Concrete mathematics : a foundation for computer science

4.4 FACTORIAL FACTORS 113
contributions of those ten numbers; this calculation corresponds to summing
the columns of the following array:
11 23456789101powersof2
divisible by 2 x x x x x 5 = [10/2J
divisible by 4 X X 2 = [10/4]
divisible by 8 X 1 = [10/S]
powersof 010201030 1 ( 8
(The column sums form what’s sometimes called the ruler function p(k),
because of their similarity to ‘m ‘, the lengths of lines marking
fractions of an inch.) The sum of these ten sums is 8; hence 2* divides lo!
but 29 doesn’t.
There’s also another way: We can sum the contributions of the rows.
The first row marks the numbers that contribute a power of 2 (and thus are
divisible by 2); there are [10/2J = 5 of them. The second row marks those
that contribute an additional power of 2; there are L10/4J = 2 of them. And
the third row marks those that contribute yet another; there are [10/S] = 1 of
them. These account for all contributions, so we have ~2 (1 O!) = 5 + 2 + 1 = 8.
For general n this method gives
ez(n!) =
This sum is actually finite, since the summand is zero when 2k > n. Therefore
it has only [lgn] nonzero terms, and it’s computationally quite easy. For
instance, when n = 100 we have
q(lOO!) = 50+25+12+6+3+1 = 97.
Each term is just the floor of half the previous term. This is true for all n,
because as a special case of (3.11) we have lr~/2~+‘J = Lln/2k] /2]. It’s especially
easy to see what’s going on here when we write the numbers in binary:
100 = (1100100)~ =lOO
L100/2] = (110010)~ = 50
L100/4] = (11001)2 = 2 5
1100/8] = (1100)2 = 12
[100/16J = (110)2 = 6
1100/32] = (1l)z = 3
[100/64J = (I)2 = 1
We merely drop the least significant bit from one term to get the next.