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%I #25 May 07 2021 08:44:05
%S 0,1,1,17,1,98,1,273,82,642,1,1650,1,2418,707,4369,1,7955,1,10898,
%T 2483,14658,1,26482,626,28578,6643,41090,1,62644,1,69905,14723,83538,
%U 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 Sum of 4th powers of proper divisors of n.
%H Amiram Eldar, <a href="/A279363/b279363.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/ProperDivisor.html">Proper divisors</a>.
%H <a href="/index/Su#sums_of_divisors">Index entries for sequences related to sums of divisors</a>
%F a(n) = 1 if n is prime.
%F a(p^k) = (p^(4*k) - 1)/(p^4 - 1) for p is prime.
%F Dirichlet g.f.: zeta(s-4)*(zeta(s) - 1).
%F a(n) = A001159(n) - A000583(n).
%F 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 Sum_{k=1..n} a(k) ~ (Zeta(5) - 1)*n^5 / 5. - _Vaclav Kotesovec_, Feb 02 2019
%e a(10) = 1^4 + 2^4 + 5^4 = 642, because 10 has 3 proper divisors {1,2,5}.
%e a(11) = 1^4 = 1, because 11 has 1 proper divisor {1}.
%t Table[DivisorSigma[4, n] - n^4, {n, 60}]
%o (PARI) for(n=1, 60, print1(sigma(n, 4) - n^4,", ")) \\ _Indranil Ghosh_, Mar 18 2017
%o (Python)
%o from sympy.ntheory import divisor_sigma
%o print([divisor_sigma(n,4) - n**4 for n in range(1,61)]) # _Indranil Ghosh_, Mar 18 2017
%Y Cf. A000583, A001159.
%Y Cf. A001065, A067558, A276634, A279364.
%K nonn,easy
%O 1,4
%A _Ilya Gutkovskiy_, Dec 10 2016
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