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Statement

Setting (pp. 311--312). $P(n)$ is the largest prime factor of $n\ge2$. If $n>1$ has canonical factorization $n=\prod p_i^{a_i}$, then $f(n)=\sum a_ip_i$, the sum of the prime factors of $n$ counted with multiplicity, and $f(1)=0$.

Theorem 2 (p. 312, quoted). "For every $\epsilon>0$, there is a $\delta>0$ such that for sufficiently large $x$ there are at least $(1-\epsilon)x$ choices for $n\le x$ such that

$$P(n)<f(n)<(1+x^{-\delta})P(n). \tag{2}$$

"

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: the statement on p. 312, the proof in §4 (p. 316).

Read depth. Claims checked: the statement was read clause by clause on the printed page. The proof was read for the pointer below and not checked step by step; nothing here is independently reviewed.

Proof pointer

§4 (p. 316). For large $x$ and composite $n\le x$, $f(n)\le P(n)+P(n/P(n))\log x/\log 2$, so outside $o(x)$ exceptions a failure of (2) forces $P(n/P(n))>x^{-2\delta}P(n)$, display (10). After discarding the $n$ with $P(n)<x^{\delta_0}$ by Dickman's theorem, the pairs of primes $p=P(n)$, $q=P(n/P(n))$ with $x^{-2\delta}p<q\le p$ are counted with Lemmas 1 and 2, and $\delta=\delta_0\epsilon/8$ suffices.

Depends on. Theorem A (Dickman) and Lemmas 1 and 2 of the paper (pp. 311, 313); none is recorded here.

Bears on

No problem page of this corpus. With Theorem 1 it gives, on p. 312, that the Aaron numbers ($f(n)=f(n+1)$) have density $0$, which Theorem 3 sharpens.