10/18/2021, 07:06 AM
This is very god damned interesting!
A quick suggestion I would make, is to look at this first for prime numbers. So let's choose \(p\) prime and let's look at:
\[
f^{p}_q(x,n) = u\\
\]
such that,
\[
^\infty \Big{[}q^{p^x\cdot a}\Big{]} =\, ^u\Big{[}q^{p^x\cdot a}\Big{]}\,\mod p^n\\
\]
Rather than dealing with \(2^x\cdot 5^y\). I'd start with prime numbers first. And then, as I see it, you've done a kind of "chinese remainder thing" where you've created the min/max result for different valuations across different primes. I suggest starting with one prime; and getting it to work for all primes, then generalizing to something like \(p_1 p_2 \cdots p_m\); and deriving an even more complex min/max formula off of valuations that works arbitrarily.
I think you'll find it's a lot easier too, to deal with one prime rather than two primes. The mod \(10^n\) is actually harder than \(2^n\), and in the simplicity some kernel of truth may come out.
Also, quick question to gauge your understanding, you are aware that for primes \(p\) that,
\(
m^{p-1} = 1\,\mod p\\
\)
This seems like it would speed up some of your proofs...
But still a very interesting paper, Luknik
A quick suggestion I would make, is to look at this first for prime numbers. So let's choose \(p\) prime and let's look at:
\[
f^{p}_q(x,n) = u\\
\]
such that,
\[
^\infty \Big{[}q^{p^x\cdot a}\Big{]} =\, ^u\Big{[}q^{p^x\cdot a}\Big{]}\,\mod p^n\\
\]
Rather than dealing with \(2^x\cdot 5^y\). I'd start with prime numbers first. And then, as I see it, you've done a kind of "chinese remainder thing" where you've created the min/max result for different valuations across different primes. I suggest starting with one prime; and getting it to work for all primes, then generalizing to something like \(p_1 p_2 \cdots p_m\); and deriving an even more complex min/max formula off of valuations that works arbitrarily.
I think you'll find it's a lot easier too, to deal with one prime rather than two primes. The mod \(10^n\) is actually harder than \(2^n\), and in the simplicity some kernel of truth may come out.
Also, quick question to gauge your understanding, you are aware that for primes \(p\) that,
\(
m^{p-1} = 1\,\mod p\\
\)
This seems like it would speed up some of your proofs...
But still a very interesting paper, Luknik
