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%I #41 Jul 12 2022 08:30:18
%S 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,
%T 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,
%U 103,53,107,109,23,113,7,59,3,61,3,5,127,17,131,67,3,137,139,71,3,73,3,149
%N Largest prime factor of the denominator of the Bernoulli number B(2*n) (A002445).
%C Least k such that k!*B(2n) is an integer where B(2n) denotes the 2n-th Bernoulli number.
%C 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
%C 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
%H Amiram Eldar, <a href="/A073409/b073409.txt">Table of n, a(n) for n = 1..10000</a>
%t Table[FactorInteger[Denominator[BernoulliB[2n]]][[ -1, 1]], {n, 100}]
%o (PARI)
%o a(n)=
%o {
%o my(bd=1);
%o forprime (p=2, 2*n+1, if( (2*n)%(p-1)==0, bd=p ) );
%o return(bd);
%o }
%o /* _Joerg Arndt_, May 06 2012 */
%o (PARI) a(n)=my(p); fordiv(n,d, if(isprime(p=2*n/d+1), return(p))) \\ _Charles R Greathouse IV_, Jun 08 2020
%Y Cf. A000367, A002445, A005097.
%K easy,nonn
%O 1,1
%A _Benoit Cloitre_, Aug 23 2002
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