# 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?

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 $$a and $$\text{lpf}(a)=\text{lpf}(b)$$. Is there always an $$n\in\mathbb N$$ such that $$a and $$\text{lpf}(n)>\text{lpf}(a)$$?

• oeis.org/A020639 Jul 27, 2019 at 1:33
• Note a fairly closed related question is In a given sequence of consecutive integers, finding the count of integers with a least prime factor greater than $p$. Jul 27, 2019 at 1:53
• @Trev I wrote & ran a C++ program which tested all integers up to $8 \times 10^{8}$ and didn't find any exceptions, i.e., there's always an $n$ where it's lpf is larger between any $2$ consecutive integers with the same lpf. Jul 27, 2019 at 5:03
• @Trev A very closely related question is MO's Least Prime Factor in a sequence of 2n consecutive integers. The comments indicate using A058989, which points to the associated A049300. What these $2$ show is the least prime $p$ where at least $2p$ consecutive integers have a lpf $\le p$ is $43$, for $89$ values, starting at $14,478,292,443,584$. As this is more than $1.4\times 10^{13}$, it shows counter-examples, if any, are very large. Jul 27, 2019 at 5:34
• @JohnOmielan Thanks! It also seems to hold for arithmetical progressions, but seems to be untrue for quadratic functions. I tried it on $f(x)=x^2+k$ for small $k$, and there are generally small counterexamples. Weirdly, $x^2+1$ has a counterexample at $x=846$ and that's the highest you see until $x^2+138$, and more surprisingly, $x^2-2$ has its first at $x=116279$, with nothing else nearby remotely as high. Jul 27, 2019 at 20:44