[go: up one dir, main page]

login
A061704 revision #48

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
OFFSET
1,8
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
KEYWORD
nonn,mult
AUTHOR
Henry Bottomley, Jun 18 2001
STATUS
editing