%I #14 Nov 11 2022 04:34:03
%S 1,4095,531442,16773119,244140626,2176254990,13841287202,68702695423,
%T 282430067923,999755863470,3138428376722,8913939907598,23298085122482,
%U 56680071092190,129746582562692,281406240452607,582622237229762,1156551128144685,2213314919066162,4094999772632494
%N a(n) = Sum_{d|n} (-1)^(n/d+1)*d^12.
%H Seiichi Manyama, <a href="/A321557/b321557.txt">Table of n, a(n) for n = 1..10000</a>
%H J. W. L. Glaisher, <a href="https://books.google.com/books?id=bLs9AQAAMAAJ&pg=RA1-PA1">On the representations of a number as the sum of two, four, six, eight, ten, and twelve squares</a>, Quart. J. Math. 38 (1907), 1-62 (see p. 4 and p. 8).
%H <a href="/index/Ge#Glaisher">Index entries for sequences mentioned by Glaisher</a>.
%F G.f.: Sum_{k>=1} k^12*x^k/(1 + x^k). - _Seiichi Manyama_, Nov 25 2018
%F From _Amiram Eldar_, Nov 11 2022: (Start)
%F 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.
%F Sum_{k=1..n} a(k) ~ c * n^13, where c = 315*zeta(13)/4096 = 0.0769137... . (End)
%t 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 *)
%o (PARI) apply( A321557(n)=sumdiv(n, d, (-1)^(n\d-1)*d^12), [1..30]) \\ _M. F. Hasler_, Nov 26 2018
%Y Cf. A321543 - A321565, A321807 - A321836 for similar sequences.
%Y Cf. A013671.
%K nonn,mult
%O 1,2
%A _N. J. A. Sloane_, Nov 23 2018