Pisano Periods - Fibonacci Numbers mod p

[LDP]

Let F_n be the n^th number of a Fibonacci sequence.
We know that F_nmod(p) forms a periodic sequence (http://en.wikipedia.org/wiki/Pisano_period) called the Pisano Period.
Let p = a prime such that p$\equiv${2,3}mod 5 so that h(p)$\mid$ 2 p + 2.
Let h(p) denote of the length of the Pisano period.

If D = {d₁,d₂,d₃$\cdots$d_k} is the non-empty set of k divisors of 2 p + 2
Then:
h(p) = min[d_i] such that F_{d(i + 1)}$\equiv$ 1 mod p
and

1. d_i ~$\mid$$\frac{1}{2}$ p (p + 1)
2. d_i ~$\mid$ p + 1
3. d_i ~$\mid$ 3 (p - 1)

Now let p = a prime such that p$\equiv${1,4}mod 5 so that h(p)$\mid$ p - 1.
If p has a primitive root such that g²$\equiv$ g + 1 mod(p) then h(p) = p - 1.
Note that g²$\equiv$ g + 1 mod(p) has two roots: 1.618033988 and -0.618033988 - variants of the Golden Ratio.
If p has no primitive root then D = {d₁,d₂,d₃$\cdots$d_k} is the non-empty set of k divisors of p - 1.
Let h(p) = min[d_i] such that F_{d(i + 1)}$\equiv$ 1 mod p
and
d_i ~$\mid$ p + 1 and d_i ~$\mid$ floor [ p/2]].

If m is any positive integer > 3 we can write F_n mod F_m where h(F_m) is given by

1. h(F_m) = 2m ↔ m is even
2. h(F_m) = 4m ↔ m is odd

 

[ramsey2879]

Let F_n be the n^th number of a Fibonacci sequence.
We know that F_nmod(p) forms a periodic sequence (http://en.wikipedia.org/wiki/Pisano_period) called the Pisano Period.
Let p = a prime such that p$\equiv${2,3}mod 5 so that h(p)$\mid$ 2 p + 2.
Let h(p) denote of the length of the Pisano period.

If D = {d₁,d₂,d₃$\cdots$d_k} is the non-empty set of k divisors of 2 p + 2
Then:
h(p) = min[d_i] such that F_{d(i + 1)}$\equiv$ 1 mod p
and

1. d_i ~$\mid$$\frac{1}{2}$ p (p + 1)
2. d_i ~$\mid$ p + 1
3. d_i ~$\mid$ 3 (p - 1)

Now let p = a prime such that p$\equiv${1,4}mod 5 so that h(p)$\mid$ p - 1.
If p has a primitive root such that g²$\equiv$ g + 1 mod(p) then h(p) = p - 1.
Note that g²$\equiv$ g + 1 mod(p) has two roots: 1.618033988 and -0.618033988 - variants of the Golden Ratio.
If p has no primitive root then D = {d₁,d₂,d₃$\cdots$d_k} is the non-empty set of k divisors of p - 1.
Let h(p) = min[d_i] such that F_{d(i + 1)}$\equiv$ 1 mod p
and
d_i ~$\mid$ p + 1 and d_i ~$\mid$ floor [ p/2]].

If m is any positive integer > 3 we can write F_n mod F_m where h(F_m) is given by

1. h(F_m) = 2m ↔ m is even
2. h(F_m) = 4m ↔ m is odd

Interesting to say the least. I am curious as to what is meant by

1. d_i ~$\mid$$\frac{1}{2}$ p (p + 1)
2. d_i ~$\mid$ p + 1
3. d_i ~$\mid$ 3 (p - 1)

I think you are saying d_i approximately divides the expressions on the right, but I don't know what that means. Can you give an example?

 

[LDP]

Thanks.

No, I was trying to find - does not divide - but couldn't so I sort of made that up.
But perhaps I should clarify it, because it is not the standard notation.

 

[ramsey2879]

Thanks.

No, I was trying to find - does not divide - but couldn't so I sort of made that up.
But perhaps I should clarify it, because it is not the standard notation.

Clarification noted. Only problem that I see is for p = 2 (for which you say h(p) = 2p + 2 = 6). Since h(2) = 3, I guess you meant odd primes that = 2,3, mod 5 (whose last digit is either a 3 or 7). PS, I noted that for the composites ending in 3 or 7 which I checked, that h(p) <> 2p + 2. Could this be a test for primes ending in 3 or 7?

 

[LDP]

Clarification noted. Only problem that I see is for p = 2 (for which you say h(p) = 2p + 2 = 6). Since h(2) = 3, I guess you meant odd primes that = 2,3, mod 5 (whose last digit is either a 3 or 7). PS, I noted that for the composites ending in 3 or 7 which I checked, that h(p) <> 2p + 2. Could this be a test for primes ending in 3 or 7?

Indeed, I should have said "only odd primes".
Good catch.
I will add some more observations on this topic as time permits.

 

[ramsey2879]

Clarification noted. Only problem that I see is for p = 2 (for which you say h(p) = 2p + 2 = 6). Since h(2) = 3, I guess you meant odd primes that = 2,3, mod 5 (whose last digit is either a 3 or 7). PS, I noted that for the composites ending in 3 or 7 which I checked, that h(p) <> 2p + 2. Could this be a test for primes ending in 3 or 7?

I checked and found that the following composites ending in 7 have Pisano periods that divide 2p + 2 and do not divide (1/2)*p(p+1) or p+1 or 3p-3. (All primes ending in 7 less than 100,000 meet this test also). The composites ending in 7 and less than 100,000 meeting the test are 377, 3827, 5777, 10877, 25877, 60377, 75077 and 90287.