I am currently examining the sequence A198438. It is defined as:
"Common differences in triples of squares in arithmetic progression, that are not a multiples of other triples in (A198384, A198385, A198386)."
Here are some terms,
24, 120, 240, 336, 840, 840, 720, 1320, 2520, 2016, 3696, 2184, 5280, 5544, 6240, 3960, 9360, 10920, 10296, 4896, 6864, 14280, 18480, 6840, 22440, 17160, 10920, 20064
I'm fairly certain there is no way to generate this algebraically (if I'm mistaken that would be awesome and please let me know) but what I'd like to know is some way to bound it algebraically. For example I'd like to say something like this,
Let $a_n = A198438$,
$a_n < 40n^2$ (This may actually be true)
Or better yet bounded between something like,
$n^2 < a_n < 40n^2$
To summarize, is there a way to bound this sequence in a closed form like above or generate it algebraically? If so, what is the least upper bound you can find and the greatest lower bound?
Edit: To make it a little more general I'm interested in what we can say about the $n$th term in the sequence. Also when I say algebraic I mean it a little more loosely for example, is there an algorithm that can compute this sequence exactly?
Edit2: Will Jagy gave some phenomenal insight to this but I'd still like to know: How do I go from my triples to the sequence and is there a way to chose m and n so it comes out in that order?
Pythagorean triples (primitive) are $$ m^2 - n^2, 2mn, m^2 + n^2. $$
Your triples are $$ m^2 + 2mn - n^2 \; , \; \; m^2 - 2mn - n^2 \; , \; \; m^2 + n^2 $$
For the common difference of the squares, I get, with $m > n$ and $m+n$ odd, and $\gcd(m,n) = 1,$ $$ 4mn \left( m^2 - n^2 \right) $$
Putting these in order is not immediate: