Is it true that for any two integers with the same least prime factor, there must be an integer with a higher lpf between them?

Question

I'm sure something like this has been asked, but I can't seem to find it.

Let $a,b$ be two natural numbers such that $a<b$ and $\text{lpf}(a)=\text{lpf}(b)$. Is there always an $n\in\mathbb N$ such that $a<n<b$ and $\text{lpf}(n)>\text{lpf}(a)$?

Answer 1

I am kind of astonished, but I've eventually discovered this to be false. There is a counterexample in the interval $[a,a+262],$ $a = 724968762211953720363081773921156853174119094876349$, which both have lpf of $131$, with intervening values all lower.

Answer 2

This answer expands on my comment here as I was asked to.

A smaller counterexample is the interval $[a,a+142],$ $a=7310131732015251470110369$.

Suppose $\operatorname{lpf}(a)=\operatorname{lpf}(b)=p$ and no such $n$ as in the original post exists. Then $a-b \ge 2p$ as $p \not = 2$ and so both $a,b$ are odd. Thus if $p$ is the $k$th prime and $h(k)$ is the Jacobsthal function applied to the product of the first $k$ primes, $$h(k-1) \ge 2p.$$ Going off the table for $h(k)$ at OEIS A048670, we have $k \ge 20$ and so $p \ge 71$.

Given this, I decided to look for a counterexample $a,b$ with $a-b = 2 \cdot 71$, and $a \equiv 0 \pmod{71}.$ It should be noted that in fact $h(19) \ge 2 \cdot 73$, so there are probably similarly sized examples with $p=73$ instead. Examples using intervals longer than $h(19)$ are likely to be larger than the one in this post, but I haven't proved that none are smaller.

I wrote a short script to choose congruences for $a$ to satisfy modulo each prime less than $71$ such that every $n$ with $a < n < b$ is divisible by some such prime. It found many such systems of congruences, the one giving the answer in the post is given by \begin{align*} a&\equiv{1} \pmod{2}\\ a&\equiv{1} \pmod{3}\\ a&\equiv{4} \pmod{5}\\ a&\equiv{2} \pmod{7}\\ a&\equiv{7} \pmod{11}\\ a&\equiv{5} \pmod{13}\\ a&\equiv{4} \pmod{17}\\ a&\equiv{5} \pmod{19}\\ a&\equiv{4} \pmod{23}\\ a&\equiv{15} \pmod{29}\\ a&\equiv{4} \pmod{31}\\ a&\equiv{27} \pmod{37}\\ a&\equiv{23} \pmod{41}\\ a&\equiv{21} \pmod{43}\\ a&\equiv{23} \pmod{47}\\ a&\equiv{12} \pmod{53}\\ a&\equiv{16} \pmod{59}\\ a&\equiv{44} \pmod{61}\\ a&\equiv{39} \pmod{67}\\ a&\equiv{0} \pmod{71} \end{align*} I then combined these in the standard way to give the example. Note that I had to enforce that $a$ is not $0$ or $-2 \cdot 71$ modulo any primes less than $71$, so that indeed $\operatorname{lpf}(a) = \operatorname{lpf}(a + 2 \cdot 71) = 71$. It is likely that slightly smaller examples exist even with $p=71$, this was just the smallest my laptop found in a short run.