A328745 Dirichlet g.f.: Product_{p prime} 1 / (1 - p^(-s))^p.

1, 2, 3, 3, 5, 6, 7, 4, 6, 10, 11, 9, 13, 14, 15, 5, 17, 12, 19, 15, 21, 22, 23, 12, 15, 26, 10, 21, 29, 30, 31, 6, 33, 34, 35, 18, 37, 38, 39, 20, 41, 42, 43, 33, 30, 46, 47, 15, 28, 30, 51, 39, 53, 20, 55, 28, 57, 58, 59, 45, 61, 62, 42, 7, 65, 66, 67, 51, 69, 70, 71, 24, 73, 74, 45

Offset: 1

Author

Ilya Gutkovskiy, Oct 26 2019

Keywords

• nonn
• mult

Comments

Number of ways to factor n into 2 kinds of 2, 3 kinds of 3, 5 kinds of 5, ... , p kinds of p.

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 1..10000
• Vaclav Kotesovec, Plot of Sum_{k=1..n} a(k) / n^2 for n = 1..1000000

Crossrefs

Cf. A008480, A050367, A059481, A135323, A182938.

Programs

Maple

a:= n-> mul(binomial(i[1]+i[2]-1, i[2]), i=ifactors(n)[2]):
seq(a(n), n=1..100);  # Alois P. Heinz, Oct 26 2019

Mathematica

a[n_] := Times @@ (Binomial[#[[1]] + #[[2]] - 1, #[[2]]] & /@ FactorInteger[n]); Table[a[n], {n, 1, 75}]

PARI

for(n=1, 100, print1(direuler(p=2, n, 1/(1 - X)^p)[n], ", ")) \\ Vaclav Kotesovec, Aug 22 2021

Formula

If n = Product (p_j^k_j) then a(n) = Product (binomial(p_j + k_j - 1, k_j)).

Conjecture: Sum_{k=1..n} a(k) ~ c * n^2, where c = 0.40373... - Vaclav Kotesovec, Mar 28 2025