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Statement

Setting (p. 481).

$$t_k(n)=\min\{t\ge1:n\mid t(t+1)\cdots(t+k-1)\}.$$

The paper notes that for primes $p\ge k$ one has $t_k(p)=p+1-k$, so the maximal order is settled and the average and normal orders are the questions of interest.

Theorem 3 (p. 481).

$$\frac1x\sum_{n\le x}t_2(n)\ll x\,\frac{\log\log\log x}{\log\log x}.$$

Conjecture after the theorem (p. 481). The authors conjecture that the right side can be replaced by $x(\log x)^{-\alpha}$ for some fixed $\alpha>0$, and say it is likely that any fixed $\alpha<\log2$ will do; since $t_2(p)=p-1$, $\alpha>1$ is impossible.

Question (3) (p. 481). The paper asks whether

$$\sum_{n=1}^{x}t_{i+1}(n)=o\Bigl(\sum_{n=1}^{x}t_i(n)\Bigr),$$

and states that it has not proved this even for $i=2$.

Source. P. Erdős and R. R. Hall, On some unconventional problems on the divisors of integers, J. Austral. Math. Soc. Ser. A 25 (1978), no. 4, 479-485: the setting, Theorem 3, the conjecture and question (3) on p. 481, the proof on pp. 483-484. The edition read is identified on the source card.

Read depth. Claims checked: the definition, the statement, the conjecture and question (3) were read clause by clause on the printed page. The proof was read but not checked step by step. A second reader checked the statement, hypotheses, label and page against the print.

Proof pointer

Pages 483-484. For squarefree $q$, a residue class $h$ modulo $q$ is called $\varepsilon$-good when some $d\mid q$ and some $r$ with $1\le r\le\varepsilon d$, $(r,d)=1$, satisfy $h\equiv-r^{-1}(q/d)^{-1}\pmod d$. Write $n\le x$ as $mq$ with the prime factors of $q$ in $(z,y]$ and $m$ free of primes in that range; the $n$ with $q$ not squarefree contribute $O(x^2/z)$ to the sum. If $m$ lies in an $\varepsilon$-good class, then $t=rmq/d$ gives $n\mid t(t+1)$ and $t_2(n)\le\varepsilon n$. The $\varepsilon$-bad classes are counted with the Chinese remainder theorem and summed over $q$, with $y=x^{1/10}$; the choice $z=\log x$ and $\varepsilon=2(\log\log\log x)/\log\log x$ gives the theorem.

Dependencies

Elementary sieve counting and the Chinese remainder theorem; no other result of the paper.

Bears on

• Problem 394: the problem's $t_k(n)$ is the least $m$ with $n\mid m(m+1)\cdots(m+k-1)$, matching the paper's $t_k$ with $t\ge1$. Theorem 3 gives $\sum_{n\le x}t_2(n)\ll x^2\log\log\log x/\log\log x$, which is weaker than the bound $x^2/(\log x)^c$ the problem's first question asks for; the conjecture after the theorem is that question, and question (3), which the print states with no range for $i$, is the problem's second question with $i$ in place of $k$ (the problem takes $k\ge2$). The paper proves neither.