OFFSET
0,6
COMMENTS
a(n)=5 if and only if n is in A017329. - Robert Israel, Feb 09 2020
From Chai Wah Wu, Feb 10 2020: (Start)
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, ...
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.
(End)
LINKS
Chai Wah Wu, Table of n, a(n) for n = 0..191 (n = 0..103 from Amiram Eldar)
S. S. Wagstaff, Jr., Factors of Bernoulli numbers.
EXAMPLE
a(10) = 283, since Bernoulli(2*10) = -174611/330, and 283 is the least prime factor of its numerator, 174611 = 283 * 617.
MATHEMATICA
Array[FactorInteger[Abs @ Numerator @ BernoulliB[2*#]][[1, 1]] &, 30, 0]
PROG
(Magma) [n le 4 select 1 else Min(PrimeDivisors(Abs(Numerator(Bernoulli(2*n))))):n in [0..48]]; // Marius A. Burtea, Feb 09 2020
(PARI) a(n) = my(x=abs(numerator(bernfrac(2*n)))); if (x==1, 1, vecmin(factor(x)[, 1])); \\ Michel Marcus, Feb 09 2020
(Python)
from sympy import bernoulli, primefactors
def A332300(n):
x = abs(bernoulli(2*n).p)
return 1 if x == 1 else min(primefactors(x)) # Chai Wah Wu, Feb 10 2020
KEYWORD
nonn
AUTHOR
Amiram Eldar, Feb 09 2020
STATUS
approved