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A279363 Sum of 4th powers of proper divisors of n.

%I A279363 #29 Feb 16 2025 08:33:37
%S A279363 0,1,1,17,1,98,1,273,82,642,1,1650,1,2418,707,4369,1,7955,1,10898,
%T A279363 2483,14658,1,26482,626,28578,6643,41090,1,62644,1,69905,14723,83538,
%U A279363 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
%N A279363 Sum of 4th powers of proper divisors of n.
%H A279363 Amiram Eldar, <a href="/A279363/b279363.txt">Table of n, a(n) for n = 1..10000</a>
%H A279363 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/ProperDivisor.html">Proper divisors</a>.
%H A279363 <a href="/index/Su#sums_of_divisors">Index entries for sequences related to sums of divisors</a>
%F A279363 a(n) = 1 if n is prime.
%F A279363 a(p^k) = (p^(4*k) - 1)/(p^4 - 1) when p is prime.
%F A279363 Dirichlet g.f.: zeta(s-4)*(zeta(s) - 1).
%F A279363 a(n) = A001159(n) - A000583(n).
%F A279363 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
%F A279363 Sum_{k=1..n} a(k) ~ (Zeta(5) - 1)*n^5 / 5. - _Vaclav Kotesovec_, Feb 02 2019
%e A279363 a(10) = 1^4 + 2^4 + 5^4 = 642, because 10 has 3 proper divisors {1,2,5}.
%e A279363 a(11) = 1^4 = 1, because 11 has 1 proper divisor {1}.
%t A279363 Table[DivisorSigma[4, n] - n^4, {n, 60}]
%o A279363 (PARI) for(n=1, 60, print1(sigma(n, 4) - n^4,", ")) \\ _Indranil Ghosh_, Mar 18 2017
%o A279363 (Python)
%o A279363 from sympy.ntheory import divisor_sigma
%o A279363 print([divisor_sigma(n,4) - n**4 for n in range(1,61)]) # _Indranil Ghosh_, Mar 18 2017
%Y A279363 Cf. A000583, A001159.
%Y A279363 Cf. A001065, A067558, A276634, A279364.
%K A279363 nonn,easy
%O A279363 1,4
%A A279363 _Ilya Gutkovskiy_, Dec 10 2016