Math 2.0

Number theory Theorem

Recurrence for Stirling numbers of the first kind

Builds on Stirling numbers of the first kind

Statement

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

Remarks

Proof idea: look at the element n+1n + 1 in a permutation of n+1n + 1 elements. Either it is a fixed point, and the rest has kk cycles. Or it sits right after one of the nn other elements in its cycle. Example: c(4,2)=3 c(3,2)+c(3,1)=3⋅3+2=11c(4, 2) = 3\, c(3, 2) + c(3, 1) = 3 \cdot 3 + 2 = 11.

Reading path

To understand this page from the ground up, read these 3 pages in order. The foundations come first.

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Permutation, definition, no evidence levelPermutationdefinitionStirling numbers of the first kind, definition, no evidence levelStirling numbers of t…definitionRecurrence for Stirling numbers of the first kind, statement, no evidence levelRecurrence for Stirli…statement
Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.
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  1. Permutation definition
  2. Stirling numbers of the first kind definition
  3. Recurrence for Stirling numbers of the first kind statement

Further reading

  1. Web https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind, Recurrence relation
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Kindstatement, theorem
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MSC11B73
StatusThis copy of the site has no build data. Run m2 status 0C5BFX94J9.
UsesStirling numbers of the first kind

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