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%I #76 Apr 15 2023 15:55:12
%S 1,2,10,244,554,202084,2162212,1594887848,7756604858,9619518701764,
%T 59259390118004,554790995145103208,954740563911205348,
%U 32696580074344991138888,105453443486621462355224,7064702291984369672858925136,4176926860695042104392112698
%N a(n) = numerator of (Zeta(2*n+1,1/4) - Zeta(2*n+1,3/4))/Pi^(2*n+1) where Zeta is the Hurwitz zeta function.
%C The function (Zeta(2*n+1,1/4) - Zeta(2*n+1,3/4))/Pi^(2*n+1) is rational for every positive integer n.
%C For denominators see A360966.
%C (Zeta(2*n+1,1/4) + Zeta(2*n+1,3/4))/Zeta(2*n+1) = 4*16^n - 2*4^n; see A193475.
%C For numerators of the function (Zeta(2*n,1/4) + Zeta(2*n,3/4))/Pi^(2*n) see A361007.
%C For denominators of the function (Zeta(2*n,1/4) + Zeta(2*n,3/4))/Pi^(2*n) see A036279.
%C (Zeta(2*n,1/4) - Zeta(2*n,3/4))/beta(2*n) = 16^n (see A001025) where beta is the Dirichlet beta function.
%C From the above formulas we can express Zeta(k,1/4) and Zeta(k,3/4) for every positive integer k.
%F a(n) = A046982(2*n).
%F (Zeta(2*n + 1, 1/4) - Zeta(2*n + 1, 3/4))/(Pi^(2*n + 1)) = A000364(n)*(2*n + 1)*2^(2*n)/(2*n + 1)!.
%e a(0) = 1 because lim_{n->0} (Zeta(2*n+1,1/4) - Zeta(2*n+1,3/4))/Pi^(2*n+1) = 1.
%e a(3) = 244 because (Zeta(2*3+1,1/4) - Zeta(2*3+1,3/4))/Pi^(2*3+1) = 244/45.
%t Table[(Zeta[2*n + 1, 1/4] - Zeta[2*n + 1, 3/4]) / Pi^(2*n + 1), {n, 1, 25}] // FunctionExpand // Numerator (* _Vaclav Kotesovec_, Feb 27 2023 *)
%t t[0, 1] = 1; t[0, _] = 0;
%t t[n_, k_] := t[n, k] = (k-1) t[n-1, k-1] + (k+1) t[n-1, k+1];
%t a[n_] := Sum[t[2n, k]/(2n)!, {k, 0, 2n+1}] // Numerator;
%t Table[a[n], {n, 0, 16}] (* _Jean-François Alcover_, Mar 15 2023 *)
%t a[n_] := SeriesCoefficient[Tan[x+Pi/4], {x, 0, 2n}] // Numerator;
%t Table[a[n], {n, 0, 16}] (* _Jean-François Alcover_, Apr 15 2023 *)
%o (PARI) a(n) = numerator(abs(eulerfrac(2*n))*(2*n + 1)*2^(2*n)/(2*n + 1)!); \\ _Michel Marcus_, Apr 11 2023
%Y Cf. A000364, A046982, A173945, A173947, A173948, A173949, A173953, A173954, A173955, A173982, A173983, A173984, A173987, A360966, A361007, A361007.
%K nonn,frac
%O 0,2
%A _Artur Jasinski_, Feb 26 2023
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