cp's OEIS Frontend

A385005 The sum of the cubefull divisors of n.

%I A385005 #8 Jun 15 2025 12:54:09
%S A385005 1,1,1,1,1,1,1,9,1,1,1,1,1,1,1,25,1,1,1,1,1,1,1,9,1,1,28,1,1,1,1,57,1,
%T A385005 1,1,1,1,1,1,9,1,1,1,1,1,1,1,25,1,1,1,1,1,28,1,9,1,1,1,1,1,1,1,121,1,
%U A385005 1,1,1,1,1,1,9,1,1,1,1,1,1,1,25,109,1,1,1
%N A385005 The sum of the cubefull divisors of n.
%C A385005 The sum of the terms in A036966 that divide n.
%C A385005 The number of these divisors is A190867(n), and the largest of them is A360540(n).
%H A385005 Amiram Eldar, <a href="/A385005/b385005.txt">Table of n, a(n) for n = 1..10000</a>
%F A385005 Multiplicative with a(p^e) = 1 if e <= 2, and a(p^e) = ((p^(e+1)-1) / (p-1)) - p - p^2 if e >= 3.
%F A385005 Dirichlet g.f.: zeta(s-1) * zeta(s) * Product_{p prime} (1 - p^(s-1) + 1/p^(3*s-3)).
%t A385005 f[p_, e_] := (p^(e+1)-1)/(p-1) - p - If[e == 1, 0, p^2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
%o A385005 (PARI) a(n) = {my(f = factor(n), p, e); prod(i = 1, #f~, p = f[i,1]; e = f[i,2]; (p^(e+1)-1)/(p-1) - p - if(e == 1, 0, p^2));}
%Y A385005 Cf. A036966, A190867, A183097, A360540.
%Y A385005 The sum of divisors d of n such that d is: A000593 (odd), A033634 (exponentially odd), A035316 (square), A038712 (power of 2), A048250 (squarefree), A072079 (3-smooth), A073185 (cubefree), A113061 (cube), A162296 (nonsquarefree), A183097 (powerful), A186099 (5-rough), A353900 (exponentially 2^n), this sequence (cubefull), A385006 (biquadratefree).
%K A385005 nonn,easy,mult
%O A385005 1,8
%A A385005 _Amiram Eldar_, Jun 15 2025