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 be the n-th Pisano period of the k-Fibonacci sequence Fk(n) (k can be any natural number, these sequences are defined as Fk(0) = 0, Fk(1) = 1, and for any natural number n > 1, Fk(n) = kFk(n−1) + Fk(n−2)). If m and n are coprime, then , 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, and so Thus it suffices to compute Pisano periods for prime powers (Usually, , 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, since 2412 divides F3(242).) For prime numbers p, these can be analyzed by using Binet's formula: where is the kth metallic mean If k2 + 4 is a quadratic residue modulo p (where p > 2 and p does not divide k2 + 4), then and 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 , since any power (such as ) has period dividing 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 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