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  "body": "\n## Statement\n\nLet $n$ and $k$ be natural numbers, and let $S(n, k)$ be the [[Stirling numbers of the second kind|0ZN07RQ3JS]].\nThen\n$$S(n + 1, k + 1) = (k + 1)\\, S(n, k + 1) + S(n, k).$$\n\n## Remarks\n\nProof idea: look at the element $n + 1$. Either it forms a block by itself, and the rest is a partition into $k$ blocks. Or it joins one of the $k + 1$ blocks of a partition of the rest.\nExample: $S(4, 2) = 2\\, S(3, 2) + S(3, 1) = 2 \\cdot 3 + 1 = 7$.\n\n## Explanations\n\n### L1 human\n\nAdd a new element to a partition problem.\nThe new element is alone, or it joins one of the existing $k + 1$ groups, and these two cases give the two terms.\n",
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  "name": "Recurrence for Stirling numbers of the second kind",
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