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本帖最后由 愚工688 于 2019-7-1 04:56 编辑
光说不练,只是空话。就以今天日期的百倍为随机数,实际计算一下连续的几个偶数素对数量(单记)吧:
S(m)≥inf(M)= (A-2)*0.5π(1- 2/r )* π[(p1-1)/(p1- 2)] /(1+ .148 )= infS(m) * k(m)
G(2019070100) = 4293540;
inf( 2019070100 )≈ 4266679.4 , Δ≈-0.00626,infS(m) = 3185974.4 , k(m)= 1.33921
G(2019070102) = 3847609;
inf( 2019070102 )≈ 3823169.3 , Δ≈-0.00635,infS(m) = 3185974.4 , k(m)= 1.2
G(2019070104) = 6439971;
inf( 2019070104 )≈ 6397539 , Δ≈-0.00659,infS(m) = 3185974.41 , k(m)= 2.00803
G(2019070106) = 3230917;
inf( 2019070106 )≈ 3209574.2 , Δ≈-0.00661,infS(m) = 3185974.41 , k(m)= 1.00741
G(2019070108) = 3562420;
inf( 2019070108 )≈ 3539971.6 , Δ≈-0.00630,infS(m) = 3185974.41 , k(m)= 1.11111
G(2019070110) = 9276262;
inf( 2019070110 )≈ 9215926 , Δ≈-0.00650,infS(m) = 3185974.42 , k(m)= 2.89266
另外用我的在哈李计算式基础上改进的对数素对计算式 Xi(M)也计算一下:
Xi(M)=t1*c1*M/(logM)^2 , ( t1=1.358-log(M)^(2.045/3)*.03178 )
(注:c1—— 类似拉曼扭杨系数C,但是我只计算√M 内的素数 )
S( 2019070100 ) = 4293540; Xi(N)≈ 4282364.69 δxi( 2019070100 )≈-0.003069;
S( 2019070102 ) = 3847609; Xi(N)≈ 3837224.08 δxi( 2019070102 )≈-0.002699;
S( 2019070104 ) = 6439971; Xi(N)≈ 6421057.64 δxi( 2019070104 )≈-0.002937;
S( 2019070106 ) = 3230917; Xi(N)≈ 3221373.36 δxi( 2019070106 )≈-0.002954;
S( 2019070108 ) = 3562420; Xi(N)≈ 3552985.25 δxi( 2019070108 )≈-0.002648;
S( 2019070110 ) = 9276262; Xi(N)≈ 9249805.68 δxi( 2019070110 )≈-0.002852;
S( 2019070112 ) = 3395628; Xi(N)≈ 3386153.96 δxi( 2019070112 )≈-0.002790;
time start =12:35:45 end time =12:36:06
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