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Statement

Setting (pp. 311, 320). $P(n)$ is the largest prime factor of $n\ge2$, and

$$\epsilon_n=\begin{cases}1,&\text{if }P(n)>P(n+1),\\0,&\text{if }P(n)<P(n+1).\end{cases}$$

Theorem (unnumbered, §7, p. 320). The number $\sum_{n=2}^{\infty}\epsilon_n/2^n$ is irrational; equivalently, the sequence $(\epsilon_n)$ is not eventually periodic.

Consequence (p. 320). For each $k$ let $h(k)$ be the number of distinct patterns of $k$ consecutive terms of $(\epsilon_n)$ that occur infinitely often. Then $h(k)\ge k+1$ for every $k$. The paper says surely $h(k)=2^k$, that this is easy for $k=1$, that for $k=2$ it can prove only $h(2)\ge3$ (and $h(2)=4$ if (20), $P(n)>P(n+1)>P(n+2)$, holds for infinitely many $n$), and on p. 321 that $h(k)=2^k$ follows from the prime $k$-tuples conjecture.

Source. P. Erdős, C. Pomerance, On the largest prime factors of $n$ and $n+1$, Aequationes Math. 17 (1978), 311--321, read in the edition named on the source card: §7, pp. 320--321.

Read depth. Claims checked: the statements and the remarks were read clause by clause on the printed pages, and the argument below was read in full. Nothing here is independently reviewed.

Proof pointer

p. 320. Suppose $(\epsilon_n)$ is eventually periodic with period $K$, and fix a prime $p>K$. By a theorem of Pólya, the set $M=\{n:P(n)\le p\}$ contains only finitely many pairs of consecutive integers. The numbers $p^i,2p^i,\ldots,Kp^i$ lie in $M$ for every $i$, so for large $i$ their successors do not, and $\epsilon_m=0$ at each of these $K$ numbers $m$. They form a complete residue system modulo $K$, so $\epsilon_n=0$ for all large $n$, which is absurd. For $h(k)\ge k+1$: $h(1)=2$, and $h$ is strictly increasing, since $h(k)=h(k+1)$ would make each late term determined by the previous $k$ and the sequence eventually periodic.

Depends on. Pólya's theorem on consecutive integers with only small prime factors (the paper cites it without a reference, noting Baker's work makes the largest such pair effectively computable); not recorded here.

Bears on

• Problem 251: context only. The problem asks about $\sum p_n/2^n$ with $p_n$ the $n$th prime; this is a different $0/1$ series and the result says nothing about that sum.
• Problem 372: context only. The paper notes that infinitely many $n$ with $P(n)>P(n+1)>P(n+2)$ would give $h(2)=4$.