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A329649 Let D = A042948(n) be the n-th positive integer congruent to 0 or 1 mod 4, then a(n) = b(D) := Sum_{i=1..D} Kronecker(D,i)*i^2, where Kronecker(D,i) is the Kronecker symbol.

%I A329649 #20 Nov 19 2019 07:29:03
%S A329649 1,10,4,16,159,48,52,680,136,48,168,288,4150,448,348,64,792,5196,740,
%T A329649 1120,1312,1232,144,192,33565,624,1484,2240,3192,2880,2684,43680,4160,
%U A329649 -544,3312,576,6424,5776,3696,192,118071,2016,6120,8096,9256,7360,6696,1152,13192
%N A329649 Let D = A042948(n) be the n-th positive integer congruent to 0 or 1 mod 4, then a(n) = b(D) := Sum_{i=1..D} Kronecker(D,i)*i^2, where Kronecker(D,i) is the Kronecker symbol.
%C A329649 Note that {Kronecker(D,i)} is a Dirichlet character mod |D| if and only if D == 0, 1 (mod 4).
%C A329649 Conjecture (a): There are no 0's in this sequence.
%C A329649 Conjecture (b): if D is not of the form k^2 or 5*k^2 or 2^t (t odd, t > 1), then b(D) is divisible by 4*D; if D = 5*k^2, then b(D) is divisible by 4*k^2 = 4*(D/5).
%C A329649 Conjecture (c): let D be a fundamental discriminant > 1 (A003658), then Sum_{k>=1} Kronecker(D,k)/k^2 = Pi^2*b(D)/D^(5/2) = (4*Pi^2)/D^(3/2) * (b(D)/(4*D)), which is an integer times (4*Pi^2)/D^(3/2) if D > 8 by the conjecture (b). Here Sum_{k>=1} Kronecker(D,k)/k^2 is the value of the Dirichlet L-series of a non-principal character modulo D at s=2.
%H A329649 Jianing Song, <a href="/A329649/b329649.txt">Table of n, a(n) for n = 1..10000</a>
%F A329649 If D is a square, then b(D) = A053818(D), so b(D) is divisible by D if and only if sqrt(D) is not in A316860.
%F A329649 If D is not a square, D is divisible by p^e, where e >= 4 if p = 2, e >= 3 if p > 2, then it is easy to see that b(D) = p^2*b(D/p^2). So we only need to consider the value of b(D) where D is a cubefree number or 8 times a cubefree odd number. Specially, for odd t, we have b(2^t) = b(8)*(2^(t-3)) = 2^(t+1) for t > 1; b(5^t) = b(5)*(5^(t-1)) = 4*5^(t-1); b(13^t) = b(13)*(13^(t-1)) = 4*13^t and so on.
%e A329649 For n = 3, D = 5, b(5) = 1^2 - 2^2 - 3^2 + 4^2 = 4. Here 5 = 5*1^2, we have b(5)/(4*1^2) = 1 is an integer. Also, Sum_{k>=1} Kronecker(5,k)/k^2 = Pi^2/(25*sqrt(5)) = Pi^2*b(5)/5^(5/2).
%e A329649 For n = 4, D = 8, b(8) = 1^2 - 3^2 - 5^2 + 7^2 = 16. We have Sum_{k>=1} Kronecker(8,k)/k^2 = Pi^2/(8*sqrt(2)) = Pi^2*b(8)/8^(5/2).
%e A329649 For n = 6, D = 12, b(12) = 1^2 - 5^2 - 7^2 + 11^2 = 48. Here 12 is not of the form k^2 or 5*k^2 or 2^t (t odd, t > 1), we have b(12)/(4*12) = 1 is an integer. Also, Sum_{k>=1} Kronecker(12,k)/k^2 = Pi^2/(6*sqrt(3)) = Pi^2*b(12)/12^(5/2).
%t A329649 b[n_] = Sum[KroneckerSymbol[n, i]*i^2, {i, 1, n}];
%t A329649 a[n_] = b[2 n - Mod[n, 2]]
%o A329649 (PARI) b(n) = sum(i=1, n, kronecker(n,i)*i^2)
%o A329649 a(n) = b(2*n - (n%2))
%Y A329649 Cf. A042948, A003658, A053818, A316860.
%K A329649 sign
%O A329649 1,2
%A A329649 _Jianing Song_, Nov 18 2019