# Looking for a $q$-analogue of a binomial identity

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)?

• 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. – Sam Hopkins May 18 at 19:53
• You're right and I was aware of too. But, what would be the $q$-analogue g.f.? – T. Amdeberhan May 18 at 19:55
• 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... – Sam Hopkins May 18 at 19:56

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 $$\sum_{j+i=n} \tau^{j} {2j\choose j}_{\tau}{2i\choose i}_\tau \frac{(\tau)_j^2(\tau)_i^2}{(\tau^2)_j^2(\tau^2)_i^2}=1.$$