A q-Clausen-type product identity for terminating basic hypergeometric sums

Let nn and rr be nonnegative integers with rnr\leqslant n, and let q,xq,x be parameters for which the displayed expressions are defined. The q-Clausen-type conjecture.

(k=rn(qn;q)k(x;q2)kqk(q;q)kr(q;q)k+r)(k=rn(qn;q)k(x;q2)kq(n+1)k(k2)(q;q)kr(q;q)k+rxk)=(1)r(q;q)n2(x;q2)rqr(q;q)nr(q;q)n+r(q2/x;q2)rxrk=rn(qn;q)k(qn+1;q)k(x;q2)k(q2/x;q2)kqk(q;q)kr(q;q)k+r(q;q)2k.\begin{aligned} &\left(\sum_{k=r}^{n}\frac{(q^{-n};q)_k(x;q^2)_kq^k}{(q;q)_{k-r}(q;q)_{k+r}}\right)\left(\sum_{k=r}^{n}\frac{(q^{-n};q)_k(x;q^2)_kq^{(n+1)k-{k\choose2}}}{(q;q)_{k-r}(q;q)_{k+r}x^k}\right)\\\\ &=\frac{(-1)^r(q;q)_n^2(x;q^2)_rq^r}{(q;q)_{n-r}(q;q)_{n+r}(q^2/x;q^2)_rx^r}\sum_{k=r}^{n}\frac{(q^{-n};q)_k(q^{n+1};q)_k(x;q^2)_k(q^2/x;q^2)_kq^k}{(q;q)_{k-r}(q;q)_{k+r}(q;q)_{2k}}. \end{aligned}

The text calls this a numerical-experiment companion to a theorem; no proof or resolution is supplied.

Sources & referencesView supporting material

Primary source

Victor J. W. Guo and Jiang Zeng, “Some q-analogues of (super)congruences of Beukers, Van Hamme and Rodriguez-Villegas”, arXiv:1408.0512 (2014).

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