Combination of positive integers in terms of primes (sophisticated version 2)

Question

Here comes a second sophisticated version of my conjecture, as critics came up the first version was trivial.

Teorem 2

for a given prime $p$ and a given power $m$ the representation of any positive integer $n\in \Bbb N$ in the form: $$ n=(a_u p - b_u) \; p^m$$ is unique with the coefficient pairs (OEIS: A226233, A226236) $$ \left\{ \begin{array}{l l} \langle a_u \rangle=1+\left\lfloor\frac{u-1}{p-1}\right\rfloor=\frac{(p-1)+u-1-((u-1)mod(p-1))}{p-1}\\ \langle b_u \rangle=u-(p-1)\left\lfloor\frac{u-1}{p-1}\right\rfloor=1+((u-1)mod(p-1)) \end{array} \right.$$

while $a_u,b_u,u\in \Bbb N$ and $m\in \Bbb N_0$.

Can anyone help raising the proof? Or connect to another existing unsolved conjecture?

Note: $\lfloor\cdot \rfloor$ denotes the floor function.

2: Theorem to be cited Vaseghi 2013

Answer 1

If I have understood the question correctly, the claim can be seen to be correct as follows.

The definition of the integer $b_u$ tells us that $b_u$ is the unique integer in the range $1\le b_u\le p-1$ that is congruent to $u$ modulo $p-1$. So choosing $u$ from the correct residue class modulo $p-1$ allows us to choose $b$ to be anything we wish in that range.

On the other hand, if $u\le p-1$, then $a_u=1$. But also if we replace $u$ with $u'=u+k(p-1)$ we replace $a_u$ with $a_u+k$. As we saw in the preceding paragraph, $u$ and $u'$ give rise to the same value of $b_u$. All this means that a choice of $u$ allows us to assign the parameter $a_u$ to be any positive integer that we wish, and to assign the parameter $b_u$ any integer in the range $1\le b_u\le p-1$. The mapping $u\mapsto (a_u,b_u)$ is clearly a bijection from $\mathbb{Z}_+$ to $\mathbb{Z}_+\times \{1,2,\ldots,p-1\}$.

The claim follows from this. We are given the values of $n$ and $p$. As the factor $a_up-b_u$ is never divisible by $p$, we are forced to select $m$ in such a way that $p^m$ is the highest power of $p$ dividing $m$. So we write $$ n=n'p^m, $$ where $n'$ is not a multiple of $p$. This determines $n'$ and $m$ uniquely. Then as $n'$ is not a multiple of $p$, there exists a unique integer $r$ in the range $1\le r \le p-1$ such that $n'+r$ is a multiple of $p$. So $n'+r=\ell p$ for some positive integer $\ell$. Now the previous paragraph says that there exists a unique $u$ such that $a_u=\ell$ and $b_u=r$. Clearly $a_up-b_u=\ell p-r=n'$ and the claimed equation then holds. No freedom remains in the choice of $u$.

Illustrating this with an example $p=3$. I table some values of $a_u,b_u,a_up-b_u$ as functions of $u$ $$ \begin{array}{c|c|c|c} u&a_u&b_u&a_up-b_u\\ \hline 1&1&1&2\\ 2&1&2&1\\ 3&2&1&5\\ 4&2&2&4\\ 5&3&1&8\\ 6&3&2&7\\ 7&4&1&11\\ 8&4&2&10 \end{array} $$ It is clear that the last column will contain all the positive integers that are not multiples of $p=3$. All such numbers appear exactly once.

So for example when $n=63=7\cdot9=7\cdot3^2$, we are forced to select $m=2$ and $u=6$ (see the table).

Answer 2

I don't know why you subscript $a,b$ with $u$ or where $u$ comes from. If you say $p^m$ is the highest power of $p$ that divides $n$ this is essentially the Euclidean division algorithm. $b_u$ is the remainder term, here in the range $[1,p-1]$. Since you do not say $p^m$ is the largest power of $p$ dividing $n$, we can decrement $m$ by $1$ and multiply $a_u, b_u$ by $p$ and get another representation.