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 k-th Bernoulli number.

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

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 * A350704 A157342 A230226

Adjacent sequences: A090822 A090823 A090824 * A090826 A090827 A090828

Keyword

nonn

Author

Benoit Cloitre, Feb 11 2004

Status

approved