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A060305
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Pisano periods for primes: period of Fibonacci numbers mod prime(n).
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14
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3, 8, 20, 16, 10, 28, 36, 18, 48, 14, 30, 76, 40, 88, 32, 108, 58, 60, 136, 70, 148, 78, 168, 44, 196, 50, 208, 72, 108, 76, 256, 130, 276, 46, 148, 50, 316, 328, 336, 348, 178, 90, 190, 388, 396, 22, 42, 448, 456, 114, 52, 238, 240, 250, 516, 176, 268, 270, 556
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OFFSET
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1,1
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COMMENTS
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Assuming Wall's conjecture (which is still open) allows one to calculate A001175(m) when m is a prime power since for any k >= 1: A001175(prime(n)^k) = a(n)*prime(n)^(k-1). For example: A001175(2^k) = 3*2^(k-1) = A007283(k-1).
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LINKS
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FORMULA
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a(n) = (3 - L(p))/2 * (p - L(p)) / A296240(n) for n >= 4, where p = prime(n) and L(p) = Legendre(p|5); so a(n) <= p-1 if p == +- 1 mod 5, and a(n) <= 2*p+2 if p == +- 2 mod 5. See Wall's Theorems 6 and 7. - Jonathan Sondow, Dec 10 2017
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MAPLE
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a:= proc(n) option remember; local F, k, p;
F:=[1, 1]; p:=ithprime(n);
for k while F<>[0, 1] do
F:=[F[2], irem(F[1]+F[2], p)]
od: k
end:
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MATHEMATICA
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Table[p=Prime[n]; a={1, 0}; a0=a; k=0; While[k++; s=Mod[Plus@@a, p]; a=RotateLeft[a]; a[[2]]=s; a!=a0]; k, {n, 100}] (* T. D. Noe, Jun 12 2006 *)
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PROG
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(PARI) for(n=1, 100, s=1; while(sum(i=n, n+s, abs(fibonacci(i)%prime(n)-fibonacci(i+s)%prime(n)))+sum(i=n+1, n+1+s, abs(fibonacci(i)%prime(n)-fibonacci(i+s)%prime(n)))>0, s++); print1(s, ", "))
(Python)
from itertools import count
from sympy import prime
x, p = (1, 1), prime(n)
for k in count(1):
if x == (0, 1):
return k
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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Louis Mello (mellols(AT)aol.com), Mar 26 2001
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EXTENSIONS
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STATUS
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approved
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