

A090825


Nonprimes n such that (3/2)*(1/n)*(2*n+1)*(3^n+1)*B(2*n) is an integer, where B(k) denotes the kth Bernoulli number.


0



1, 49, 91, 119, 133, 169, 217, 221, 247, 259, 289, 301, 323, 329, 335, 343, 361, 403, 413, 427, 469, 481, 497, 511, 517, 527, 553, 559, 589, 611, 629, 637, 679, 703, 707, 721, 731, 749
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OFFSET

1,2


COMMENTS

Conjecture: composite numbers with all prime factors in A053176 are in the sequence. For p prime (3/2)*(1/p)*(2*p+1)*(3^p+1)*B(2*p) == 1 (mod p). There are few terms n with (3/2)*(1/n)*(2*n+1)*(3^n+1)*B(2*n) == 1 (mod n): 91,247,....Is this subsequence finite?


LINKS

Table of n, a(n) for n=1..38.


PROG

(PARI) for(n=1, 750, if(frac( (3/2)*(1/n)*(2*n+1)*(3^n+1)*bernfrac(2*n))==0, if(isprime(n)==0, print1(n, ", "))))


CROSSREFS

Sequence in context: A326257 A231275 A158725 * A157342 A230226 A178705
Adjacent sequences: A090822 A090823 A090824 * A090826 A090827 A090828


KEYWORD

nonn


AUTHOR

Benoit Cloitre, Feb 11 2004


STATUS

approved



