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%I #31 Sep 08 2022 08:46:25
%S 1,1,1,1,1,5,691,7,3617,43867,283,11,103,13,7,5,37,17,
%T 26315271553053477373,19,137616929,1520097643918070802691,11,23,653,5,
%U 13,39409,7,29,2003,31,1226592271,11,17,5,3112655297839,37,19,13,631,41,233,43,11,5,23,47,7823741903
%N The least prime factor of the numerator of Bernoulli(2*n), or 1 if the numerator is 1.
%C a(n)=5 if and only if n is in A017329. - _Robert Israel_, Feb 09 2020
%C From _Chai Wah Wu_, Feb 10 2020: (Start)
%C For n > 1, clearly if a(n) = n, then n is prime. However, the converse is not true. Prime numbers p such that a(p) != p are: 2, 3, 109, 167, 211, 227, 271, ...
%C Conjecture: for prime p > 3, p is a prime factor of the numerator of Bernoulli(2*p), thus the conjecture implies that a(p) <= p for prime p.
%C (End)
%H Chai Wah Wu, <a href="/A332300/b332300.txt">Table of n, a(n) for n = 0..191</a> (n = 0..103 from Amiram Eldar)
%H S. S. Wagstaff, Jr., <a href="http://www.cerias.purdue.edu/homes/ssw/bernoulli/bnum">Factors of Bernoulli numbers</a>.
%F a(n) = A020639(abs(A000367(n))).
%e a(10) = 283, since Bernoulli(2*10) = -174611/330, and 283 is the least prime factor of its numerator, 174611 = 283 * 617.
%t Array[FactorInteger[Abs @ Numerator @ BernoulliB[2*#]][[1, 1]] &, 30, 0]
%o (Magma) [n le 4 select 1 else Min(PrimeDivisors(Abs(Numerator(Bernoulli(2*n))))):n in [0..48]]; // _Marius A. Burtea_, Feb 09 2020
%o (PARI) a(n) = my(x=abs(numerator(bernfrac(2*n)))); if (x==1, 1, vecmin(factor(x)[,1])); \\ _Michel Marcus_, Feb 09 2020
%o (Python)
%o from sympy import bernoulli, primefactors
%o def A332300(n):
%o x = abs(bernoulli(2*n).p)
%o return 1 if x == 1 else min(primefactors(x)) # _Chai Wah Wu_, Feb 10 2020
%Y Cf. A000367, A017329, A020639, A079294, A090947, A242193, A326727.
%K nonn
%O 0,6
%A _Amiram Eldar_, Feb 09 2020
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