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A061704 Number of cubes dividing n. 23
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,8
LINKS
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
a(n) = A000005(A053150(n)).
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
EXAMPLE
a(128) = 3 since 128 is divisible by 1^3 = 1, 2^3 = 8 and 4^3 = 64.
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
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 *)
PROG
(PARI) a(n) = sumdiv(n, d, ispower(d, 3)); \\ Michel Marcus, Jan 31 2015
CROSSREFS
Sequence in context: A307428 A368248 A362852 * A325837 A366902 A050361
KEYWORD
nonn,mult
AUTHOR
Henry Bottomley, Jun 18 2001
STATUS
approved

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Last modified April 24 19:51 EDT 2024. Contains 371963 sequences. (Running on oeis4.)