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%I #49 May 10 2023 07:37:05
%S 1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,1,2,1,1,1,1,2,1,1,
%T 1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,1,1,1,1,2,1,2,1,1,1,1,1,1,1,3,1,1,1,1,
%U 1,1,1,2,1,1,1,1,1,1,1,2,2,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1
%N Number of cubes dividing n.
%H Antti Karttunen, <a href="/A061704/b061704.txt">Table of n, a(n) for n = 1..10000</a>
%H Vaclav Kotesovec, <a href="/A061704/a061704.jpg">Graph - the asymptotic ratio (100000 terms)</a>
%H <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>
%F Multiplicative with a(p^e) = floor(e/3) + 1. - _Mitch Harris_, Apr 19 2005
%F G.f.: Sum_{n>=1} x^(n^3)/(1-x^(n^3)). - _Joerg Arndt_, Jan 30 2011
%F a(n) = A000005(A053150(n)).
%F Dirichlet g.f.: zeta(3*s)*zeta(s). - _Geoffrey Critzer_, Feb 07 2015
%F Sum_{k=1..n} a(k) ~ zeta(3)*n + zeta(1/3)*n^(1/3). - _Vaclav Kotesovec_, Dec 01 2020
%F a(n) = Sum_{k=1..n} (1 - ceiling(n/k^3) + floor(n/k^3)). - _Wesley Ivan Hurt_, Jan 28 2021
%e a(128) = 3 since 128 is divisible by 1^3 = 1, 2^3 = 8 and 4^3 = 64.
%p N:= 1000: # to get a(1)..a(N)
%p G:= add(x^(n^3)/(1-x^(n^3)),n=1..floor(N^(1/3))):
%p S:= series(G,x,N+1):
%p seq(coeff(S,x,j),j=1..N); # _Robert Israel_, Jul 28 2017
%p # alternative
%p A061704 := proc(n)
%p local a,pe ;
%p a := 1 ;
%p for pe in ifactors(n)[2] do
%p op(2,pe) ;
%p a := a*(1+floor(%/3)) ;
%p end do:
%p a ;
%p end proc:
%p seq(A061704(n),n=1..80) ; # _R. J. Mathar_, May 10 2023
%t nn = 100; f[list_, i_]:= list[[i]]; Table[ DirichletConvolve[ f[ Boole[ Map[ IntegerQ[#] &, Map[#^(1/3) &, Range[nn]]]], n],f[Table[1, {nn}], n], n, m], {m, 1, nn}] (* _Geoffrey Critzer_, Feb 07 2015 *)
%t Table[DivisorSum[n, 1 &, IntegerQ[#^(1/3)] &], {n, 105}] (* _Michael De Vlieger_, Jul 28 2017 *)
%t f[p_, e_] := 1 + Floor[e/3]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* _Amiram Eldar_, Sep 15 2020 *)
%o (PARI) a(n) = sumdiv(n, d, ispower(d, 3)); \\ _Michel Marcus_, Jan 31 2015
%Y Cf. A000005, A000578, A046951, A053150.
%K nonn,mult
%O 1,8
%A _Henry Bottomley_, Jun 18 2001
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