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高木くんがアクセプトされるまで見守るスレ ★4 (1002レス)
高木くんがアクセプトされるまで見守るスレ ★4 http://rio2016.5ch.net/test/read.cgi/math/1688037294/
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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
751: ◆pObFevaelafK [sage] 2023/07/29(土) 20:18:58.79 ID:3mRZQXR+ >>750 訂正 ×this 〇these http://rio2016.5ch.net/test/read.cgi/math/1688037294/751
752: ◆pObFevaelafK [sage] 2023/07/29(土) 21:17:57.50 ID:3mRZQXR+ >>750 訂正 ×since 2m+1<log(p_n) 〇since 2m+1<(log(p_n))^2 http://rio2016.5ch.net/test/read.cgi/math/1688037294/752
753: ◆pObFevaelafK [sage] 2023/07/30(日) 09:14:15.26 ID:usyTwEio >>750 訂正 ×there are 〇there is http://rio2016.5ch.net/test/read.cgi/math/1688037294/753
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