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

0,6

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.

Formula

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

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

Crossrefs

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

Sequence in context: A203925 A198597 A180315 * A242193 A326727 A090947

Adjacent sequences: A332297 A332298 A332299 * A332301 A332302 A332303

Keyword

nonn

Author

Amiram Eldar, Feb 09 2020

Status

approved