A061200 tau_5(n) = number of ordered 5-factorizations of n.

1, 5, 5, 15, 5, 25, 5, 35, 15, 25, 5, 75, 5, 25, 25, 70, 5, 75, 5, 75, 25, 25, 5, 175, 15, 25, 35, 75, 5, 125, 5, 126, 25, 25, 25, 225, 5, 25, 25, 175, 5, 125, 5, 75, 75, 25, 5, 350, 15, 75, 25, 75, 5, 175, 25, 175, 25, 25, 5, 375, 5, 25, 75, 210, 25, 125, 5, 75, 25, 125, 5

Offset: 1

Author

Vladeta Jovovic, Apr 21 2001

Keywords

• easy
• nonn
• mult

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Enrique Pérez Herrero)

Crossrefs

Cf. tau_2(n): A000005, tau_3(n): A007425, tau_4(n): A007426, tau_6(n): A034695, (unordered) 2-factorization of n: A038548, (unordered) 3-factorization of n: A034836, A001055, A006218, A061201, A061202, A061203 (partial sums), A061204.

Column k=5 of A077592.

Programs

Mathematica

tau[n_, 1] = 1; tau[n_, k_] := tau[n, k] = Plus @@ (tau[ #, k - 1] & /@ Divisors[n]); Table[ tau[n, 5], {n, 77}] (* Robert G. Wilson v *)
tau[1, k_] := 1; tau[n_, k_] := Times @@ (Binomial[Last[#]+k-1, k-1]& /@ FactorInteger[n]); Table[tau[n, 5], {n, 1, 100}] (* Amiram Eldar, Sep 13 2020 *)

PARI

for(n=1,100,print1(sumdiv(n,k,sumdiv(k,x,sumdiv(x,y,numdiv(y)))),","))

PARI

a(n)=sumdivmult(n,k,sumdivmult(k,x,sumdivmult(x,y,numdiv(y)))) \\ Charles R Greathouse IV, Sep 09 2014

PARI

a(n, f=factor(n))=f=f[, 2]; prod(i=1, #f, binomial(f[i]+4, 4)) \\ Charles R Greathouse IV, Oct 28 2017

PARI

for(n=1, 100, print1(numerator(direuler(p=2, n, 1/(1-X)^5)[n]), ", ")) \\ Vaclav Kotesovec, May 06 2025

Python

from math import prod, comb
from sympy import factorint
def A061200(n): return prod(comb(4+e,4) for e in factorint(n).values()) # Chai Wah Wu, Dec 22 2024

Formula

tau_k(n) = |{(x_1,x_2,...,x_k): x_1*x_2*...*x_k=n}|, number of ordered k-factorizations of n.

tau_k(p^m) = (-1)^(k-1)*binomial(-m-1,k-1), p prime.

limit(tau_k(n)/n^epsilon, n=infinity) = 0, for any epsilon>0.

tau_k(n) = Sum_{d|n} tau_(k-1)(d), tau_1(n)=1.

Dirichlet g.f.: (zeta(s))^k.

For explicit formula, see A007425.

G.f.: Sum_{k>=1} tau_4(k)*x^k/(1 - x^k). - Ilya Gutkovskiy, Oct 30 2018