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  "body": "\n## Statement\n\nLet $A$ be the [[shift operator|91QMF4R1CY]] on sequences, so that $(Af)(n) = f(n+1)$. Let $r$ be a constant and let $k \\ge 1$ be an integer.\n\nConsider the recurrence $(A - r)^k f = 0$. Its [[characteristic polynomial|0BWDJGQYST]] is $(x - r)^k$, so $r$ is its only root, with multiplicity $k$.\n\nThen the general solution is\n$$f(n) = c_1 r^n + c_2 n r^n + c_3 n^2 r^n + \\dots + c_k n^{k-1} r^n,$$\nwhere $c_1, \\dots, c_k$ are arbitrary constants. That is, each sequence of this form solves the recurrence, and each solution has this form for some constants.\n\n## Remarks\n\nFor $k = 1$ the recurrence is $f(n+1) = r f(n)$, and the solutions are $c_1 r^n$.\n\nFor $k = 2$ the solutions are $(c_1 + c_2 n) r^n$.\n\nThe shape matches the differential equation case, where a root of multiplicity $m$ gives $x^j e^{rx}$ for $0 \\le j \u003c m$.\n",
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