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A279364
Sum of 5th powers of proper divisors of n.
4
0, 1, 1, 33, 1, 276, 1, 1057, 244, 3158, 1, 9076, 1, 16840, 3369, 33825, 1, 67101, 1, 104182, 17051, 161084, 1, 290676, 3126, 371326, 59293, 555688, 1, 870552, 1, 1082401, 161295, 1419890, 19933, 2206525, 1, 2476132, 371537, 3336950, 1, 4646784, 1, 5315740, 821793, 6436376, 1, 9301876, 16808, 9868783
OFFSET
1,4
FORMULA
a(n) = 1 if n is prime.
a(p^k) = (p^(5*k) - 1)/(p^5 - 1) for p is prime.
Dirichlet g.f.: zeta(s-5)*(zeta(s) - 1).
a(n) = A001160(n) - A000584(n).
G.f.: -x*(1 + 26*x + 66*x^2 + 26*x^3 + x^4)/(1 - x)^6 + Sum_{k>=1} k^5 x^k/(1 - x^k). - Ilya Gutkovskiy, Mar 18 2017
EXAMPLE
a(10) = 1^5 + 2^5 + 5^5 = 3158, because 10 has 3 proper divisors {1,2,5}.
a(11) = 1^5 = 1, because 11 has 1 proper divisor {1}.
MATHEMATICA
Table[DivisorSigma[5, n] - n^5, {n, 50}]
PROG
(PARI) for(n=1, 50, print1(sigma(n, 5) - n^5, ", ")) \\ Indranil Ghosh, Mar 18 2017
(Python)
from sympy.ntheory import divisor_sigma
print([divisor_sigma(n, 5) - n**5 for n in range(1, 51)]) # Indranil Ghosh, Mar 18 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Ilya Gutkovskiy, Dec 10 2016
STATUS
approved