A332300 The least prime factor of the numerator of Bernoulli(2*n), or 1 if the numerator is 1.

1, 1, 1, 1, 1, 5, 691, 7, 3617, 43867, 283, 11, 103, 13, 7, 5, 37, 17, 26315271553053477373, 19, 137616929, 1520097643918070802691, 11, 23, 653, 5, 13, 39409, 7, 29, 2003, 31, 1226592271, 11, 17, 5, 3112655297839, 37, 19, 13, 631, 41, 233, 43, 11, 5, 23, 47, 7823741903

Offset: 0

Author

Amiram Eldar, Feb 09 2020

Keywords

• nonn

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)

Examples

			a(10) = 283, since Bernoulli(2*10) = -174611/330, and 283 is the least prime factor of its numerator, 174611 = 283 * 617.
		

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.

Crossrefs

Cf. A000367, A017329, A020639, A079294, A090947, A242193, A326727.

Programs

Magma

[n le 4 select 1 else Min(PrimeDivisors(Abs(Numerator(Bernoulli(2*n))))):n in [0..48]]; // Marius A. Burtea, Feb 09 2020

Mathematica

Array[FactorInteger[Abs @ Numerator @  BernoulliB[2*#]][[1, 1]] &, 30, 0]

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

Formula

a(n) = A020639(abs(A000367(n))).