为什么the probability that g_c_d(u, v) = d is equal to l/d times l/d times p, namely p/d^2?
楼上说的为什么
(u/d)*1 (u/d)*2 ... (u/d)*(d-1) (u/d)*d m (d个中选一个) m要在(u/d)*1~(u/d)*d中选?
[Theorem](G. Lejeune Dirichlet, Abhandlungen Kbniglich PreuB. Akad. Wiss.(1849), 69-83). If u and v are integers chosen at random, the probability that g_c_d(u, v) = 1 is 6/π^2 ≈.60793.
A precise formulation of this theorem, which carefully defines what is meant by being “chosen at random,” appears in exercise 10 with a rigorous proof. Let us content ourselves here with a heuristic argument that shows why the theorem is plausible.
If we assume, without proof, the existence of a well-defined probability p that g_c_d(u, v) equals unity, then we can determine the probability that g_c_d(u, TJ) = d for any positive integer d, because g_c_d(u, v) = d if and only if u is a multiple of d and v is a multiple of d and g_c_d(u/d, v/d) = 1. Thus the probability that g_c_d(u, v) = d is equal to l/d times l/d times p, namely p/d^2. Now let us sum these probabilities over all possible values of d; we should get
1 = ∑ p/d^2 = p(1 + 1/4 + 1/9 + 1/16 + . . . ).
d>l
Since the sum 1 + 1/4 + 1/9 + . . . = H∞(2) is equal to π^2/6 , we need p = 6/π^2 in order to make this equation come out right.