A351308 Sum of the cubes of the square divisors of n.
1, 1, 1, 65, 1, 1, 1, 65, 730, 1, 1, 65, 1, 1, 1, 4161, 1, 730, 1, 65, 1, 1, 1, 65, 15626, 1, 730, 65, 1, 1, 1, 4161, 1, 1, 1, 47450, 1, 1, 1, 65, 1, 1, 1, 65, 730, 1, 1, 4161, 117650, 15626, 1, 65, 1, 730, 1, 65, 1, 1, 1, 65, 1, 1, 730, 266305, 1, 1, 1, 65, 1, 1, 1, 47450, 1
Offset: 1
Examples
a(16) = 4161; a(16) = Sum_{d^2|16} (d^2)^3 = (1^2)^3 + (2^2)^3 + (4^2)^3 = 4161.
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
-
Mathematica
f[p_, e_] := (p^(6*(1 + Floor[e/2])) - 1)/(p^6 - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Feb 07 2022 *)
-
PARI
a(n) = sumdiv(n, d, if (issquare(d), d^3)); \\ Michel Marcus, Mar 24 2023
Formula
a(n) = Sum_{d^2|n} (d^2)^3.
Multiplicative with a(p) = (p^(6*(1+floor(e/2))) - 1)/(p^6 - 1). - Amiram Eldar, Feb 07 2022
From Amiram Eldar, Sep 19 2023: (Start)
Dirichlet g.f.: zeta(s) * zeta(2*s-6).
Sum_{k=1..n} a(k) ~ (zeta(7/2)/7) * n^(7/2). (End)
G.f.: Sum_{k>=1} k^6 * x^(k^2) / (1 - x^(k^2)). - Ilya Gutkovskiy, Jun 05 2024
a(n) = Sum_{d|n} d^3 * c(d), where c = A010052. - Wesley Ivan Hurt, Jun 20 2024
a(n) = Sum_{d|n} lambda(d)*d^3*sigma_3(n/d), where lambda = A008836. - Ridouane Oudra, Jul 18 2025
Comments