6
$\begingroup$

The following identity is well-known and there are a few proofs to it (see Bijective proof problems, by R P Stanley, for this and similar formulae): $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}=4^n \qquad \iff \qquad \sum_{j+i=n}\binom{2j}j\binom{2i}i\frac1{4^n}=1.\tag1$$

QUESTION. Is there a $q$-analogue to this "innocent-looking" identity in equation (1)?

$\endgroup$
3
  • 1
    $\begingroup$ The identity is essentially equivalent to the g.f. identity $\sum_{k \geq 0} \binom{2k}{k} x^k = 1/\sqrt{1-4x}$, so if you have a q-analog of that g.f. you'll get what you want. $\endgroup$ – Sam Hopkins May 18 at 19:53
  • 1
    $\begingroup$ You're right and I was aware of too. But, what would be the $q$-analogue g.f.? $\endgroup$ – T. Amdeberhan May 18 at 19:55
  • 1
    $\begingroup$ Yes, I have no idea about, maybe I am just restating your problem. Of course you can try to find an expression for $\sum_{k\geq 0} \binom{2k}{k}_q x^k$, the usual q-binomial... $\endgroup$ – Sam Hopkins May 18 at 19:56
3
$\begingroup$

We start with $q$-binomial theorem $$ (x+y)(x+qy)\cdots(x+q^{n-1}y)=\sum q^{k\choose 2}{n\choose k}_qy^k x^{n-k}.\quad\quad\quad(\heartsuit) $$ Put $n=a+b$ in $(\heartsuit)$ and consider separately the first $a$ multiples in LHS and the last $b$ multiples. We get $$ \left(\sum q^{j\choose 2}{a\choose j}_qy^j x^{a-j}\right)\cdot \left(\sum q^{{i\choose 2}+ai}{b\choose i}_qy^i x^{b-j}\right)= \sum q^{k\choose 2}{a+b\choose k}_qy^k x^{a+b-k}. $$ Taking coefficients of $x^ky^{n-k}$ we get $$ \sum_{j+i=k} q^{{j\choose 2}+{i\choose 2}+ai}{a\choose j}_q {b\choose i}_q=q^{k\choose 2}{a+b\choose k}_q.\quad\quad\quad\quad\quad\quad\quad\quad(\clubsuit) $$ For fixed $k$ (and $q$) both LHS and RHS of $(\clubsuit)$ are polynomials in $q^a$ and $q^b$. Thus we may substitute $q^a=q^b=q^{-1/2}$ (I do this because in the ordinary version the identity is equivalent to the $a=b=-1/2$ version of the Vandermonde--Chu convolution $\sum_{a+b=k} {a\choose i}{b\choose j}={a+b\choose k}$). We have $${-1\choose k}_q=\frac{(q^{-1}-1)(q^{-2}-1)\cdots(q^{-k}-1)}{(q^k-1)\cdots(q-1)}=q^{-{k+1\choose 2}}(-1)^k.$$ Next, denote $q=\tau^2$, then $${-1/2\choose j}_q=\frac{(q^{-1/2}-1)(q^{-3/2}-1)\cdots(q^{-(2j-1)/2}-1)}{(q^j-1)\cdots(q-1)}\\=\tau^{-j^2}(-1)^j\frac{(\tau-1)(\tau^3-1)\cdots (\tau^{2j-1}-1)}{(\tau^2-1)(\tau^4-1)\cdots (\tau^{2j}-1)}\\=\tau^{-j^2}(-1)^j{2j\choose j}_{\tau}\frac1{(1+\tau)^2(1+\tau^2)^2\cdots (1+\tau^j)^2}.$$ So $(\clubsuit)$ reads as $$ \sum_{j+i=k} \tau^{j} {2j\choose j}_{\tau}{2i\choose i}_\tau\frac1{(1+\tau)^2(1+\tau^2)^2\cdots (1+\tau^j)^2\cdot (1+\tau)^2(1+\tau^2)^2\cdots (1+\tau^i)^2}=1. $$

This is an analogue of the identity you ask about in the sense that for $\tau=1$ we get it.

Note by OP: If we write the $j$-factorial analogue as $(q)_j=\prod_{k=1}^j\frac{1-q^k}{1-q}$, then the last identity takes the form $$ \sum_{j+i=n} \tau^{j} {2j\choose j}_{\tau}{2i\choose i}_\tau \frac{(\tau)_j^2(\tau)_i^2}{(\tau^2)_j^2(\tau^2)_i^2}=1. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.