According to ingham's theorem $p_{n+1}-p_n< p_n+Ap_n^{5/8}$ in which $A$ is constant number.

Now let $p_n$ be largest prime less than or equal than $N$ so

$p_n\le N< p_{n+1}< p_n+Ap_n^{5/8}\le N+AN^{5/8}$,

let $p_{n+1}= q_1$,then:

$N< q_1< N+AN^{5/8}$,with above method we have below inequalities:

$N+AN^{5/8}< q_2< (N+AN^{5/8})+A(N+AN^{5/8})^{5/8}< N+A(2^{2}-1)N^{5/8}$ in which

$q_2=p_{n+s_1}$

$N+3AN^{5/8}< q_3< (N+3AN^{5/8})+A(N+3AN^{5/8})^{5/8}< N+A(2^{3}-1)N^{5/8}$

in which $q_3=p_{n+s_2}$

if we continue so:

$N+A(2^{k-1}-1)N^{5/8}< q_k< N+A(2^{k-1}-1)N^{5/8})+A(N+A(2^{k-1}-1)N^{5/8})^{5/8}< N+A(2^{k}-1)N^{5/8}$

in which $q_k=p_{n+s_k-1}$q_k=p_{n+s_{k-1}}$then after k step we reach to$2N$so$N+A(2^{k}-1)N^{5/8}\le 2N< N+A(2^{k+1}-1)N^{5/8}$we have$m\ge k>(log(N^{3/8}/{A}+1)/log2-1$,for a large$N$8 deleted 4 characters in body According to ingham's theorem$p_{n+1}-p_n< p_n+Ap_n^{5/8}$in which$A$is constant number. Now let$p_n$be largest prime less than or equal than$N$so$p_n\le N< p_{n+1}< p_n+Ap_n^{5/8}\le N+AN^{5/8}$, let$p_{n+1}= q_1$,then:$N< q_1< N+AN^{5/8}$,with above method we have below inequalities:$N+AN^{5/8}< q_2< (N+AN^{5/8})+A(N+AN^{5/8})^{5/8}< N+A(2^{2}-1)N^{5/8}$in which$q_2=p_{n+s_1}N+3AN^{5/8}< q_3< (N+3AN^{5/8})+A(N+3AN^{5/8})^{5/8}< N+A(2^{3}-1)N^{5/8}$in which$q_3=p_{n+s_2}$if we continue so:$N+A(2^{k-1}-1)N^{5/8}< q_k< N+A(2^{k-1}-1)N^{5/8})+A(N+A(2^{k-1}-1)N^{5/8})^{5/8}< N+A(2^{k}-1)N^{5/8}$in which$q_k=p_{n+s_k-1}$then$N+A(2^{k}-1)N^{5/8}\le 2N< N+A(2^{k+1}-1)N^{5/8}$we have$m\ge k>(log(N^{3/8}/{A}+1)/log2-1$,for a large$N$7 deleted 1 characters in body According to ingham's theorem$p_{n+1}-p_n< p_n+Ap_n^{5/8}$in which$A$is constant number. Now let$p_n$be largest prime less than or equal than$N$so$p_n\le N< p_{n+1}< p_n+Ap_n^{5/8}\le N+AN^{5/8}$, let$p_{n+1}= q_1$,then:$N< q_1< N+AN^{5/8}$,with above method we have below inequalities:$N+AN^{5/8}< q_2< (N+AN^{5/8})+A(N+AN^{5/8})^{5/8}< N+A(2^{2}-1)N^{5/8}$in which$q_2=p_{n+s_1}N+3AN^{5/8}< q_3< (N+3AN^{5/8})+A(N+3AN^{5/8})^{5/8}< N+A(2^{3}-1)N^{5/8}$in which$q_3=p_{n+s_2}$if we continue so:$N+A(2^{k-1}-1)N^{5/8}< q_3q_k< N+A(2^{k-1}-1)N^{5/8})+A(N+A(2^{k-1)-1)N^{5/8})^{5/8N+A(2^{k-1}-1)N^{5/8})+A(N+A(2^{k-1}-1)N^{5/8})^{5/8}

< N+A{2^{k}-1}N^{5/8}$N+A(2^{k}-1)N^{5/8}$

in which $q_k=p_{n+s_{k-1}$q_k=p_{n+s_k-1}$then$N+A{2^{k}-1}N^{5/8}\le N+A(2^{k}-1)N^{5/8}\le 2N< N+A{2^{k+1}-1}N^{5/8}$N+A(2^{k+1}-1)N^{5/8}$

we have $m\ge k>(log(N^{3/8}/{A}+1)/log2-1$,for a large $N$

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