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A352447
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Numbers k such that BarnesG(k) is divisible by Gamma(k).
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0
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1, 2, 7, 9, 10, 11, 13, 15, 16, 17, 19, 21, 22, 23, 25, 26, 27, 28, 29, 31, 33, 34, 35, 36, 37, 39, 40, 41, 43, 45, 46, 47, 49, 50, 51, 52, 53, 55, 56, 57, 58, 59, 61, 63, 64, 65, 66, 67, 69, 70, 71, 73, 75, 76, 77, 78, 79, 81, 82, 83, 85, 86, 87, 88, 89, 91, 92, 93, 94, 95
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OFFSET
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1,2
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COMMENTS
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These are k such that G(k)/Gamma(k) = 1!*2!*3!*...*(k-2)!/(k-1)! = 1!*2!*3!*...*(k-3)!/(k-1) are integer. Let k=1+c, so require 1!*2!*3!*...*(c-2)!/c to be integer. If c is composite, take any factorization of c=c_1*c_2 with 2<=c_1<=c_2<=c/2; then matching factors exist in the product 1!*2!*3!*...*(c-2)! that cancel this factor [either c_1! and c_2! if c_1 <> c_2, or c_1! and (c_1+1)! if c_1=c_2 and c-2 >= c_1+1]. If c is prime, none of the 1!*2!*..(c-2)! contains a factor matching that prime. So this shows that the sequence is (apart from offset and at c=4) the same as A079696. - R. J. Mathar, Mar 25 2022
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LINKS
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Eric Weisstein's World of Mathematics, Divisible.
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FORMULA
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EXAMPLE
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BarnesG(7) = 34560, Gamma(7) = 720, 34560 is divisible by 720, so 7 is in this sequence.
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MATHEMATICA
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Table[If[Divisible[BarnesG[k], Gamma[k]], k, Nothing], {k, 115}]
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PROG
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(Python)
from itertools import count, islice
from collections import Counter
from sympy import factorint
def A352447_gen(): # generator of terms
yield 1
a = Counter()
for k in count(2):
b = Counter(factorint(k-1))
if all(b[p] <= a[p] for p in b):
yield k
a += b
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CROSSREFS
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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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