A279395 a(n) = Sum_{ d >= 1, d divides n} (-1)^(n-d)*d^4.

1, 15, 82, 271, 626, 1230, 2402, 4367, 6643, 9390, 14642, 22222, 28562, 36030, 51332, 69903, 83522, 99645, 130322, 169646, 196964, 219630, 279842, 358094, 391251, 428430, 538084, 650942, 707282, 769980, 923522, 1118479, 1200644, 1252830, 1503652, 1800253, 1874162, 1954830, 2342084, 2733742

Offset: 1

Author

Wolfdieter Lang, Jan 09 2017

Keywords

• nonn
• mult
• easy

Comments

This is the k=4 member of the family sigma^*_k(n), defined in the Hardy reference, which is sigma_k(2*j+1) if n = 2*j+1 and sigma_k^e(2*j) - sigma_k^o(2*j) if n=2*j, where the superscript e and o stands for a restriction to even and odd divisors in the sum of their k-th powers, respectively.

References

• G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, AMS Chelsea Publishing, Providence, Rhode Island, 2002, p. 142.

Links

• Amiram Eldar, 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

Cf. A112329 (k=0), A113184 (k=1), A064027 (k=2), A008457(k=3).

Programs

Magma

[&+[(-1)^(n-d)*d^4:d in Divisors(n)]:n in [1..40]]; // Marius A. Burtea, Aug 17 2019

Maple

# A version with signs - N. J. A. Sloane, Nov 23 2018
zet1:=(n,i)->add((-1)^(d-1)*d^i, d in divisors(n));
szet1:=i->[seq(zet1(n,i),n=1..120)];
szet1(4);

Mathematica

f[p_, e_] := If[p == 2, (2^(4*(e + 1)) - 31)/15, (p^(4*(e + 1)) - 1)/(p^4 - 1)]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 40] (* Amiram Eldar, Aug 17 2019 *)

PARI

a(n) = sumdiv(n, d, (-1)^(n-d)*d^4); \\ Michel Marcus, Jan 09 2017

Formula

a(n) = Sum_{ d >= 1, d divides n} (-1)^(n-d)*d^4.

Bisection: a(2*j-1) = A001159(2*j-1), a(2*j) = 16*A001159(j) - A051001(j), j >= 1. See the comment above for k=4, and the Hardy reference.

G.f.: Sum_{k>=1} k^4*x^k/(1-(-x)^k).

Multiplicative with a(2^k) = 2^4*(2^(4*k)-1)/(2^4-1) - 1 = (2^(4*(k+1)) - 31)/15 and a(p^k) = (p^(4*(k+1))-1)/(p^4-1) for primes p > 2 (see A001159).