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 $N$
