Pisano period

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In number theory, the nth Pisano period, written π(n), is the period with which the sequence of Fibonacci numbers, modulo n repeats. For example, the Fibonacci numbers modulo 3 are 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, etc., with the first eight numbers repeating, so π(3) = 8.

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][2]

Tables[edit]

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 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, 16, 30, 48, 24, 100, 84, 72, 48, 14, 120, 30, 48, 40, 36, 80, 24, 76, 18, 56, 60, 40, 48, 88, 30, 120, 48, 32, 24, 112, 300, ...

For n > 2 the period is even, because alternatingly F(n)2 is one more and one less than F(n − 1)F(n + 1) (Cassini's identity).

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

Generalizations[edit]

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

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

1, 1, 3, 1, 24, 3, 48, 1, 9, 24, 10, 3, 12, 48, 24, 1, 144, 9, 180, 24, 48, 10, 22, 3, 120, 12, 27, 48, 840, 24, 320, 1, 30, 144, 48, 9, 36, 180, 12, 24, ... (sequence A175289 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, ... (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, ... (sequence A106295 in OEIS)

Number theory[edit]

Pisano periods can be analyzed using algebraic number theory.

Let \pi_k(n) be the n-th Pisano period of the k-Fibonacci sequence, 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)

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}-(-k/\varphi_k)^{n}} \over {\sqrt {k^2+4}}},\, where \varphi_k\, is the kth expensive metal ratio
\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 n = 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\left(k\right) \equiv 3\cdot \left(8^k - 4^k\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 π(3) = 8 which equals 32 − 1 = 8; for p = 7, one has π(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. For p divides the squarefree part of k2+4, the Pisano period is π(k2+4) = p2-p = p(p − 1), which does not divide p − 1 or p2 − 1.

For the n-Fibonacci sequence, we should check whether the n-th term of this sequence

5, 2, 13, 5, 29, 10, 53, 17, 85, 26, 5, 37, 173, 2, 229, 65, 293, 82, 365, 101, 445, 122, 533, 145, 629, 170, 733, 197, 5, 226, ... (sequence A013946 in OEIS)

is a quadratic residue mod p or not, if so, the period must divide p − 1, if not, it must divide 2p+2, and this analysis fails for p = 2 or while p divides the n-th term of last sequence.

Sums[edit]

Using:

\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:

\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:

\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}

Powers of 10[edit]

The Pisano periods when n is a power of 10 are 60, 300, 1500, 15000, 150000, ... (sequence A096363 in OEIS). Dov Jarden proved that for n greater than 2 the periodicity mod 10n is 15·10n−1.[1]

Cultural references[edit]

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[edit]

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

References[edit]

  1. ^ a b Weisstein, Eric W., "Pisano Period", MathWorld.
  2. ^ On Arithmetical functions related to the Fibonacci numbers. Acta Arithmetica XVI (1969). Retrieved 22 September 2011.

External links[edit]