Heres a fairly useless but interesting tidbit:

if i = n
then
r = (abs(((((p)-(9**i)-9)+1))-((((9**i)-(p)-9)-2)))-p+1+1)

then r%a will (almost*) always return 0. when n = floor(a/2) for the lowest non-trivial factor of a two factor product.

Thats not really the interesting bit though. The interesting bit is the result of r will always be some product with a *larger* factor tree that includes the factor A of p, but not p's other larger factor, B.

So, useless from what I can see. But its an interesting function on its own, simply because of what it does.

I wrote a script to test it. For all two-factor products of the first 1000 primes, (with no repeating combinations, so if we calculated say, 23*31, we skip 31*23), only 3262 products failed this little formula, out of half a million.

All others reliably returned 0 for the following..
~~~
i = floor(a/2)

r = (abs(((((p)-(9**i)-9)+1))-((((9**i)-(p)-9)-2)))-p+1+1)

r%a
~~~

The distribution of failures was *very* early on in the set of factors, and once fixed at the value of 3262, stopped increasing for the rest of the run.

I didn't calculate if some primes were more likely to cause a product to fail or not. Nor the factor trees, nor if the factor trees had any factors in common between products, or anything of that nature.

All in all I count this as a worthwhile experiment.

If you want to run the code yourself, I posted it to pastebin here:

https://pastebin.com/Q4LFKBjB

edit:

Tried wolfram alpha just to see what it says, but apparently not much. Wish it could tell me more.