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发表于 2020-9-21 11:13
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本帖最后由 愚工688 于 2020-9-21 03:16 编辑
今天是2020-09-21日,以日期的千倍为起点的连续偶数素对数量的计算,看看计算值精度如何?
G(20200921000) = 37719257;
inf( 20200921000 )≈ 37697158.9 , jd ≈ 0.999414 ,infS(m) = 25875117.26 , k(m)= 1.45689;
G(20200921002) = 51784335;
inf( 20200921002 )≈ 51760120 , jd ≈ 0.999532 ,infS(m) = 25875117.26 , k(m)= 2.00038;
G(20200921004) = 28838337;
inf( 20200921004 )≈ 28822915.4 , jd ≈ 0.999465 ,infS(m) = 25875117.26 , k(m)= 1.11392;
G(20200921006) = 31955193;
inf( 20200921006 )≈ 31937287.6 , jd ≈ 0.999440 ,infS(m) = 25875117.26 , k(m)= 1.23429;
G(20200921008) = 51778467;
inf( 20200921008 )≈ 51750234.5 , jd ≈ 0.999455 ,infS(m) = 25875117.27 , k(m)= 2;
G(20200921010) = 35287693;
inf( 20200921010 )≈ 35266826.5 , jd ≈ 0.999408 ,infS(m) = 25875117.27 , k(m)= 1.36296;
G(20200921012) = 25890610;
inf( 20200921012 )≈ 25875117.3 , jd ≈ 0.999402 ,infS(m) = 25875117.27 , k(m)= 1;
G(20200921014) = 51785957;
inf( 20200921014 )≈ 51750234.6 , jd ≈ 0.999310 ,infS(m) = 25875117.27 , k(m)= 2;
G(20200921016) = 25893239;
inf( 20200921016 )≈ 25875117.3 , jd ≈ 0.999300 ,infS(m) = 25875117.28 , k(m)= 1;
G(20200921018) = 25949913;
inf( 20200921018 )≈ 25937167.9 , jd ≈ 0.999509 ,infS(m) = 25875117.28 , k(m)= 1.0024;
G(20200921020) = 82848474;
inf( 20200921020 )≈ 82800375.3 , jd ≈ 0.999419 ,infS(m) = 25875117.28 , k(m)= 3.2;
G(20200921022) = 27129681;
inf( 20200921022 )≈ 27112885.1 , jd ≈ 0.999381 ,infS(m) = 25875117.28 , k(m)= 1.04784;
time start =09:40:58 ,time end =09:45:05 ,time use =
计算式如下:
inf( 20200921000 ) = 1/(1+ .1535 )*( 20200921000 /2 -2)*p(m) ≈ 37697158.9
inf( 20200921002 ) = 1/(1+ .1535 )*( 20200921002 /2 -2)*p(m) ≈ 51760120
inf( 20200921004 ) = 1/(1+ .1535 )*( 20200921004 /2 -2)*p(m) ≈ 28822915.4
inf( 20200921006 ) = 1/(1+ .1535 )*( 20200921006 /2 -2)*p(m) ≈ 31937287.6
inf( 20200921008 ) = 1/(1+ .1535 )*( 20200921008 /2 -2)*p(m) ≈ 51750234.5
inf( 20200921010 ) = 1/(1+ .1535 )*( 20200921010 /2 -2)*p(m) ≈ 35266826.5
inf( 20200921012 ) = 1/(1+ .1535 )*( 20200921012 /2 -2)*p(m) ≈ 25875117.3
inf( 20200921014 ) = 1/(1+ .1535 )*( 20200921014 /2 -2)*p(m) ≈ 51750234.6
inf( 20200921016 ) = 1/(1+ .1535 )*( 20200921016 /2 -2)*p(m) ≈ 25875117.3
inf( 20200921018 ) = 1/(1+ .1535 )*( 20200921018 /2 -2)*p(m) ≈ 25937167.9
inf( 20200921020 ) = 1/(1+ .1535 )*( 20200921020 /2 -2)*p(m) ≈ 82800375.3
inf( 20200921022 ) = 1/(1+ .1535 )*( 20200921022 /2 -2)*p(m) ≈ 27112885.1
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