%I #11 Mar 30 2016 02:30:06
%S 2,3,5,7,13,11,17,19,29,23,37,47,41,61,31,89
%N Start of the permutation of the primes in the order in which p^2 first appears as a factor of a number in the Fibonacci sequence.
%C 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
%Y Cf. A065069, A065106.
%Y Cf. A001602 (smallest m such that prime(n) divides Fibonacci(m)).
%K nonn,more
%O 1,1
%A _Len Smiley_, Nov 21 2001
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