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Problem #233: Let $d_n=p_{n+1}-p_n$, where $p_n$ is the $n$th prime

Let $d_n=p_{n+1}-p_n$, where $p_n$ is the $n$th prime. Prove that\[\sum_{1\leq n\leq N}d_n^2 \ll N(\log N)^2.\]

Problem Statement

Let $d_n=p_{n+1}-p_n$, where $p_n$ is the $n$th prime. Prove that\[\sum_{1\leq n\leq N}d_n^2 \ll N(\log N)^2.\]
Categories: Number Theory Primes

Progress

Cramer [Cr36] proved an upper bound of $O(N(\log N)^4)$ conditional on the Riemann hypothesis. Selberg [Se43] improved this slightly (still assuming the Riemann hypothesis) to\[\sum_{1\leq n\leq N}\frac{d_n^2}{n}\ll (\log N)^4.\]The prime number theorem immediately implies a lower bound of\[\sum_{1\leq n\leq N}d_n^2\gg N(\log N)^2.\]The values of the sum are listed at A074741 on the OEIS.

This is discussed in problem A8 of Guy's collection [Gu04].

Source: erdosproblems.com/233 | Last verified: January 14, 2026

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