

A116514


a(1) = 1; thereafter a(n) = (p  (5p)) divided by the smallest m such that p divides Fibonacci(m), where p is the nth prime and (5p) is the Legendre symbol.


2



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, 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, 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
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OFFSET

1,6


COMMENTS

Lucas showed that A001602 divides p1 or p+1, according as (5p) = 1 or 1 respectively. This is the quotient.


LINKS

Patrick McKinley, Table of n, a(n) for n = 1..10000


FORMULA

a(n) = (prime(n)  (5prime(n))) / A001602(n).


EXAMPLE

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.


CROSSREFS

Cf. A001602.
Sequence in context: A074295 A331348 A236479 * A334028 A124767 A319443
Adjacent sequences: A116511 A116512 A116513 * A116515 A116516 A116517


KEYWORD

easy,nonn


AUTHOR

Nick Krempel, Mar 24 2006


EXTENSIONS

a(1)=1 added by N. J. A. Sloane, Dec 07 2020


STATUS

approved



