A284926 a(n) = Sum_{d|n} (-1)^(n/d+1)*d^5.

1, 31, 244, 991, 3126, 7564, 16808, 31711, 59293, 96906, 161052, 241804, 371294, 521048, 762744, 1014751, 1419858, 1838083, 2476100, 3097866, 4101152, 4992612, 6436344, 7737484, 9768751, 11510114, 14408200, 16656728, 20511150, 23645064, 28629152, 32472031, 39296688

Offset: 1

Author

Seiichi Manyama, Apr 06 2017

Keywords

• nonn
• mult

Comments

Multiplicative because this sequence is the Dirichlet convolution of A000584 and A062157 which are both multiplicative. - Andrew Howroyd, Jul 20 2018

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

Sum_{d|n} (-1)^(n/d+1)*d^k: A000593 (k=1), A078306 (k=2), A078307 (k=3), A284900 (k=4), this sequence (k=5), A284927 (k=6), A321552 (k=7), A321553 (k=8), A321554 (k=9), A321555 (k=10), A321556 (k=11), A321557 (k=12).

Cf. A000584, A013664, A062157.

Programs

Mathematica

Table[Sum[(-1)^(n/d + 1)*d^5, {d, Divisors[n]}], {n, 50}] (* Indranil Ghosh, Apr 06 2017 *)
f[p_, e_] := (p^(5*e + 5) - 1)/(p^5 - 1); f[2, e_] := (15*2^(5*e + 1) + 1)/31; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Nov 11 2022 *)

PARI

a(n) = sumdiv(n, d, (-1)^(n/d + 1)*d^5); \\ Indranil Ghosh, Apr 06 2017

Python

from sympy import divisors
print([sum((-1)**(n//d + 1)*d**5 for d in divisors(n)) for n in range(1, 51)]) # Indranil Ghosh, Apr 06 2017

Formula

G.f.: Sum_{k>=1} k^5*x^k/(1 + x^k). - Ilya Gutkovskiy, Apr 07 2017

From Amiram Eldar, Nov 11 2022: (Start)

Multiplicative with a(2^e) = (15*2^(5*e+1)+1)/31, and a(p^e) = (p^(5*e+5) - 1)/(p^5 - 1) if p > 2.

Sum_{k=1..n} a(k) ~ c * n^6, where c = 31*zeta(6)/192 = 0.164258... . (End)

Extensions

Keyword:mult added by Andrew Howroyd, Jul 23 2018