_T. D. Noe (noe(AT)sspectra.com), _, May 08 2010

Least k>0 such that (p*2^k-1)/3 is prime, or zero if no k exists, where p=prime(n).

3, 0, 1, 4, 1, 2, 1, 4, 3, 1, 2, 4, 3, 4, 1, 9107, 3, 6, 2, 1, 2, 4, 7, 1, 6, 1, 2, 1, 32, 11, 4, 3, 45, 24, 3, 6, 8, 16, 21, 3, 29, 2, 1, 2, 1, 4, 2, 66, 1, 8, 7, 5, 10, 1, 5, 3, 1, 14, 18, 13, 6, 59, 2, 3, 4, 1, 18, 2, 5, 4, 3, 1, 6, 5016, 8, 3, 15, 14, 3, 12, 3, 46, 5, 2, 4, 3, 5, 4, 1, 2, 1, 3

1,1

When a(n) is not zero, a(n) is even if p=1 (mod 6); a(n) is odd if p=5 (mod 6). If we let q=(p*2^k-1)/3 be a prime generated by p for some k>0, then the first prime number after q in the Collatz iteration of q is p. When k=1, q is less than p. The primes, other than 3, for which a(n)=0 are in A177331.

Table[p=Prime[n]; If[p==3, k=0, k=1; While[q=(p*2^k-1)/3; k<10000 && !PrimeQ[q], k++ ]]; k, {n, 100}]

Cf. A177000

nonn,new

T. D. Noe (noe(AT)sspectra.com), May 08 2010

approved