A361794 Sum of the cubes d^3 of the divisors d satisfying d^2|n.
1, 1, 1, 9, 1, 1, 1, 9, 28, 1, 1, 9, 1, 1, 1, 73, 1, 28, 1, 9, 1, 1, 1, 9, 126, 1, 28, 9, 1, 1, 1, 73, 1, 1, 1, 252, 1, 1, 1, 9, 1, 1, 1, 9, 28, 1, 1, 73, 344, 126, 1, 9, 1, 28, 1, 9, 1, 1, 1, 9, 1, 1, 28, 585, 1, 1, 1, 9, 1, 1, 1, 252, 1, 1, 126, 9, 1, 1, 1, 73
Offset: 1
Links
- Winston de Greef, Table of n, a(n) for n = 1..10000
- A. Dixit, B. Maji, and A. Vatwani, Voronoi summation formula for the generalized divisor function sigma_z^k(n), arXiv:2303.09937 [math.NT], 2023, sigma(z=3,k=2,n).
Programs
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Maple
gsigma := proc(n,z,k) local a,d ; a := 0 ; for d in numtheory[divisors](n) do if modp(n,d^k) = 0 then a := a+d^z ; end if ; end do: a ; end proc: seq( gsigma(n,3,2),n=1..80) ; -
Mathematica
f[p_, e_] := (p^(3*(Floor[e/2] + 1)) - 1)/(p^3 - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Mar 24 2023 *)
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PARI
a(n) = sumdiv(n, d, if (issquare(d), sqrtint(d)^3)); \\ Michel Marcus, Mar 24 2023
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Python
from math import prod from sympy import factorint def A361794(n): return prod((p**(3*(e>>1)+3)-1)//(p**3-1) for p, e in factorint(n).items()) # Chai Wah Wu, Mar 24 2023
Formula
a(n) = Sum_{d^2|n} d^3.
Multiplicative with a(p^e) = (p^(3*(floor(e/2) + 1)) - 1)/(p^3 - 1). - Amiram Eldar, Mar 24 2023
G.f.: Sum_{k>=1} k^3 * x^(k^2) / (1 - x^(k^2)). - Ilya Gutkovskiy, Jun 05 2024
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