Hoffman's conjecture for symmetric triple series

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Let A=3A=3 and let n≥0n\geq 0 be an integer. Define

K1=k1+5n2+2,K2=k2+3n2+1,K3=k3+n2.K_1=k_1+\frac{5n}{2}+2,\qquad K_2=k_2+\frac{3n}{2}+1,\qquad K_3=k_3+\frac{n}{2}.

For a polynomial P(K1,K2,K3)P(K_1,K_2,K_3) in any of the four families specified in Theorem

ofthesource,withthestatedparityconditionsonitsnon−negativeexponents,considertheseriesof the source, with the stated parity conditions on its non-negative exponents, consider the series

\sum_{k_1\geq k_2\geq k_3\geq 1}\frac{P(K_1,K_2,K_3)}{(k_1+2n+2){n+1}^{A}(k_2+n+1){n+1}^{A}(k_3)_{n+1}^{A}}.

∗∗Hoffman′ssymmetric−seriesconjecture.∗∗Wheneverthisseriesisconvergent,Theorem**Hoffman's symmetric-series conjecture.** Whenever this series is convergent, Theorem

holds: it is a rational linear combination of multiple zeta values of depth at most 33, with neither ζ(2,2)\zeta(2,2) nor ζ(2,2,2)\zeta(2,2,2) appearing; and, if the additional condition (8)(8) holds, it is a rational linear combination of the explicitly listed values 11, ζ(2)\zeta(2), ζ(3)\zeta(3), ζ(2,3)\zeta(2,3), ζ(3,2)\zeta(3,2), ζ(3,3)\zeta(3,3), ζ(2,2,3)\zeta(2,2,3), ζ(2,3,2)\zeta(2,3,2), ζ(2,3,3)\zeta(2,3,3), ζ(3,2,2)\zeta(3,2,2), ζ(3,2,3)\zeta(3,2,3), ζ(3,3,2)\zeta(3,3,2), and ζ(3,3,3)\zeta(3,3,3). The source says this was checked computationally for n≤2n\leq 2 and leaves the general case open; in particular, the open problem is to prove that ζ(2,2)\zeta(2,2) does not appear.

References

Primary source

Stéphane Fischler, “Multiple series connected to Hoffman's conjecture on multiple zeta values”, arXiv:math/0609799 (2007).

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