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?
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
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Feb 11 2004
STATUS
approved