A279363 Sum of 4th powers of proper divisors of n.
0, 1, 1, 17, 1, 98, 1, 273, 82, 642, 1, 1650, 1, 2418, 707, 4369, 1, 7955, 1, 10898, 2483, 14658, 1, 26482, 626, 28578, 6643, 41090, 1, 62644, 1, 69905, 14723, 83538, 3027, 133923, 1, 130338, 28643, 174994, 1, 236692, 1, 249170, 57893, 279858, 1, 423794, 2402, 401267, 83603, 485810, 1, 644372, 15267, 659842, 130403, 707298, 1, 1053636
Offset: 1
Examples
a(10) = 1^4 + 2^4 + 5^4 = 642, because 10 has 3 proper divisors {1,2,5}.
a(11) = 1^4 = 1, because 11 has 1 proper divisor {1}.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
- Eric Weisstein's World of Mathematics, Proper divisors.
- Index entries for sequences related to sums of divisors
Programs
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Mathematica
Table[DivisorSigma[4, n] - n^4, {n, 60}] -
PARI
for(n=1, 60, print1(sigma(n, 4) - n^4,", ")) \\ Indranil Ghosh, Mar 18 2017
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Python
from sympy.ntheory import divisor_sigma print([divisor_sigma(n,4) - n**4 for n in range(1,61)]) # Indranil Ghosh, Mar 18 2017
Formula
a(n) = 1 if n is prime.
a(p^k) = (p^(4*k) - 1)/(p^4 - 1) when p is prime.
Dirichlet g.f.: zeta(s-4)*(zeta(s) - 1).
G.f.: -x*(1 + 11*x + 11*x^2 + x^3)/(1 - x)^5 + Sum_{k>=1} k^4 x^k/(1 - x^k). - Ilya Gutkovskiy, Mar 18 2017
Sum_{k=1..n} a(k) ~ (Zeta(5) - 1)*n^5 / 5. - Vaclav Kotesovec, Feb 02 2019