---

Statement

Udrescu's result, as the paper states it (p. 2): if $\varepsilon$ is a real number with $0<\varepsilon\le1$ and $n$ satisfies $\varepsilon m\le n\le m$, then $\pi(m+n)\le\pi(m)+\pi(n)$ for every sufficiently large positive integer $m$. Dusart had shown that it holds for every integer $m\ge e^{3.1/\log(1+\varepsilon)}$.

Theorem 1.2 (p. 2, quoted). "Udrescu's result holds for every integer

$$m\ge e^{\sqrt{0.3426/\log(1+\varepsilon)}}.$$

"

The proof (p. 4) works with integers $m,n\ge2$. For each fixed $\varepsilon$ this is a threshold in $m$ alone; it grows without bound as $\varepsilon\to0$.

Proof pointer

Section 4, pp. 4--5. For $\varepsilon\in[1/1950,1]$ the result is Theorem 1.1. For $\varepsilon<1/1950$ the threshold forces $m\ge168\,527\,259\,431$ and, since $x\mapsto xe^{\sqrt{0.3426/\log(1+x)}}$ decreases on $(0,1/1950)$, $n\ge86\,424\,235$. The paper then bounds $\pi(m)$ and $\pi(n)$ from below, displays (4.1) and (4.3), and $\pi(m+n)$ from above, all with the common denominator $\log(m+n)-1-1/\log(m+n)-3.15/\log^2(m+n)-14.25/\log^3(m+n)$, using explicit bounds from Axler's earlier papers (its references [1] and [2]); adding the two lower bounds gives the result.

Read depth

Claims checked: the statement and the description of Udrescu's and Dusart's results were read clause by clause on the pages of the copy named on the source card. The proof was read but not checked. Nothing here is independently reviewed.

Dependencies

• Theorem 1.1, for $\varepsilon\ge1/1950$.
• Explicit bounds for $\pi(x)$ from C. Axler, Integers 16 (2016), Paper No. A22, Corollary 3.5, and C. Axler, Integers 18 (2018), Paper No. A52, Corollaries 1 and 3 (see its card).

Source. Christian Axler, "Some Results on a Conjecture of Hardy and Littlewood," arXiv:1909.12625v2 (2019), the edition read for the source card.

Bears on

• Problem 855: for each fixed ratio bound $\varepsilon$, the theorem proves the problem's inequality for all integer pairs with $\varepsilon X\le Y\le X$ and $X$ beyond an explicit threshold. The threshold is not uniform as $\varepsilon\to0$, so it gives no single bound beyond which the inequality holds for all large $X$ and $Y$, and it does not decide the problem.