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A116515
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a(n) = the period of the Fibonacci numbers modulo p divided by the smallest m such that p divides Fibonacci(m), where p is the n-th prime.
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1
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1, 2, 4, 2, 1, 4, 4, 1, 2, 1, 1, 4, 2, 2, 2, 4, 1, 4, 2, 1, 4, 1, 2, 4, 4, 1, 2, 2, 4, 4, 2, 1, 4, 1, 4, 1, 4, 2, 2, 4, 1, 1, 1, 4, 4, 1, 1, 2, 2, 1, 4, 1, 2, 1, 4, 2, 4, 1, 4, 2, 2, 4, 2, 1, 4, 4, 1, 4, 2, 1, 4, 1, 2, 4, 1, 2, 4, 4, 2, 2, 1, 4, 1, 4, 1, 2, 2, 4, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 4, 2, 2, 1
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Conditions on p_n mod 4 and mod 5 restrict possible values of a(n). The unknown (?) case is p = 1 mod 4 and (5|p) = 1, equivalently, p = 1 or 9 mod 20, where {1, 2, 4} all occur.
Number of zeros in fundamental period of Fibonacci numbers mod prime(n). [From T. D. Noe (noe(AT)sspectra.com), Jan 14 2009]
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FORMULA
| a(n) = A060305(n) / A001602(n). a(n) is always one of {1, 2, 4}.
a(n) = A001176(prime(n)) [From T. D. Noe (noe(AT)sspectra.com), Jan 14 2009]
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EXAMPLE
| a(4) = 2, as 7 is the 4th prime, the Fibonacci numbers mod 7 have period 16, the first Fibonacci number divisible by 7 is F(8) = 21 = 3*7 and 16 / 8 = 2.
One period of the Fibonacci numbers mod 7 is 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1, 0, which has two zeros. Hence a(4)=2. [From T. D. Noe (noe(AT)sspectra.com), Jan 14 2009]
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CROSSREFS
| Cf. A060305, A001602.
Cf. A112860, A053027, A053028 (primes producing 1, 2 and 4 zeros) [From T. D. Noe (noe(AT)sspectra.com), Jan 14 2009]
Sequence in context: A079045 A021417 A105791 * A037178 A077748 A152753
Adjacent sequences: A116512 A116513 A116514 * A116516 A116517 A116518
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KEYWORD
| easy,nonn
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AUTHOR
| Nick Krempel (ndkrempel(AT)blueyonder.co.uk), Mar 24 2006
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