Addendum I thank David Fowler for pointing out that John Wallis exhibited “Stirling numbers of the first kind” in several formulas for figurate numbers, for example on page 162 of his Arithmetica Infinitorum …

7/6, etc. These coefficients seem to grow at an alarming rate, but eventually they settle down somewhat; Stirling's approximation tells us that am+1000 = (m + 1000)m/m! ~ (1+ 1000/m)mem/V2mm ~ 100

Stirling subset number \becomes Stirling partition number \endchange \amendpage 4a.829 new entry in the rightmost column (23.05.05) Algorithm 7.2.1.5P, 428. \endchange \amendpage 4a.831 in the …

P152 On the inversion of y-to-the-alpha times e-to-the-y by means of associated Stirling numbers.

The basic recurrence for Stirling partition numbers, m +1 d +1 =( d +1) + , (3 . 4) is based on the fact that a partition of m girls into d +1 nonempty blocks either puts the oldest girl into a ( d + 1)-block partition …

(See section 3 of my unpublication ``Poly-Bernoulli Bijections.'') @d maxn 25 /* Stirling partition numbers will be less than $2^{61}$ */ @c #include <stdio.h> #include <stdlib.h> int m,n; /* command-line …

they settle down somewhat; Stirling's approximation tells us that $a_{m+1000} =(m+1000)^m\!/m!\approx(1+1000/m)^me^m\!/\sqrt{2\pi m} \approx e^{1000+m}\!/\sqrt{2\pi m}$. So …