A050471 a(n) = Sum_{d|n, n/d=1 mod 4} d^3 - Sum_{d|n, n/d=3 mod 4} d^3.

1, 8, 26, 64, 126, 208, 342, 512, 703, 1008, 1330, 1664, 2198, 2736, 3276, 4096, 4914, 5624, 6858, 8064, 8892, 10640, 12166, 13312, 15751, 17584, 18980, 21888, 24390, 26208, 29790, 32768, 34580, 39312, 43092, 44992, 50654, 54864, 57148

Offset: 1

Author

N. J. A. Sloane, Dec 23 1999

Keywords

• nonn
• easy
• mult

Comments

Multiplicative because it is the Dirichlet convolution of A000578 = n^3 and A101455 = [1 0 -1 0 1 0 -1 ...], which are both multiplicative. - Christian G. Bower, May 17 2005

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. A000578, A101455, A175572.

Glaisher's E'_i (i=0..12): A002654, A050469, A050470, this sequence, A050468, A321829, A321830, A321831, A321832, A321833, A321834, A321835, A321836.

Programs

Mathematica

max = 40; s = Sum[n^3*x^(n-1)/(1+x^(2*n)), {n, 1, max}] + O[x]^max; CoefficientList[s, x] (* Jean-François Alcover, Dec 02 2015, after Vladeta Jovovic *)
s[n_] := If[OddQ[n], (-1)^((n-1)/2), 0]; (* A101455 *)
f[p_, e_] := (p^(3*e+3) - s[p]^(e+1))/(p^3 - s[p]); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Nov 04 2023 *)

PARI

a(n) = sumdiv(n, d, d^3*(((n/d) % 4)==1)) - sumdiv(n, d, d^3*(((n/d) % 4)==3)); \\ Michel Marcus, Feb 16 2015

Formula

G.f.: Sum_{n>=1} n^3*x^n/(1+x^(2*n)). - Vladeta Jovovic, Oct 16 2002

From Amiram Eldar, Nov 04 2023: (Start)

Multiplicative with a(p^e) = (p^(3*e+3) - A101455(p)^(e+1))/(p^3 - A101455(p)).

Sum_{k=1..n} a(k) ~ c * n^4 / 4, where c = A175572. (End)

a(n) = Sum_{d|n} (n/d)^3*sin(d*Pi/2). - Ridouane Oudra, Sep 26 2024

Extensions

Offset changed from 0 to 1 by R. J. Mathar, Jul 15 2010