A347042 Number of divisors d > 1 of n such that bigomega(d) divides bigomega(n), where bigomega = A001222.
0, 1, 1, 2, 1, 3, 1, 2, 2, 3, 1, 3, 1, 3, 3, 3, 1, 3, 1, 3, 3, 3, 1, 5, 2, 3, 2, 3, 1, 4, 1, 2, 3, 3, 3, 6, 1, 3, 3, 5, 1, 4, 1, 3, 3, 3, 1, 3, 2, 3, 3, 3, 1, 5, 3, 5, 3, 3, 1, 8, 1, 3, 3, 4, 3, 4, 1, 3, 3, 4, 1, 3, 1, 3, 3, 3, 3, 4, 1, 3, 3, 3, 1, 8, 3, 3, 3
Offset: 1
Keywords
Examples
The a(n) divisors for selected n: n = 1: 2: 4: 6: 24: 30: 36: 60: 96: 144: 210: 216: 240: 360: --------------------------------------------------------------------- {} 2 2 2 2 2 2 2 2 2 2 2 2 2 4 3 3 3 3 3 3 3 3 3 3 3 6 4 5 4 4 4 4 5 4 4 4 6 30 6 5 6 6 6 6 5 5 24 9 6 8 8 7 8 6 6 36 10 12 9 10 9 8 8 15 96 12 14 12 10 9 60 18 15 18 12 10 144 21 27 15 12 35 216 20 15 210 30 18 240 20 30 45 360
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
Table[Length[Select[Rest[Divisors[n]],IntegerQ[PrimeOmega[n]/PrimeOmega[#]]&]],{n,100}]
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PARI
a(n) = my(bn=bigomega(n)); sumdiv(n, d, if (d>1, !(bn % bigomega(d)))); \\ Michel Marcus, Aug 18 2021
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Python
from sympy import divisors, primeomega def a(n): bigomegan = primeomega(n) return sum(bigomegan%primeomega(d) == 0 for d in divisors(n)[1:]) print([a(n) for n in range(1, 88)]) # Michael S. Branicky, Aug 18 2021
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Python
from sympy import factorint, divisors from sympy.utilities.iterables import multiset_combinations def A347042(n): fs = factorint(n,multiple=True) return sum(len(list(multiset_combinations(fs,d))) for d in divisors(len(fs),generator=True)) # Chai Wah Wu, Aug 21 2021