Stirling numbers of the first kind


The Stirling numbers of the first kind, frequently denoted as


are the integer coefficients of the falling factorialMathworldPlanetmath polynomials. To be more precise, the defining relation for the Stirling numbers of the first kind is:


Here is the table of some initial values.

n\k 1 2 3 4 5
1 1
2 -1 1
3 2 -3 1
4 -6 11 -6 1
5 24 -50 35 -10 1

Recurrence Relation.

The evident observation that


leads to the following equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath characterization of the s(n,k), in terms of a 2-place recurrence formula:


subject to the following initial conditions:


Generating Function.

There is also a strong connection with the generalized binomial formula, which furnishes us with the following generating functionMathworldPlanetmath:


This generating function implies a number of identitiesPlanetmathPlanetmathPlanetmath. Taking the derivative of both sides with respect to t and equating powers, leads to the recurrence relation described above. Taking the derivative of both sides with respect to x gives


This is because the derivative of the left side of the generating funcion equation with respect to x is


The relationMathworldPlanetmathPlanetmath


yields the following family of summation identities. For any given k1,k2,d1 we have


Enumerative interpretation.

The absolute valueMathworldPlanetmathPlanetmathPlanetmath of the Stirling number of the first kind, s(n,k), counts the number of permutationsMathworldPlanetmath of n objects with exactly k orbits (equivalently, with exactly k cycles). For example, s(4,2)=11, corresponds to the fact that the symmetric groupMathworldPlanetmathPlanetmath on 4 objects has 3 permutations of the form

(**)(**) — 2 orbits of size 2 each,

and 8 permutations of the form

(***) — 1 orbit of size 3, and 1 orbit of size 1,

(see the entry on cycle notation for the meaning of the above expressions.)

Let us prove this. First, we can remark that the unsigned Stirling numbers of the first are characterized by the following recurrence relation:


To see why the above recurrence relation matches the count of permutations with k cycles, consider forming a permutation of n+1 objects from a permutation of n objects by adding a distinguished object. There are exactly two ways in which this can be accomplished. We could do this by forming a singleton cycle, i.e. leaving the extra object alone. This accounts for the s(n,k-1) term in the recurrence formula. We could also insert the new object into one of the existing cycles. Consider an arbitrary permutation of n object with k cycles, and label the objects a1,,an, so that the permutation is represented by

(a1aj1)(aj1+1aj2)(ajk-1+1an)k cycles.

To form a new permutation of n+1 objects and k cycles one must insert the new object into this array. There are, evidently n ways to perform this insertion. This explains the ns(n,k) term of the recurrence relation. Q.E.D.

Title Stirling numbers of the first kind
Canonical name StirlingNumbersOfTheFirstKind
Date of creation 2013-03-22 12:33:44
Last modified on 2013-03-22 12:33:44
Owner rmilson (146)
Last modified by rmilson (146)
Numerical id 6
Author rmilson (146)
Entry type Definition
Classification msc 05A15
Synonym Stirling numbers
Related topic StirlingNumbersSecondKind