the Pisano period of any other prime. The periods of powers of these primes are as follows: If n = 2k, then (n) = 3·2k–1 = = . if n = 5k, then (n) = 20·5k–1 = = 4n. From these it follows that if n = 2·5k then (n) = 6n. The remaining primes all lie in the residue classes or . If p is a prime different from 2 and 5, then the modulo p analogue of Binet's formula implies that (p) is the multiplicative order of a root of modulo p. If , these roots belong to (by quadratic reciprocity). Thus their order, (p) is a divisor of p − 1. For example, (11) = 11 − 1 = 10 and (29) = (29 − 1)/2 = 14. If the roots modulo p of do not belong to (by quadratic reciprocity again), and belong to the finite field As the Frobenius automorphism exchanges these roots, it follows that, denoting them by r and s, we have r p = s, and thus r p+1 = –1. That is r 2(p+1) = 1, and the Pisano period, which is the order of r, is the quotient of 2(p+1) by an odd divisor. This quotient is always a multiple of 4. The first examples of such a p, for which (p) is smaller than 2(p+1), are (47) = 2(47 + 1)/3 = 32, (107) = 2(107 + 1)/3 = 72 and (113) = 2(113 + 1)/3 = 76. (See the table below) It follows from above results, that if n = pk is an odd prime power such that (n) > n, then (n)/4 is an integer that is not greater than n. The multiplicative property of Pisano periods imply thus that (n) ≤ 6n, with equality if and only if n = 2 · 5r, for r ≥ 1. The first examples are (10) = 60 and (50) = 300. If n is not of the form 2 · 5r, then (n) ≤ 4n. Tables The first twelve Pisano periods and their cycles (with spaces before the zeros for readability) are (using hexadecimal cyphers A and B for ten and eleven, respectively): The first 144 Pisano periods are shown in the following table: Pisano periods of Fibonacci numbers If n = F(2k) (k ≥ 2), then π(n) = 4k; if n = F(2k + 1) (k ≥ 2),