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x/(lnx)^γ函数与筛法
因为 x/(lnx)^0=x,从p1p2p3...pk中筛去所有p1,p2,p3,...,pk的合数及p1,p2,p3,...,pk的数的个数就是φ(p1p2p3...pk)=x/(lnx)^0+|γ1|,再筛去p>pk的所有合数且加上p1,p2,p3,...,pk再减去整数1的数的个数就是π(x)=x/(lnx)^0+|γ1|+|γ2|,再筛去p≡q-2 (modq)|q≤x0.5的数的个数就是T(x)= x/(lnx)^0+|γ1|+|γ2|+|γ3|,例如:
6/(ln6)^0=6
φ(6)=2=6/(ln6)^0+1.883772126...
π(6)=3=6/(ln6)^0+1.883772126...-0.695244243...=3/6^1.188527883...
T(6)=1=6/(ln6)^0+1.883772126...-0.695244243...+1.883772127...=6/(ln6)^3.07230001...
30/(ln30)^0=30
φ(30)=8=30/(ln30)^0+1.079753379...
π(30)=10=30/(ln30)^0+1.079753379...-0.182287836...=30/(ln30)^0.897465544...
T(30)=4=30/(ln30)^0+1.079753379...-0.182287836...+0.748525542...=30/(ln30)^1.645991086...
210/(ln210)^0=210
φ(210)=48=210/(ln210)^0+0.880320565...
π(210)=46=210/(ln210)^0+0.880320565...+0.025385147...=210/(ln210)^0.905705712...
T(210)=15=210/(ln210)^0+0.880320565...+0.025385147...+0.668388859...=210/(ln210)^1.574094571...
2310/(ln2310)^0=2310
φ(2310)=480=2310/(ln2310)^0+0.767552507...
π(2310)=343=2310/(ln2310)^0+0.767552507...0.164166001...=2310/(ln2310)^0.931718508...
T(2310)=69=2310/(ln2310)^0+0.767552507...+0.164166001...+0.783383717...=2310/(2310)^1.715102225...
30030/(ln30030)^0=30030
φ(30030)=5760=30030/(ln30030)^0+0.707750448...
π(30030)=3248=30030/(ln30030)^0+0.707750448...+0.245551276...=30030/(ln30030)^0.953301724...
T(30030)=468=30030/(ln30030)^0+0.707750448...+0.245551276...+0.830362345...=30030/(ln30030)^1.783664069...
φ(510510)=22275=510510/(ln510510)^0.664576564
φ(9699690)=378675=9699690/(ln9699690)^0.635680366
φ(223092870)=7952175=223092870/(ln223092879)^0.612426943
φ(6469693230)=214708725=6469693230/(ln6469693230)^0.591974227641516
φ(200560490130)=6226553025=200560490130/(ln200560490130)^0.5763315871587100791
φ(7420738134810)=217929355875=7420738134810/(ln7420738134810)^0.562319897477908
同li↓λ(x)函数一样,φ(6)=2=6/(ln6)^0+1.883772126...<π(6)=3,所以有limπ(p1p2p3...pk+m)>li↓1.883772126(p1p2p3...pk+m), m→∞
我们可以看到γ1在不断的减少,但γ2且在不断的增加,其结果是γ1+γ2=1-ε,而ε在不断的减少,使limγ1+λ2=1 (x→∞).并且我们已经证明 limλ1+λ2+λ3<1.978905066.,而limγ1+γ2+γ3<limλ1+λ2+λ3.
作者施承忠 2009.12.6
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