Geodesic
On the Difference between the Number of Prime Divisors from Subsets for Consecutive Integers
Matematičeskie zametki, Tome 68 (2000) no. 5, pp. 725-738 
N. M. Timofeev; M. B. Khripunova. On the Difference between the Number of Prime Divisors from Subsets for Consecutive Integers. Matematičeskie zametki, Tome 68 (2000) no. 5, pp. 725-738. http://geodesic.mathdoc.fr/item/MZM_2000_68_5_a8/
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Voir la notice de l'article provenant de la source Math-Net.Ru

Suppose that $E_1$, $E_2$ are arbitrary subsets of the set of primes and $g_1(n)$, $g_2(n)$ are additive functions taking integer values such that $g_i(p)=1$, if $p\in E_i$ and $g_i(p)=0$ otherwise, $i=1,2$. Set $$ E_i(x)=\sum_{\substack{p\le x,\\p\in E_i}}\frac 1p,\quad i=1,2. $$ It is proved in this paper that if $R(x)=\max(E_1(x),E_2(x))$, $a\ne0$ is an integer, then $$ \sup_m|\{n:n\le x, g_2(n+a)-g_1(n)=m\}| \ll\frac x{\sqrt{R(x)}}. $$ If, moreover, $E_i(x)\ge T$ for $x\ge x_0$, where $T$ is a sufficiently large constant and $$ |m-(E_2(x)-E_1(x))|\le\mu\sqrt{R(x)}, $$ then there exists a constant $c(\mu,a,T)>0$ such that for $x\ge x_0$ we have $$ \sum_{i=0}^3|\{n:n\le x,g_2(n+a)-g_1(n)=m+i\}|\ge c(\mu,a,T)\frac x{\sqrt{R(x)}}. $$

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