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)
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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

749: ◆pObFevaelafK  [sage] 2023/07/29(土) 20:09:17.49 ID:3mRZQXR+

>>748 推敲版
Another proof of Legendre's conjecture

Let p_n be the largest prime number among the primes smaller than m^2.
We assume that the following inequalities hold.
p_n<m^2<(m+1)^2<p_(n+1)

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

The following inequality must hold when this inequalities hold.
2m+1<log(p_n)

√p_n<m
2√(p_n)+1<2m+1
Since according to Dusart's inequality,
p_n>n(log(n)+log(log(n))-1) holds for n≧2,
2√(n(log(n)+log(log(n))-1))+1<2m+1
holds.

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.

From the above,
2√(n(log(n)+log(log(n))-1))+1<log(n(log(n)+log(log(n))))
holds. However, this inequality does not hold for n≧3.

Therefore, there are at least one prime between m^2 and (m+1)^2 for m≧3
since 2m+1<log(p_n) does not hold for n≧6 and the assumption is false.

http://rio2016.5ch.net/test/read.cgi/math/1688037294/749

750: ◆pObFevaelafK  [sage] 2023/07/29(土) 20:17:22.73 ID:3mRZQXR+

>>749 再推敲
Another proof of Legendre's conjecture

Let p_n be the largest prime number among the primes smaller than m^2.
We assume that the following inequalities hold.
p_n<m^2<(m+1)^2<p_(n+1)

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

The following inequality must hold when this inequalities hold.
2m+1<(log(p_n))^2

√p_n<m
2√(p_n)+1<2m+1
Since according to Dusart's inequality,
p_n>n(log(n)+log(log(n))-1) holds for n≧2,
2√(n(log(n)+log(log(n))-1))+1<2m+1
holds.

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.

From the above,
2√(n(log(n)+log(log(n))-1))+1<(log(n(log(n)+log(log(n)))))^2
holds. However, it is confirmed that this inequality does not hold for
n≧58 by numeric computation.

Therefore, there are at least one prime between m^2 and (m+1)^2 for m≧16
since 2m+1<log(p_n) does not hold for n≧58 and the assumption is false.

http://rio2016.5ch.net/test/read.cgi/math/1688037294/750

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