P-adic valuation expression for a given natural number

[DaTario]

TL;DR Summary

Hi All, is there a closed expression that yields the p-adic valuation of a natural number n ?

Hi All,

If p is a prime, the p-adic valuation, $v_p(n)$, of a positive natural number is defined as the highest power of p that divides n. For instance, $v_3(45) = 2$.
My question is: Is there a mathematical expression for $v_p(n)$ ?

 

[mathman]

doubtful - primes are too scattered around

 

[DaTario]

Thank you, mathman. Could you please elaborate a bit more on this comment?

 

[DaTario]

I have just found this paper:
https://arxiv.org/ftp/arxiv/papers/1907/1907.11902.pdf
where the author derives a formula for $v_p(n)$. The approach goes like this:

$v_p (n) = v_p (n!) – v_p ((n –1)!)$
$= (n – s_p (n))/(p –1) – (n –1– s_p (n–1))/(p –1)$
$= (1 – s_p (n) + s_p (n–1))/(p –1)$
$= (1 – Δ s_p (n–1))/(p –1).$

where $s_p (n)$ denotes the sum of the digits in the base-p expansion of n.

Example: $v_3(36) = 2$
$36 = 1 . 9 + 1 . 27 \Rightarrow s_3(36) = 1 + 1 = 2$
$35 = 2 . 1 + 2 . 3 + 1 . 27 \Rightarrow s_3(35) = 2 + 2 + 1 = 5$
Thus,
$v_3(36) = \frac{1 - \Delta s_3(36) + \Delta s_3(35)}{3-1} = \frac{1 - 2 + 5}{2} = \frac{4}{2} = 2$.

Thak you all.

 

[DaTario]

Editing the last post: In the last equation there is no $\Delta$ anymore.

 

[mathwonk]

is there some formula for the sum of the digits in the base p expansion? this is something that seems to require an iterative application of the euclidean algorithm.

 

[DaTario]

Yes, mathwonk, it seems to require such algorithmic approach. Do you think using instead functions as Floor[ ], Ceiling[ ] and FractionalPart[ ], to implement $v_p(n)$, will demand the same algorithmic structure?

 

[DaTario]

I found a simple and alternative formula for $v_p(n)$, given by:
$$
v_p(n) = \lfloor \log_p (n) \rfloor - \sum_{j=1}^{\lfloor \log_p (n) \rfloor} \left \lceil \left \{ \frac{n}{p^j}\right \} \right\rceil,
$$

But it does not seem to be of any practical use.

 

[mathman]

Thank you, mathman. Could you please elaborate a bit more on this comment?

No deep study. Just a gut feeling, since prime factoring seems to be "random" as a function of n.

 

[DaTario]

Thank you, mathman.