A116515 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.

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

Offset

1,2

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, Jan 14 2009]

Links

Table of n, a(n) for n=1..105.

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, Jan 14 2009]

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, Jan 14 2009]

Crossrefs

Cf. A060305, A001602.

Cf. A112860, A053027, A053028 (primes producing 1, 2 and 4 zeros) [From T. D. Noe, Jan 14 2009]

Sequence in context: A355346 A364608 A105791 * A303118 A037178 A352650

Adjacent sequences: A116512 A116513 A116514 * A116516 A116517 A116518

Keyword

easy,nonn

Author

Nick Krempel, Mar 24 2006

Status

approved