高木くんがアクセプトされるまで見守るスレ ★4

[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)
上下前次1-新
通常表示 512ﾊﾞｲﾄ分割ﾚｽ栞
抽出解除ﾚｽ栞

このｽﾚｯﾄﾞは過去ﾛｸﾞ倉庫に格納されています｡
次ｽﾚ検索歴削→次ｽﾚ栞削→次ｽﾚ過去ﾛｸﾞﾒﾆｭｰ

ﾘﾛｰﾄﾞ規制です｡10分ほどで解除するので､他のﾌﾞﾗｳｻﾞへ避難してください｡

746: ◆pObFevaelafK  [sage] 2023/07/29(土) 19:24:59.09 ID:3mRZQXR+

Another proof of Legendre's (Oppermann's) 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
p_(n+1)-p_n<2m+1

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(n)+log(log(n))))
holds.
p_(n+1)-p_n<(log(n(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+1)(log(n+1)+log(log(n+1))-1))
m+1>√((n+1)(log(n+1)+log(log(n+1))-1))
2m+1>2√((n+1)(log(n+1)+log(log(n+1))-1))-1
holds. 2m+1 is the distance between m^2 and (m+1)^2.

We consider the following inequality.
(log(n(log(n)+log(log(n)))))^2<2√((n+1)(log(n+1)+log(log(n+1))-1))-1
It is confirmed that this inequality holds for n≧73 by numeric computation.
Therefore, p_(n+1)-p_n is smaller than 2m+1 for m≧19.

http://rio2016.5ch.net/test/read.cgi/math/1688037294/746

上下前次1-新書関写板覧索設栞歴

ｽﾚ情報赤ﾚｽ抽出画像ﾚｽ抽出歴の未読ｽﾚ AAｻﾑﾈｲﾙ

ぬこの手ぬこTOP 0.034s