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Statement

Theorem 1.1 (p. 2, quoted). "Let $m$ and $n$ be integers satisfying $m,n\ge2$ and $m/1950\le n\le m$. Then we have

$$\pi(m+n)\le\pi(m)+\pi(n).$$

"

So the inequality of the second Hardy--Littlewood conjecture holds on the cone in which the smaller argument is at least $1/1950$ of the larger. The paper presents it as an improvement on the cone $x/109\le y\le x$ for real $x,y\ge3$ that it cites from Dusart (its display (1.5), p. 1).

Proof pointer

Section 3, pp. 3--4. Proposition 3.1 (p. 3) is a criterion for real $x\ge y\ge3$ in a ratio band $x/r\le y\le x/s$ with $r\ge s\ge1$: given $b\in(1,2)$ and a $B$ with $\pi(t)\le t/(\log t-1-b/\log t)$ for $t\ge B$, the inequality $\pi(x+y)\le\pi(x)+\pi(y)$ holds once $x$ passes an explicit threshold (3.3) built from $r$, $s$, $b$ and $B$. The proof of Theorem 1.1 (p. 4) takes $b=1.15$ and $B=38\,284\,442\,297$ (from Axler's earlier paper, its reference [2]) and covers $m/1950\le n\le m/109$ by fifteen bands tabulated on p. 4, each sharing an endpoint with the next. Every band's threshold is at most $38\,284\,440\,640$. Below that threshold the band gives $m+n\le(1+1/109)m\le39\,708\,229\,123$, which Proposition 2.4 covers. The remaining cone $m/109\le n\le m$ is the cited results (1.4) and (1.5) of p. 1.

Read depth

Claims checked: the statement, Proposition 3.1 and the table were read clause by clause on the pages of the copy named on the source card. The proof was read but not checked, and the table's thresholds were not recomputed. Nothing here is independently reviewed.

Dependencies

• Proposition 2.4 (p. 2), the computation up to $m+n\le39\,708\,229\,123$.
• Proposition 3.1 (p. 3), described above.
• The cone $x/109\le y\le x$ of Dusart, the paper's reference [4], Proposition 3 (see its card), and Gordon and Rodemich's range $2\le\min(m,n)\le1731$, display (1.4).

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: with $X\ge Y$ the two arguments, the theorem proves the problem's inequality for all integers $X,Y\ge2$ with $Y\ge X/1950$. It leaves open the pairs with $Y<X/1950$, so it does not decide the problem.