A284900 a(n) = Sum_{d|n} (-1)^(n/d+1)*d^4.

1, 15, 82, 239, 626, 1230, 2402, 3823, 6643, 9390, 14642, 19598, 28562, 36030, 51332, 61167, 83522, 99645, 130322, 149614, 196964, 219630, 279842, 313486, 391251, 428430, 538084, 574078, 707282, 769980, 923522, 978671, 1200644, 1252830, 1503652, 1587677

Offset: 1

Author

Seiichi Manyama, Apr 05 2017

Keywords

• nonn
• mult

Comments

Multiplicative because this sequence is the Dirichlet convolution of A000583 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), this sequence (k=4), A284926 (k=5), A284927 (k=6), A321552 (k=7), A321553 (k=8), A321554 (k=9), A321555 (k=10), A321556 (k=11), A321557 (k=12).

Cf. A000583, A013663, A062157, A386729.

Programs

Mathematica

Table[Sum[(-1)^(n/d + 1)*d^4, {d, Divisors[n]}], {n, 50}] (* Indranil Ghosh, Apr 05 2017 *)
f[p_, e_] := (p^(4*e + 4) - 1)/(p^4 - 1); f[2, e_] := (7*2^(4*e + 1) + 1)/15; 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^4); \\ Indranil Ghosh, Apr 05 2017

Python

from sympy import divisors
print([sum([(-1)**(n//d + 1)*d**4 for d in divisors(n)]) for n in range(1, 51)]) # Indranil Ghosh, Apr 05 2017

Formula

G.f.: Sum_{k>=1} k^4*x^k/(1 + x^k). - Ilya Gutkovskiy, Apr 07 2017

From Amiram Eldar, Nov 11 2022: (Start)

Multiplicative with a(2^e) = (7*2^(4*e+1)+1)/15, and a(p^e) = (p^(4*e+4) - 1)/(p^4 - 1) if p > 2.

Sum_{k=1..n} a(k) ~ c * n^5, where c = 3*zeta(5)/16 = 0.194423... . (End)