---

Statement

Display (5) of the paper (p. 27) writes the product of $n$ consecutive integers as

$$\prod_{i=1}^{n}(x+i)=\prod\nolimits_1 p_i^{\alpha_i(x,n)}\prod\nolimits_2 q_j^{\beta_j(x,n)}, \tag{5}$$

where $\prod_1$ runs over the primes $p_i\le n$ and $\prod_2$ over the primes $q_j>n$, and display (6) names the family of all these exponents, $\{\alpha_i(x,n),\beta_j(x,n)\}$.

On p. 27 Erdős first remarks that it is easy to see that for every $n$ there are infinitely many $x$ for which the $\alpha_i(x,n)$ are all distinct, and that surely the $\alpha_i(x,n)$ are subject only to the condition $\alpha_i(x,n)\ge\alpha_i(n)$, the exponent of $p_i$ in $n!$ (he did not carry out the details). He does not believe that, for large $n$, all the exponents (6) can be distinct, and states:

Conjecture (p. 27). For every $k$ there is an $n_0$ such that for $n>n_0$ at least $k$ of the exponents (6) equal $1$.

The print states no condition on $x$; the conjecture is read as holding for every $x$ for which (5) is a product of positive integers. Erdős calls the conjecture "no doubt unattainable at present" (p. 27). The paper gives no proof or evidence for it.

Source. P. Erdős, Miscellaneous problems in number theory, Proceedings of the Eleventh Manitoba Conference on Numerical Mathematics and Computing (Winnipeg, Man., 1981), Congr. Numer. 34 (1982), 25--45; displays (5) and (6) and the conjecture on p. 27. The edition read is identified on the source card.

Read depth. Claims checked: the passage was read clause by clause on the page image. There is no proof to check.

Dependencies

None. The weaker conjecture with $k=1$ is the Erdős--Selfridge conjecture of p. 28, recorded on its own page.

Bears on

• Problem 137: an exponent equal to $1$ is a prime dividing the product exactly once, so the conjecture with $k=1$ would make the product of $n$ consecutive positive integers never powerful for every $n>n_0$; it would leave the problem open for $3\le n\le n_0$. The paper poses the conjecture and records no result on it.