A060305 Pisano periods for primes: period of Fibonacci numbers mod prime(n).

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

Offset: 1

Author

Louis Mello (mellols(AT)aol.com), Mar 26 2001

Keywords

• nonn

Comments

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).

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
• B. Avila and T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and J. Int. Seq. 17 (2014), # 14.8.5.
• A. Elsenhans and J. Jahnel, The Fibonacci sequence modulo p^2 -- An investigation by computer for p < 10^14, (2004).
• D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly, 67 (1960), 525-532.
• Wikipedia, Pisano period.

Crossrefs

Cf. A001175, A000961, A071774, A003147, A296240.

Programs

Maple

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:
seq(a(n), n=1..70);  # Alois P. Heinz, Oct 16 2015

Mathematica

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 *)

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
def A060305(n):
    x, p = (1,1), prime(n)
    for k in count(1):
        if x == (0,1):
            return k
        x = (x[1], (x[0]+x[1]) % p) # Chai Wah Wu, May 31 2022

Formula

a(n) = A001175(prime(n)). - Jonathan Sondow, Dec 09 2017

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

Extensions

Corrected by Benoit Cloitre, Jun 04 2002

Name clarified by Jonathan Sondow, Dec 09 2017