Quote:
Originally Posted by carpetpool
and find that 2^n1 is prime for n = 2, 3, 5, 7, 11, 13, 17, 19, 31, 61, 89. The first 6 terms are a trivial result of dividing 2^n1 by all prime factors of the form 2*k*n+1, with k > l, and 2*l*n+1 > 2^n1. The others are a result of a primality test after 'sieving'. Other large sequences should work the same way, leaving less primes indices n to test in S(n).

2^111 = 23 * 89 is composite