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%I #13 Oct 15 2012 22:15:46
%S 1,49,91,119,133,169,217,221,247,259,289,301,323,329,335,343,361,403,
%T 413,427,469,481,497,511,517,527,553,559,589,611,629,637,679,703,707,
%U 721,731,749
%N 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.
%C 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?
%o (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,","))))
%K nonn
%O 1,2
%A _Benoit Cloitre_, Feb 11 2004
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