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%I #38 Jun 06 2022 08:16:25
%S 1,90,113400,681080400,12504636144000,548828480360160000,
%T 49229914688306352000000,8094874872198213459360000000,
%U 2252447502438386084347676160000000,997586474354936812896742294502400000000,669959124447288464805194190141921792000000000
%N a(n) = (4n+2)!/2^(2n+1).
%H J. M. Borwein, D. M. Bradley, and D. J. Broadhurst, <a href="https://arxiv.org/abs/hep-th/9611004">Evaluations of k-fold Euler/Zagier sums: a compendium of results for arbitrary k</a>, arXiv:hep-th/9611004, 1996.
%H Roudy El Haddad, <a href="https://arxiv.org/abs/2102.00821">Multiple Sums and Partition Identities</a>, arXiv:2102.00821 [math.CO], 2021.
%H Roudy El Haddad, <a href="https://doi.org/10.7546/nntdm.2022.28.2.200-233">A generalization of multiple zeta value. Part 2: Multiple sums</a>. Notes on Number Theory and Discrete Mathematics, 28(2), 2022, 200-233, DOI: 10.7546/nntdm.2022.28.2.200-233.
%F sin(x)*sinh(x) = Sum_{n>=0} (-1)^n*x^(4n+2)/a(n). - _Benoit Cloitre_, Feb 02 2002
%F 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
%F From _Amiram Eldar_, Feb 25 2022: (Start)
%F Sum_{n>=0} 1/a(n) = (cosh(sqrt(2)) - cos(sqrt(2)))/2.
%F Sum_{n>=0} (-1)^n/a(n) = sin(1)*sinh(1). (End)
%t Table[(4n+2)!/2^(2n+1), {n, 0, 10}] (* _Amiram Eldar_, Feb 25 2022 *)
%o (PARI) a(n) = (4*n+2)!/2^(2*n+1); \\ _Michel Marcus_, Feb 20 2022
%Y Cf. A002432 (denominators of zeta(2*n)/Pi^(2*n)).
%Y Cf. A068447, A067912, A013662 (zeta(4)).
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_, Feb 05 2000
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