Stirling numbers, including negative arguments

Stirling numbers are something like binomial coefficients. They come in two varieties, imaginatively called the first kind and second kind. Unfortunately it is the second kind that are simpler to describe and that come up more often in applications, so we’ll start there.

Stirling numbers of the second kind

The Stirling number of the second kind

S_2(n,k) = \left\{ {n \atop k} \right\}

counts the number of ways to partition a set of n elements into k non-empty subsets. The notation on the left is easier to use inline, and the subscript reminds us that we’re talking about Stirling numbers of the second kind. The notation on the right suggests that we’re dealing with something analogous to binomial coefficients, and the curly braces suggest this thing might have something to do with counting sets.

Since the nth Bell number counts the total number of ways to partition a set of n elements into any number of non-empty subsets, we have

B_n = \sum_{k=1}^n \left\{ {n \atop k}\right\}

Another connection to Bell is that S2(n, k) is the sum of the coefficients in the partial exponential Bell polynomial Bn, k.

Stirling numbers of the first kind

The Stirling numbers of the first kind

S_1(n,k) = \left[ {n \atop k} \right]

count how many ways to partition a set into cycles rather than subsets. A cycle is a sort of ordered subset. The order of elements matters, but only a circular way. A cycle of size k is a way to place k items evenly around a circle, where two cycles are considered the same if you can rotate one into the other. So, for example, [1, 2, 3] and [2, 3, 1] represent the same cycle, but [1, 2, 3] and [1, 3, 2] represent different cycles.

Since a set with at least three elements can be arranged into multiple cycles, Stirling numbers of the first kind are greater than or equal to Stirling numbers of the second kind, given the same arguments.

Addition identities

We started out by saying Stirling numbers were like binomial coefficients, and here we show that they satisfy addition identities similar to binomial coefficients.

For binomial coefficients we have

{n \choose k} = {n - 1 \choose k} + {n-1 \choose k-1}.

To see this, imagine we start with the set of the numbers 1 through n. How many ways can we select a subset of k items? We have selections that exclude the number 1 and selections that include the number 1. These correspond to the two terms on the right side of the identity.

The analogous identities for Stirling numbers are

\left\{{n \atop k} \right\}= k \left\{ {n-1 \atop k} \right\} + \left\{ {n-1 \atop k-1} \right\}

\left[ {n \atop k} \right]= (n-1) \left[ {n-1 \atop k} \right] + \left[ {n-1 \atop k-1} \right]
The combinatorial proofs of these identities are similar to the argument above for binomial coefficients. If you want to partition the numbers 1 through n into k subsets (or cycles), either 1 is in a subset (cycle) by itself or not.

More general arguments

Everything above has implicitly assumed n and k were positive, or at least non-negative, numbers. Let’s look first at how we might handle zero arguments, then negative arguments.

It works out best if we define S1(0, k) and S2(0, k) to be 1 if k is 0 and zero otherwise. Also we define S1(n, 0) and S2(n, 0) to be 1 if n is 0 and zero otherwise. (See Concrete Mathematics.)

When either n or k is zero the combinatoric interpretation is strained. If we let either be negative, there is no combinatorial interpretation, though it can still be useful to consider negative arguments, much like one does with binomial coefficients.

We can take the addition theorems above, which are theorems for non-negative arguments, and treat them as definitions for negative arguments. When we do, we get the following beautiful dual relationship between Stirling numbers of the first and second kinds:

\left\{{n \atop k} \right\}= \left[ {-k \atop -n} \right]

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One thought on “Stirling numbers, including negative arguments

  1. Thanks for this very interesting article. Suspense was intense and the conclusion is mind blowing!
    Did you mean [1,2,3] instead of [1,3,3] ?

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