answers 2

up vote 26 down vote accepted

It is possible to give a direct combinatorial proof, but it is quite difficult to find it.

One possibility is to use paths between points with integer coordinates and steps $(1,1)$ and $(1,-1)$.

1) $\binom{2n}{n}$ counts all paths from $(0,0)$ to $(2n,0)$.

2) $2^{2n}$ counts all paths starting from $(0,0)$ with $2n$ steps.

3) $\binom{2n}{n}$ counts all paths with $2n$ steps that never touch the $x$-axis again after the start. (This one is not obvious, but can be proved with a bijection.)

Now you can conclude that all paths are a concatenation of a path that returns a certain number of times to the $x$-axis and a path that never does.

Note that the main difficulty here was that the two binomial coefficients are interpreted differently.

Edited to add reference: In Richard P. Stanley: Enumerative Combinatorics Volume 1, Chapter 1, Solution to exercice 2c the following reference is given:

The problem of giving a combinatorial proof was raised by P. Veress and solved by G. Hajos in the 1930s. A recent proof appears in D.J. Kleitman, Studies in Applied Math. 54 (1975), 289 - 292. See also M. Sved, Math. Intelligencer, vol.6, no. 4 (1984), 44-45.

But I have not looked to check which article gives the proof I have outlined above.

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Here is the Sved article; I can't seem to find an electronic version of Kleitman's. –  J. M. May 9 '11 at 13:01
There's also another combinatorial proof by de Angelis in Amer. Math. Montly, Aug-Sept 2006, and a probabilistic proof, which generalizes to $\prod_{i_1+\dots+i_m=n} \binom{2i_1}{i_1} \dots \binom{2i_m}{i_m} = (4^n/n!) \cdot \Gamma(m/2+n) / \Gamma(m/2)$, by Chang & Xu in Amer. Math. Montly, Feb 2011. –  Hans Lundmark May 9 '11 at 15:13
@Hand Lundmark: The product should be a sum. Thanks to all for the references. –  Phira May 9 '11 at 15:20
Oops! Of course it was meant to be a sum. Nice answer, by the way (even though it made me waste way too much time today trying to figure out what that bijection is...). –  Hans Lundmark May 9 '11 at 17:35

Here is another proof, one that I slightly prefer. I'll start with the hardest part.

Lemma. The number of all words of length $n$ in the alphabet $\{A,B\}$ such that no prefix (left factor) of it contains more letters $B$ than $A$, is $\binom n{\lceil n/2\rceil}$.

Instead of these words one may also take, interpreting $A$ as an up-step and $B$ as a down-step, paths as in the answer by Phira that never go below the horizontal axis; or one can formulate as ballot sequences as in Bertand's ballot problem, with the difference that we allow $B$ to catch up with $A$ without overtaking, and that the (non-negative) size of the eventual lead of $A$ is not fixed.

Proof. The following step can be applied to any word for which some prefix does contain more letters $B$ than$~A$: find the smallest prefix for which the majority of its letters $B$ over its letters $A$ is maximal among all prefixes, and change its last letter (which is a $B$) into $A$. There is an inverse step that can be applied to any word with more letters $A$ than letters $B$ (or more generally to a word for which some suffix (right factor) has this property): find the smallest suffix for which the majority of its letters $A$ over its letters $B$ is maximal among all suffixes, and change its first letter (which is an $A$) into $B$. The easiest way to see that these are inverse operations is that the presence of subwords in the Dyck language for $\{A,B\}$ has no effect on these operations (in particular they will never change inside such words), and that what remains when ignoring such subwords is of the form $BB\ldots BAA\ldots A$, where the last $B$ respectively first $A$ will be changed. Now given a word of length $n$ with $\lceil n/2\rceil$ letters $A$ and $\lfloor n/2\rfloor$ letters $B$, one can iterate the first operation until no prefix contains more letters $B$ than $A$, and conversely given a word of length $n$ satisfying that condition, if there are $d\geq0$ more letters $A$ than $B$ in all, one can iterate the reverse operation $\lfloor d/2\rfloor$ times to obtain a word of length $n$ with $\lceil n/2\rceil$ letters $A$ and $\lfloor n/2\rfloor$ letters $B$. This bijection proves the lemma. QED

Now to prove the identity of the question, consider the words of length $2n+1$ in which the letters $A$ are in the majority; their number is $2^{2n+1}/2=2^{2n}$. Consider the longest prefix (possibly empty) in which there are as many letters $A$ as $B$; it has an even length $2k$, and given that length there are $\binom{2k}k$ possibilities for this prefix. The next letter is necessarily an $A$, and after that there is a suffix of length $2n-2k$ in which no prefix (of that suffix) contains more letters $B$ than $A$. By the lemma there are $\binom{2n-2k}{n-k}$ of them, whence the result.

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Instead of "subwords in the Dyck language for $\left\{A,B\right\}$", can't you just say "subwords of the form $AB$"? –  darij grinberg Feb 10 at 19:20
I would also define what you mean by "majority of its letters $B$ over its letters $A$" (you seem to use it for "number of its letters $B$ minus number of its letters $A$", but it also could be mistaken for a ratio by someone not used to combinatorics). Otherwise, this is a very nicely written half-page introduction into crystal operators; I didn't expect to see a bijective proof of this identity that short! –  darij grinberg Feb 10 at 19:27

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