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%I #27 Oct 27 2019 12:01:04
%S 1,2,1407,1444,4540261
%N a(n) = b(n) if b(n) := (Sum_{k>=1}(k/5)/k^(2n))*sqrt(5)^(floor((n+1)/2)*floor((n+11)/2)-1)/(24*zeta(2n)) is an integer, and a(n) = 0 otherwise, where (k/5) is the Legendre symbol.
%C Conjecture: For each n = 1,2,3,... the number (Sum_{k>=1}(k/5)/k^(2n))*sqrt(5)/Pi^(2n) is rational, where (k/5) is the Legendre symbol. Moreover, a(n) is a nonzero integer for every positive integer n.
%C A famous formula of Ramanujan states that Sum_{k>=1}(k/5)q^k/(1-q^k)^2 = q*Product_{n>=1}(1-q^(5n))^5/(1-q^n) for |q| < 1. This can be viewed as a q-analog of the identity Sum_{k>=1}(k/5)/k^2 = 4*Pi^2/(25*sqrt(5)).
%C A well-known formula of Euler asserts that zeta(2n) = (-1)^(n-1)*2^(2n-1)*Pi^(2n)*B_{2n}/(2n)! for all n = 1,2,3,..., where B_{2n} is the 2n-th Bernoulli number. We find that Mathematica 10 could evaluate the value of s(n) = Sum_{k>=1}(k/5)/k^(2n) for n = 1,2,3,4,5; namely, s(1) = (4*Pi^2)/(25*sqrt(5)), s(2) = 8*Pi^4/(375*sqrt(5)), s(3) = 536*Pi^6/(234375*sqrt(5)), s(4) = 5776*Pi^8/(24609375*sqrt(5)) and s(5) = 3302008*Pi^10/(138427734375*sqrt(5)).
%D B. C. Berndt, Number Theory in the Spirit of Ramanujan, Amer. Math. Soc., Providence, RI, 2006, p. 107.
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1802.01473">A q-analogue of Euler's formula zeta(2) = pi^2/6</a>, arXiv:1802.01473 [math.NT], 2018.
%e a(1) = 1 since Sum_{k>=1}((k/5)/k^2)*sqrt(5)^(floor((1+1)/2)*floor((1+11)/2)-1)/(24*zeta(2)) = 4*Pi^2/(25*sqrt(5))*sqrt(5)^5/(24*Pi^2/6) = 1.
%t a[n_]:=a[n]=FullSimplify[Sum[JacobiSymbol[k,5]/(k^(2n)),{k,1,Infinity}]]Sqrt[5]^(Floor[(n+1)/2]Floor[(n+11)/2]-1)/(24*Zeta[2n]);
%t Table[If[IntegerQ[a[n]],a[n],0],{n,1,5}]
%Y Cf. A000796, A080891, A027641, A027642.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Feb 09 2018
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