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

%I #2 Mar 30 2012 17:22:56

%S 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,

%T 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,

%U 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

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

%C 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.

%t 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}]

%Y Cf. A177000

%K nonn

%O 1,1

%A _T. D. Noe_, May 08 2010