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Statement

Setting. $\mathcal U$ is the set of positive odd integers not of the form $p+2^k$ with $p$ prime and $k$ a positive integer (pp. 1--2).

Definitions (p. 3). An infinite arithmetic progression $\{mh+a:h=0,1,\ldots\}$ is quasi-non-representable if $a>0$ and $\{mh+a:h=0,1,\ldots\}\setminus\mathcal U$ has asymptotic density zero. A quasi-non-representable progression is longest if it is a proper subset of no quasi-non-representable infinite arithmetic progression $\{m'h+a':h=0,1,\ldots\}$.

Theorem 1.5 (p. 3). $\{11184810h+a:h=0,1,\ldots\}$ is a longest quasi-non-representable infinite arithmetic progression if and only if $a$ belongs to the following list, the paper's (1.1):

509203, 762701, 992077, 1247173, 1254341, 1330207, 1330319, 1730653, 1730681, 1976473, 2313487, 2344211, 2554843, 3177553, 3292241, 3419789, 3423373, 3661529, 3661543, 3784439, 4384979, 4442323, 4506097, 4507889, 4626967, 5049251, 5050147, 6610811, 7117807, 7576559, 7629217, 8086751, 8101087, 8252819, 8253043, 8643209, 9053711, 9053767, 9545351, 9560713, 9666029, 10219379, 10280827, 10581097, 10609769, 10702091, 10913233, 10913681.

Corollary 1.6 (p. 3). For an integer $b$, $\{11184810h+b:h=0,1,\ldots\}\subseteq\mathcal U$ if and only if $b\ge0$ and $b\equiv a\pmod{11184810}$ for some $a$ in the list (1.1).

The list was checked here by recomputing the odd $a$ with $0\le a<11184810$ and $\gcd(a-2^k,11184810)>1$ for $1\le k\le24$, which the paper's proof (p. 23) identifies with (1.1): the computation gives exactly these 48 numbers.

Source. Yong-Gao Chen, A conjecture of Erdős on $p+2^k$, arXiv:2312.04120v3 (2024). Labels and pages are those of arXiv v3: the definitions and statements on p. 3, the proof of Theorem 1.5 on pp. 23--24, the proof of Corollary 1.6 on pp. 24--25. The edition read is identified on the source card.

Read depth. Claims checked: the definitions and statements were read clause by clause on the printed pages, and the list (1.1) was recomputed as described above. The finite verification in the proof of Corollary 1.6 was rerun: no $p+2^{k'}$ with $p\in\{3,5,7,13,17,241\}$ and $1\le k'\le24$ is congruent modulo $11184810$ to a number in (1.1). Nothing here is independently reviewed.

Proof pointer

Pages 23--25. For a longest quasi-non-representable $\{11184810h+b\}$, Sun's positive-proportion result (Lemma 4.4) forces $\gcd(b-2^k,11184810)>1$ for all $k\ge1$; since $2^{24}\equiv1$ modulo $5592405$ this is a condition on $k\le24$, and the reduced residue $a$ lies in (1.1). Conversely, for $a$ in (1.1) every $n=p+2^k$ in the progression has $p\in\{3,5,7,13,17,241\}$, so the exceptions have density zero, and maximality follows from Theorem 1.3. Corollary 1.6 adds a finite check that no $p+2^{k'}$ with these $p$ and $1\le k'\le24$ is congruent to a listed $a$.

Dependencies

Theorem 1.3; Lemma 4.4 (Sun's positive-proportion theorem, p. 23).

Bears on

• Problem 16: Theorem 1.5 supplies the second ingredient of the paper's third disproof (p. 25); see Theorem 3.1.