1,2

If the canonical factorization of n into prime powers is the product of p^e(p) then sigma_k(n) = Product_p ((p^((e(p)+1)*k))-1)/(p^k-1).

Sum_{d|n} 1/d^k is equal to sigma_k(n)/n^k. So sequences A017665-A017712 also give the numerators and denominators of sigma_k(n)/n^k for k = 1..24. The power sums sigma_k(n) are in sequences A000203 (k=1), A001157-A001160 (k=2,3,4,5), A013954-A013972 for k = 6,7,...,24. - comment from Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 05 2001.

sigma_4(n) is the sum of the 4th powers of the divisors of n (A001159).

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math.Series 55, Tenth Printing, December 1972, p. 827.

T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 38.

T. D. Noe, Table of n, a(n) for n = 1..10000

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, December 1972 [alternative scanned copy].

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math.Series 55, Tenth Printing, December 1972, p. 827.

Multiplicative with a(p^e) = (p^(4e+4)-1)/(p^4-1). - David W. Wilson (davidwwilson(AT)comcast.net), Aug 01, 2001.

G.f. sum(k>=1, k^4*x^k/(1-x^k)). - Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 21 2003

(PARI from Joerg Arndt (arndt(AT)jjj.de), May 03, 2008)

N=17; default(seriesprecision, N); x=z+O(z^(N+1))

c=sum(j=1, N, j*x^j); \\ log case

s=-log(prod(j=1, N, (1-x^j)^(j^3))); \\ A001159 sigma_4(n)

s=serconvol(s, c)

v=Vec(s)

Cf. A000005, A000203, A001157, A001158.

Sequence in context: A034678 A065960 A017671 this_sequence A053820 A142059 A041554

Adjacent sequences: A001156 A001157 A001158 this_sequence A001160 A001161 A001162

nonn,easy,mult

njas