Dietzfelbinger M. Primality testing in polynomial time ... - tiera.ru

3.6 Chebychev’s Theorem on the Density of Prime Numbers 47
� �
2n
= p
n
k1
1 ···pkr r .
The crucial observation we will make is that for no ps can the factor pks s in this
product be larger than 2n. To get the big number � � 2n
n as a product of such
small contributions requires that the prime decomposition of � � 2n
n contains
many different primes — namely, Ω(2n/ log(2n)) many — all of them ≤ 2n,
of course. To extend the estimate to odd numbers 2n + 1 is only a technical
matter.
Next, we will fill in the details of this sketch.
�
.Wehave
First, we orient ourselves about the order of magnitude of � 2n
n
2 2n
2n ≤
� �
2n
< 2
n
2n , for all n ≥ 1. (3.6.11)
Roughly, this is because � � � 2n 2n
0≤i≤2n i = 2 by the binomial theorem,
and because � � 2n
n is the largest term in the sum. (For the details, see
Lemma A.1.2(c) in Appendix A.1.)
Foranumbermand a prime p we denote the exact power to which p
appears in the prime factorization of m by νp(m). Thus νp(m) isthelargest
k ≥ 0sothatpk | m, and
m = �
p νp(m) ,
p | m
where the product extends over all prime factors of m.
For example, ν3(18) = ν3(2·3 2 )=2,ν2(10 k )=ν5(10 k )=k. Interestingly,
it is almost trivial to calculate to which power a prime divides the number
n!. This is most easily expressed using the “floor function” (defined in Ap-
pendix A.2). To give some intuitive sense to the following formula, note that
for integers a ≥ 0andb≥1thenumber⌊a b ⌋ = a div b equals the number of
multiples b, 2b, 3b,... of b that do not exceed a.
Lemma 3.6.4 (Legendre). For all n ≥ 1 and all primes p we have
νp(n!) = �
�
n
pk �
.
k≥1
Proof. The proof is a typical example for a simple, but very helpful counting
technique used a lot in combinatorics as well as in the amortized analysis of
algorithms. Consider the set
Rp,n = {(i, k) | 1 ≤ i ≤ n and p k divides i }.
Table 3.6 depicts an example for this set (p =2andn = 20) as a matrix with
log p (n) rowsandn columns, and entries 1 (•) and 0 (empty). We obtain