
COMMENTS

Assuming that there are no square primitive factors in the Fibonacci sequence (an open question), then this sequence continues 53, 43, 113, 73, 109, 233, 59, 107, 199, 67, 97, 71, 101, 149, 79, 139, 83, 151, 281, 421, 211, 137, 103, 157, 307, 521. This is obtained by sorting the pairs (prime(n)*A001602(n), prime(n)) by the first position and noting the order of the primes in the second position.  T. D. Noe, Apr 15 2004
