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 103X.[1][2]

## 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, 2484, 4636 ...

defined by the recurrence relation

${\displaystyle F_0 = 0}$
${\displaystyle F_1=1}$
${\displaystyle F_{i}=F_{i-1}+F_{i-2}.}$

For any integer n, the sequence of Fibonacci numbers Fi taken modulo n is periodic. The Pisano period, denoted Template:Pi(n), is the length of the period of this sequence. For example, the sequence of Fibonacci numbers modulo 10 begins:

0, 1, 1, 2, 3, 5, 8, 1, 9, X, 7, 5, 0, 5, 5, X, 3, 1, 4, 5, 9, 2, E, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, X, 7, 5, 0, 5, 5, X, 3, 1, 4, 5, 9, 2, E, 1, 0, ...

This sequence has period 20, so Template:Pi(10) = 20.

## Properties

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

${\displaystyle \mathbf {Q} ={\begin{bmatrix}1&1\\1&0\end{bmatrix}}}$

in the general linear group GL2(ℤn) of invertible 2 by 2 matrices in the finite ring ℤn of integers modulo n. Since Q has determinant -1, the determinant of QTemplate:Pi(n) is (-1)Template:Pi(n), and since this must = 1 in ℤn, either n≤2 or Template:Pi(n) is even.[3]

If m and n are coprime, then Template:Pi(mn) is the least common multiple of Template:Pi(m) and Template:Pi(n), by the Chinese remainder theorem. For example, Template:Pi(3) = 8 and Template:Pi(4) = 6 imply Template:Pi(10) = 20. Thus the study of Pisano periods may be reduced to that of Pisano periods of prime powers q = pk, for k ≥ 1.

If p is prime, Template:Pi(pk) divides pk–1 Template:Pi(p). It is conjectured that

${\displaystyle \pi (p^{k})=p^{k-1}\pi (p)}$ for every prime p and integer k > 1. Any prime p providing a counterexample would necessarily be a Wall-Sun-Sun prime, and such primes are also conjectured not to exist.[4]

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 = 2k, then Template:Pi(n) = 3·2k–1 = Template:Sfrac = Template:Sfrac.
• if n = 5k, then Template:Pi(n) = 18·5k–1 = Template:Sfrac = 4n.

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

The remaining primes all lie in the conjugacy classes

${\displaystyle p\equiv \pm 1{\pmod {\mathcal {X}}}}$ or

${\displaystyle p\equiv \pm 3{\pmod {\mathcal {X}}}}$ . If p is a prime different from 2 and 5, then the modulo p analogue of Binet's formula implies that Template:Pi(p) is the multiplicative order of the roots of Template:Math modulo p. If

${\displaystyle p\equiv \pm 1{\pmod {\mathcal {X}}}}$ , these roots belong to

${\displaystyle \mathbb {F} _{p}=\mathbb {Z} /p\mathbb {Z} }$ (by quadratic reciprocity). Thus their order, Template:Pi(p) is a divisor of p – 1. For example, Template:Pi(E) = E – 1 = X and Template:Pi(25) = (25 – 1)/2 = 12. If

${\displaystyle p\equiv \pm 3{\pmod {\mathcal {X}}},}$ the roots modulo p of Template:Math do not belong to

${\displaystyle \mathbb{F}_p }$ (by quadratic reciprocity again), and belong to the finite field

${\displaystyle \mathbb {F} _{p}[x]/(x^{2}-x-1).}$

As the Frobenius automorphism

${\displaystyle x\mapsto x^{p}}$ exchanges these roots, it follows that, denoting them by r and s, we have rp = s, and thus rp+1 = –1. That is r2(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 Template:Pi(p) is smaller than 2(p+1), are Template:Pi(3E) = 2(3E + 1)/3 = 28, Template:Pi(8E) = 2(8E + 1)/3 = 60 and Template:Pi(95) = 2(95 + 1)/3 = 64. (See the table below)

It follows from above results, that if n = pk is an odd prime power such that Template:Pi(n) > n, then Template:Pi(n)/4 is an integer that is not greater than n. The multiplicative property of Pisano periods imply thus that

Template:Pi(n) ≤ 6n,

with equality if and only if n = 2 · 5r, for r ≥ 1.[5] The first examples are Template:Pi(X) = 50 and Template:Pi(42) = 210. If n is not of the form 2 · 5r, then Template:Pi(n) ≤ 4n.

## Tables

The first 10 Pisano periods Template:OEIS and their cycles (with spaces before the zeros for readability) are:[6]

n π(n) number of zeros in the cycle (Template:Oeis) cycle (Template:Oeis) OEIS sequence for the cycle
1 1 1 0 Template:OEIS link
2 3 1 011 Template:OEIS link
3 8 2 0112 0221 Template:OEIS link
4 6 1 011231 Template:OEIS link
5 18 4 01123 03314 04432 02241 Template:OEIS link
6 20 2 011235213415 055431453251 Template:OEIS link
7 14 2 01123516 06654261 Template:OEIS link
8 10 2 011235 055271 Template:OEIS link
9 20 2 011235843718 088764156281 Template:OEIS link
X 50 4 011235831459437 077415617853819 099875279651673 033695493257291 Template:OEIS link
E X 1 01123582X1 Template:OEIS link
10 20 2 011235819X75 055X314592E1 Template:OEIS link

The first 100 Pisano periods are shown in the following table:

π(n) +1 +2 +3 +4 +5 +6 +7 +8 +9 +X +E +10
0+ 1 3 8 6 18 20 14 10 20 50 X 20
10+ 24 40 34 20 30 20 16 50 14 26 40 20
20+ 84 70 60 40 12 X0 26 40 34 30 68 20
30+ 64 16 48 50 34 40 74 26 X0 40 28 20
40+ 94 210 60 70 90 60 18 40 60 36 4X X0
50+ 50 26 40 80 E8 X0 E4 30 40 180 5X 20
60+ 104 170 148 16 68 120 66 X0 160 X0 120 40
70+ 130 1X0 48 50 38 X0 94 40 X0 80 130 40
80+ 144 240 X0 210 42 60 154 70 68 90 60 60
90+ 90 50 108 40 64 60 180 36 120 126 100 X0
X0+ 92 50 34 26 358 40 194 140 74 2E0 XX X0
E0+ 100 2X0 260 30 1E0 40 3X 180 28 156 E8 20

## 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 2 zeros. If the base is a Fibonacci number (≥5) with an odd index, the period is 4 times the index and the cycle has 4 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 18 0, 1, 1, 2, 3, (0, 3, 3, 1, 4)
6 8 10 0, 1, 1, 2, 3, 5, (0, 5, 5, 2, 7, 1)
7 11 24 0, 1, 1, 2, 3, 5, 8, (0, 8, 8, 3, E, 1, 10)
8 19 14 0, 1, 1, 2, 3, 5, 8, 11, (0, 11, 11, 5, 16, 2, 18, 1)
9 2X 30 0, 1, 1, 2, 3, 5, 8, 11, 19, (0, 19, 19, 8, 25, 3, 28, 1, 29)
X 47 18 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, (0, 2X, 2X, 11, 3E, 5, 44, 2, 46, 1)
E 75 38 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, (0, 47, 47, 19, 64, 8, 70, 3, 73, 1, 74)
10 100 20 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, (0, 75, 75, 2X, X3, 11, E4, 5, E9, 2, EE, 1)
11 175 44 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100
12 275 24 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175
13 42X 50 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275
14 6X3 28 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X
15 E11 58 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3
16 15E4 30 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11
17 2505 64 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4
18 3XE9 34 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505
19 6402 70 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505, 3XE9
1X X2EE 38 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505, 3XE9, 6402
1E 14701 78 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505, 3XE9, 6402, X2EE
20 22X00 40 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505, 3XE9, 6402, X2EE, 14701

## 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 4 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 14 0, 1, 1, 2, (3, 5, 1, 6)
5 E X 0, 1, 1, 2, 3, (5, 8, 2, X, 1)
6 16 20 0, 1, 1, 2, 3, 5, (8, 11, 3, 14, 1, 15)
7 25 12 0, 1, 1, 2, 3, 5, 8, (11, 19, 5, 22, 2, 24, 1)
8 3E 28 0, 1, 1, 2, 3, 5, 8, 11, (19, 2X, 8, 36, 3, 39, 1, 3X)
9 64 16 0, 1, 1, 2, 3, 5, 8, 11, 19, (2X, 47, 11, 58, 5, 61, 2, 63, 1)
X X3 34 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, (47, 75, 19, 92, 8, 9X, 3, X1, 1, X2)
E 147 1X 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, (75, 100, 2X, 12X, 11, 13E, 5, 144, 2, 146, 1)
10 22X 40 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, (100, 175, 47, 200, 19, 219, 8, 225, 3, 228, 1, 229)
11 375 22 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100
12 5X3 48 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175
13 958 26 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275
14 133E 54 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X
15 2097 2X 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3
16 3416 60 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11
17 54E1 32 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4
18 8907 68 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505
19 121E8 36 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505, 3XE9
1X 1XE03 74 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505, 3XE9, 6402
1E 310EE 3X 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505, 3XE9, 6402, X2EE
20 50002 80 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505, 3XE9, 6402, X2EE, 14701

For even k, the cycle has 2 zeros. For odd k, the cycle has only 1 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.

## Pisano periods of powers of 10

The period of the last n digits of the Fibonacci numbers is π(10n), and π(10n) for n = 1, 2, 3, ... are 20, 20, 200, 2000, 20000, 200000, 2000000, 20000000, 200000000, 2000000000, 20000000000, 200000000000, .... For n > 1, π(10n) = 2×10n−1. (thus, π(10n) = 2×10max(n−1, 1) for all n ≥ 1)

## 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.

The number of zeros in the cycle for Fibonacci sequence mod n are:

+1 +2 +3 +4 +5 +6 +7 +8 +9 +X +E +10
0+ 1 1 2 1 4 2 2 2 2 4 1 2
10+ 4 2 2 2 4 2 1 2 2 1 2 2
20+ 4 4 2 2 1 2 1 2 2 4 2 2
30+ 4 1 2 2 2 2 2 1 2 2 2 2
40+ 2 4 2 2 4 2 2 2 2 1 1 2
50+ 4 1 2 2 4 2 2 2 2 2 1 2
60+ 4 4 2 1 2 2 1 2 2 2 2 2
70+ 4 2 2 2 4 2 2 2 2 2 2 2
80+ 4 2 2 2 1 2 2 2 2 4 2 2
90+ 4 2 2 2 4 2 2 1 2 1 2 2
X0+ 1 4 2 1 4 2 2 2 2 4 1 2
E0+ 2 2 2 2 4 2 1 2 2 1 2 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 +2 +3 +4 +5 +6 +7 +8 +9 +X +E +10
0+ 1 3 4 6 5 10 8 6 10 13 X 10
10+ 7 20 18 10 9 10 16 26 8 26 20 10
20+ 21 19 30 20 12 50 26 20 18 9 34 10
30+ 17 16 24 26 18 20 38 26 50 20 14 10
40+ 48 63 30 36 23 30 X 20 30 36 4X 50
50+ 13 26 20 40 2E 50 58 16 20 X0 5X 10
60+ 31 49 84 16 34 70 66 50 90 50 70 20
70+ 39 E0 24 26 E 50 48 20 50 40 76 20
80+ 41 120 50 106 42 30 88 36 34 23 30 30
90+ 23 26 64 20 17 30 X0 36 70 126 60 50
X0+ 92 13 18 26 X5 20 X8 80 38 89 XX 50
E0+ 60 150 130 16 59 20 3X X0 14 156 5X 10

In Renault's paper the number of zeros is called the "order" of F mod m, denoted

${\displaystyle \omega (m)}$ , and the "rank of apparition" is called the "rank" and denoted

${\displaystyle \alpha (m)}$ .[7] According to Wall's conjecture,

${\displaystyle \alpha (p^{e})=p^{e-1}\alpha (p)}$ . If

${\displaystyle m}$ has prime factorization

${\displaystyle m=p_{1}^{e_{1}}p_{2}^{e_{2}}\dots p_{n}^{e_{n}}}$ then

${\displaystyle \alpha (m)=\operatorname {lcm} (\alpha (p_{1}^{e_{1}}),\alpha (p_{2}^{e_{2}}),\dots ,\alpha (p_{n}^{e_{n}}))}$ [7]

## Number theory

Pisano periods can be analyzed using algebraic number theory.

If m and n are coprime, then

${\displaystyle \pi (m\cdot n)=\mathrm {lcm} (\pi (m),\pi (n)),}$ 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,

${\displaystyle \pi (3)=8}$ and

${\displaystyle \pi (4)=6,}$ so

${\displaystyle \pi (10=3\cdot 4)=\mathrm {lcm} (\pi (3),\pi (4))=\mathrm {lcm} (8,6)=20.}$ Thus it suffices to compute Pisano periods for prime powers

${\displaystyle q=p^{k}.}$

For prime numbers p, these can be analyzed by using Binet's formula:

${\displaystyle F\left(k\right)={{\varphi ^{k}-(1-\varphi )^{k}} \over {\sqrt {5}}}={{\varphi ^{k}-(-1/\varphi )^{k}} \over {\sqrt {5}}},\,}$where${\displaystyle \varphi\,}$ is the golden ratio
${\displaystyle \varphi ={\frac {1+{\sqrt {5}}}{2}}.}$

If 5 is a quadratic residue modulo p (and

${\displaystyle p>2}$ ), then

${\displaystyle {\sqrt {5}},1/2,}$ and

${\displaystyle 1/{\sqrt {5}}}$ 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

${\displaystyle \phi (p)=p-1}$ , since any power (such as

${\displaystyle \varphi^k}$ ) has period dividing

${\displaystyle \phi (p),}$ as this is the order of the group of units modulo p.

This first occurs for n = E, where 42 = 14 ≡ 5 (mod E) and 2 · 6 = 10 ≡ 1 (mod E) and 4 · 3 = 10 ≡ 1 (mod E) so 4 = √5, 6 = 1/2 and 1/√5 = 3, yielding φ = (1 + 4) · 6 = 30 ≡ 8 (mod E) and the congruence

${\displaystyle F\left(k\right)\equiv 3\cdot \left(8^{k}-4^{k}\right){\pmod {\mathcal {E}}}.}$

Another example, which shows that the period can properly divide p − 1, is π(25) = 12.

If 5 is not a quadratic residue (and p ≠ 2, 5), then Binet's formula is instead defined over the quadratic extension field (Z/p)[√5], 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 π(3) = 8 which equals 32 − 1 = 8; for p = 7, one has π(7) = 14, which properly divides 72 − 1 = 40.

This analysis fails for p = 2 and p = 5 since in these cases 2 and 5 are zero divisors, so one must be careful in interpreting 1/2 or √5. For p = 2, 5 is congruent to 1 mod 2, but the Pisano period is not p − 1 = 1, but rather 3. For p = 5, the Pisano period is π(5) = 18 = 5(5 − 1), which does not divide p − 1 = 4 or p2 − 1 = 20.

## Sums

Using:

${\displaystyle \sum _{i=n}^{n}F_{i}=F_{n+2}-1}$,

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

${\displaystyle \sum _{i=n}^{\pi (n)}F_{i}=nk}$

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

${\displaystyle \sum _{i=n}^{n+5}F_{i}=4F_{n+4}}$
${\displaystyle \sum _{i=n}^{n+9}F_{i}={\mathcal {E}}F_{n+6}}$
${\displaystyle \sum _{i=n}^{n+11}F_{i}=25F_{n+8}}$
${\displaystyle \sum _{i=n}^{n+15}F_{i}=64F_{n+{\mathcal {X}}}}$
${\displaystyle \sum _{i=n}^{n+19}F_{i}=147F_{n+10}}$

## 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 = X 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)

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
X 0, 02246 06628 08864 04482, 055, 2684 134718976392 6
E 0 02246X5492, 0336942683, 044819X874, 055X437X65, 0661784156, 0773X21347, 0885279538, 0997516729, 0XX986391X, 14593, 18964X3257, 28X76 12
10 0, 02246X42682X 0XX8628X64X2, 033693, 0448 0884, 066, 099639 07729E873X1E 0EEX974E3257, 1347E65E437X538E761783E2, 156E5491XE98516718952794 X

Number of Fibonacci integer cycles mod n are:

1, 2, 2, 4, 3, 4, 4, 8, 5, 6, 12, X, 7, 8, 10, 14, 9, 14, 1X, 14, 25, 24, 10, 26, 11, 12, 12, 1X, 53, 20, 2X, 28, 33, 2X, 26, 4X, 17, 72, 28, 44, 37, 4X, 1X, 66, 33, 3X, 5X, 86, ... Template:OEIS

## Generalization

The Pisano periods of Pell numbers (or 2-Fibonacci numbers) are

+1 +2 +3 +4 +5 +6 +7 +8 +9 +X +E +10
0+ 1 2 8 4 10 8 6 8 20 10 20 8
10+ 24 6 20 14 14 20 34 10 20 20 1X 8
20+ 50 24 60 10 18 20 26 28 20 14 10 20
30+ 64 34 48 20 X 20 74 20 20 1X 3X 14
40+ 36 50 14 24 90 60 20 20 34 18 34 20
50+ X4 26 20 54 70 20 E4 14 74 10 5X 20
60+ 60 64 X0 34 20 48 22 40 160 X 120 20
70+ 40 74 34 20 74 20 70 38 X0 3X X0 28
80+ 80 36 20 50 150 14 2X 48 20 90 160 60
90+ 164 20 108 40 48 34 E0 18 120 34 40 20
X0+ 1X0 X4 34 50 210 20 X6 X8 74 70 1X0 20
E0+ X0 E4 60 14 58 74 1E4 10 134 5X 120 40

The Pisano periods of 3-Fibonacci numbers are

+1 +2 +3 +4 +5 +6 +7 +8 +9 +X +E +10
0+ 1 3 2 6 10 6 14 10 6 10 8 6
10+ 44 40 10 20 14 6 34 10 14 20 1X 10
20+ 50 110 16 40 24 10 54 40 8 40 40 6
30+ 64 X0 44 10 24 40 36 20 10 56 80 20
40+ 94 50 14 110 22 16 20 40 34 70 20 10
50+ 26 140 40 80 110 20 E4 40 1X 40 100 10
60+ 104 170 50 X0 14 110 22 20 46 70 120 40
70+ 40 36 24 20 130 10 154 56 54 80 X0 40
80+ 144 240 20 50 42 40 86 110 40 66 8X 16
90+ 18 20 64 40 94 X0 E0 70 110 20 14 10
X0+ 74 26 24 140 210 40 X6 140 36 110 X 20
E0+ 68 2X0 30 40 78 56 E6 40 80 100 88 20

The Pisano periods of 4-Fibonacci numbers are

+1 +2 +3 +4 +5 +6 +7 +8 +9 +X +E +10
0+ 1 2 8 2 18 8 14 4 8 18 X 8
10+ 24 14 34 8 10 8 6 18 14 X 14 8
20+ 84 24 20 14 12 34 X 14 34 10 68 8
30+ 64 6 48 18 34 14 74 X 34 14 28 8
40+ 94 84 20 24 30 20 18 14 20 12 4X 34
50+ 18 X 14 28 E8 34 E4 10 14 68 5X 8
60+ 104 64 148 6 68 48 22 34 60 34 48 14
70+ 50 74 48 18 38 34 94 14 34 28 50 14
80+ 144 94 34 84 42 20 154 24 68 30 20 20
90+ 30 18 108 14 64 20 68 12 48 4X 40 34
X0+ 92 18 34 X 358 14 194 54 74 E8 XX 34
E0+ 40 E4 X0 10 78 14 3X 68 28 5X E8 8

The Pisano periods of Jacobsthal numbers (or (1,2)-Fibonacci numbers) are

+1 +2 +3 +4 +5 +6 +7 +8 +9 +X +E +10
0+ 1 1 6 2 4 6 6 2 16 4 X 6
10+ 10 6 10 2 8 16 16 4 6 X 1X 6
20+ 18 10 46 6 24 10 X 2 26 8 10 16
30+ 30 16 10 4 18 6 12 X 30 1X 3X 6
40+ 36 18 20 10 44 46 18 6 16 24 4X 10
50+ 50 X 16 2 10 26 56 8 56 10 5X 16
60+ 16 30 50 16 26 10 66 4 116 18 6X 6
70+ 8 12 70 X 1X 30 10 1X 26 3X 30 6
80+ 40 36 76 18 84 20 86 10 10 44 8X 46
90+ 30 18 30 6 24 16 38 24 30 4X 20 10
X0+ 92 50 50 X 84 16 12 2 36 10 XX 26
E0+ 16 56 90 8 58 56 E6 10 E6 5X 50 16

The Pisano periods of (1,3)-Fibonacci numbers are

+1 +2 +3 +4 +5 +6 +7 +8 +9 +X +E +10
0+ 1 3 1 6 20 3 20 6 3 20 X0 6
10+ 110 20 20 10 14 3 76 20 20 X0 1X 6
20+ X0 110 9 20 24 20 180 20 X0 40 20 6
30+ 123 76 110 20 240 20 36 X0 20 56 514 10
40+ 120 X0 14 110 44 9 X0 20 76 70 2020 20
50+ 18 180 20 40 220 X0 524 40 1X 20 2E00 6
60+ 620 123 X0 76 X0 110 33 20 23 240 13E4 20
70+ 40 36 24 X0 1640 20 220 56 180 1340 260 20
80+ 480 120 X0 X0 84 40 2X 110 20 110 8X 16
90+ 3530 X0 123 20 94 76 1X0 70 110 2020 40 20
X0+ 920 50 240 180 420 20 53 80 36 220 XX X0
E0+ 260 1370 60 40 3754 56 E6 20 514 2E00 XX0 10

The Pisano periods of (1,4)-Fibonacci numbers are

+1 +2 +3 +4 +5 +6 +7 +8 +9 +X +E +10
0+ 1 1 8 1 6 8 40 2 20 6 X0 8
10+ 10 40 20 4 E4 20 16 6 40 X0 380 8
20+ 26 10 60 40 2E 20 228 8 X0 E4 40 20
30+ 123 16 20 6 89 40 36 X0 20 380 3X 8
40+ 240 26 E4 10 22 60 X0 40 60 2E 4X 20
50+ 329 228 40 14 10 X0 56 E4 380 40 E80 20
60+ 930 123 X0 16 180 20 3740 10 160 89 6X 40
70+ 2X0 36 1E4 X0 74 20 40 380 228 3X 16 8
80+ 1440 240 X0 26 84 E4 86 10 40 22 6760 60
90+ 6X6 X0 960 40 293 60 380 2E 20 4X 580 20
X0+ 920 329 5X0 228 106 40 X6 28 120 10 9E20 X0
E0+ 100 56 60 E4 E4 380 22X0 40 134 E80 X0 20

The Pisano periods of (3,−1)-Fibonacci numbers are

+1 +2 +3 +4 +5 +6 +7 +8 +9 +X +E +10
0+ 1 3 4 3 X 10 8 6 10 26 5 10
10+ 12 20 18 10 16 10 9 26 8 13 20 10
20+ 42 36 30 20 7 50 13 20 18 16 34 10
30+ 32 9 24 26 18 20 38 13 50 20 14 10
40+ 48 106 30 36 46 30 X 20 30 19 25 50
50+ 26 13 20 40 5X 50 58 16 20 X0 2E 10
60+ 62 96 84 9 34 70 33 50 90 50 70 20
70+ 76 E0 24 26 1X 50 48 20 50 40 76 20
80+ 82 120 50 106 21 30 88 36 34 46 30 30
90+ 46 26 64 20 32 30 X0 19 70 73 60 50
X0+ 47 26 18 13 18X 20 X8 80 38 156 55 50
E0+ 60 150 130 16 E6 20 1E X0 14 89 5X 10

The Pisano periods of (4,−1)-Fibonacci numbers are

+1 +2 +3 +4 +5 +6 +7 +8 +9 +X +E +10
0+ 1 2 6 4 3 6 8 4 16 6 X 10
10+ 10 8 6 8 16 16 5 10 20 X E 10
20+ 13 10 46 8 13 6 28 14 26 16 20 30
30+ 30 X 10 10 12 20 E 18 16 1X 1E 20
40+ 48 26 16 10 9 46 26 8 26 26 4X 10
50+ 50 28 60 28 10 26 2X 30 56 20 7 30
60+ 30 30 26 18 34 10 68 20 116 12 6X 20
70+ 16 1X 26 18 76 16 20 38 80 3X 13 40
80+ 14 48 76 50 15 16 88 10 20 16 8X 90
90+ 90 26 30 8 96 26 29 50 30 4X 60 10
X0+ 92 50 36 28 63 60 X8 54 56 10 XX 50
E0+ 34 2X 46 30 E6 56 5X 20 E6 12 50 60

The Pisano periods of Tribonacci numbers (or 3-step Fibonacci numbers) are

+1 +2 +3 +4 +5 +6 +7 +8 +9 +X +E +10
0+ 1 4 11 8 27 44 40 14 33 X4 92 88
10+ 120 40 297 28 80 110 260 188 440 164 3X1 154
20+ 10E 120 99 40 E8 E24 237 54 9E2 80 X40 220
30+ 331 260 1320 354 3X8 440 218 308 849 1344 3X 2X8
40+ 240 438 880 120 44 330 1E82 40 2860 E8 2071 1X48
50+ 10E0 924 440 X8 3020 17X4 X67 80 41E1 X40 2E61 440
60+ 3100 1104 11EE 260 1640 1320 1980 6X8 253 3X8 1EE 440
70+ 1880 218 1078 614 4777 2970 240 2688 25X7 78 6560 594
80+ 1X01 240 2596 874 488 880 43 240 E240 44 8X0 660
90+ 6X6 3E44 3641 80 7557 2860 9E07 1E4 1320 8244 80 3894
X0+ 84X 10E0 4268 1648 547 440 3140 194 2398 3020 3388 3388
E0+ 500 3624 2123 80 XE37 14784 22X0 X40 41X EX04 5420 880