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%I #14 Dec 07 2020 02:32:34
%S 1,1,1,1,1,2,2,1,1,2,1,2,2,1,3,2,1,4,1,1,2,1,1,8,2,2,1,3,4,6,1,1,2,3,
%T 4,3,2,1,1,2,1,2,1,2,2,9,5,1,1,2,18,1,2,1,2,3,4,1,2,10,1,2,7,1,2,2,3,
%U 2,3,2,6,1,1,2,1,1,4,2,4,2,1,20,1,2,1,1,2,2,10,1,1,1,1,1,1,1,2,20,1,6,1,18,3
%N a(1) = 1; thereafter a(n) = (p - (5|p)) divided by the smallest m such that p divides Fibonacci(m), where p is the n-th prime and (5|p) is the Legendre symbol.
%C Lucas showed that A001602 divides p-1 or p+1, according as (5|p) = 1 or -1 respectively. This is the quotient.
%H Patrick McKinley, <a href="/A116514/b116514.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = (prime(n) - (5|prime(n))) / A001602(n).
%e a(6) = 2, as 13 is the 6th prime, 5 is not a quadratic residue mod 13, 13 first occurs as a prime factor of Fibonacci(7) and (13 - (-1)) / 7 = 2.
%Y Cf. A001602.
%K easy,nonn
%O 1,6
%A _Nick Krempel_, Mar 24 2006
%E a(1)=1 added by _N. J. A. Sloane_, Dec 07 2020
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