|
在大偶数区域,如果既想计算偶数的素数对下界数量,又想下界计算值的相对误差更小一些,那么我们就不能使用统一的下界修正系数(1/(1+μ))的μ值,而要依据偶数所处区域的不同选取适宜的μ值,才能得到比较高的计算精度。
比如:在1000亿——1500亿区域,我们可以取μ=0.162的值对连乘式计算值的相对误差进行修正。
G(100000000000) = 149091160;
inf( 100000000000 )≈ 148863296.6 , Δ≈-0.001528 ,infS( 100000000000 )= 111647472.43 ,
G(100000000002) = 268556111;
inf( 100000000002 )≈ 268127817.0 , Δ≈-0.001595 ,infS( 100000000002 )= 111647472.43 ,
G(100000000004) = 111836359;
inf( 100000000004 )≈ 111653826.5 , Δ≈-0.001632 ,infS( 100000000004 )= 111647472.43 ,
G(100000000006) = 111843604;
inf( 100000000006 )≈ 111675773.4 , Δ≈-0.001501 ,infS( 100000000006 )= 111647472.43 ,
G(100000000008) = 223655943;
inf( 100000000008 )≈ 223294944.9 , Δ≈-0.001614 ,infS( 100000000008 )= 111647472.43 ,
G(100000000010) = 150645060;
inf( 100000000010 )≈ 150414834.4 , Δ≈-0.001528,infS( 100000000010 )= 111647472.44 ,
G(100000000012) = 128533939;
inf( 100000000012 )≈ 128330428.1 , Δ≈-0.001583,infS( 100000000012 )= 111647472.44 ,
G(100000000014) = 238586864;
inf( 100000000014 )≈ 238209773.7 , Δ≈-0.001581,infS( 100000000014 )= 111647472.44 ,
G(100000000016) = 134188011;
inf( 100000000016 )≈ 133976966.9 , Δ≈-0.001573,infS( 100000000016 )= 111647472.44 ,
G(100000000018) = 111942653;
inf( 100000000018 )≈ 111774488.9 , Δ≈-0.001502,infS( 100000000018 )= 111647472.45 ,
G(100000000020) = 298192310;
inf( 100000000020 )≈ 297726593.2 , Δ≈-0.001562,infS( 100000000020 )= 111647472.45 ,
G(100000000022) = 124402721;
inf( 100000000022 )≈ 124210930.6 , Δ≈-0.001542,infS( 100000000022 )= 111647472.45 ,
Sp( 100000000000 ) = 1/(1+ .162 )*( 100000000000 /2 -2)*p(m) ≈ 148863296.6 , k(m)= 1.33333
Sp( 100000000002 ) = 1/(1+ .162 )*( 100000000002 /2 -2)*p(m) ≈ 268127817 , k(m)= 2.40156
Sp( 100000000004 ) = 1/(1+ .162 )*( 100000000004 /2 -2)*p(m) ≈ 111653826.5 , k(m)= 1.00006
Sp( 100000000006 ) = 1/(1+ .162 )*( 100000000006 /2 -2)*p(m) ≈ 111675773.4 , k(m)= 1.00025
Sp( 100000000008 ) = 1/(1+ .162 )*( 100000000008 /2 -2)*p(m) ≈ 223294944.9 , k(m)= 2
Sp( 100000000010 ) = 1/(1+ .162 )*( 100000000010 /2 -2)*p(m) ≈ 150414834.4 , k(m)= 1.34723
Sp( 100000000012 ) = 1/(1+ .162 )*( 100000000012 /2 -2)*p(m) ≈ 128330428.1 , k(m)= 1.14943
Sp( 100000000014 ) = 1/(1+ .162 )*( 100000000014 /2 -2)*p(m) ≈ 238209773.7 , k(m)= 2.13359
Sp( 100000000016 ) = 1/(1+ .162 )*( 100000000016 /2 -2)*p(m) ≈ 133976966.9 , k(m)= 1.2
Sp( 100000000018 ) = 1/(1+ .162 )*( 100000000018 /2 -2)*p(m) ≈ 111774488.9 , k(m)= 1.00114
Sp( 100000000020 ) = 1/(1+ .162 )*( 100000000020 /2 -2)*p(m) ≈ 297726593.2 , k(m)= 2.66667
Sp( 100000000022 ) = 1/(1+ .162 )*( 100000000022 /2 -2)*p(m) ≈ 124210930.6 , k(m)= 1.11253
G(120000000000) = 352503092;
inf( 120000000000 )≈ 352131790.3 , Δ≈-0.001053 ,infS( 120000000000 )= 132049421.35 ,
G(120000000002) = 137230841;
inf( 120000000002 )≈ 137072275.3 , Δ≈-0.001155 ,infS( 120000000002 )= 132049421.35 ,
G(120000000004) = 132188594;
inf( 120000000004 )≈ 132049421.4 , Δ≈-0.001053 ,infS( 120000000004 )= 132049421.35 ,
G(120000000006) = 280130367;
inf( 120000000006 )≈ 279807448.7 , Δ≈-0.001153 ,infS( 120000000006 )= 132049421.35 ,
G(120000000008) = 158634730;
inf( 120000000008 )≈ 158459305.6 , Δ≈-0.001106 ,infS( 120000000008 )= 132049421.35 ,
G(120000000010) = 209105088;
inf( 120000000010 )≈ 208865513.7 , Δ≈-0.001146 ,infS( 120000000010 )= 132049421.36 ,
计算式:
inf( 120000000000 ) = 1/(1+ .162 )*( 120000000000 /2 -2)*p(m) ≈ 352131790.3 ,
inf( 120000000002 ) = 1/(1+ .162 )*( 120000000002 /2 -2)*p(m) ≈ 137072275.3 ,
inf( 120000000004 ) = 1/(1+ .162 )*( 120000000004 /2 -2)*p(m) ≈ 132049421.4 ,
inf( 120000000006 ) = 1/(1+ .162 )*( 120000000006 /2 -2)*p(m) ≈ 279807448.7 ,
inf( 120000000008 ) = 1/(1+ .162 )*( 120000000008 /2 -2)*p(m) ≈ 158459305.6 ,
inf( 120000000010 ) = 1/(1+ .162 )*( 120000000010 /2 -2)*p(m) ≈ 208865513.7 ,
G(130000000000) = 206957741;
inf( 130000000000 )≈ 206780555 , Δ≈-0.000856 ,infS( 130000000000 )= 142161631.58 ,
G(130000000002) = 291494087;
inf( 130000000002 )≈ 291257976.9 , Δ≈-0.000810 ,infS( 130000000002 )= 142161631.59 ,
G(130000000004) = 170724988;
inf( 130000000004 )≈ 170593957.9 , Δ≈-0.000767 ,infS( 130000000004 )= 142161631.59 ,
G(130000000006) = 142661257;
inf( 130000000006 )≈ 142542144.6 , Δ≈-0.000835 ,infS( 130000000006 )= 142161631.59 ,
G(130000000008) = 303509249;
inf( 130000000008 )≈ 303278147.4 , Δ≈-0.000761 ,infS( 130000000008 )= 142161631.59 ,
G(130000000010) = 189710906;
inf( 130000000010 )≈ 189562218 , Δ≈-0.000784 ,infS( 130000000010 )= 142161631.59 ,
计算式:
inf( 130000000000 ) = 1/(1+ .162 )*( 130000000000 /2 -2)*p(m) ≈ 206780555 ,
inf( 130000000002 ) = 1/(1+ .162 )*( 130000000002 /2 -2)*p(m) ≈ 291257976.9 ,
inf( 130000000004 ) = 1/(1+ .162 )*( 130000000004 /2 -2)*p(m) ≈ 170593957.9 ,
inf( 130000000006 ) = 1/(1+ .162 )*( 130000000006 /2 -2)*p(m) ≈ 142542144.6 ,
inf( 130000000008 ) = 1/(1+ .162 )*( 130000000008 /2 -2)*p(m) ≈ 303278147.4 ,
inf( 130000000010 ) = 1/(1+ .162 )*( 130000000010 /2 -2)*p(m) ≈ 189562218 ,
G(150000000000) = 432693233;
inf( 150000000000 )≈ 432611673 , Δ≈-0.0001885,infS( m )= 162229377.38 , k(m)= 2.66667
G(150000000002) = 162281514;
inf( 150000000002 )≈ 162229377.4 , Δ≈-0.000321,infS( m )= 162229377.38 , k(m)= 1
G(150000000004) = 173090450;
inf( 150000000004 )≈ 173052270.7 , Δ≈-0.0002206,infS( m )= 162229377.38 , k(m)= 1.06671
G(150000000006) = 324533701;
inf( 150000000006 )≈ 324477220.4 , Δ≈-0.0001740,infS( m )= 162229377.39 , k(m)= 2.00011
G(150000000008) = 163640122;
inf( 150000000008 )≈ 163599942.2 , Δ≈-0.0002455,infS( m )= 162229377.39 , k(m)= 1.00845
G(150000000010) = 259646691;
inf( 150000000010 )≈ 259567003.8 , Δ≈-0.0003069,infS( m )= 162229377.39 , k(m)= 1.6
G(150000000012) = 324534559;
inf( 150000000012 )≈ 324458754.8 , Δ≈-0.0002336,infS( m )= 162229377.39 , k(m)= 2
G(150000000014) = 166666276;
inf( 150000000014 )≈ 166627941.1 , Δ≈-0.0002300,infS( m )= 162229377.4 , k(m)= 1.02711
G(150000000016) = 162262009;
inf( 150000000016 )≈ 162229377.4 , Δ≈-0.0002011,infS( m )= 162229377.4 , k(m)= 1
G(150000000018) = 373009121;
inf( 150000000018 )≈ 372941097.5 , Δ≈-0.0001824,infS( m )= 162229377.4 , k(m)= 2.29885
G(150000000020) = 237083721;
inf( 150000000020 )≈ 237037741.1 , Δ≈-0.0001939,infS( m )= 162229377.4 , k(m)= 1.46113
G(150000000022) = 162255812;
inf( 150000000022 )≈ 162229377.4 , Δ≈-0.0001629,infS( 150000000022 )= 162229377.4 , k(m)= 1
计算式:
inf( 150000000000 ) = 1/(1+ .162 )*( 150000000000 /2 -2)*p(m) ≈ 432611673 ,
inf( 150000000002 ) = 1/(1+ .162 )*( 150000000002 /2 -2)*p(m) ≈ 162229377.4 ,
inf( 150000000004 ) = 1/(1+ .162 )*( 150000000004 /2 -2)*p(m) ≈ 173052270.7 ,
inf( 150000000006 ) = 1/(1+ .162 )*( 150000000006 /2 -2)*p(m) ≈ 324477220.4 ,
inf( 150000000008 ) = 1/(1+ .162 )*( 150000000008 /2 -2)*p(m) ≈ 163599942.2 ,
inf( 150000000010 ) = 1/(1+ .162 )*( 150000000010 /2 -2)*p(m) ≈ 259567003.8 ,
inf( 150000000012 ) = 1/(1+ .162 )*( 150000000012 /2 -2)*p(m) ≈ 324458754.8 ,
inf( 150000000014 ) = 1/(1+ .162 )*( 150000000014 /2 -2)*p(m) ≈ 166627941.1 ,
inf( 150000000016 ) = 1/(1+ .162 )*( 150000000016 /2 -2)*p(m) ≈ 162229377.4 ,
inf( 150000000018 ) = 1/(1+ .162 )*( 150000000018 /2 -2)*p(m) ≈ 372941097.5 ,
inf( 150000000020 ) = 1/(1+ .162 )*( 150000000020 /2 -2)*p(m) ≈ 237037741.1 ,
inf( 150000000022 ) = 1/(1+ .162 )*( 150000000022 /2 -2)*p(m) ≈ 162229377.4 ,
再大一些的偶数的 1600亿区域,虽然样本偶数的相对误差仍然都是负值,但是绝对值太小了,μ=0.162的值不适宜再作下界计算式的修正系数,否则样本外的偶数容易出现正相对误差,就违反下界计算式的定义了。
G(160000000000) = 229574132;
inf( 160000000000 )≈ 229559235.1 , Δ≈-0.0000649,infS( 160000000000 )= 172169426.33 ,
G(160000000002) = 367315420;
inf( 160000000002 )≈ 367295743.5 , Δ≈-0.0000536,infS( 160000000002 )= 172169426.33 ,
G(160000000004) = 187842530;
inf( 160000000004 )≈ 187821192.4 , Δ≈-0.0001136,infS( 160000000004 )= 172169426.34 ,
G(160000000006) = 233415788;
inf( 160000000006 )≈ 233400374.2 , Δ≈-0.00006604,infS( 160000000006 )= 172169426.34 ,
G(160000000008) = 364844031;
inf( 160000000008 )≈ 364820136.6 , Δ≈-0.00006549,infS( 160000000008 )= 172169426.34 ,
G(160000000010) = 229594896;
inf( 160000000010 )≈ 229576603.6 , Δ≈-0.00007968,infS( 160000000010 )= 172169426.34 ,
计算式:
inf( 160000000000 ) = 1/(1+ .162 )*( 160000000000 /2 -2)*p(m) ≈ 229559235.1
inf( 160000000002 ) = 1/(1+ .162 )*( 160000000002 /2 -2)*p(m) ≈ 367295743.5
inf( 160000000004 ) = 1/(1+ .162 )*( 160000000004 /2 -2)*p(m) ≈ 187821192.4
inf( 160000000006 ) = 1/(1+ .162 )*( 160000000006 /2 -2)*p(m) ≈ 233400374.2
inf( 160000000008 ) = 1/(1+ .162 )*( 160000000008 /2 -2)*p(m) ≈ 364820136.6
inf( 160000000010 ) = 1/(1+ .162 )*( 160000000010 /2 -2)*p(m) ≈ 229576603.6
|
|