A321830 a(n) = Sum_{d|n, n/d==1 mod 4} d^6 - Sum_{d|n, n/d==3 mod 4} d^6.

1, 64, 728, 4096, 15626, 46592, 117648, 262144, 530713, 1000064, 1771560, 2981888, 4826810, 7529472, 11375728, 16777216, 24137570, 33965632, 47045880, 64004096, 85647744, 113379840, 148035888, 190840832, 244156251, 308915840, 386889776

Offset: 1

Author

N. J. A. Sloane, Nov 24 2018

Keywords

• nonn
• easy
• mult

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000
• J. W. L. Glaisher, On the representations of a number as the sum of two, four, six, eight, ten, and twelve squares, Quart. J. Math. 38 (1907), 1-62 (see p. 4 and p. 8).
• Index entries for sequences mentioned by Glaisher.

Crossrefs

Cf. A101455, A258814.

Cf. A321543 - A321565, A321807 - A321836 for similar sequences.

Glaisher's E'_i (i=0..12): A002654, A050469, A050470, A050471, A050468, A321829, this sequence, A321831, A321832, A321833, A321834, A321835, A321836.

Programs

Mathematica

s[n_,r_] := DivisorSum[n, #^6 &, Mod[n/#,4]==r &]; a[n_] := s[n,1] - s[n,3]; Array[a, 30] (* Amiram Eldar, Nov 26 2018 *)
s[n_] := If[OddQ[n], (-1)^((n-1)/2), 0]; (* A101455 *)
f[p_, e_] := (p^(6*e+6) - s[p]^(e+1))/(p^6 - s[p]); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Nov 04 2023 *)

PARI

apply( A321830(n)=factorback(apply(f->f[1]^(6*f[2]+6)\/(f[1]^6+f[1]%4-2),Col(factor(n)))), [1..30]) \\ M. F. Hasler, Nov 26 2018

Formula

G.f.: Sum_{k>=1} k^6*x^k/(1 + x^(2*k)). - Ilya Gutkovskiy, Nov 26 2018

Multiplicative with a(p^e) = round(p^(6e+6)/(p^6 + p%4 - 2)), where p%4 is the remainder of p modulo 4. (Following R. Israel in A321833.) - M. F. Hasler, Nov 26 2018

Sum_{k=1..n} a(k) ~ c * n^7 / 7, where c = 61*Pi^7/184320 (A258814). - Amiram Eldar, Nov 04 2023

a(n) = Sum_{d|n} (n/d)^6*sin(d*Pi/2). - Ridouane Oudra, Sep 27 2024