A061704 Number of cubes dividing n.
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, 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, 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
Offset: 1
Examples
a(128) = 3 since 128 is divisible by 1^3 = 1, 2^3 = 8 and 4^3 = 64.
Links
Programs
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Maple
N:= 1000: # to get a(1)..a(N) G:= add(x^(n^3)/(1-x^(n^3)),n=1..floor(N^(1/3))): S:= series(G,x,N+1): seq(coeff(S,x,j),j=1..N); # Robert Israel, Jul 28 2017 # alternative A061704 := proc(n) local a,pe ; a := 1 ; for pe in ifactors(n)[2] do op(2,pe) ; a := a*(1+floor(%/3)) ; end do: a ; end proc: seq(A061704(n),n=1..80) ; # R. J. Mathar, May 10 2023
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Mathematica
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 *) Table[DivisorSum[n, 1 &, IntegerQ[#^(1/3)] &], {n, 105}] (* Michael De Vlieger, Jul 28 2017 *) f[p_, e_] := 1 + Floor[e/3]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 15 2020 *) -
PARI
a(n) = sumdiv(n, d, ispower(d, 3)); \\ Michel Marcus, Jan 31 2015
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Python
from math import prod from sympy import factorint def A061704(n): return prod(e//3+1 for e in factorint(n).values()) # Chai Wah Wu, Jun 05 2025
Formula
Multiplicative with a(p^e) = floor(e/3) + 1. - Mitch Harris, Apr 19 2005
G.f.: Sum_{n>=1} x^(n^3)/(1-x^(n^3)). - Joerg Arndt, Jan 30 2011
Dirichlet g.f.: zeta(3*s)*zeta(s). - Geoffrey Critzer, Feb 07 2015
Sum_{k=1..n} a(k) ~ zeta(3)*n + zeta(1/3)*n^(1/3). - Vaclav Kotesovec, Dec 01 2020
a(n) = Sum_{k=1..n} (1 - ceiling(n/k^3) + floor(n/k^3)). - Wesley Ivan Hurt, Jan 28 2021