Math 2.0

statement theorem

Recurrence for 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.

Explanations

L1 human

Insert a new element into a permutation written as cycles. It forms a new cycle of length 11, or it goes into an existing cycle right after one of the nn old elements.

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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

Sources

  1. Web https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind, Recurrence relation