A048785 a(0) = 0; a(n) = tau(n^3), where tau = number of divisors (A000005).
0, 1, 4, 4, 7, 4, 16, 4, 10, 7, 16, 4, 28, 4, 16, 16, 13, 4, 28, 4, 28, 16, 16, 4, 40, 7, 16, 10, 28, 4, 64, 4, 16, 16, 16, 16, 49, 4, 16, 16, 40, 4, 64, 4, 28, 28, 16, 4, 52, 7, 28, 16, 28, 4, 40, 16, 40, 16, 16, 4, 112, 4, 16, 28, 19, 16, 64, 4, 28, 16
Offset: 0
Examples
a(6) = 16 because there are 16 divisors of 6^3 = 216: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216. Also there are 16 ordered triples of divisors of 6 that are pairwise relatively prime: (1,1,1), (1,1,2), (1,1,3), (1,1,6), (1,2,1), (1,2,3), (1,3,1), (1,3,2), (1,6,1), (2,1,1), (2,1,3), (2,3,1), (3,1,1), (3,1,2), (3,2,1), (6,1,1).
Links
- Antti Karttunen, Table of n, a(n) for n = 0..10000
Programs
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Maple
seq(numtheory:-tau(n^3), n=0..100); # Robert Israel, Jan 11 2015
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Mathematica
Join[{0,1},Table[Product[3 k + 1, {k, FactorInteger[n][[All, 2]]}], {n, 2, 69}]] (* Geoffrey Critzer, Jan 11 2015 *) Join[{0},DivisorSigma[0,Range[70]^3]] (* Harvey P. Dale, Jan 23 2016 *) -
PARI
A048785(n) = if(!n,n,numdiv(n^3)); \\ Antti Karttunen, May 19 2017
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PARI
print1("0, "); for(n=1, 100, print1(direuler(p=2, n, (1 + 2*X)/(1 - X)^2)[n], ", ")) \\ Vaclav Kotesovec, May 15 2021 print1("0, "); for(n=1, 100, print1(direuler(p=2, n, (1 - 3*X^2 + 2*X^3)/(1 - X)^4)[n], ", ")) \\ Vaclav Kotesovec, Aug 20 2021 -
Python
from math import prod from sympy import factorint def A048785(n): return 0 if n == 0 else prod(3*e+1 for e in factorint(n).values()) # Chai Wah Wu, May 10 2022
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Python
from sympy import divisor_count def A048785(n): return divisor_count(n**3) # Karl-Heinz Hofmann, May 10 2022
Formula
a(n) = Sum_{d|n} 3^omega(d), where omega(x) is the number of distinct prime factors in the factorization of x. - Benoit Cloitre, Apr 14 2002
Multiplicative with a(p^e) = 3e+1. - Mitch Harris, Jun 09 2005
L.g.f.: -log(Product_{k>=1} (1 - x^k)^(3^omega(k)/k)) = Sum_{n>=1} a(n)*x^n/n. - Ilya Gutkovskiy, May 26 2018
For n>0, a(n) = Sum_{d|n} mu(d)^2*tau(d)*tau(n/d). - Ridouane Oudra, Nov 18 2019
Dirichlet g.f.: zeta(s)^2 * Product_{primes p} (1 + 2/p^s). - Vaclav Kotesovec, May 15 2021
Dirichlet g.f.: zeta(s)^4 * Product_{primes p} (1 - 3/p^(2*s) + 2/p^(3*s)). - Vaclav Kotesovec, Aug 20 2021
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