Math 2.0

Number theory Theorem

Recurrence for Stirling numbers of the second kind

Builds on Stirling numbers of the second kind

Statement

Let nn and kk be natural numbers, and let S(n,k)S(n, k) be the Stirling numbers of the second kind. Then S(n+1,k+1)=(k+1) S(n,k+1)+S(n,k).S(n + 1, k + 1) = (k + 1)\, S(n, k + 1) + S(n, k).

Remarks

Proof idea: look at the element n+1n + 1. Either it forms a block by itself, and the rest is a partition into kk blocks. Or it joins one of the k+1k + 1 blocks of a partition of the rest. Example: S(4,2)=2 S(3,2)+S(3,1)=2⋅3+1=7S(4, 2) = 2\, S(3, 2) + S(3, 1) = 2 \cdot 3 + 1 = 7.

Reading path

To understand this page from the ground up, read these 2 pages in order. The foundations come first.

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Stirling numbers of the second kind, definition, no evidence levelStirling numbers of t…definitionRecurrence for Stirling numbers of the second kind, statement, no evidence levelRecurrence for Stirli…statement
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  1. Stirling numbers of the second kind definition
  2. Recurrence for Stirling numbers of the second kind statement

Further reading

  1. Web https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind, Recurrence relation
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Tag0C019SDTER
URIm2://0C019SDTER
Kindstatement, theorem
WritingW0 stub
MSC11B73
StatusThis copy of the site has no build data. Run m2 status 0C019SDTER.
UsesStirling numbers of the second kind

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