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  "body": "\n## Definition\n\nLet $k \\ge 1$, let $S$ be a set, and let $F \\colon S^k \\to S$ be a map.\nConsider the autonomous difference equation of order $k$\n$$x_{n+k} = F(x_n, x_{n+1}, \\dots, x_{n+k-1}), \\qquad n \\ge 0.$$\nLet $p \\ge 1$ be an integer.\nA solution $(x_n)_{n \\ge 0}$ is *periodic with period $p$* if it is a [[periodic sequence|Q3B6K9HYRG]] with period $p$, that is, $x_{n+p} = x_n$ for all $n \\ge 0$.\nIts *prime period* is its least period.\n\n## Remarks\n\nExample: for $x_{n+2} = x_{n+1} / x_n$ on $S = (0, \\infty)$ with $x_0 = a$ and $x_1 = b$, the solution is $a, b, b/a, 1/a, 1/b, a/b, a, b, \\dots$, so every solution is periodic with period $6$.\n\nNon-example: for $x_{n+1} = 2 x_n$ on $\\R$ with $x_0 = 1$, the solution $x_n = 2^n$ is strictly increasing, so $x_{n+p} \\ne x_n$ for all $p \\ge 1$ and it is not periodic.\n\nFor $k = 1$ and $F = f$, the solution starting at $x_0$ is periodic with period $p$ if and only if $f^p(x_0) = x_0$, that is, $x_0$ is a [[periodic point|T75CZ5H47N]] of $f$.\nA periodic solution of period $1$ is an equilibrium.\nFor a non-autonomous equation, one asks for periodic solutions when the equation itself is periodic in $n$.\n",
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