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A073409
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Largest prime factor of the denominator of the Bernoulli number B(2*n) (A002445).
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2
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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
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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MATHEMATICA
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Table[FactorInteger[Denominator[BernoulliB[2n]]][[ -1, 1]], {n, 100}]
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PROG
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(PARI)
a(n)=
{
my(bd=1);
forprime (p=2, 2*n+1, if( (2*n)%(p-1)==0, bd=p ) );
return(bd);
}
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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