A321557 a(n) = Sum_{d|n} (-1)^(n/d+1)*d^12.
1, 4095, 531442, 16773119, 244140626, 2176254990, 13841287202, 68702695423, 282430067923, 999755863470, 3138428376722, 8913939907598, 23298085122482, 56680071092190, 129746582562692, 281406240452607, 582622237229762, 1156551128144685, 2213314919066162, 4094999772632494
Offset: 1
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.
Programs
-
Mathematica
f[p_, e_] := (p^(12*e + 12) - 1)/(p^12 - 1); f[2, e_] := (2047*2^(12*e + 1) + 1)/4095; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Nov 11 2022 *)
-
PARI
apply( A321557(n)=sumdiv(n, d, (-1)^(n\d-1)*d^12), [1..30]) \\ M. F. Hasler, Nov 26 2018
Formula
G.f.: Sum_{k>=1} k^12*x^k/(1 + x^k). - Seiichi Manyama, Nov 25 2018
From Amiram Eldar, Nov 11 2022: (Start)
Multiplicative with a(2^e) = (2047*2^(12*e+12)+1)/4095, and a(p^e) = (p^(12*e+12) - 1)/(p^12 - 1) if p > 2.
Sum_{k=1..n} a(k) ~ c * n^13, where c = 315*zeta(13)/4096 = 0.0769137... . (End)