A321548 a(n) = Sum_{d|n} (-1)^(d-1)*d^9.
1, -511, 19684, -262655, 1953126, -10058524, 40353608, -134480383, 387440173, -998047386, 2357947692, -5170101020, 10604499374, -20620693688, 38445332184, -68853957119, 118587876498, -197981928403, 322687697780, -512998309530, 794320419872, -1204911270612, 1801152661464, -2647111858972
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
Table[Total[(-1)^(#-1) #^9&/@Divisors[n]],{n,30}] (* Harvey P. Dale, Sep 07 2020 *) f[p_, e_] := (p^(9*e + 9) - 1)/(p^9 - 1); f[2, e_] := 2 - (2^(9*e + 9) - 1)/511; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 25] (* Amiram Eldar, Nov 04 2022 *)
-
PARI
apply( a(n)=sumdiv(n, d, (-1)^(d-1)*d^9), [1..30]) \\ M. F. Hasler, Nov 26 2018
Formula
G.f.: Sum_{k>=1} (-1)^(k-1)*k^9*x^k/(1 - x^k). - Ilya Gutkovskiy, Dec 23 2018
Multiplicative with a(2^e) = 2 - (2^(9*e + 9) - 1)/511, and a(p^e) = (p^(9*e + 9) - 1)/(p^9 - 1) for p > 2. - Amiram Eldar, Nov 04 2022