# Pisano period

In number theory, the nth Pisano period, written as 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 the 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 of Fibonacci numbers F$_i$ 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 the OEIS)
This sequence has period 8, so π(3) = 8.

## Properties
### Parity

With the exception of π(2) = 3, the Pisano period π(n) is always even. A proof of this can be given by observing that π(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 $\mathbb {Z} _{n}$of integers modulo n. Since Q has determinant −1, the determinant of Q$^{π(n)}$ is (−1)$^{π(n)}$, and since this must equal 1 in $\mathbb {Z} _{n}$, either n ≤ 2 or π(n) is even.
Since $F_{0}=0$ and $F_{1}=1$ we have that $n$ divides $F_{\pi (n)}$ and $(F_{\pi (n)+1}-1)$.

### Reducing the study to prime numbers

If m and n are coprime, then π(mn) is the least common multiple of π(m) and π(n), by the Chinese remainder theorem. For example, π(3) = 8 and π(4) = 6 imply π(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, π(p$^k$) divides p$^{k–1}$ π(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).

### Study of prime numbers

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 π(n) = 3·2$^{k–1}$ = ⁠3·2$^k$/2⁠ = ⁠3n/2⁠.
* if n = 5$^k$, then π(n) = 20·5$^{k–1}$ = ⁠20·5$^k$/5⁠ = 4n.
From these it follows that if n = 2 · 5$^k$ then π(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 π(p) is the multiplicative order of a root of x$^2$ − x − 1 modulo p.
If $p\equiv \pm 1{\pmod {10}}$, these roots belong to $\mathbb {F} _{p}=\mathbb {Z} /p\mathbb {Z}$ (by quadratic reciprocity). Thus their order, π(p) is a divisor of p − 1. For example, π(11) = 11 − 1 = 10 and π(29) = (29 − 1)/2 = 14.
If $p\equiv \pm 3{\pmod {10}},$ the roots modulo p of x$^2$ − x − 1 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. This quotient is always a multiple of 4. The first examples of such a p, for which π(p) is smaller than 2(p+1), are π(47) = 2(47 + 1)/3 = 32, π(107) = 2(107 + 1)/3 = 72 and π(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 π(n) > n, then π(n)/4 is an integer that is not greater than n. The multiplicative property of Pisano periods imply thus that
    π(n) ≤ 6n, with equality if and only if n = 2 · 5$^r$, for r ≥ 1.
The first examples are π(10) = 60 and π(50) = 300. If n is not of the form 2 · 5$^r$, then π(n) ≤ 4n.

## Tables

The first twelve Pisano periods (sequence A001175 in the OEIS) and their cycles (with spaces before the zeros for readability) are (using X and E for ten and eleven, respectively):
| n | π(n) | number of zeros in the cycle (OEIS: A001176) | cycle (OEIS: A161553) | OEIS sequence for the cycle |
| 1 | 1 | 1 | 0 | A000004 |
| 2 | 3 | 1 | 011 | A011655 |
| 3 | 8 | 2 | 0112 0221 | A082115 |
| 4 | 6 | 1 | 011231 | A079343 |
| 5 | 20 | 4 | 01123 03314 04432 02241 | A082116 |
| 6 | 24 | 2 | 011235213415 055431453251 | A082117 |
| 7 | 16 | 2 | 01123516 06654261 | A105870 |
| 8 | 12 | 2 | 011235 055271 | A079344 |
| 9 | 24 | 2 | 011235843718 088764156281 | A007887 |
| 10 | 60 | 4 | 011235831459437 077415617853819 099875279651673 033695493257291 | A003893 |
| 11 | 10 | 1 | 01123582X1 | A105955 |
| 12 | 24 | 2 | 011235819X75 055X314592E1 | A089911 |
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 p$^2$ ≡ 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, ... (sequence A001177 in the OEIS)
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, ... (sequence A106291 in the OEIS)
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, ... (sequence A175181 in the 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, ... (sequence A175182 in the 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, ... (sequence A175286 in the 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, ... (sequence A175291 in the 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, ... (sequence A046738 in the 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, ... (sequence A106295 in the 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, p$^2$ 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 241$^2$ divides F$_3$(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 k$^2$ + 4 is a quadratic residue modulo p (where p > 2 and p does not divide k$^2$ + 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 4$^2$ = 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 π$_1$(29) = 14.
If k$^2$ + 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 p$^2$ elements and whose group of units thus has order p$^2$ − 1, and thus the Pisano period divides p$^2$ − 1. For example, for p = 3 one has π$_1$(3) = 8 which equals 3$^2$ − 1 = 8; for p = 7, one has π$_1$(7) = 16, which properly divides 7$^2$ − 1 = 48.
This analysis fails for p = 2 and p is a divisor of the squarefree part of k$^2$ + 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, k$^2$ + 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 k$^2$ + 4, the Pisano period is π$_k$(k$^2$ + 4) = p$^2$ − p = p(p − 1), which does not divide p − 1 or p$^2$ − 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, ... (sequence A015134 in the OEIS)