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.
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| 格式: | Journal article |
| 語言: | English |
| 出版: |
Oxford University Press
2019
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| 總結: | 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. |
|---|