Andrews–Merca–Guo–Zeng coefficient non-negativity conjecture for averaged Jacobi truncations

Let (a;q)=j0(1aqj)(a;q)_\infty=\prod_{j\ge 0}(1-aq^j) and (a,b,c;q)=(a;q)(b;q)(c;q)(a,b,c;q)_\infty=(a;q)_\infty(b;q)_\infty(c;q)_\infty. For positive integers m,k,R,Sm,k,R,S with 1S<R/21\le S<R/2, consider the coefficient of qmq^m in

(1)k1(qS,qRS,qR;qR)0n<k(1)nq(n+12)RnS(1q(2n+1)S).\frac{(-1)^{k-1}}{(q^S,q^{R-S},q^{R};q^{R})_{\infty}} \sum_{0\le n<k}(-1)^nq^{\binom{n+1}{2}R-nS}(1-q^{(2n+1)S}).

Andrews–Merca–Guo–Zeng conjecture. This coefficient is non-negative. This conjecture generalizes non-negativity results for averaged truncations of Euler's pentagonal number theorem and Jacobi's triple product identity. Its resolution is not specified in the source.

Sources & referencesView supporting material

Primary source

Michael J. Schlosser and Nian Hong Zhou, “Expansions of averaged truncations of basic hypergeometric series”, arXiv:2307.10821 (2023).

Additional references

2 papers in this index state this conjecture (2021–2023). The statement above is taken from the most recent of them; the others are arXiv:2108.04148.

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.