| Version current |
Version 2 |
| For $n \in \mathbb{N}$ and any prime number $p$, $\pfac{n}$ is the product of numbers $1 \le m \le n$, where $p \not\vert m$. |
\newcommand{\pfac}[1]{\left(#1!\right)_p} |
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For $n \in \mathbb{N}$ and prime number $p$, $\pfac{n}$ is the product of numbers $1 \le m \le n \vert p \not\vert m$. |
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For natural numbers $n, s$ and a given prime number $p$, we have the congruence
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For $n, s \in \mathbb{N}$ and prime number $p$, we have the congruence
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\begin{displaymath} \frac{n!}{p^{\sum\limits_{i=0}^d \left\lfloor
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\begin{displaymath} \frac{n!}{p^{\sum_{i=0}^d \left\lfloor
|
| \frac{n}{p^i}\right\rfloor}} \equiv\prod\limits_{i=0}^d\left((\pm |
\frac{n}{p^i}\right\rfloor}} \equiv\prod\limits_{i=0}^d\left((\pm |
| 1)^{\left\lfloor\frac{n}{p^s+i}\right\rfloor} \pfac{N_i}\right) \pmod{p^s}, |
1)^{\left\lfloor\frac{n}{p^s+i}\right\rfloor} \pfac{N_i}\right) \pmod{p^s}, |
| \end{displaymath} |
\end{displaymath} |
| where $N_i$ is the least non-negative residue of $\left\lfloor \frac{n}{p^i}\right\rfloor |
where $N_i$ is the least non-negative residue of $\left\lfloor \frac{n}{p^i}\right\rfloor |
| \pmod{p^s}$. Here $d+1$ denotes the number of digits in the representation |
\pmod{p^s}$. $d+1$ denotes the number of digits in the $p$-adic representation |
| of $n$ in base $p$. More precisely, $\pm 1$ is $-1$ unless $p=2, s \ge 3$. |
of $n$. More preciesly, $\pm 1$ is $-1$ unless $p=2, s \ge 3$. |
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\textbf{Proof:} Let $i \ge 0$. Then the set of numbers between 1 and |
| \begin{proof} |
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| Let $i \ge 0$. Then the set of numbers between 1 and |
|
| $\left\lfloor \frac{n}{p^i}\right\rfloor$ is |
$\left\lfloor \frac{n}{p^i}\right\rfloor$ is |
| \[\left\{kp, k \ge 1, k \le \left\lfloor\frac{n}{p^{i+1}}\right\rfloor\right\}.\] |
\[\left\{kp, k \ge 1, k \le \left\lfloor\frac{n}{p^{i+1}}\right\rfloor\right\}.\] |
| This is true for every integer $i$ with $p^{i+1} \le n$. So we have |
This is true for every integer $i$ with $p^{i+1} \le n$. So we have |
| \begin{equation} |
\begin{equation} |
| \label{F1} |
\label{F1} |
| \frac{\left\lfloor \frac{n}{p^i}\right\rfloor!}{\left\lfloor \frac{n}{p^{i+1}}\right\rfloor |
\frac{\left\lfloor \frac{n}{p^i}\right\rfloor!}{\left\lfloor \frac{n}{p^{i+1}}\right\rfloor |
| p^{\left\lfloor\frac{n}{p^{i+1}}\right\rfloor}} |
p^{\left\lfloor\frac{n}{p^{i+1}}\right\rfloor}} |
| =\pfac{\left\lfloor\frac{n}{p^i}\right\rfloor}. |
=\pfac{\left\lfloor\frac{n}{p^i}\right\rfloor}. |
| \end{equation} |
\end{equation} |
| Multiplying all terms with $0 \le i \le d$, where $d$ is the largest power of |
Multiplying all terms with $0 \le i \le d$, where $d$ is the largest power of |
| $p$ not greater than $n$, the statement follows from the generalization of |
$p$ not greater than $n$, the statement follows from the generalization of |
| Anton's congruence. |
Anton's congruence. |
| \end{proof} |
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