A073409 Largest prime factor of the denominator of the Bernoulli number B(2*n) (A002445).

3, 5, 7, 5, 11, 13, 3, 17, 19, 11, 23, 13, 3, 29, 31, 17, 3, 37, 3, 41, 43, 23, 47, 17, 11, 53, 19, 29, 59, 61, 3, 17, 67, 5, 71, 73, 3, 5, 79, 41, 83, 43, 3, 89, 31, 47, 3, 97, 3, 101, 103, 53, 107, 109, 23, 113, 7, 59, 3, 61, 3, 5, 127, 17, 131, 67, 3, 137, 139, 71, 3, 73, 3, 149

Offset: 1

Author

Benoit Cloitre, Aug 23 2002

Keywords

• easy
• nonn

Comments

Least k such that k!*B(2n) is an integer where B(2n) denotes the 2n-th Bernoulli number.

a((p-1)/2) = p, where p is odd prime. All odd primes appear in this sequence. The very first appearance of odd prime p is a((p-1)/2). - Alexander Adamchuk, Jul 31 2006

Conjecture: a(n) is the largest prime p <= 2n+1 such that p * A000367(n) == - A002445(n) (mod p^2) for n > 0. Note that 2^(2n) == 1 (mod a(n)), since a(n) is the largest prime p such that b^(2n)== 1 (mod p) for every b coprime to p; i.e., a(n) is the largest prime p such that p-1 | 2n. - Thomas Ordowski, May 17 2020

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000367, A002445, A005097.

Programs

Mathematica

Table[FactorInteger[Denominator[BernoulliB[2n]]][[ -1, 1]], {n, 100}]

PARI

a(n)=
{
    my(bd=1);
    forprime (p=2, 2*n+1, if( (2*n)%(p-1)==0, bd=p ) );
    return(bd);
}
/* Joerg Arndt, May 06 2012 */

PARI

a(n)=my(p); fordiv(n,d, if(isprime(p=2*n/d+1), return(p))) \\ Charles R Greathouse IV, Jun 08 2020