Pisano period explained

In number theory, the nth Pisano period, written as (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.^[1]

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, ... 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 F_i 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

With the exception of (2) = 3, the Pisano period (n) is always even. A proof of this can be given by observing that (n) is equal to the order of the Fibonacci matrix

Q = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} in the general linear group GL _{2(Z} _{n)} of invertible 2 by 2 matrices in the finite ring Z _{n} of integers modulo n. Since Q has determinant −1, the determinant of Q⁽ⁿ⁾ is (−1)⁽ⁿ⁾, and since this must equal 1 in Z _{n} , either n ≤ 2 or (n) is even.^[2]

Since

F _{0} = 0 and F _{1} = 1 we have that n divides F _{\pi(n)} and (F _{\pi(n)+1} - 1 ) .

If m and n are coprime, then (mn) is the least common multiple of (m) and (n), by the Chinese remainder theorem. For example, (3) = 8 and (4) = 6 imply (12) = 24. Thus the study of Pisano periods may be reduced to that of Pisano periods of prime powers q = p^k, for k ≥ 1.

If p is prime, (p^k) divides p^k–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).

So the study of Pisano periods may be further reduced to that of Pisano periods of primes. In this regard, two primes are anomalous. The prime 2 has an odd Pisano period, and the prime 5 has period that is relatively much larger than the Pisano period of any other prime. The periods of powers of these primes are as follows:

• If n = 2^k, then (n) = 3·2^k–1 = = .
• if n = 5^k, then (n) = 20·5^k–1 = = 4n.

From these it follows that if n = 2·5^k then (n) = 6n.

The remaining primes all lie in the residue classes

p \equiv \pm 1 \pmod { 10 } or p \equiv \pm 3 \pmod { 10 } . 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 p \equiv \pm 1 \pmod { 10 } , these roots belong to F _{p} = Z /p Z (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

p \equiv \pm 3 \pmod { 10 }, the roots modulo p of do not belong to F _{p} (by quadratic reciprocity again), and belong to the finite fieldF _{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. 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 = p^k 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 · 5^r, for r ≥ 1.

The first examples are (10) = 60 and (50) = 300. If n is not of the form 2 · 5^r, then (n) ≤ 4n.

Tables

The first twelve Pisano periods and their cycles (with spaces before the zeros for readability) are^[3] (using hexadecimal cyphers A and B for ten and eleven, respectively):

n	π(n)	number of zeros in the cycle	cycle	OEIS sequence for the 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
11	10	1	01123582A1
12	24	2	011235819A75 055A314592B1

The first 144 Pisano periods are shown in the following table:

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

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.

k	F(k)	π(F(k))	first half of cycle (for even k ≥ 4) or first quarter of cycle (for odd k ≥ 4) or all cycle (for k ≤ 3) (with selected second halves or second quarters)
1	1	1	0
2	1	1	0
3	2	3	0, 1, 1
4	3	8	0, 1, 1, 2, (0, 2, 2, 1)
5	5	20	0, 1, 1, 2, 3, (0, 3, 3, 1, 4)
6	8	12	0, 1, 1, 2, 3, 5, (0, 5, 5, 2, 7, 1)
7	13	28	0, 1, 1, 2, 3, 5, 8, (0, 8, 8, 3, 11, 1, 12)
8	21	16	0, 1, 1, 2, 3, 5, 8, 13, (0, 13, 13, 5, 18, 2, 20, 1)
9	34	36	0, 1, 1, 2, 3, 5, 8, 13, 21, (0, 21, 21, 8, 29, 3, 32, 1, 33)
10	55	20	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, (0, 34, 34, 13, 47, 5, 52, 2, 54, 1)
11	89	44	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, (0, 55, 55, 21, 76, 8, 84, 3, 87, 1, 88)
12	144	24	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, (0, 89, 89, 34, 123, 13, 136, 5, 141, 2, 143, 1)
13	233	52	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144
14	377	28	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233
15	610	60	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377
16	987	32	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610
17	1597	68	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987
18	2584	36	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597
19	4181	76	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584
20	6765	40	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181
21	10946	84	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765
22	17711	44	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946
23	28657	92	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711
24	46368	48	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657

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.

k	L(k)	π(L(k))	first half of cycle (for odd k ≥ 2) or first quarter of cycle (for even k ≥ 2) or all cycle (for k = 1) (with selected second halves or second quarters)
1	1	1	0
2	3	8	0, 1, (1, 2)
3	4	6	0, 1, 1, (2, 3, 1)
4	7	16	0, 1, 1, 2, (3, 5, 1, 6)
5	11	10	0, 1, 1, 2, 3, (5, 8, 2, 10, 1)
6	18	24	0, 1, 1, 2, 3, 5, (8, 13, 3, 16, 1, 17)
7	29	14	0, 1, 1, 2, 3, 5, 8, (13, 21, 5, 26, 2, 28, 1)
8	47	32	0, 1, 1, 2, 3, 5, 8, 13, (21, 34, 8, 42, 3, 45, 1, 46)
9	76	18	0, 1, 1, 2, 3, 5, 8, 13, 21, (34, 55, 13, 68, 5, 73, 2, 75, 1)
10	123	40	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, (55, 89, 21, 110, 8, 118, 3, 121, 1, 122)
11	199	22	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, (89, 144, 34, 178, 13, 191, 5, 196, 2, 198, 1)
12	322	48	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, (144, 233, 55, 288, 21, 309, 8, 317, 3, 320, 1, 321)
13	521	26	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144
14	843	56	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233
15	1364	30	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377
16	2207	64	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610
17	3571	34	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987
18	5778	72	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597
19	9349	38	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584
20	15127	80	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181
21	24476	42	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765
22	39603	88	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946
23	64079	46	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711
24	103682	96	0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657

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) .^[4]

According to Wall's conjecture,

\alpha(p ^{e)} = p ^{e-1} \alpha(p) . If m has prime factorization m = p ^{e _{1}} _{1} p ^{e _{2}} _{2} ... p ^{e _{n}} _{n} then \alpha(m) = \operatorname{lcm}(\alpha(p ^{e _{1}} _{1} ), \alpha(p ^{e _{2}} _{2} ), ... , \alpha(p ^{e _{n}} _{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 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 ⋅} n) = 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 ⋅} 4 ) = lcm (\pi _{1(3),\pi} _{1(4))=} lcm ( 8 , 6 ) =24 . Thus it suffices to compute Pisano periods for prime powers q = p ^{n.} (Usually, \pi ^{n)} _{k(p} = p ^{n-1} ⋅ \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 ^{n}} _{k)} \over {\sqrt {k ^{n} _{k} - ( -1 /\varphi ^{n} _{k)} } \over },\, where \varphi _{k} is the kth metallic mean\varphi _{k} = (k + \sqrt{k ^{2+4})/() }.

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 ^{n} _{k} ) 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 ⋅ \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

( 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 π_k(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)

n	multiples	other cycles	number of cycles (including the original Fibonacci cycles)
1			1
2	0		2
3	0		2
4	0, 022	033213	4
5	0	1342	3
6	0, 0224 0442, 033		4
7	0	02246325 05531452, 03362134 04415643	4
8	0, 022462, 044, 066426	033617 077653, 134732574372, 145167541563	8
9	0, 0336 0663	022461786527 077538213472, 044832573145 055167426854	5
10	0, 02246 06628 08864 04482, 055, 2684	134718976392	6
11	0	02246X5492, 0336942683, 044819X874, 055X437X65, 0661784156, 0773X21347, 0885279538, 0997516729, 0XX986391X, 14593, 18964X3257, 28X76	14
12	0, 02246X42682X 0XX8628X64X2, 033693, 0448 0884, 066, 099639	07729E873X1E 0EEX974E3257, 1347E65E437X538E761783E2, 156E5491XE98516718952794	10

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, ...

External links

• The Fibonacci sequence modulo m
• A research for Fibonacci numbers
• Fibonacci sequence starts with q, r modulo m
  • , a video with Dr. James Grime and the University of Nottingham

Notes and References

1. http://matwbn.icm.edu.pl/ksiazki/aa/aa16/aa1621.pdf On Arithmetical functions related to the Fibonacci numbers
2. http://www.theoremoftheday.org/Binomial/PeriodicFib/TotDPeriodic.pdf A Theorem on Modular Fibonacci Periodicity
3. A001175. graph. Graph of the cycles modulo 1 to 24. Each row of the image represents a different modulo base n, from 1 at the bottom to 24 at the top. The columns represent the Fibonacci numbers mod n, from F(0) mod n at the left to F(59) mod n on the right. In each cell, the brightness indicates the value of the residual, from dark for 0 to near-white for n−1. Blue squares on the left represent the first period; the number of blue squares is the Pisano number.
4. Web site: The Fibonacci Sequence Modulo M, by Marc Renault. webspace.ship.edu. 2018-08-22.