On the difference between the number of prime divisors from subsets for consecutive integers (Q5959482)

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scientific article; zbMATH DE number 1729017
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On the difference between the number of prime divisors from subsets for consecutive integers
scientific article; zbMATH DE number 1729017

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    On the difference between the number of prime divisors from subsets for consecutive integers (English)
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    26 October 2003
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    Let \(E_1,E_2\) be sets of primes and let \(R(x)=\max(E_1(x),E_2(x))\), where \(E_i(x)\) counts the primes \(p\leq x\) in~\(E_i\) with the weight~\(1/p\), for \(i=1,2\). Let \(g_1,g_2\) be additive functions taking integer values such that \(g_i(p)=1\) or~0 depending on whether \(p\in E_i\) or~not, for \(i=1,2\). By employing a sieve method the authors prove that, for integer \(a\neq 0\), \[ \sup_m|\{n:n\leq x, g_2(n+a)-g_1(n)=m\}|\ll{x\over\sqrt{R(x)}}. \] Moreover, if \(T\) is large, \(E_i(x)\geq T\) for \(x\geq x_0\) and \(|m-(E_2(x)-E_1(x))|\leq\mu\sqrt{R(x)}\), then there exists \(c(\mu,a,T)>0\) such that \[ \sum_{0\leq i\leq 3}|\{n:n\leq x, g_2(n+a)-g_1(n)=m+i\}|\geq c(\mu,a,T){x\over\sqrt{R(x)}}. \]
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    additive functions
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    sieve method
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