a(3)=2,因为对于每一个素数p>=素数(3+1)=7,p^2 mod(4*2*3*5=120)是2个值{1,49}之一:
7^2 mod 120=49 mod 120=49
11^2模块120=121模块120=1
13^2模块120=169模块120=49
17^2模120=289模120=49
19^2模块120=361模块120=1
23^2模块120=529模块120=49
29^2模块120=841模块120=1
...
.
q=(n+1)stb=残留物p^2 mod b
p>=qa(n)的n素数4*素数(n)
= ========= =============== ======================= ====
0 2 4 = 4 {0,1} 2
1 3 4*2 = 8 {1} 1
2 5 4*2*3 = 24 {1} 1
3 7 4*2*3*5 = 120 {1,49} 2
4 11 4*2*3*5*7 = 840 {1,121,169,289,361,529} 6