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.

1, 10, 4, 16, 159, 48, 52, 680, 136, 48, 168, 288, 4150, 448, 348, 64, 792, 5196, 740, 1120, 1312, 1232, 144, 192, 33565, 624, 1484, 2240, 3192, 2880, 2684, 43680, 4160, -544, 3312, 576, 6424, 5776, 3696, 192, 118071, 2016, 6120, 8096, 9256, 7360, 6696, 1152, 13192

Offset: 1

Author

Jianing Song, Nov 18 2019

Keywords

• sign

Comments

Note that {Kronecker(D,i)} is a Dirichlet character mod |D| if and only if D == 0, 1 (mod 4).

Conjecture (a): There are no 0's in this sequence.

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).

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.

Examples

			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).
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).
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).
		

Links

• Jianing Song, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A042948, A003658, A053818, A316860.

Programs

Mathematica

b[n_] = Sum[KroneckerSymbol[n, i]*i^2, {i, 1, n}];
a[n_] = b[2 n - Mod[n, 2]]

PARI

b(n) = sum(i=1, n, kronecker(n,i)*i^2)
a(n) = b(2*n - (n%2))

Formula

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.

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.