bibkey: maynardpandeyradziwill2026primes
authors: James Maynard; Mayank Pandey; Maksym Radziwiłł
year: 2026
title: Exponential sums over primes
doi: null
url: https://arxiv.org/abs/2608.14777v1
claim: "Theorem 1.1 replaces the classical 4/5 exponent by 19/24 for additive von Mangoldt sums under its rational-approximant conditions. Harmonic averaging gives a uniform centered golden-interval allowance; the common untwisted prime error and same-root Robin sign remain uncontrolled."
strata_touched: []
license: citation-only
triage: anchor

Additive prime sums and the golden interval interface

The inspected primary is the authors’ preprint arXiv:2608.14777v1, submitted 14 August 2026. Locators refer to that version: Theorem 1.1, printed p.2, and the introduction’s comparison with the classical bound on pp.1–2. The source statement and the parameter application below were checked, not the complete analytic proof. This note makes no peer-review, originality or Lean-verification claim. The new exponential-sum theorem belongs to its authors and is reused here.

The source keeps both denominator and frequency offset

Let $e(z)=\exp (2\pi iz)$ and let $\Lambda$ be the von Mangoldt function, including proper prime powers. For $N\ge 1$, write

$$\alpha =\frac{a}{q}+\xi ,(a,q)=1,q\le N^{1/2},|\xi |\le \frac{1}{q{N}^{1/2}},B=\max (q,qN|\xi |).$$

Theorem 1.1 states

$$\left|\sum _{n<N}\Lambda (n)e(n\alpha )\right|\le N^{o(1)}\left(\frac{N}{\sqrt{B}}+N^{19/24}\right).\text{\hspace{1em}(M1)}$$

The source compares this with the classical $N^{4/5}$ term and notes $4/5-19/24=1/120$. Its statement is unconditional, with all the displayed parameters retained. At zero frequency, $q=1,\xi =0$ gives $B=1$ and a bound of order $N^{1+o(1)}$; it is not a centered square-root estimate for $\psi (N)-N$.

Average the approximant parameter over golden harmonics

Put $\phi =(1+\sqrt{5})/2$. Its quadratic irrationality supplies a fixed $c>0$ with $\Vert m\phi \Vert \ge c/m$ for all integers $m\ge 1$. For each $h\ge 1$, choose the reduced Dirichlet approximant to $h\phi$ at scale $N^{1/2}$ required in (M1), and call its parameter $B_h$. This gives

$$B_h\le N^{1/2},B_h\ge \max \left(q_h,\frac{cN}{h{q}_h}\right)\gg \sqrt{N/h}.$$

It is necessary to average $B_h^{-1/2}$, rather than substitute the smallest allowed $B_h$ for every harmonic. In a block $H<h\le 2H$, if $B_h\le Q$, then

$$m=h{q}_h\le 2HQ,\Vert m\phi \Vert \le Q/N.$$

Distinct such integers have circle separation at least $c/(2HQ)$. The union of the two arcs adjacent to zero has total length $2Q/N$, so it contains $O(1+H{Q}^2/N)$ such integers. Each integer $m$ has at most $\tau (m)$ possible divisors $h$. For $H\le N^{5/24}$ and $Q\le N^{1/2}$, the standard divisor bound absorbs this multiplicity in $N^{o(1)}$. Hence

$$\#\{H<h\le 2H:B_h\le Q\}\ll N^{o(1)}(1+H{Q}^2/N).\text{\hspace{1em}(M2)}$$

Split $Q$ into dyadic scales between a constant times $\sqrt{N/H}$ and $\sqrt{N}$. Multiplying (M2) by $Q^{-1/2}$ and summing gives

$$\sum _{H<h\le 2H}{B}_h^{-1/2}\ll N^{o(1)}\left((H/N{)}^{1/4}+H{N}^{-1/4}\right)\ll H{N}^{-1/4+o(1)}.$$

The first block containing $h=1$ obeys the same bound. Applying (M1) block by block therefore yields, with $L=\lfloor N^{5/24}\rfloor$,

$$\sum _{1\le h\le L}\frac{1}{h}\left|\sum _{n<N}\Lambda (n)e(h\phi n)\right|\ll N^{19/24+o(1)}.$$

Use the interval approximants in DMS v2, Theorem 3.3, printed p.25, whose main coefficient is the interval length, nonconstant coefficients are bounded by $1/(\pi |h|)$, and positive error coefficients by $1/(L+1)$. Nonnegativity of $\Lambda$ and the classical bound $\psi (N)=O(N)$ pay the zero coefficient of that error by $O(N/L)$. The nonconstant error coefficients are paid by the same harmonic sum since $1/(L+1)\le 1/h$. Consequently, uniformly over all circle intervals $I$,

$$\sum _{n<N}\Lambda (n)(1_I(\{n\phi \})-|I|)=O_{ϵ}(N^{19/24+ϵ})\text{for every fixed }ϵ>0.$$

Remove prime powers before reading a support-prime label

The preceding indicator reads the digits of $n=p^k$, not of its base prime. Remove all terms with $k\ge 2$ before identifying a support-prime class. Their total weight is $O(N^{1/2}(\log N{)}^2)$, uniformly in $I$, so it is smaller than the displayed allowance. Define

$${\Theta }_I(X)=\sum _{\begin{array}{c}p\le X\\ \{p\phi \}\in I\end{array}}\log p,{\Pi }_I(X)=\#\{p\le X:\{p\phi \}\in I\}.$$

Endpoint changes between $n<N$ and $p\le X$ cost only $O(\log X)$. Abel summation of the centered logarithmic prime count then gives, for each fixed $0<ϵ<5/24$, uniformly in $I$ and $X\to \infty$,

$${\Theta }_I(X)-|I|\vartheta (X)=O_{ϵ}(X^{19/24+ϵ}),{\Pi }_I(X)-|I|\pi (X)=O_{ϵ}(X^{19/24+ϵ}/\log X).\text{\hspace{1em}(M3)}$$

Finite endpoint conventions and unions of at most two circle intervals preserve these uniform constants. This is a parameter application of (M1) and standard interval approximation, not a new prime-distribution theorem. DMS’s exact low-address detection transports (M3) to the same actual CA correction, including a sufficient range of growing Fibonacci prefixes.

The centering in (M3) is the actual total prime count and weight. It does not supply a power-saving bound for their ordinary PNT error, a single CA integer’s digit law, or a sign for the full-support Robin residual.