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A347042
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Number of divisors d > 1 of n such that bigomega(d) divides bigomega(n), where bigomega = A001222.
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6
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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
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OFFSET
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1,4
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LINKS
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EXAMPLE
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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
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MATHEMATICA
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Table[Length[Select[Rest[Divisors[n]], IntegerQ[PrimeOmega[n]/PrimeOmega[#]]&]], {n, 100}]
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PROG
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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
(Python)
from sympy import divisors, primeomega
def a(n):
bigomegan = primeomega(n)
return sum(bigomegan%primeomega(d) == 0 for d in divisors(n)[1:])
(Python)
from sympy import factorint, divisors
from sympy.utilities.iterables import multiset_combinations
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
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CROSSREFS
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The smallest of these divisors is A020639
The case of divisors with half bigomega is A345957 (rounded: A096825).
A001221 counts distinct prime factors.
A001222 counts all prime factors, also called bigomega.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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