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Source. Lemma 2.1, Section 2, p. 2 of arXiv:2406.19491v1 (27 June 2024), the edition named on the source card; the construction on p. 2, the proof on pp. 2-3. Read on the PDF page images.

Statement

Construction (p. 2). For each $n\in\mathbb N$ let $m=m(n)$ be an integer with

$$p_{m+1}\equiv\cdots\equiv p_{m+n}\equiv1\pmod{2^n},$$

where $p_j$ is the $j$-th prime; by Shiu's theorem such an $m$ exists, and for $n$ sufficiently large one with $m(n)<\exp_4(n)$, the four-fold iterated exponential. Put $n_0=1$ and, for $k\ge0$,

$$m_k=m(n_k),\qquad \pi_k=p_{m_k+n_k},\qquad n_{k+1}=4\pi_k , \qquad\text{and}\qquad \alpha=\sum_{k=0}^{\infty}2^{-n_k}.$$

Lemma 2.1 (p. 2). "The number $\alpha$ is transcendental, and hence is irrational."

Read depth. Claims checked: the construction and the lemma were read clause by clause on the page images, and the proof was read in full. Nothing here is independently reviewed.

Proof pointer

pp. 2-3. With $q_k=2^{n_k}$ and $a_k=2^{n_k}\sum_{l\le k}2^{-n_l}$, the fractions $a_k/q_k$ are in lowest terms. Since $\pi_k>2^{n_k}$, the exponents grow at least exponentially ($n_{k+1}>2^{n_k}$), which gives $|q_k\alpha-a_k|<q_k^{-k}$ for all large $k$; Liouville's theorem then rules out $\alpha$ being algebraic.

Dependencies

Shiu's theorem (Theorem 1(i) of Shiu, Strings of congruent primes, J. London Math. Soc. (2) 61, 2000) for the existence of $m(n)$; Liouville's theorem.

Bears on

• Problem 997: the lemma supplies the irrationality of the $\alpha$ for which Theorem 1.1 shows that $(\alpha p_n)$ is not well-distributed modulo $1$. On its own it says nothing about well-distribution.