The differences between consecutive primes, V

We show that ∑ pn≤x pn+1−pn≥ √ pn (pn+1−pn)≪εx3/5+ε or any fixed ε>0⁠. This improves a result of Matomäki, in which the exponent was 2/3⁠.

Bibliográfalaš dieđut
Váldodahkki: Heath-Brown, D
Materiálatiipa: Journal article
Giella:English
Almmustuhtton: Oxford University Press 2019
Govvádus
Čoahkkáigeassu:We show that ∑ pn≤x pn+1−pn≥ √ pn (pn+1−pn)≪εx3/5+ε or any fixed ε>0⁠. This improves a result of Matomäki, in which the exponent was 2/3⁠.