OFFSET
1,3
COMMENTS
For n from 1 to 2000, a(n)/prime(n) is always < 1.8.
As N increases, (Sum_{n=1..N} a(n)) / (Sum_{n=1..N} prime(n)) tends to log(2)/3; this is consistent with the prime number theorem as the probability that x*2^n-1 is prime with odd x divisible by 3 is ~ 3/(n*log(2)) and after n*log(2)/3 try (n*log(2)/3)*(3/(n*log(2)) = 1.
LINKS
Pierre CAMI, Table of n, a(n) for n = 1..2000
MATHEMATICA
sk[n_]:=Module[{k=1, t=2^Prime[n]}, While[!PrimeQ[(6k-3)*t-1], k++]; k]; Array[ sk, 70] (* Harvey P. Dale, Nov 14 2018 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Pierre CAMI, May 22 2017
STATUS
approved