|
楼主 |
发表于 2019-7-29 21:42
|
显示全部楼层
大一些的偶数,其素对数量的下界计算值的相对误差绝对值依然随偶数增大而缓慢地缩小:
5千亿的偶数系列:
G( 500000000000 ) = 655630055;
inf( 500000000000 )≈ 631936977.1 , Δ≈-0.036138,infS(m) = 473952732.79 , k(m)= 1.33333
G( 500000000002 ) = 530781937;
inf( 500000000002 )≈ 511599914 , Δ≈-0.036139,infS(m) = 473952732.79 , k(m)= 1.07943
G( 500000000004 ) = 984045373;
inf( 500000000004 )≈ 948474778.2 , Δ≈-0.036147,infS(m) = 473952732.79 , k(m)= 2.0012
计算式:
inf( 500000000000 ) = 1/(1+ .21 )*( 500000000000 /2 -2)*p(m) ≈ 631936977.1
inf( 500000000002 ) = 1/(1+ .21 )*( 500000000002 /2 -2)*p(m) ≈ 511599914
inf( 500000000004 ) = 1/(1+ .21 )*( 500000000004 /2 -2)*p(m) ≈ 948474778.2
1万亿的偶数系列:
G( 1000000000000 )= 1243722370 ;
inf( 1000000000000 )≈1201359378.5 , Δ≈-0.034061 , infS(m) = 901019533.87 , k(m)= 1.33333
G( 1000000000002 )= 1865594604 ;
inf( 1000000000002 )≈ 1802039067.8 ,Δ≈-0.034067 , infS(m) = 901019533.87 , k(m)= 2
G( 1000000000004 )= 1006929938;
inf( 1000000000004 )≈ 972589636.4 , Δ≈-0.034104 ,infS(m) = 901019533.88 , k(m)= 1.07943
计算式:
inf( 1000000000000 ) = 1/(1+ .21 )*( 1000000000000 /2 -2)*p(m) ≈ 1201359378.5
inf( 1000000000002 ) = 1/(1+ .21 )*( 1000000000002 /2 -2)*p(m) ≈ 1802039067.8
inf( 1000000000004 ) = 1/(1+ .21 )*( 1000000000004 /2 -2)*p(m) ≈ 972589636.4
p(m)—— 即素数连乘式,
|
|