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A329649
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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.
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1
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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FORMULA
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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.
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EXAMPLE
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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).
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MATHEMATICA
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b[n_] = Sum[KroneckerSymbol[n, i]*i^2, {i, 1, n}];
a[n_] = b[2 n - Mod[n, 2]]
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PROG
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(PARI) b(n) = sum(i=1, n, kronecker(n, i)*i^2)
a(n) = b(2*n - (n%2))
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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