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Statement

Setting (p. 312). If $n>1$ has canonical factorization $n=\prod p_i^{a_i}$, then $f(n)=\sum a_ip_i$, and $f(1)=0$. An $n$ with $f(n)=f(n+1)$ is an Aaron number, the paper's term, with a pointer to Nelson, Penney and Pomerance (its [13]), where the density question is raised.

Theorem 3 (p. 312, quoted). "For every $\epsilon>0$, the number of $n\le x$ for which $f(n)=f(n+1)$ is $O(x/(\log x)^{1-\epsilon})$."

In particular the Aaron numbers have density $0$, which the paper first derives from Theorem 1 and Theorem 2.

On p. 312 the authors say they can prove the sharper bound $O(x/\log x)$ by a harder argument they do not give; that they suspect $O(x/(\log x)^k)$ for every $k$ but cannot prove it for any $k>1$, nor even $o(x/\log x)$; that they cannot prove there are infinitely many Aaron numbers, which would follow from Schinzel's Hypothesis H; and that they believe the count up to $x$ is $\Omega(x^{1-\epsilon})$ for every $\epsilon>0$. On p. 313 they record that the least $n$ with $f(n)=f(n+1)=f(n+2)$ is $n=417162$, found by David E. Penney in a computer search, say they cannot prove that the number of such $n\le x$ is $o(x/\log x)$, and conjecture that for every $k$ there are integers $n$ with $f(n)=f(n+1)=\cdots=f(n+k)$.

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 and remarks on pp. 312--313, the proof in §5 (pp. 317--319).

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

Proof pointer

§5 (pp. 317--319). Split by whether $P(n)>x^{1/2}$. In the first case $f(n)=f(n+1)$ and Lemma 3 ($f(n)\le P(n)\log n/\log P(n)$ when $P(n)\ge5$) give $P(n+1)>P(n)/3$ and $|P(n)-P(n+1)|<4x/P(n)$, and a congruence modulo $P(n)$ then forces $P(n)<3x^{2/3}$; a Hardy--Littlewood bound for primes in an interval and Lemma 1 bound the count by $O(x/\log x)$. In the second case, after discarding $O(x/\log x)$ integers with $P(n)\le x^{1/(3\log\log x)}$, the equation gives $P(n)/(4\log\log x)<P(n+1)<3P(n)\log\log x$, and Lemmas 1 and 2 bound the count by $O(x\log\log x\log\log\log x/\log x)$.

Depends on. Lemmas 1, 2 and 3 of the paper (p. 313); none is recorded here.

Bears on

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