I found the following problem while working through Richard Stanley's Bijective Proof Problems (Page 5, Problem 16). It asks for a combinatorial proof of the following: $$ \sum_{i+j+k=n} \binom{i+j}{i}\binom{j+k}{j}\binom{k+i}{k} = \sum_{r=0}^{n} \binom{2r}{r}$$ where $n \ge 0$, and $i,j,k \in \mathbb{N}$, though any proof would work for me.

I also found a similar identity in Concrete Mathematics, which was equivalent to this one, but I could not see how the identity follows from the hint provided in the exercises.

My initial observation was to note that the ordinary generating function of the right hand side is $\displaystyle \frac {1}{1-x} \frac{1}{\sqrt{1-4x}}$, but couldn't think of any way to establish the same generating function for the left hand side.

  • $\begingroup$ LHS is $\sum\binom{i+j}i\binom{n-i}j\binom{n-j}i$. So the question about $\sum\binom{n-i}j\binom{n-j}i$ looks (somewhat) related. $\endgroup$
    – Grigory M
    Dec 31, 2014 at 19:02
  • $\begingroup$ also triple product looks superficially similar to 3-variable form of Dixon / Strehl $\endgroup$
    – Grigory M
    Jan 4, 2015 at 12:47
  • $\begingroup$ I now suspect that both sides count 00-avoiding $3n$-periodic binary sequences with exactly $n$ zeroes — maybe someone can prove it? $\endgroup$
    – Grigory M
    Jan 7, 2015 at 20:35
  • $\begingroup$ I've also asked a (different but) related question @ MO $\endgroup$
    – Grigory M
    Jan 9, 2015 at 14:12
  • 1
    $\begingroup$ @GrigoryM I believe the 00-avoiding $3n$-periodic binary sequences are a different sequence, counted by ${2n \choose n} + {2n-1 \choose n-1}$. In particular there are $30$ (not $29$) such sequences when $n = 3$. $\endgroup$ Jan 9, 2015 at 19:38

4 Answers 4


Restating your question, you are seeking to find the generating function of the left-hand-side: $$ g(x) = \sum_{n=0}^\infty x^n \sum_{i+j+k=n}\binom{i+j}{i} \binom{j+k}{j} \binom{k+i}{k} = \sum_{i=0}^\infty \sum_{j=0}^\infty \sum_{k=0}^\infty x^{i+j+k} \frac{(i+j)! (i+k)! (j+k)!}{i!^2 j!^2 k!^2} $$ First, carry out summation over $i$: $$ g(x) = \sum_{j=0}^\infty \sum_{k=0}^\infty x^{j+k} \frac{(j+k)!}{j!\cdot k!} {}_2F_1\left(1+j, 1+k; 1; x\right) $$ Now use Euler's transformation ${}_2F_1\left(1+j, 1+k; 1; x\right) = (1-x)^{-j-k-1} \, {}_2F_1\left(-j, -k; 1, x\right)$, which gives $$ g(x) = \frac{1}{1-x} \sum_{j=0}^\infty \sum_{k=0}^\infty \left(\frac{x}{1-x}\right)^{j+k} \frac{(j+k)!}{j!\cdot k!} {}_2F_1\left(-j, -k; 1; x\right) = \\ \frac{1}{1-x} \sum_{j=0}^\infty \sum_{k=0}^\infty \left(\frac{x}{1-x}\right)^{j+k} \frac{(j+k)!}{j!\cdot k!} \sum_{r=0}^{\min(j,k)} \binom{j}{r}\binom{k}{r} x^r = \\ \frac{1}{1-x} \sum_{r=0}^\infty x^r \sum_{j=r}^\infty \sum_{k=r}^\infty \binom{k+j}{k} \binom{j}{r}\binom{k}{r} \left(\frac{x}{1-x}\right)^{j+k} $$ Using $$ \sum_{j=r}^\infty \sum_{k=r}^\infty \binom{k+j}{k} \binom{j}{r}\binom{k}{r} z^{j+k} = \sum_{j=r}^\infty \binom{j}{r} z^{j+r} \sum_{k=0}^\infty \frac{(k+j+r)!}{j! r! k!} z^k =\\ \sum_{j=r}^\infty \binom{j}{r} z^{j+r} \binom{j+r}{j} \sum_{k=0}^\infty \frac{(j+r+1)_k}{k!} z^k = \sum_{j=r}^\infty \binom{j}{r} z^{j+r} \binom{j+r}{j} \left(1-z\right)^{-j-r-1} = \\ \frac{z^{2r}}{(1-z)^{2r+1}} \frac{1}{r!^2} \sum_{j=0}^\infty \frac{(j+2r)!}{j!} \left(\frac{z}{1-z}\right)^j = \frac{z^{2r}}{(1-z)^{2r+1}} \binom{2r}{r} \left(1-\frac{z}{1-z}\right)^{-1-2r} = \binom{2r}{r} z^{2r} \left(1-2z\right)^{-2r-1} $$ we continue: $$ g(x) = \frac{1}{1-x} \sum_{r=0}^\infty x^r \binom{2r}{r} \left(\frac{x}{1-x}\right)^{2r} \left(1 - 2 \frac{x}{1-x} \right)^{-1-2r} = \\ \frac{1}{1-x} \sum_{r=0}^\infty \binom{2r}{r} \frac{1-x}{1-3x} \left(\frac{x^3}{(1-3x)^2}\right)^r = \\ \frac{1}{1-3x} \sum_{r=0}^\infty \binom{2r}{r}\left(\frac{x^3}{(1-3x)^2}\right)^r = \frac{1}{1-3x} \left(1 - 4 \frac{x^3}{(1-3x)^2}\right)^{-1/2} = \frac{1}{1-3x} \left( \frac{(1-4x)(1-x)^2}{(1-3x)^2}\right)^{-1/2} = \frac{1}{1-x} \frac{1}{\sqrt{1-4x}} $$ which is exactly the generating function of the right-hand-side: $$ \sum_{n=0}^\infty x^n \sum_{r=0}^n \binom{2r}{r} \stackrel{n=r+k}{=} \sum_{k=0}^\infty x^r \sum_{r=0}^\infty \binom{2r}{r} x^r = \frac{1}{1-x} \cdot \frac{1}{\sqrt{1-4x}} $$

  • 3
    $\begingroup$ Awesome solution! $\endgroup$ Jul 31, 2012 at 19:21

Short summary: On the right we are summing the number of words of $r$ $a$s and $r$ $b$s over $0\le r\le n.$ Denote the set of words with $r$ $a$s and $r$ $b$s by $U_r.$ On the left we are computing the number of triples of words, the first with $i$ $a$s and $j$ $b$s, the second with $j$ $a$s and $k$ $b$s, the third with $k$ $a$s and $i$ $b$s, for all $i+j+k=n.$ Denote this set of triples $V_n.$

Let $(x,y,z)\in V_n.$ Define $(x,y,z)$ to be $0$-padded if $x$ does not end in $b$ or $y$ does not end in $a.$ Define $(x,y,z)$ to be $r$-padded if $(x,y,z)=(x'b^r,y'a^r,z)$ where $(x',y',z)\in V_{n-r}$ is $0$-padded.

Then any word $w\in U_r$ can be written in a unique way as $w=x'y'z$ where $(x',y',z)\in V_r$ is $0$-padded. (This is proved below.) Then $(x'b^{n-r},y'a^{n-r},z)\in V_n$ is $(n-r)-padded.$ This establishes a bijection between $V_n$ on the left, and $U_0\cup U_1\cup\ldots\cup U_n$ on the right, thereby proving the identity. The elements of $U_n$ correspond to $0$-padded elements of $V_n,$ the elements of $U_{n-1}$ correspond to $1$-padded elements of $V_n,$ and so on.

Detailed answer: The expression $\binom{2r}{r}$ counts words constructed from $r$ $a$s and $r$ $b$s. The sum on the right counts all such words of lengths ranging from $0$ to $2n.$

On the left, we have the product $\binom{i+j}{i}\binom{j+k}{j}\binom{k+i}{k},$ which counts all words constructed by concatenating a word with $i$ $a$s and $j$ $b$s to a word with $j$ $a$s and $k$ $b$s to a word with $k$ $a$s and $i$ $b$s. Such words contain $i+j+k$ $a$s and $i+j+k$ $b$s. The sum is over all $i+j+k=n,$ so these words all have $n$ $a$s and $n$ $b$s.

These observations hint at the possibility of a bijection, but a few questions arise:

  1. Can any word of $n$ $a$s and $n$ $b$s be constructed by concatenating a word with $i$ $a$s and $j$ $b$s to a word with $j$ $a$s and $k$ $b$s to a word with $k$ $a$s and $i$ $b$s, for suitable $i,$ $j,$ $k$?
  2. Is it possible for a word to be constructed in more than one way by such a procedure?
  3. What about words of length less than $2n$?

In answering the first question, we will actually answer all three.

Given a word $w$ of $n$ $a$s and $n$ $b$s, let's see if we can find $i+j+k=n,$ a word $x$ of $i$ $a$s and $j$ $b$s, a word $y$ of $j$ $a$s and $k$ $b$s, and a word $z$ of $k$ $a$s and $i$ $b$s such that $xyz=w.$ Observe that if such words can be found then $\lvert x\rvert=i+j\le i+j+2k=(j+k)+(k+i)=\lvert y\rvert+\lvert z\rvert.$ So $\lvert x\rvert\le n.$

We proceed by splitting $w$ into three, possibly empty, parts as follows: let $X$ consist of the first $P$ letters of $w,$ with $0\le P\le n,$ and let $I$ equal the number of $a$s in $X$ and $J$ the number of $b$s in $X$ (so that $I+J=P$). Let $K=n-I-J.$ Let $Z$ equal the string of letters from position $Q+1$ to $2n$ of $w,$ with $Q$ chosen so that $Z$ has $K$ $a$s. Note that $Q\ge P$ automatically holds: since the number of $a$s in $X$ is $I,$ the number of $a$s in the remainder of $w$ is $n-I\ge n-P=K.$ If $n-I>K,$ then there are $a$s to the right of $X$ and to the left of $Z$ and so $Q>P.$ If $n-I=K,$ then $J=0,$ and $X$ is simply a string of $I$ $a$s. So $Z$ contains all of the remaining $a$s and possibly some $b$s as well. Since $X$ contains no $b$s, we have $Q\ge P.$ So we may let $Y$ equal the string of letters from position $P+1$ to $Q$ of $w,$ giving $w=XYZ.$

So far we have that $I+J+K=n,$ that $X$ contains $I$ $a$s and $J$ $b$s, that $Z$ contains $K$ $a$s and $B:=2n-Q-K$ $b$s, and that $Y$ contains $n-I-K=J$ $a$s and $n-J-B$ $b$s. In order to obtain our desired partition of $w,$ we need $B=I.$ We claim that with suitable choices of $P$ and $Q,$ this can always be achieved.

To verify that this is so, we need to understand what happens as $P$ increases in steps of $1$ from $0$ to $n.$ To emphasize that $I,$ $J,$ and $K$ are functions of $P,$ we write $I(P),$ etc. We have already seen that there is always at least one possible value of $Q$ (and hence $B$) compatible with a given value of $P.$ There may, however, be more than one such compatible value, so we write $Q_l(P)$ and $Q_u(P)$ for the lower and upper bounds on $Q$ for a given $P.$ We have corresponding lower and upper bounds on $B,$ $$B_l(P)=2n-Q_u(P)-K(P)=n+P-Q_u(P)$$ and $$B_u(P)=2n-Q_l(P)-K(P)=n+P-Q_l(P).$$

When $P$ increases by $1,$ $K$ decreases by $1$: $K(P+1)=K(P)-1.$ If the letter of $w$ at position $P+1$ is $a$ then $I(P+1)=I(P)+1$ and $J(P+1)=J(P)$; if that letter is $b$ then $I(P+1)=I(P)$ and $J(P+1)=J(P)+1.$ Since $K(P)=n-P,$ the $K(P)^\text{th}$ $a$ from the right in $w$ is the $(P+1)^\text{st}$ $a$ from the left. Let $u$ be the position of the $P^\text{th}$ $a$ from the left in $w$ ($u=0$ if $P=0$) and let $v$ be the position of the $(P+1)^\text{st}$ $a$ from the left in $w$ ($v=2n+1$ if $P=n$). Then $Q_l(P)=u$ and and $Q_u(P)=v-1.$ In other words, if $\lvert X\rvert=P$ then $Z$ may start anywhere between the position immediately following the $P^\text{th}$ $a$ and the position of the $(P+1)^\text{st}$ $a.$ Notice that $Q_l(P+1)=v$ and therefore, that $Q_l(P+1)=Q_u(P)+1.$ This implies that $$B_u(P+1)=n+P+1-Q_l(P+1)=n+P+1-1-Q_u(P)=n+P-Q_u(P)=B_l(P).$$ Therefore, if $I(P)<B_l(P),$ then, since $I(P+1)$ is at most $I(P)+1,$ we have $I(P+1)\le B_u(P+1).$ On the other hand, $B_l(P)$ is non-increasing as $P$ increases from $0$ to $n,$ and ultimately equals $0.$ Therefore there must eventually be a $P$ such that $B_l(P)\le I(P)\le B_u(P).$ Once we find such a $P,$ we set $i=I(P),$ $j=J(P),$ $k=n-P,$ $x=X,$ $y=Y,$ $z=Z.$

Will this be the only solution? Not in general. Let there be a solution $(i,j,k,x,y,z)$ corresponding to a particular $P$ and $Q.$ If $y$ begins with $b$ and $z$ begins with $a$ then we get an additional solution by increasing both $P$ and $Q$ by $1.$ This results in the first $b$ of $y$ becoming the last letter of $x$ and the first $a$ of $z$ becoming the last letter of $y,$ which increases $j$ by $1$ and decreases $k$ by $1$ while leaving $i$ unchanged. The process obviously also works in reverse. There are no other ways to obtain additional solutions, for if the first letter of $y$ is $a,$ adding it to $x$ would force us to add a $b$ to $z,$ but $z$ cannot increase in length when $x$ increases in length.

In summary, all solutions for a particular word $w$ have the same $i.$ Equivalently, $\lvert y\rvert=j+k=n-i$ is the same for all solutions. The solution with minimum $j$ must be such that $x$ ends in $a$ or $y$ ends in $b.$ The solution with maximum $j$ must be such that $y$ begins with $a$ or $z$ begins with $b.$

Consider an example. Let $w=baabbbaaabab.$ Here $n=6.$ Then we have $$ \begin{array}{ccc|ccc} P & I(P) & J(P) & K(P) & [B_l(P),B_u(P)] & [Q_l(P),Q_u(P)]\\ \hline 0 & 0 & 0 & 6 & [5,6] & [0,1]\\ 1 & 0 & 1 & 5 & [5,5] & [2,2]\\ 2 & 1 & 1 & 4 & [2,5] & [3,6]\\ 3 & 2 & 1 & 3 & [2,2] & [7,7]\\ 4 & 2 & 2 & 2 & [2,2] & [8,8]\\ 5 & 2 & 3 & 1 & [1,2] & [9,10]\\ 6 & 2 & 4 & 0 & [0,1] & [11,12] \end{array} $$ We get three solutions, corresponding to $(P,Q)=(3,7),$ $(4,8),$ $(5,9).$ These are $$ \begin{aligned} &(i,j,k,x,y,x)=(2,1,3,baa,bbba,aabab),\\ &(i,j,k,x,y,x)=(2,2,2,baab,bbaa,abab),\\ &(i,j,k,x,y,x)=(2,3,1,baabb,baaa,bab). \end{aligned} $$

We have answered questions $1$ and $2$ in the affirmative. This suggests a way of producing a bijective proof of the identity. Let $W$ be the set of words of length at most $2n$ containing equal numbers of $a$s and $b$s. Let $S$ be the set of sextuples $(i,j,k,x,y,z)$ where $i,$ $j,$ $k$ are nonnegative integers satisfying $i+j+k=n,$ $x$ is a word consisting of $i$ $a$s and $j$ $b$s, $y$ is a word consisting of $j$ $a$s and $k$ $b$s, and $z$ is a word consisting of $k$ $a$s and $i$ $b$s. The map from $S$ to $W$ that simply concatenates $x,$ $y,$ and $z$ is not onto, since it only produces words of length $2n,$ and not one-to-one, as we have seen above. By modifying this simple concatenation map, we obtain a map that is both onto and one-to-one.

The map: Let $s=(i,j,k,x,y,z)\in S.$ If the last letter of $x$ is $b$ and the last letter of $y$ is $a,$ delete the last letter of both words. Repeat until $x$ or $y$ is empty or the last letter of $x$ is not $b$ or the last letter of $y$ is not $a.$ Denote the resulting words $x',$ $y'.$ We define $f:S\to W$ by $f(s)=x'y'z.$

The inverse map, applied to a word $w\in W$ with $N$ $a$s and $N$ $b$s, $0\le N\le n,$ is computed

  1. by applying the algorithm described above to write $w=x'y'z,$ where $x'$ is a word with $i$ $a$s and $j'$ $b$s, $y'$ is a word with $j'$ $a$s and $k$ $b$s, $z$ is a word with $k$ $a$s and $i$ $b$s, $i+j'+k=N,$ and $j'$ is as small as possible,
  2. by appending $n-N$ $b$s to $x'$ to form $x,$ appending $n-N$ $a$s to $y'$ to form $y,$ and
  3. by forming the sextuple $(i,j'+n-N,k,x,y,z).$

Example from original answer: I've replaced my original answer with what I hope is a more convincing presentation, but this example from that answer may be helpful, so I leave it.

Suppose that $n=3.$ The right hand side enumerates the union of the following sets $$ \begin{aligned} &\{e\},\\ &\{ab,ba\},\\ &\{aabb,abab,abba,baab,baba,bbaa\},\\ &\{aaabbb,aababb,aabbab,aabbba,abaabb,ababab,ababba,abbaab,abbaba,abbbaa,\\ &\ \ baaabb,baabab,baabba,babaab,bababa,babbaa,bbaaab,bbaaba,bbabaa,bbbaaa\}. \end{aligned} $$ Here $e$ denotes the empty word. The left hand side enumerates words constructed as follows.

  • Let $A$ be the set of words containing $i$ $a$s and $j$ $b$s. ($\lvert A\rvert=\binom{i+j}{i}$)
  • Let $B$ be the set of words containing $j$ $a$s and $k$ $b$s. ($\lvert B\rvert=\binom{j+k}{j}$)
  • Let $C$ be the set of words containing $k$ $a$s and $i$ $b$s. ($\lvert C\rvert=\binom{k+i}{k}$)

$$ \begin{aligned} &(i,j,k)=(0,0,3),\ A=\{e\},B=\{bbb\},C=\{aaa\}:\\ &\ \longrightarrow\{bbbaaa\}\\ &(i,j,k)=(0,1,2),\ A=\{b\},B=\{abb,bab,bba\},C=\{aa\}:\\ &\ \longrightarrow\{babbaa,bbabaa,\dot{b}bb\dot{a}aa=bbaa\}\\ &(i,j,k)=(0,2,1),\ A=\{bb\},B=\{aab,aba,baa\},C=\{a\}:\\ &\ \longrightarrow\{bbaaba,b\dot{b}ab\dot{a}a=baba,\dot{b}\dot{b}b\dot{a}\dot{a}a=ba\}\\ &(i,j,k)=(0,3,0),\ A=\{bbb\},B=\{aaa\},C=\{e\}:\\ &\ \longrightarrow\{\dot{b}\dot{b}\dot{b}\dot{a}\dot{a}\dot{a}=e\}\\ &(i,j,k)=(1,0,2),\ A=\{a\},B=\{bb\},C=\{aab,aba,baa\}:\\ &\ \longrightarrow\{abbaab,abbaba,abbbaa\}\\ &(i,j,k)=(1,1,1),\ A=\{ab,ba\},B=\{ab,ba\},C=\{ab,ba\}:\\ &\ \longrightarrow\{ababab,ababba,a\dot{b}b\dot{a}ab=abab,a\dot{b}b\dot{a}ba=abba,baabab,baabba,babaab,bababa\}\\ &(i,j,k)=(1,2,0),\ A=\{abb,bab,bba\},B=\{aa\},C=\{b\}:\\ &\ \longrightarrow\{a\dot{b}\dot{b}\dot{a}\dot{a}b=ab,ba\dot{b}a\dot{a}b=baab,bbaaab\}\\ &(i,j,k)=(2,0,1),\ A=\{aa\},B=\{b\},C=\{abb,bab,bba\}:\\ &\ \longrightarrow\{aababb,aabbab,aabbba\}\\ &(i,j,k)=(2,1,0),\ A=\{aab,aba,baa\},B=\{a\},C=\{bb\}:\\ &\ \longrightarrow\{aa\dot{b}\dot{a}bb=aabb,abaabb,baaabb\}\\ &(i,j,k)=(3,0,0),\ A=\{aaa\},B=\{e\},C=\{bbb\}:\\ &\ \longrightarrow\{aaabbb\} \end{aligned} $$ A dot over a letter indicates that the letter is to be deleted.

  • $\begingroup$ Thank you! I'm awarding the bounty now — and will try to understand the proof later. $\endgroup$
    – Grigory M
    Jan 9, 2015 at 21:02
  • $\begingroup$ I'm sorry to keep revising my answer. In the end, I think the bijection is fairly straightforward, and am a bit frustrated that the explanation is so cumbersome. Perhaps someone can produce a better explanation. $\endgroup$ Jan 13, 2015 at 15:11
  • $\begingroup$ When I see these combinatorial proofs of algebraic identities, they often seem like magic to me and I do not see how the combinatorial proof could have been derived without knowing the identity. $\endgroup$ Mar 2, 2022 at 3:52
  • $\begingroup$ @martycohen Can you elaborate on what you mean? Isn't it the case that with any proof method you need to know the identity you're trying to prove? I suppose that if you had two distinct methods for evaluating a generating function, you might have the proof before having the identity. But the same would be true if you somehow knew the bijection before knowing the identity. In physics, combinatorial identities often arise in the course of complicated calculations that are subject to some consistency condition. If the calculation and the consistency condition are known to be true, then the... $\endgroup$ Mar 2, 2022 at 18:50
  • $\begingroup$ ...identity follows. But such complicated proofs are often unsatisfying, and one seeks a simpler proof. If the identity can be interpreted combinatorially, it seems natural to look for a combinatorial explanation. In this case, the bijection was easier to find than it was to explain. $\endgroup$ Mar 2, 2022 at 18:51

Turns out, this indeed follows (relatively) directly from Strehl's identity.

Let's rewrite LHS in terms of $i$, $k$ and $l=i+j$: $$ \text{LHS}= \sum_{l+k=n}\sum_{i=0}^l\binom li\binom{k+l-i}{l-i}\binom{k+i}i= \sum_{l+k=n}\sum_{i=0}^l(-1)^l\binom li\binom{-k-1}{l-i}\binom{-k-1}i; $$ application of Strehl's identity to the internal sum (in that post's notation, for $n=-k-1$, $m=l$) yields $$ \sum_{l+k=n}\sum_i(-1)^l\binom{-k-1}i\binom li\binom{2i}l= \sum(-1)^{l-i}\binom{k+i}i\binom li\binom{2i}l= \sum(-1)^{l-i}\binom{k+i}i\binom i{l-i}\binom{2i}i; $$ the $\binom{2i}i$ terms suggests that we're almost there — and indeed what we've got is $$ \sum_{i=0}^n\binom{2i}i\sum_j(-1)^j\binom{n-j}i\binom ij $$ and the internal sum is (always) equal to 1 by Vandermonde convolution. QED.

Remark. In principle, this is not that far from being bijective, since we have a bijective proof of Strehl's identity — but that proof works only for positive parameters...

  • $\begingroup$ It would help if you would link to a version of Strehl's identity general enough for your use of it! $\endgroup$ Sep 19, 2015 at 1:08
  • $\begingroup$ @darij why, I already do (see par. 3 of 'Comments and thoughts' there) $\endgroup$
    – Grigory M
    Sep 19, 2015 at 1:26
  • $\begingroup$ Ah, that's the one you're referencing! $\endgroup$ Sep 19, 2015 at 1:33

Recall that $\sum_{n \ge 0} \binom nm x^n = \frac{x^m}{(1-x)^{m+1}}$.

We have \begin{align*} \sum_{i+j+k=n}\binom{i+j}{i} \binom{j+k}{j} \binom{k+i}{k} &= \sum_{i,j}\binom{i+j}{i} \binom{n-i}{j} \binom{n-j}{i} \\ &= \sum_{i,j}\binom{i+j}{i} \left([x^{n-i}] \frac{x^j}{(1-x)^{j+1}}\right) \left([y^{n-j}] \frac{y^i}{(1-y)^{i+1}}\right) \\ &= [x^ny^n] \sum_{i,j}\binom{i+j}{i} \frac{x^{i+j}}{(1-x)^{j+1}} \frac{y^{i+j}}{(1-y)^{i+1}} \end{align*}

Now, \begin{align*} \sum_{i,j}\binom{i+j}{i} \frac{x^{i+j}}{(1-x)^{j+1}} \frac{y^{i+j}}{(1-y)^{i+1}} &= \frac 1{(1-x)(1-y)} \sum_{p} x^p y^p \sum_{i+j = p} \binom pi \frac 1{(1-x)^{j}} \frac 1{(1-y)^{i}} \\ &= \frac 1{(1-x)(1-y)} \sum_{p} x^py^p \left(\frac 1{1-x} + \frac 1{1-y} \right)^p \\ &= \frac 1{(1-x)(1-y)} \frac 1 {1 - \frac{xy(2-x-y)}{(1-x)(1-y)}} \\ &= \frac 1 {(1-x-y)(1-xy)} \end{align*}

We have $[x^ry^r] \frac 1 {(1-x-y)} = \binom {2r} r$, so $$[x^ny^n] \frac 1 {(1-x-y)(1-xy)} = \sum_{r=0}^n \binom {2r} r.$$

  • $\begingroup$ The only proof that matches the beauty of the formula. Bravo. $\endgroup$ Apr 30, 2023 at 17:05

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