Pisano Period - Tables

Tables

The first Pisano periods (sequence A001175 in OEIS) and their cycles (with spaces before the zeros for readability) are:

Onward the Pisano periods are 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, 16, 30, 48, 24, 100 ...

For n > 2 the period is even, because alternatingly F(n)2 is one more and one less than F(n − 1)F(n + 1) (Cassini's identity).

The period is relatively small, 4k + 2, for n = F (2k) + F (2k + 2), i.e. Lucas number L (2k + 1), with k a positive integer. This is because F(−2k − 1) = F (2k + 1) and F(−2k) = −F (2k), and the latter is congruent to F(2k + 2) modulo n, showing that the period is a divisor of 4k + 2; the period cannot be 2k + 1 or less because the first 2k + 1 Fibonacci numbers from 0 are less than n.

The second half of the cycle, which is of course equal to the part on the left of 0, consists of alternatingly numbers F(2m + 1) and n − F(2m), with m decreasing.

Furthermore, the period is 4k for n = F(2k), and 8k + 4 for n = F(2k + 1).

The number of occurrences of 0 per cycle is 1, 2, or 4. Let p be the number after the first 0 after the combination 0, 1. Let the distance between the 0s be q.

• There is one 0 in a cycle, obviously, if p = 1. This is only possible if q is even or n is 1 or 2.
• Otherwise there are two 0s in a cycle if p2 ≡ 1. This is only possible if q is even.
• Otherwise there are four 0s in a cycle. This is the case if q is odd and n is not 1 or 2.

For generalized Fibonacci sequences (satisfying the same recurrence relation, but with other initial values, e.g. the Lucas numbers) the number of occurrences of 0 per cycle is 0, 1, 2, or 4. Also, it can be proven that

π(n) ≤ 6n,

with equality if and only if n = 2 · 5k, for k ≥ 1, the first examples being π(10) = 60 and π(50) = 300.