# Is this a meaningful approach to primes?

## 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)

• No, $J(6,3)=2$, and $3\leq 6< 3^2$. – Thomas Andrews Apr 27 '15 at 19:57
• You are right, I had only inspected in the case of $i=2$ and I have accordingly edited it ... – drewdles Apr 27 '15 at 20:09
• There's a well known relationship between the log of the factorial and the log the primorial. – Larry Freeman Apr 27 '15 at 21:04
• Your equation is known as Legendre's Theorem for Prime Factorization of Factorials. Here's an article about it. – Larry Freeman Apr 28 '15 at 2:30

## 2 Answers

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 $p$ ;)

I hope the question becomes clear for you this way:

1) How many multiples of 5 are there in 103!? these multiples are 5,10,15,....95,100 i.e 20 hence floor function acting on 103/5.

2)The same question with $5^2$ gives 25,50,75,100 i.e 4 hence floor function acting on 103/25.

3) The same with $5^3$ gives 0 and so is for greater powers of 5 till the infinite.

Consequently the power with wchich appears the prime 5 in 103! is 24.

Remember $a^0$ = 1 so your formula is just to consider in a concise (and elegant) way all natural n since for all prime floor function must give constantly 0 from a certain rank.

P.S. Your (go easy on me) in your P.S. maked me to do what I realized now it is unnecessary. Anyway....