A052277 a(n) = (4n+2)!/2^(2n+1).
1, 90, 113400, 681080400, 12504636144000, 548828480360160000, 49229914688306352000000, 8094874872198213459360000000, 2252447502438386084347676160000000, 997586474354936812896742294502400000000, 669959124447288464805194190141921792000000000
Offset: 0
Links
- J. M. Borwein, D. M. Bradley, and D. J. Broadhurst, Evaluations of k-fold Euler/Zagier sums: a compendium of results for arbitrary k, arXiv:hep-th/9611004, 1996.
- Roudy El Haddad, Multiple Sums and Partition Identities, arXiv:2102.00821 [math.CO], 2021.
- Roudy El Haddad, A generalization of multiple zeta value. Part 2: Multiple sums. Notes on Number Theory and Discrete Mathematics, 28(2), 2022, 200-233, DOI: 10.7546/nntdm.2022.28.2.200-233.
Crossrefs
Programs
-
Mathematica
Table[(4n+2)!/2^(2n+1), {n, 0, 10}] (* Amiram Eldar, Feb 25 2022 *) -
PARI
a(n) = (4*n+2)!/2^(2*n+1); \\ Michel Marcus, Feb 20 2022
Formula
sin(x)*sinh(x) = Sum_{n>=0} (-1)^n*x^(4n+2)/a(n). - Benoit Cloitre, Feb 02 2002
a(n) = Pi^(4n)/Zeta({4}n) where ({4}_n) is the standard multiple zeta values notation for (4, ..., 4) where the multiplicity of 4 is n. - _Roudy El Haddad, Feb 19 2022
From Amiram Eldar, Feb 25 2022: (Start)
Sum_{n>=0} 1/a(n) = (cosh(sqrt(2)) - cos(sqrt(2)))/2.
Sum_{n>=0} (-1)^n/a(n) = sin(1)*sinh(1). (End)