%I #65 May 02 2020 13:15:04
%S 2,28,1488,182880,41207040,14856307200,7847004211200,5713142135500800,
%T 5484741986820096000,6713362606110031872000,
%U 10204325758699297505280000,18857600746080668455403520000,41637586170036526348967608320000,108257726461843735266949595136000000,327371366649945523117538738700288000000
%N a(n) = (4^(n + 1) - 2)*(2*n)!.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PolygammaFunction.html">Polygamma Function</a>
%F a(n) = -(polygamma(2n, 1/4)/2^(2n-1) + abs(Euler(2n))*Pi^(2n+1))/zeta(2n+1) for n > 0.
%F a(n) = (2n)!*(2^(2n + 2) - 2).
%F a(n+1) = (2n)!*Stirling2(2n, 2)/binomial(2n, 2).
%F a(n) = A010050(n)*(A000302(n+1) - 2). - _Omar E. Pol_, May 02 2020
%e polygamma(2, 1/4) = -2^1*(28*zeta(3) + Pi^3),
%e polygamma(4, 1/4) = -2^3*(1488*zeta(5) + 5*Pi^5),
%e polygamma(6, 1/4) = -2^5*(182880*zeta(7) + 61*Pi^7),
%e polygamma(8, 1/4) = -2^7*(41207040*zeta(9) + 1385*Pi^9),
%e etc.
%t Table[(2 n + 2)!*StirlingS2[2 n + 2, 2]/Binomial[2 n + 2, 2], {n, 0, 17}]
%t Prepend[FullSimplify[Table[-(PolyGamma[2 n, 1/4]/2^(2 n - 1) + Abs[EulerE[2 n]] Pi^(2 n + 1))/Zeta[2 n + 1], {n, 1, 16}]], 2]
%Y Cf. A000364, A000302, A010050.
%K nonn
%O 0,1
%A _Artur Jasinski_, Jan 29 2020
%E a(0) = 2 and new name by _Peter Luschny_, May 02 2020.