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Can some explain me how the period of Fibonacci mod $1000033$ is $4684$.

As we know if $n$ mod $5$ is $2$ or $3$ then period is $2n + 2$ so the period should me $2\times1000033 + 2$ but why it is $4684$.

Thanks.

usermath
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Luckymaster
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  • No, we don't. I am pretty sure that the period modulo $1$ is $1$, and $1$ is not $4$. – mercio Jun 02 '14 at 16:48
  • Sorry ( n mod 5) is 2 or 3 thanks for your comment. – Luckymaster Jun 02 '14 at 16:51
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    I think what we know is that the period divides $2n+2$, rather than always being equal to it. (Also, I think this only holds when $n$ is prime.) – Barry Cipra Jun 02 '14 at 17:01
  • You must be mistaken. The period mod 8 is 12, the period mod 18 is 24, the period mod anything ending in 8 doesn't fit your pattern. Similarly, the period mod 33 is 40, and so on. Your assumed pattern is incorrect. – davidlowryduda Jun 02 '14 at 17:01
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    Note that $2000068=2^2\times 7 \times 61 \times 1171$ and $4684=2^2 \times 1171$ – Mark Bennet Jun 02 '14 at 17:04
  • But if n=1000000007 then period is 2000000016 which is form of n mod 5 = 2 and period 2*1000000007+2 , If I am wrong can some one explain me how to find the period. Thanks. – Luckymaster Jun 02 '14 at 17:09
  • See http://math.stackexchange.com/questions/217285/applications-of-abstract-algebra-to-elementary-mathematics/217289#217289 – Will Jagy Jun 02 '14 at 17:21
  • see also table of Pisano periods $\pmod n$ for all $n \leq 10000$ in text format at http://oeis.org/A001175/b001175.txt – Will Jagy Jun 02 '14 at 17:47

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The period is actually $ \pi(n) = k\cdot\text{ord}_n(F_{k+1}) $ where k is the smallest positive integer such that $n\mid F_k$.

So first find the first Fibonacci number where 1000033 divides $F_k$, This happens at $F _{1171}$, so $F_{1171}/1000033$ does not give any remainder when dividing. Every fibonacci number before that does. Thus $k=1171$.

The second part: $\text{ord}_n(F_{k+1})$ is the multiplicative order. You can look that up yourself, but for fibonacci numbers its either 1, 2, or 4. In this case its 4.

(Wolfram Alpha proof: https://www.wolframalpha.com/input/?i=multiplicative+order%28fib%281172%29%2C1000033%29)

So: $1171*4=4684$

I think you are talking about $2*n+2$ with primes. That refers to the fact that all primes "land" on that, but there period can be shorter, meaning $F_{2*1000033+2}\equiv 0 \mod 1000033$. Thus for some primes $F_{2n+2}\equiv 0 \mod 1000033$.

"For p≡±1(mod5) and ±2(mod5) the periods for Fibonacci sequences modulo p divide p−1 and 2p+2 respectively."

Dane Bouchie
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