# Stirling numbers of the first kind

This page is a solution to the task Stirling numbers of the first kind in the Rosetta Code, written in Fōrmulæ.

## Contents

### Description (from Rosetta Code)

 Stirling numbers of the first kind, or Stirling cycle numbers, count permutations according to their number of cycles (counting fixed points as cycles of length one). They may be defined directly to be the number of permutations of n elements with k disjoint cycles. Stirling numbers of the first kind express coefficients of polynomial expansions of falling or rising factorials. Depending on the application, Stirling numbers of the first kind may be "signed" or "unsigned". Signed Stirling numbers of the first kind arise when the polynomial expansion is expressed in terms of falling factorials; unsigned when expressed in terms of rising factorials. The only substantial difference is that, for signed Stirling numbers of the first kind, values of S1(n, k) are negative when n + k is odd. Stirling numbers of the first kind follow the simple identities: ``` S1(0, 0) = 1 S1(n, 0) = 0 if n > 0 S1(n, k) = 0 if k > n S1(n, k) = S1(n - 1, k - 1) + (n - 1) * S1(n - 1, k) # For unsigned or S1(n, k) = S1(n - 1, k - 1) - (n - 1) * S1(n - 1, k) # For signed ``` Task Write a routine (function, procedure, whatever) to find Stirling numbers of the first kind. There are several methods to generate Stirling numbers of the first kind. You are free to choose the most appropriate for your language. If your language has a built-in, or easily, publicly available library implementation, it is acceptable to use that. Using the routine, generate and show here, on this page, a table (or triangle) showing the Stirling numbers of the first kind, S1(n, k), up to S1(12, 12). it is optional to show the row / column for n == 0 and k == 0. It is optional to show places where S1(n, k) == 0 (when k > n). You may choose to show signed or unsigned Stirling numbers of the first kind, just make a note of which was chosen. If your language supports large integers, find and show here, on this page, the maximum value of S1(n, k) where n == 100. See also Related Tasks

### Solution

A single solution for signed and unsigned versions is presented. Unsigned version uses 1 as factor, signed version used -1.

#### Recursive

Table up to S1(12, 12), unsigned version:

Table up to S1(12, 12), signed version:

#### Non recursive

A faster, non recursive version is presented:

Table up to S1(12, 12), unsigned version:

(The result is the same as recursive version)

Table up to S1(12, 12), signed version:

(The result is the same as recursive version)

#### Extra

Maximum value of S1(100, k) for k ≤ 100, unsigned version:

Maximum value of S1(100, k) for k ≤ 100, signed version: