Concrete mathematics : a foundation for computer science

6.1 STIRLING NUMBERS 249
We can go the other way too, because Stirling cycle numbers are the
coefficients of ordinary powers that yield factorial powers:
xiT = xo.
Xi = xiI
xi = x2 + x’ ;
x” - x3 +3x2 +2x';
x" : x4 +6x3 +11x* +6x'.
We have (x+n- l).xk =xk+’ + (n - 1 )xk, so a proof like the one just given
shows that
(xfn-1)~~~’ - = (x+n-1); ril]xk = F [r;]xk.
This leads to a proof by induction of the general formula
integer n 3 0. (6.11)
(Setting x = 1 gives (6.9) again.)
But wait, you say. This equation involves rising factorial powers xK, while
(6.10) involves falling factorials xc. What if we want to express xn in terms of
ordinary powers, or if we want to express X” in terms of rising powers? Easy;
we just throw in some minus signs and get
This works because, for example, the formula
integer n > 0; (6.12)
x4 = x(x-1)(x-2)(x-3) = x4-6x3+11x2-6x
is just like the formula
XT = ~(~+1)(~+2)(x+3) = x4+6x3+11x2+6x
but with alternating signs. The general identity
(6.13)
x3 = (-ly-# (6.14)
of exercise 2.17 converts (6.10) to (6.12) and (6.11) to (6.13) if we negate x.