Pisano period

The Fibonacci numbers are the numbers in the integer sequence: :0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, ... defined by the recurrence relation :F_0 = 0 :F_1 = 1 :F_i = F_{i-1} + F_{i-2}. For any integer n, the sequence of Fibonacci numbers Fi taken modulo n is periodic. The Pisano period, denoted ''(n''), is the length of the period of this sequence. For example, the sequence of Fibonacci numbers modulo 3 begins: :0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, ... This sequence has period 8, so ''''(3) = 8. ==Properties==

Parity With the exception of '(2) = 3, the Pisano period '(n) is always even. This follows by observing that ''(n'') is equal to the order of the Fibonacci matrix : \mathbf Q = \begin{bmatrix} 1 & 1\\1 & 0 \end{bmatrix} in the general linear group \text{GL}_2(\mathbb{Z}_n)of invertible 2 by 2 matrices in the finite ring \mathbb{Z}_nof integers modulo n. Since Q has determinant −1, the determinant of Q''(n) is (−1)(n), which is equal to 1 when either n ≤ 2 or (n'') is even. Pisano periods of composite numbers If m and n are coprime, then ''(mn) is the least common multiple of (m) and (n''). This follows from Chinese remainder theorem. Thus the Pisano periods of composite numbers can be computed by looking at the Pisano periods of prime powers q = pk, for k ≥ 1. If p is prime, ''(p''k) divides pk–1 ''(p''). It is unknown if \pi(p^k) = p^{k-1}\pi(p) for every prime p and integer k > 1. Any prime p providing a counterexample would necessarily be a Wall–Sun–Sun prime, and conversely every Wall–Sun–Sun prime p gives a counterexample (set k = 2). For p = 2 and 5, the exact values of the Pisano periods are known. The periods of powers of these prime powers are as follows: • If n = 2k, then \pi(n) = 3 \cdot 2^{k-1} = \frac{3n}2 • if n = 5k, then \pi(n) = 4 \cdot 5^k = 4n From these it follows that if n = 2·5k then ''(n) = 6n''. Pisano periods of prime numbers If prime p is different from 2 and 5, then ''(p) is a divisor of p2 − 1. This follows from the modulo p'' analogue of Binet's formula, which implies that ''(p) is the multiplicative order of a root of modulo p''. Every p other than 2 and 5 lie in the residue classes p \equiv \pm 1\ (\mathrm{mod}\ 10) or p \equiv \pm 3\ (\mathrm{mod}\ 10). • If p \equiv \pm 1\ (\mathrm{mod}\ 10), then ''(p) divides p'' − 1. • If p \equiv \pm 3\ (\mathrm{mod}\ 10), then ''(p) divides 2(p'' + 1). The former can be proven by observing that if p \equiv \pm 1\ (\mathrm{mod}\ 10), then the roots of modulo p belong to \mathbb{F}_{p} = \mathbb{Z}/p\mathbb{Z} (by quadratic reciprocity). Thus their order, ''(p) is a divisor of p'' − 1. To prove the latter, if p \equiv \pm 3\ (\mathrm{mod}\ 10), the roots modulo p of do not belong to \mathbb{F}_{p} (by quadratic reciprocity again), and belong to the finite field \mathbb{F}_{p}[x]/(x^2 - x - 1). As the Frobenius automorphism x \mapsto x^p 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. 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. 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 X and E 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), then π(n) = 8k + 4. That is, if the modulo base is a Fibonacci number (≥ 3) with an even index, the period is twice the index and the cycle has two zeros. If the base is a Fibonacci number (≥ 5) with an odd index, the period is four times the index and the cycle has four zeros. ==Pisano periods of Lucas numbers==

If n = L(2k) (k ≥ 1), then π(n) = 8k; if n = L(2k + 1) (k ≥ 1), then π(n) = 4k + 2. That is, if the modulo base is a Lucas number (≥ 3) with an even index, the period is four times the index. If the base is a Lucas number (≥ 4) with an odd index, the period is twice the index. For even k, the cycle has two zeros. For odd k, the cycle has only one zero, and 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. ==Number of zeros in the cycle==

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. The ratio of the Pisano period of n and the number of zeros modulo n in the cycle gives the rank of apparition or Fibonacci entry point of n. That is, smallest index k such that n divides F(k). They are: :1, 3, 4, 6, 5, 12, 8, 6, 12, 15, 10, 12, 7, 24, 20, 12, 9, 12, 18, 30, 8, 30, 24, 12, 25, 21, 36, 24, 14, 60, 30, 24, 20, 9, 40, 12, 19, 18, 28, 30, 20, 24, 44, 30, 60, 24, 16, 12, ... In Renault's paper the number of zeros is called the "order" of F mod m, denoted \omega(m), and the "rank of apparition" is called the "rank" and denoted \alpha(m). According to Wall's conjecture, \alpha(p^e) = p^{e-1} \alpha(p). If m has prime factorization m = p_1^{e_1} p_2^{e_2} \dots p_n^{e_n} then \alpha(m) = \operatorname{lcm}(\alpha(p_1^{e_1}), \alpha(p_2^{e_2}), \dots, \alpha(p_n^{e_n})). == Generalizations ==

The Pisano periods of Lucas numbers are :1, 3, 8, 6, 4, 24, 16, 12, 24, 12, 10, 24, 28, 48, 8, 24, 36, 24, 18, 12, 16, 30, 48, 24, 20, 84, 72, 48, 14, 24, 30, 48, 40, 36, 16, 24, 76, 18, 56, 12, 40, 48, 88, 30, 24, 48, 32, ... The Pisano periods of Pell numbers (or 2-Fibonacci numbers) are :1, 2, 8, 4, 12, 8, 6, 8, 24, 12, 24, 8, 28, 6, 24, 16, 16, 24, 40, 12, 24, 24, 22, 8, 60, 28, 72, 12, 20, 24, 30, 32, 24, 16, 12, 24, 76, 40, 56, 24, 10, 24, 88, 24, 24, 22, 46, 16, ... The Pisano periods of 3-Fibonacci numbers are :1, 3, 2, 6, 12, 6, 16, 12, 6, 12, 8, 6, 52, 48, 12, 24, 16, 6, 40, 12, 16, 24, 22, 12, 60, 156, 18, 48, 28, 12, 64, 48, 8, 48, 48, 6, 76, 120, 52, 12, 28, 48, 42, 24, 12, 66, 96, 24, ... The Pisano periods of Jacobsthal numbers (or (1,2)-Fibonacci numbers) are :1, 1, 6, 2, 4, 6, 6, 2, 18, 4, 10, 6, 12, 6, 12, 2, 8, 18, 18, 4, 6, 10, 22, 6, 20, 12, 54, 6, 28, 12, 10, 2, 30, 8, 12, 18, 36, 18, 12, 4, 20, 6, 14, 10, 36, 22, 46, 6, ... The Pisano periods of (1,3)-Fibonacci numbers are :1, 3, 1, 6, 24, 3, 24, 6, 3, 24, 120, 6, 156, 24, 24, 12, 16, 3, 90, 24, 24, 120, 22, 6, 120, 156, 9, 24, 28, 24, 240, 24, 120, 48, 24, 6, 171, 90, 156, 24, 336, 24, 42, 120, 24, 66, 736, 12, ... The Pisano periods of Tribonacci numbers (or 3-step Fibonacci numbers) are :1, 4, 13, 8, 31, 52, 48, 16, 39, 124, 110, 104, 168, 48, 403, 32, 96, 156, 360, 248, 624, 220, 553, 208, 155, 168, 117, 48, 140, 1612, 331, 64, 1430, 96, 1488, 312, 469, 360, 2184, 496, 560, 624, 308, 440, 1209, 2212, 46, 416, ... The Pisano periods of Tetranacci numbers (or 4-step Fibonacci numbers) are :1, 5, 26, 10, 312, 130, 342, 20, 78, 1560, 120, 130, 84, 1710, 312, 40, 4912, 390, 6858, 1560, 4446, 120, 12166, 260, 1560, 420, 234, 1710, 280, 1560, 61568, 80, 1560, 24560, 17784, 390, 1368, 34290, 1092, 1560, 240, 22230, 162800, 120, 312, 60830, 103822, 520, ... See also generalizations of Fibonacci numbers. == Number theory ==

Pisano periods can be analyzed using algebraic number theory. Let \pi_k(n) be the n-th Pisano period of the k-Fibonacci sequence Fk(n) (k can be any natural number, these sequences are defined as Fk(0) = 0, Fk(1) = 1, and for any natural number n > 1, Fk(n) = kFk(n−1) + Fk(n−2)). If m and n are coprime, then \pi_k(m\cdot n) = \mathrm{lcm}(\pi_k(m),\pi_k(n)), by the Chinese remainder theorem: two numbers are congruent modulo mn if and only if they are congruent modulo m and modulo n, assuming these latter are coprime. For example, \pi_1(3)=8 and \pi_1(4)=6, so \pi_1(12=3\cdot 4) = \mathrm{lcm}(\pi_1(3),\pi_1(4))= \mathrm{lcm}(8,6)=24. Thus it suffices to compute Pisano periods for prime powers q=p^n. (Usually, \pi_k(p^n) = p^{n-1}\cdot \pi_k(p), unless p is k-Wall–Sun–Sun prime, or k-Fibonacci–Wieferich prime, that is, p2 divides Fk(p − 1) or Fk(p + 1), where Fk is the k-Fibonacci sequence, for example, 241 is a 3-Wall–Sun–Sun prime, since 2412 divides F3(242).) For prime numbers p, these can be analyzed by using Binet's formula: :F_k\left(n\right) = {{\varphi_k^n-(k-\varphi_k)^n} \over {\sqrt {k^2+4}}}={{\varphi_k^{n}-(-1/\varphi_k)^{n}} \over {\sqrt {k^2+4}}},\, where \varphi_k is the kth metallic mean :\varphi_k = \frac{k + \sqrt{k^2+4}}{2}. If k2 + 4 is a quadratic residue modulo p (where p > 2 and p does not divide k2 + 4), then \sqrt{k^2+4}, 1/2, and k/\sqrt{k^2+4} can be expressed as integers modulo p, and thus Binet's formula can be expressed over integers modulo p, and thus the Pisano period divides the totient \phi(p)=p-1, since any power (such as \varphi_k^n) has period dividing \phi(p), as this is the order of the group of units modulo p. For k = 1, this first occurs for p = 11, where 42 = 16 ≡ 5 (mod 11) and 2 · 6 = 12 ≡ 1 (mod 11) and 4 · 3 = 12 ≡ 1 (mod 11) so 4 = , 6 = 1/2 and 1/ = 3, yielding φ = (1 + 4) · 6 = 30 ≡ 8 (mod 11) and the congruence :F_1\left(n\right) \equiv 3\cdot \left(8^n - 4^n\right) \pmod{11}. Another example, which shows that the period can properly divide p − 1, is π1(29) = 14. If k2 + 4 is not a quadratic residue modulo p, then Binet's formula is instead defined over the quadratic extension field (\mathbb{Z}/p)[\sqrt{k^2+4}], which has p2 elements and whose group of units thus has order p2 − 1, and thus the Pisano period divides p2 − 1. For example, for p = 3 one has π1(3) = 8 which equals 32 − 1 = 8; for p = 7, one has π1(7) = 16, which properly divides 72 − 1 = 48. This analysis fails for p = 2 and p is a divisor of the squarefree part of k2 + 4, since in these cases are zero divisors, so one must be careful in interpreting 1/2 or \sqrt{k^2+4}. For p = 2, is congruent to 1 mod 2 (for k odd), but the Pisano period is not p − 1 = 1, but rather 3 (in fact, this is also 3 for even k). For p divides the squarefree part of k2 + 4, the Pisano period is πk(k2 + 4) = p2 − p = p(p − 1), which does not divide p − 1 or p2 − 1. ==Fibonacci integer sequences modulo n==

One can consider Fibonacci integer sequences and take them modulo n, or put differently, consider Fibonacci sequences in the ring Z/nZ. The period is a divisor of π(n). The number of occurrences of 0 per cycle is 0, 1, 2, or 4. If n is not a prime the cycles include those that are multiples of the cycles for the divisors. For example, for n = 10 the extra cycles include those for n = 2 multiplied by 5, and for n = 5 multiplied by 2. Table of the extra cycles: (the original Fibonacci cycles are excluded) (using X and E for ten and eleven, respectively) Number of Fibonacci integer cycles mod n are: :1, 2, 2, 4, 3, 4, 4, 8, 5, 6, 14, 10, 7, 8, 12, 16, 9, 16, 22, 16, 29, 28, 12, 30, 13, 14, 14, 22, 63, 24, 34, 32, 39, 34, 30, 58, 19, 86, 32, 52, 43, 58, 22, 78, 39, 46, 70, 102, ... ==Notes==