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分类计算不同偶数的素数对的误差,只能计算一些小偶数,不能分辨出总体偶数的误差(相对误差)的变化趋势。
因为大偶数时分类也将变得非常的复杂,分类的种类异常的多样,使人陷于困境。
同时由于各类分类的误差没有多少的差别,那么这样的分类还有什么意义?
我的连乘式的计算:
G(2024021500) = 4283845
G(2024021502) = 6428662
G(2024021504) = 3216019
G(2024021506) = 3564766
G(2024021508) = 6610899
G(2024021510) = 5141392
G(2024021512) = 3403963
G(2024021514) = 6428604
G(2024021516) = 3607430
G(2024021518) = 3215386
Sp( 2024021500 *)≈ 4282799.1 , jd ≈ 0.99976;
Sp( 2024021502 *)≈ 6424198.7 , jd ≈ 0.99931;
Sp( 2024021504 *)≈ 3215738.8 , jd ≈ 0.99991;
Sp( 2024021506 *)≈ 3562030.3 , jd ≈ 0.99923;
Sp( 2024021508 *)≈ 6607747.2 , jd ≈ 0.99952;
Sp( 2024021510 *)≈ 5139358.9 , jd ≈ 0.99960;
Sp( 2024021512 *)≈ 3401046.4 , jd ≈ 0.99914;
Sp( 2024021514 *)≈ 6424941.1 , jd ≈ 0.99943;
Sp( 2024021516 *)≈ 3607373.5 , jd ≈ 0.99998;
Sp( 2024021518 *)≈ 3212099.4 , jd ≈ 0.99898;
start time =14:41:10,end time=14:41:54 ,time use =
计算式:
Sp( 2024021500 *) = 1/(1+ .1411 )*( 2024021500 /2 -2)*p(m) ≈ 4282799.1 , k(m)= 1.333333
Sp( 2024021502 *) = 1/(1+ .1411 )*( 2024021502 /2 -2)*p(m) ≈ 6424198.7 , k(m)= 2
Sp( 2024021504 *) = 1/(1+ .1411 )*( 2024021504 /2 -2)*p(m) ≈ 3215738.8 , k(m)= 1.001133
Sp( 2024021506 *) = 1/(1+ .1411 )*( 2024021506 /2 -2)*p(m) ≈ 3562030.3 , k(m)= 1.108942
Sp( 2024021508 *) = 1/(1+ .1411 )*( 2024021508 /2 -2)*p(m) ≈ 6607747.2 , k(m)= 2.057143
Sp( 2024021510 *) = 1/(1+ .1411 )*( 2024021510 /2 -2)*p(m) ≈ 5139358.9 , k(m)= 1.6
Sp( 2024021512 *) = 1/(1+ .1411 )*( 2024021512 /2 -2)*p(m) ≈ 3401046.4 , k(m)= 1.058824
Sp( 2024021514 *) = 1/(1+ .1411 )*( 2024021514 /2 -2)*p(m) ≈ 6424941.1 , k(m)= 2.000231
Sp( 2024021516 *) = 1/(1+ .1411 )*( 2024021516 /2 -2)*p(m) ≈ 3607373.5 , k(m)= 1.123058
Sp( 2024021518 *) = 1/(1+ .1411 )*( 2024021518 /2 -2)*p(m) ≈ 3212099.4 , k(m)= 1
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