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Statement

Setting (p. 480). $\tau_k(n)$ is the number of divisors $d$ of $n$ of the form

$$d=t(t+1)\cdots(t+k-1).$$

The paper states that for $k=2$ this is the number of indices $i$ with $d_{i+1}-d_i=1$, so $\tau_2(n)\le f(n)$ (with $f$ as in Theorem 1), and that for $k\ge2$ the average order of $\tau_k$ is a positive constant:

$$\sum_{n\le x}\tau_k(n)=\frac{x}{(k-1)(k-1)!}+O(x^{1/k}).$$

Theorem 2 (p. 480, quoted). "For each $k\geqslant2$, and every fixed $A<e^{1/k}$, we have $\tau_k(n)>(\log n)^A$ infinitely often."

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 and Theorem 2 on p. 480, its proof on p. 483. The edition read is identified on the source card.

Read depth. Claims checked: the definition and the statement 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

Page 483. Fix $B$ with $A<B<e^{1/k}$ and take $n=\operatorname{lcm}(1,2,\ldots,y)$, so that $y=(1+o(1))\log n$. Among the integers $m<y^B$, discard those with $m\equiv-i$ modulo some prime or prime power $Q$ in $[y/k!,y^B)$ for some $i=1,\ldots,k$. Mertens' theorem leaves at least $\varepsilon y^B$ integers, with $\varepsilon=\varepsilon(B)>0$ because $B<e^{1/k}$, and for each of them $\prod_{i=1}^k(m+i)$ divides $n$; each is a divisor of the required form.

Dependencies

The prime number theorem and Mertens' theorem; no other result of the paper.

Bears on

No Erdős problem page of the corpus consumes this theorem.