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A116514
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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.
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2
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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
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1,6
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COMMENTS
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Lucas showed that A001602 divides p-1 or p+1, according as (5|p) = 1 or -1 respectively. This is the quotient.
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LINKS
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FORMULA
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a(n) = (prime(n) - (5|prime(n))) / A001602(n).
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EXAMPLE
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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.
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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