Pisano period

In number theory, the nth Pisano period, written π(n), is the period with which the sequence of Fibonacci numbers taken modulo n repeats. Pisano periods are named after Leonardo Pisano, better known as Fibonacci. The existence of periodic functions in Fibonacci numbers was noted by Joseph Louis Lagrange in 1774.

Definition

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, ... (sequence A000045 in OEIS)

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 F_i (mod n) of Fibonacci numbers taken modulo n is periodic. The Pisano period, denoted π(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, ... (sequence A082115 in OEIS)

This sequence has period 8, so π(3) = 8.

Tables

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

n	π(n)	nr. of 0s	cycle
1	1	1	0
2	3	1	011
3	8	2	0112 0221
4	6	1	011231
5	20	4	01123 03314 04432 02241
6	24	2	011235213415 055431453251
7	16	2	01123516 06654261
8	12	2	011235 055271
9	24	2	011235843718 088764156281
10	60	4	011235831459437 077415617853819 099875279651673 033695493257291

Onward the Pisano periods are shown in the following table:

π(n)	+0	+1	+2	+3	+4	+5	+6	+7	+8	+9	+10	+11
0+		1	3	8	6	20	24	16	12	24	60	10
12+	24	28	48	40	24	36	24	18	60	16	30	48
24+	24	100	84	72	48	14	120	30	48	40	36	80
36+	24	76	18	56	60	40	48	88	30	120	48	32
48+	24	112	300	72	84	108	72	20	48	72	42	58
60+	120	60	30	48	96	140	120	136	36	48	240	70
72+	24	148	228	200	18	80	168	78	120	216	120	168
84+	48	180	264	56	60	44	120	112	48	120	96	180
96+	48	196	336	120	300	50	72	208	84	80	108	72
108+	72	108	60	152	48	76	72	240	42	168	174	144
120+	120	110	60	40	30	500	48	256	192	88	420	130
132+	120	144	408	360	36	276	48	46	240	32	210	140

For n > 2 the period is even, because the Fibonacci matrix ($\begin{bmatrix}1 & 1 \\1 & 0 \end{bmatrix}$) has determinant -1 and therefore has even order in GL₂(Z_n).

If n = F (2k) (k > 1), then π(n) = 4k; If n = F (2k + 1) (k > 1), 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. If the base is a Fibonacci number (>=5) with an odd index, the period is 4 times the index.

k	F (k)	π(F (k))
1	1	1
2	1	1
3	2	3
4	3	8
5	5	20
6	8	12
7	13	28
8	21	16
9	34	36
10	55	20
11	89	44
12	144	24
13	233	52
14	377	28
15	610	60
16	987	32
17	1597	68
18	2584	36
19	4181	76
20	6765	40

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.

k	L (2k + 1)	π(L (2k + 1))	first half of cycle (with selected second halves)
1	4	6	0, 1, 1, (2, 3, 1)
2	11	10	0, 1, 1, 2, 3, (5, 8, 2, 10, 1)
3	29	14	0, 1, 1, 2, 3, 5, 8, (13, 21, 5, 26, 2, 28, 1)
4	76	18	0, 1, 1, 2, 3, 5, 8, 13, 21, (34, 55, 13, 68, 5, 73, 2, 75, 1)
5	199	22	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, (89, 144, 34, 178, 13, 191, 5, 196, 2, 198, 1)
6	521	26	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233
7	1364	30	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610
8	3571	34	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597
9	9349	38	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181

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 · 5^r, for r ≥ 1, the first examples being π(10) = 60 and π(50) = 300.

Generalizations

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, 42, 60, ... (sequence A175181 in OEIS)

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, 112, 60, ... (sequence A175182 in OEIS)

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, 42, 20, ... (sequence A175286 in OEIS)

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, 168, 120, ... (sequence A175291 in OEIS)

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, 336, 620, ... (sequence A046738 in OEIS)

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, 2394, 1560, ... (sequence A106295 in OEIS)

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 F_k(n) ( k can be any natural number, these sequences are defined as F_k(0) = 0, F_k(1) = 1, and for any natural number n > 1, F_k(n) = kF_k(n-1) + F_k(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 F_k(p-1) or F_k(p+1), where F_k 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 (and $p > 2$), 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 = √5, 6 = 1/2 and 1/√5 = 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 (and p ≠ 2, and p does not divide the squarefree part of k²+4), then Binet's formula is instead defined over the quadratic extension field (Z/p)[√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 √k^2+4. For p = 2, k²+4 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 π_k(k²+4) = p²-p = p(p − 1), which does not divide p − 1 or p² − 1.

Sums

Using:

$\sum_{i=1}^{n} F_i = F_{n+2} - 1$,

it follows that the sum of π(n) consecutive Fibonacci numbers is a multiple of n. Thus:

$\sum_{i=1}^{\pi(n)} F_i = nk$

Moreover, for the examples listed below, the sum of π(n) consecutive Fibonacci numbers is n times the (π(n)/2 + 1)th element:

$\sum_{i=n}^{n+5} F_i = 4 F_{n+4}$

$\sum_{i=n}^{n+9} F_i = 11 F_{n+6}$

$\sum_{i=n}^{n+13} F_i = 29 F_{n+8}$

$\sum_{i=n}^{n+17} F_i = 76 F_{n+10}$

Cultural references

The Fibonacci sequence modulo 5 (Pisano period 20, with 4 zeros) is featured in the episode "The Case of the Willing Parrot" of the TV series Mathnet, where the sequence is depicted as tiles on a wall.

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/n. 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:

n	multiples	other cycles
1
2	0
3	0
4	0, 022	033213
5	0	1342
6	0, 0224 0442, 033
7	0	02246325 05531452, 03362134 04415643
8	0, 022462, 044, 066426	033617 077653, 134732574372, 145167541563
9	0, 0336 0663	022461786527 077538213472, 044832573145 055167426854
10	0, 02246 06628 08864 04482, 055, 2684	134718976392

External links

• The Fibonacci sequence modulo m
• A research for Fibonacci numbers
• Pisano period calculator
• k-Fibonacci sequence modulo m
• Fibonacci Mystery - Numberphile, a YouTube video with Dr. James Grime and the University of Nottingham