设素数连乘积 p_m# = 2*3*5*...*p_m = M
区间(a,b)内,至少有一个与 p_m# 既约的自然数,则
b∏(1-1/p) - a∏(1-1/p) = (b-a)∏(1-1/p) > 1,
2 ≤ p ≤ p_m
b-a > 1 / ∏(1-1/p) = ∏[p/(p-1)]
.
实例验证:
p_m = 5,b-a > 2*3*5 / 1*2*4 = 30/8 = 3.75
p_m = 7,b-a > 2*3*5*7 / 1*2*4*6 = 210/48 = 4.375
p_m = 11,b-a > 2*3*5*7*11 / 1*2*4*6*10 = 2310/480 = 4.8125