[過去ﾛｸﾞ] 高木くんがアクセプトされるまで見守るスレ ★4 (1002ﾚｽ)
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649(1): ◆pObFevaelafK [sage] 2023/07/26(水)08:39 ID:VqCU3DJB(2/17)
>>631 推敲版
Another proof of Lengedre conjecture

Let m and n be integers. We suppose p_n is the smallest prime number
among primes greater than m^2 and consider the following inequalities hold.
m^2<p_n<p_(n+1)<(m+1)^2

According to Cramer's conjecture,
p_(n+1)-p_n<(log(p_n))^2
holds for n≧5.

Since according to Dusart's inequality,
p_n<n(log(n)+log(log(n))) holds for n≧6,
log(p_n)<log(n)+log(log(n)+log(log(n)))
holds.
p_(n+1)-p_n<(log(n)+log(log(n)+log(log(n))))^2

By the way,
m+1>√p_(n+1)
holds. Since according to Dusart's inequality,
p_n>n(log(n)+log(log(n))-1) holds for n≧2,
√p_(n+1)>√(n(log(n)+log(log(n))-1))
m+1>√(n(log(n)+log(log(n))-1))
2m+1>2√(n(log(n)+log(log(n))-1))-1
holds. 2m+1 is the distance between m^2 from (m+1)^2.

We consider the following inequality.
(log(n)+log(log(n)+log(log(n))))^2<2√(n(log(n)+log(log(n))-1))-1
It is confirmed that this inequality holds for n≧75 by numeric computation.
Therefore, p_(n+1)-p_n is smaller than 2m+1 for m≧19.

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700: ◆pObFevaelafK [sage] 2023/07/27(木)07:28 ID:Rrvg2yJA(3/11)
>>649 訂正
×between m^2 from (m+1)^2
〇between m^2 and (m+1)^2

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