Here is my current progress on the problem I described in my previous post.

​First, I observed that \(|a_{n+1}-a_n|\) is maximized when \(a_n\) is equidistant (or as close as possible to being equidistant) to consecutive multiples of \(n+1\). Hence: \[|a_{n+1}-a_n|\leq\left\lfloor \frac{n+1}{2} \right\rfloor\] Next, I considered a general term \(a_n=np\) and \(a_{n+1}=(n+1)q\) (where of course \(p,q\in\mathbb{Z}\). WLOG, I let \(a_{n+1}>a_n\). This was the case of interest anyway. Under these assumptions, I had: \[np+R=(n+1)q\Rightarrow q=\frac{np+R}{n+1}\] I set this to be less than or equal to \(p\). At the time, it just assumed this via heuristics. It wasn’t until later until I was able to actually prove it.

Lemma: Let \((n+1)q>np\) be the closest multiple of \(n+1\) to \(np\). Then \(p\geq q\)

Proof: Suppose that to the contrary, \((n+1)r>np\) was the closest multiple of \(n+1\) to \(np\), where \(r>p\). Then, the difference between the two numbers is: \[nr+r-np=n(r-p)+r\] Observe that \(r-p\geq1\Rightarrow n(r-p)\geq n\) and \(r>1\) (by implicit assumption). Adding these two inequalities yields: \[n(r-p)+r>n+1\] The maximum possible distance between a multiple of \(n\) and the closest larger multiple of \(n+1\) is \(n+1\), and this occurs when the aforementioned multiple of \(n\) is also a multiple of \(n+1\). However, we have shown that the distance between \((n+1)r\) and \(np\) is greater than \(n+1\), contradicting our initial assumption that \((n+1)r\) was the closest multiple of \(n+1\) that was larger than \(np\). Hence, \(\exists q\leq p\) such that \((n+1)q\) is the closest larger multiple of \(n+1\) to \(np\). \(\square\)

Ok that’s cool. So that means that we can safely set our expression for \(q\) to be lesser than or equal to \(p\): \[\frac{np+R}{n+1}\leq p\] We can rearrange this inequality to obtain: \[p\geq R\] The equality condition is more interesting. When \(p=R\), we must also have \(p=q\), and the sequence becomes arithmetic (\(n=k\)). This can more easily be obtained by noting that \(R=(n+1)q-np\). But when does \(p=R\)? Here is what I did know: \[0\leq R\leq\left\lfloor \frac{n+1}{2} \right\rfloor\] \(0\) wasn’t particularly interesting… but what about the upper bound? Letting \(R\) equal its upper bound seemed to match experimental results of my program for \(a_k\)! So I strongly suspect that: \[a_k=k\left\lfloor \frac{k+1}{2} \right\rfloor\] The question is, does this \(a_k\) always occur? How do I go about showing that it does? I have shown that \(q\leq p\), but how do I show that \(q=p\) will eventually occur?