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A276634
Sum of cubes of proper divisors of n.
6
0, 1, 1, 9, 1, 36, 1, 73, 28, 134, 1, 316, 1, 352, 153, 585, 1, 981, 1, 1198, 371, 1340, 1, 2556, 126, 2206, 757, 3160, 1, 4752, 1, 4681, 1359, 4922, 469, 8605, 1, 6868, 2225, 9710, 1, 12600, 1, 12052, 4257, 12176, 1, 20476, 344, 16759, 4941, 19846, 1, 26496, 1457, 25624, 6887, 24398, 1
OFFSET
1,4
COMMENTS
More generally, the Dirichlet generating function for the sum of k-th powers of proper divisors of n is zeta(s-k)*(zeta(s) - 1).
LINKS
Eric Weisstein's World of Mathematics, Proper divisors.
FORMULA
a(n) = 1 if n is prime.
a(p^k) = (p^(3*k) - 1)/(p^3 - 1) for p prime.
Dirichlet g.f.: zeta(s-3)*(zeta(s) - 1).
a(n) = A001158(n) - A000578(n).
A000035(a(n)) = A053867(n).
Sum_{n=1..k} a(n) ~ k^2*(Pi^4*k^2/90 - (k + 1)^2)/4.
G.f.: -x*(1 + 4*x + x^2)/(1 - x)^4 + Sum_{k>=1} k^3*x^k/(1 - x^k). - Ilya Gutkovskiy, Mar 17 2017
EXAMPLE
a(10) = 1^3 + 2^3 + 5^3 = 134, because 10 has 3 proper divisors {1,2,5}.
a(11) = 1^3 = 1, because 11 has 1 proper divisor {1}.
MATHEMATICA
Table[DivisorSigma[3, n] - n^3, {n, 70}]
PROG
(PARI) a(n) = sigma(n, 3) - n^3; \\ Michel Marcus, Sep 08 2016
(Magma) [DivisorSigma(3, n) - n^3: n in [1..70]]; // Vincenzo Librandi, Sep 09 2016
KEYWORD
nonn,easy
AUTHOR
Ilya Gutkovskiy, Sep 08 2016
STATUS
approved