# Lhermite's models

Lhermite's models are interesting ways to synthesize various objects that are apparently scattered.

## Prime numbers and the model of three arrows

$\mathbb {U} _{n}=\sum _{i=1}^{f\left(n\right)}{\left(\left[{\frac {1+\sum _{m=1}^{i}{\varphi \left(m\right)}}{n+1}}\right]\times \left[{\frac {n+1}{1+\sum _{m=1}^{i}{\varphi \left(m\right)}}}\right]\times i\times \varphi \left(i\right)\right)}$

$\mathbb {U} _{n}=\sum _{i=1}^{f\left(n\right)}{\left(\left[{\frac {\alpha +\sum _{m=1}^{i}{\varphi \left(m\right)}}{n+\alpha }}\right]\times \left[{\frac {n+\alpha }{\alpha +\sum _{m=1}^{i}{\varphi \left(m\right)}}}\right]\times i\times \varphi \left(i\right)\right)}$

with $f\left(n\right)\geq \mathbb {U} _{n}$  and $\alpha >0$

$P_{n}=\sum _{i=1}^{2^{2^{n}}}\left(\left\lfloor {\frac {1+\sum _{m=1}^{i}{\left(1-\left\lfloor {\frac {\left\lfloor {\frac {\left(m!\right)^{2}}{m^{3}}}\right\rfloor }{\frac {\left(m!\right)^{2}}{m^{3}}}}\right\rfloor \right)}}{n+1}}\right\rfloor \times {\left\lfloor {\frac {n+1}{1+\sum _{m=1}^{i}{\left(1-\left\lfloor {\frac {\left\lfloor {\frac {\left(m!\right)^{2}}{m^{3}}}\right\rfloor }{\frac {\left(m!\right)^{2}}{m^{3}}}}\right\rfloor \right)}}}\right\rfloor }\times {i}\times {\left({1-\left\lfloor {\frac {\left\lfloor ({\frac {\left(i!\right)^{2}}{i^{3}}}\right\rfloor }{\frac {\left(i!\right)^{2}}{i^{3}}}}\right\rfloor }\right)}\right)$
$P_{n}=\sum _{i=1}^{2^{n}}\left(\left\lfloor {\frac {1+\sum _{m=1}^{i}{\left(1-\left\lfloor {\frac {\left\lfloor {\frac {\left(m!\right)^{2}}{m^{3}}}\right\rfloor }{\frac {\left(m!\right)^{2}}{m^{3}}}}\right\rfloor \right)}}{n+1}}\right\rfloor \times {\left\lfloor {\frac {n+1}{1+\sum _{m=1}^{i}{\left(1-\left\lfloor {\frac {\left\lfloor {\frac {\left(m!\right)^{2}}{m^{3}}}\right\rfloor }{\frac {\left(m!\right)^{2}}{m^{3}}}}\right\rfloor \right)}}}\right\rfloor }\times {i}\times {\left({1-\left\lfloor {\frac {\left\lfloor ({\frac {\left(i!\right)^{2}}{i^{3}}}\right\rfloor }{\frac {\left(i!\right)^{2}}{i^{3}}}}\right\rfloor }\right)}\right)$
$P_{n}=\sum _{i=1}^{1+n!}\left(\left\lfloor {\frac {1+\sum _{m=1}^{i}{\left(1-\left\lfloor {\frac {\left\lfloor {\frac {\left(m!\right)^{2}}{m^{3}}}\right\rfloor }{\frac {\left(m!\right)^{2}}{m^{3}}}}\right\rfloor \right)}}{n+1}}\right\rfloor \times {\left\lfloor {\frac {n+1}{1+\sum _{m=1}^{i}{\left(1-\left\lfloor {\frac {\left\lfloor {\frac {\left(m!\right)^{2}}{m^{3}}}\right\rfloor }{\frac {\left(m!\right)^{2}}{m^{3}}}}\right\rfloor \right)}}}\right\rfloor }\times {i}\times {\left({1-\left\lfloor {\frac {\left\lfloor ({\frac {\left(i!\right)^{2}}{i^{3}}}\right\rfloor }{\frac {\left(i!\right)^{2}}{i^{3}}}}\right\rfloor }\right)}\right)$
$P_{n}=\sum _{i=1}^{2^{n}}\left(\left[{\frac {1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]\right)}}{n+1}}\right]\times {\left[{\frac {n+1}{1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]\right)}}}\right]}\times {i}\times {\left({1-\left[{\frac {\left[({\frac {\left(i!\right)^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]}\right)}\right)$
$P_{n}=\sum _{i=1}^{2^{2^{n}}}\left(\left[{\frac {1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]\right)}}{n+1}}\right]\times {\left[{\frac {n+1}{1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]\right)}}}\right]}\times {i}\times {\left({1-\left[{\frac {\left[({\frac {\left(i!\right)^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]}\right)}\right)$

## Red balls and blue balls and prime numbers

$P_{\left(\left(1-\left[{\frac {\left[{\frac {\left(n!\right)^{2}}{n^{3}}}\right]}{\frac {\left(n!\right)^{2}}{n^{3}}}}\right]\right)\times \left(\sum _{m=1}^{n}{\left(1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]\right)}-i\right)+i\right)}=\left(P_{i}-n\right)\times \left[{\frac {\left[{\frac {\left(n!\right)^{2}}{n^{3}}}\right]}{\frac {\left(n!\right)^{2}}{n^{3}}}}\right]+n$

## Prime numbers and the model of three arrows according to Wilson's theorem

$\forall n\in \mathbb {N'}$
$(n-1)!\equiv \ -1{\pmod {n}}\Leftrightarrow n\in \mathbb {P}$

in the same way, it is advanced that

$\forall n\in \mathbb {N'}$
$\left[{\frac {\left[{\frac {\left(n-1\right)!+1}{n}}\right]}{\frac {\left(n-1\right)!+1}{n}}}\right]=1\Leftrightarrow n\in \mathbb {P}$

It's very evident that

$\forall n\in \mathbb {N'}$
$\left[{\frac {\left[{\frac {\left(n-1\right)!+1}{n}}\right]}{\frac {\left(n-1\right)!+1}{n}}}\right]=0\Leftrightarrow n\notin \mathbb {P}$

Therefore, according to Lhermite's models and Wilson's theorem, there are two evident theorems :

$\forall n\in \mathbb {N^{*}}$
$\left[{\frac {\left[{\frac {\left(n-1\right)!+1}{n}}\right]}{\frac {\left(n-1\right)!+1}{n}}}\right]-\left[{\frac {1}{n}}\right]=1\Leftrightarrow n\in \mathbb {P}$
$\forall n\in \mathbb {N^{*}}$
$\left[{\frac {\left[{\frac {\left(n-1\right)!+1}{n}}\right]}{\frac {\left(n-1\right)!+1}{n}}}\right]-\left[{\frac {1}{n}}\right]=0\Leftrightarrow n\notin \mathbb {P}$

Therefore the following relation becomes true :

$\forall n\in \mathbb {N^{*}}$
$\left[{\frac {\left[{\frac {\left(n-1\right)!+1}{n}}\right]}{\frac {\left(n-1\right)!+1}{n}}}\right]-\left[{\frac {1}{n}}\right]=1-\left[{\frac {\left[{\frac {\left(n!\right)^{2}}{n^{3}}}\right]}{\frac {\left(n!\right)^{2}}{n^{3}}}}\right]$

Let's choose one of the formulas that are indicated in the first section :

$P_{n}=\sum _{i=1}^{2^{2^{n}}}\left(\left[{\frac {1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]\right)}}{n+1}}\right]\times {\left[{\frac {n+1}{1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]\right)}}}\right]}\times {i}\times {\left({1-\left[{\frac {\left[({\frac {\left(i!\right)^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]}\right)}\right)$

let's replace

$1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]by\left[{\frac {\left[{\frac {\left(m-1\right)!+1}{m}}\right]}{\frac {\left(m-1\right)!+1}{m}}}\right]-\left[{\frac {1}{m}}\right]$

and

$1-\left[{\frac {\left[{\frac {\left(i!\right)^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]by\left[{\frac {\left[{\frac {\left(i-1\right)!+1}{i}}\right]}{\frac {\left(i-1\right)!+1}{i}}}\right]-\left[{\frac {1}{i}}\right]$

Therefore an equivalent expression is :

$P_{n}=\sum _{i=1}^{2^{2^{n}}}{\left(\left[{\frac {1+\sum _{m=1}^{i}{\left(\left[{\frac {\left[{\frac {\left(m-1\right)!+1}{m}}\right]}{\frac {\left(m-1\right)!+1}{m}}}\right]-\left[{\frac {1}{m}}\right]\right)}}{n+1}}\right]\times \left[{\frac {n+1}{1+\sum _{m=1}^{i}{\left(\left[{\frac {\left[{\frac {\left(m-1\right)!+1}{m}}\right]}{\frac {\left(m-1\right)!+1}{m}}}\right]-\left[{\frac {1}{m}}\right]\right)}}}\right]\times i\times \left(\left[{\frac {\left[{\frac {\left(i-1\right)!+1}{i}}\right]}{\frac {\left(i-1\right)!+1}{i}}}\right]-\left[{\frac {1}{i}}\right]\right)\right)}$

## Function Ω according to Lhermite's models

$\Omega \left(n\right)=\sum _{j=1}^{n}\left({\sum _{i=1}^{n}{\left({{\left[{\frac {\left[{\frac {n}{i^{j}}}\right]}{\left({\frac {n}{i^{j}}}\right)}}\right]}\times \left(1-\left[{\frac {\left[{\frac {\left(i!\right)^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]\right)}\right)}}\right)$

## Liouville's function and Lhermite's models

$\lambda \left(n\right)=\left(-1\right)^{\left(\sum _{j=1}^{n}\left({\sum _{i=1}^{n}{\left({{\left[{\frac {\left[{\frac {n}{i^{j}}}\right]}{\left({\frac {n}{i^{j}}}\right)}}\right]}\times \left(1-\left[{\frac {\left[{\frac {\left(i!\right)^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]\right)}\right)}}\right)\right)}$

## Three Arrows or Jonatan's Arrows

There are three possibilities :$a>b$  or $b  or $a=b$  . In the same way , there are three possibilities : $V_{i}>n$  or $V_{i}  or $V_{i}=n$

with