Concrete mathematics : a foundation for computer science

6.5 BERNOULLI NUMBERS 275
Here 01’ = i, and 0~2, . . . , CX~,,,/~I are appropriate complex numbers whose
values depend on m. For example,
Ss(n) = n2(n- 1)2/4;
&t(n) = n(n-t)(n-l)(n- t + m)(n - t - fl)/5;
Ss(n) = n’(n-l)‘(n- i + m)(n- i - m)/6;
Ss(n) = n(n-$)(n-l)(n-i + (x)(n-5 - Ix)(n--t +E)(n-t --I%),
where 01= 2~5i23~‘/231’i4(~~+ i dm).
If m is odd and greater than 1, we have B, = 0; hence S,,,(n) is divisible
by n2 (and by (n - 1)‘). Otherwise the roots of S,(n) don’t seem to obey a
simple law.
Let’s conclude our study of Bernoulli numbers by looking at how they
relate to Stirling numbers. One way to compute S,(n) is to change ordinary
powers to falling powers, since the falling powers have easy sums. After doing
those easy sums we can convert back to ordinary powers:
S,(n) = x n-’ km
k=O
= 7 7 {;}l& = x{y}z kj
k=O j?O j>O k=O
t-11
j+l- k i + 1 nk
[ 1k Therefore, equating coefficients with those in (6.78), we must have the identity
;{;}[i:‘](-jy;-* = --&(mk+l)Brn+i,. (6.99)
It would be nice to prove this relation directly, thereby discovering Bernoulli
numbers in a new way. But the identities in Tables 250 or 251 don’t give
us any obvious handle on a proof by induction that the left-hand sum in
(6.99) is a constant times rnc. If k = m + 1, the left-hand sum is just
{R} [EI;]/(m+l) = l/(m+l I, so that case is easy. And if k = m, the lefthandsidesumsto~~~,~[~]m~~-~~~[“‘~~](m+1~~~
=$(m-l)-im=-i;
so that case is pretty easy too. But if k < m, the left-hand sum looks hairy.
Bernoulli would probably not have discovered his numbers if he had taken
this route.