Motivation:
I tend to be good at recognising patterns and I saw one with factorial: $$ n! = \prod_{i=1} p_i^{J(n,p_i)} $$ where $p_i$ is the $i$'th prime and $$ J(n,i)= \sum_{S=1}^\infty [n/i^S] $$ and $[x]$ is the greatest integer function/floor function acting on $x$.
Question:
After some inspection, I figured $J(n,i)$ can be re-written as:
$$ J(n,i) = \begin{cases}
0 & 1 \leq n < i \\
\vdots & \vdots \leq n < \vdots \\
\sum_{\lambda=j} a_\lambda \frac{1-i^{\lambda}}{1-i} & \sum_{j=1} a_j i^{j} \leq n < \sum_{j=1} b_j i^{j} \\
\end{cases}$$
For example:
$ J(6,3) = 2 \frac{(3-1)}{(3-1)} = 2 $ where $3< n= 2 \times 3 < 9 $
Or another example:
$ J(67,5) = 2 \times \frac{5^2 -1}{5-1} + 3 \times \frac{5-1}{5-1} = 15 $ where $ 5^2 \times 2 + 5 \times 3 < n = 67 $
Taking the $\log$ of $n$ factorial:
$$ \log(n!)= \sum_{i=1}^{p_i \leq n} J(n,p_i) \log(p_i) $$
Can some kind of meaningful integral transform (or operation) be done so as to elude further on this approach to primes?
P.S: I'm just a physics undergraduate (go easy on me)
It is well known that the $p$-adic valutation of $n!$ is exactly $$ \upsilon_p(n!)=\sum_{i \in \mathbf{N}_+}\left\lfloor \frac{n}{p^i}\right\rfloor = \frac{n-s_p(n)}{p-1}, $$ where $s_p(n)$ stands for the sum of digits of $n$ in base $p$ ;)