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%I #30 Sep 08 2022 08:46:17
%S 0,1,1,9,1,36,1,73,28,134,1,316,1,352,153,585,1,981,1,1198,371,1340,1,
%T 2556,126,2206,757,3160,1,4752,1,4681,1359,4922,469,8605,1,6868,2225,
%U 9710,1,12600,1,12052,4257,12176,1,20476,344,16759,4941,19846,1,26496,1457,25624,6887,24398,1
%N Sum of cubes of proper divisors of n.
%C 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).
%H Amiram Eldar, <a href="/A276634/b276634.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>.
%F a(n) = 1 if n is prime.
%F a(p^k) = (p^(3*k) - 1)/(p^3 - 1) for p prime.
%F Dirichlet g.f.: zeta(s-3)*(zeta(s) - 1).
%F a(n) = A001158(n) - A000578(n).
%F A000035(a(n)) = A053867(n).
%F Sum_{n=1..k} a(n) ~ k^2*(Pi^4*k^2/90 - (k + 1)^2)/4.
%F 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
%e a(10) = 1^3 + 2^3 + 5^3 = 134, because 10 has 3 proper divisors {1,2,5}.
%e a(11) = 1^3 = 1, because 11 has 1 proper divisor {1}.
%t Table[DivisorSigma[3, n] - n^3, {n, 70}]
%o (PARI) a(n) = sigma(n, 3) - n^3; \\ _Michel Marcus_, Sep 08 2016
%o (Magma) [DivisorSigma(3, n) - n^3: n in [1..70]]; // _Vincenzo Librandi_, Sep 09 2016
%Y Cf. A000035, A000578, A001065, A001158, A032741, A053867, A067558.
%K nonn,easy
%O 1,4
%A _Ilya Gutkovskiy_, Sep 08 2016
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