Search: id:A116514
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%I A116514
%S A116514 1,1,1,1,2,2,1,1,2,1,2,2,1,3,2,1,4,1,1,2,1,1,8,2,2,1,3,4,6,1,1,2,3,4,3,
%T A116514 2,1,1,2,1,2,1,2,2,9,5,1,1,2,18,1,2,1,2,3,4,1,2,10,1,2,7,1,2,2,3,2,3,2,
%U A116514 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 A116514 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 A116514 Lucas showed that A001602 divides p-1 or p+1, according as (5|p) = 1 
               or -1 respectively, this is the quotient.
%F A116514 (p_n - (5|p_n)) / A001602(n)
%e A116514 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 A116514 Cf. A001602.
%Y A116514 Sequence in context: A078880 A000002 A074295 this_sequence A124767 A112933 
               A088427
%Y A116514 Adjacent sequences: A116511 A116512 A116513 this_sequence A116515 A116516 
               A116517
%K A116514 easy,nonn
%O A116514 2,5
%A A116514 Nick Krempel (ndkrempel(AT)blueyonder.co.uk), Mar 24 2006

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