Postila Bètran

## Postila Bètran

Ant yon antye n ki pi gran ke 1 e doub antye sila a, toujou gen yon nonm premye.
se yon teyorèm ki te konjektire pa Bertrand e ki demontre an 1850 pou lapremyè fwa pa matematisyen ris Tchebycheff
Gen plizyè varyant pou demontre teyorèm oubyen postila a. Men de ladan yo ki privilejye apwòch elemantè.

## Premyè demonstrasyon
### Premye lèm
Premyè demach la ap mennen nou redemontre tankou yon lèm ke 

\frac{2^{2n}}{2 \sqrt n}
pou tout n ki siperyè ou egal ak 2 avèk

 N = C_{2n}^n=}\times{2^{2n}}

 N {>} \frac {2^{2n}}{2\sqrt{n}} 

\frac {2^{2n}}{2\sqrt{n}}  {

Inegalite goch la demontre

kidonk lè nou konbine inegalite goch la ak inegalite dwat la nou gen  premye lèm

\frac{2^{2n}}{2 \sqrt n}

### Dezyèm lèm
An nou demontre yon dezyèm pwopozisyon sa ki va sèvi kòm lèm

\theta\left(n\right) {

An soulye ke fonksyon \theta Tchebychèf la defini konsa

\theta\left(x\right) = \sum_{p\le x}^{}{Ln p}   avèk x \le 0   ...

Nan kad sa a nou sipoze ke n \in \mathbb{N}^*

 \theta\left(1\right) = 0 

 \theta\left(2\right) = Ln2 {

Inegalite a verifye pou n = 1 ak n = 2 .

An nou sipoze li vrè pou yon sèten  n>2 

kounye a an nou konsidere \frac{N}{2}

 \frac{N}{2}=\frac{1}{2} N = \frac{1}{2}=\frac{\left(\left(2n-1\right)!\right)\times2n}{2\times\left( n!\right) \times n!} =\frac{\left(\left(2n-1\right)!\right)\times n}{\left(n!\right)\times{\left(\left(n-1\right)!\right)}\times n} =\frac{\left(2n-1\right)!}{\left(\left(n-1\right)!\right)\times n!} = \frac{\left(2n-1\right)}{\left(\left(n-1\right)!\right)\times\left(\left(2n-1\right)-\left(n-1\right)\right)!} = C^_ = { {2n-1} \choose{n-1}}

pou tou n siperyè ak 1, ekspresyon sa gen yon reyalite konkrè e se yon antye : se pa egzanp kantite souzansanm de n-1 eleman diferan ke nou ka
fòme a pati de yon ansanm de 2n-1 eleman.

An nou gade byen nimeratè a ak denominatè a

\frac{\left(2n-1\right)!}{\left(n!\right)\times\left(\left(n-1\right)!\right)} = \frac{1\times2\times3\times ... \times\left(2n-2\right)\times\left(2n-1\right)}{\left(1\times2\times3\times ... \times {\left(n-1\right)}\times n\right)\times{\left(1\times2\times3\times ... \times \left(n-2\right)\times\left(n-1\right)\right)}} 

Tout nonm antye non nil ki enferyè ak 2n-1 se yon divizè de nimeratè a , an patikilye tout nonm premye ki enferyè ou egal ak 2n-1 se yon divizè nimeratè a .

Menm konsiderasyon an ka fèt pou denominatè a jiska n .

An nou konsidere nonm premye p ki siperyè ak n e ki enferyè oubyen egal ak 2n-1.

 n{

 p\nmid {\left(\left(n!\right)\times\left(\left(n-1\right)!\right) \right)}

 p\mid \left(2n-1\right)!

kidonk yon faktè premye p ki tèl ke 
 n   
pa senplifye.

 \frac{N}{2} se donk miltipl non nil chak nonm premye tèl ke  n   .
Se donk yon miltipl non nil pwodui tout nonm premye sila yo.

 \frac{N}{2} siperyè oubyen egal ak pwodui nonm sa yo 
 

 \frac{N}{2}\ge {\prod_{n

Fonksyon Ln lan kwasant, nou genyen

 {Ln{\frac{N}{2}}}\ge {Ln{\prod_{n

tout fonksyon logaritm transforme pwodui an sòm logaritm

 {Ln{\frac{N}{2}}}\ge {\sum_{n

 {Ln{\frac{N}{2}}}\ge {\sum_{1\le{p}\le{n}}{Lnp}} - {\sum_{1\le{p}\le{n}}{Lnp}} + {\sum_{n{

 {Ln{\frac{N}{2}}}\ge \left({\sum_{1\le{p}\le{n}}{Lnp}}  + {\sum_{n{

 {Ln{\frac{N}{2}}}\ge {\sum_{1\le{p}\le{2n - 1}}{Lnp}} - {\sum_{1\le{p}\le{n}}{Lnp}}

 {Ln{\frac{N}{2}}}\ge \theta\left(2n-1\right) - \theta\left(n\right)

 \theta\left(2n-1\right) - \theta\left(n\right)\le {Ln{\frac{N}{2}}}

An nou konsidere inegalite dwat premye lèn lan oubyen dezyèm inegalite premye lèm lan

 {N} {

 nou kapab ekri li 

{N} 

An nou pran logaritm Neperyen de manm yon.

Kòm se yon fonksyon kwasant, nap genyen

Ln N 

Ln N 

Ln N {

An nou konbine

\{_{\theta\left(2n - 1\right) - \theta \left(n\right) \le Ln {\frac{N}{2}}}^{Ln N 

\theta\left(2n - 1\right) - \theta\left(n\right) + Ln N{

\theta\left(2n - 1\right) - \theta\left(n\right) 

\theta\left(2n - 1\right) - \theta\left(n\right) {

\theta\left(2n - 1\right) - \theta\left(n\right) 

\theta\left(2n - 1\right) - \theta\left(n\right) 

\theta\left(2n - 1\right) - \theta\left(n\right) 

An nou pa bliye ke nou vle demontre kòm dezyèm lèm

\theta\left(n\right) {

An nou konsidere kòm Ipotèz e an montre ke li vre pou 2n - 1 ak 2n pou tout n > 2

 \theta\left(2n - 1\right) - \theta\left(n\right)+ \theta\left(n\right) {

 \theta\left(2n - 1\right) {

....

kòm  n>2 

 2n {>} {2\times2}

 2n  {>} { 2^2} 

 Ln2n {>} {Ln{2^2}}

 Ln2n {>} 2Ln2 

 \frac{1}{2}Ln2n {>} Ln2 

 - \frac{1}{2}Ln2n {

 \left(4n-1\right)Ln2 - \frac{1}{2}Ln2n {

 \left(4n-1\right)Ln2 - \frac{1}{2}Ln2n {

 \left(4n-1\right)Ln2 - \frac{1}{2}Ln2n {

 \theta\left(2n -1 \right) {

 {\theta\left(2n - 1\right)} 

si nou gen  n \ge 2  alò  2n \ge 4 

 2n \ne 2

 2n \notin \mathbb{P} 

{\{_ {n \in \mathbb{N}}^{n\ge2}}\Rightarrow {2n \notin \mathbb{P}}\Leftrightarrow \not\exists p \in \mathbb{P} {/} 2n-1{

\theta\left(2n\right) = \sum_{1\le p \le 2n}Lnp = \sum_{1\le p \le 2n - 1}Lnp + \sum_{2n - 1\le p \le 2n}Lnp = \sum_{1\le p \le 2n - 1}{Lnp}  + 0  = \sum_{1\le p \le 2n - 1}{Lnp} = \theta\left(2n\right)

kidonk

 n \ge 2 \Rightarrow \theta\left(2n-1\right) = \theta\left(2n\right) 

pou tout  n\ge 2 

si lèm lan vre pou n li vre pou 2n - 1

an nou  pwouve rapidman ke inegalite vre pou 2n tou.

\theta\left(2n-1\right) {

\theta\left(2n\right) {

\theta\left(2n\right){

\theta\left(2n\right) {

\theta\left(2n\right) {

\theta\left(2n\right) 

si inegalite a vre pou  n\ge 2  
li vre alafwa pou 2n - 1 ak 2n .

An nou montre kounye a si fòmil la vrè pou entèval ]{2^{r-1}} , 2^r]
alò li vrè pou entèval ]2^r , 2^{r+1}]

Avan nou montre sa an nou verifye ke ingalite a vrè pou entèval  ]2,4] 

Inegalite a vrè pou n = 2 selon enplikasyon avan an li vrè pou  2\times2 - 1 et pou 2\times2  ki donk li vre e pou 3 e pou 4.
Pwopriyete a vrè pou  [2 , 4 ]  oubyen ankò li vrè pou  ]2 ,4] 
Pwopriyete a vrè pou  ]2^1 , 2^2]

Ipotèz rekirans 

An nou konsidere ke inegalite sou tout entèval ]2^{r-1} , 2^r]  la e an nou montre ke li vrè pou tout entèval

]2^r , 2^{r+1}]  la tou .

An nou pran yon antye m kèlkon sou entèval 

]2^r , 2^{r+1}] 

sa vle di  m\in ]2^r , 2^{r+1}]  

 m\in ]2^r , 2^{r+1}] \Leftrightarrow {2^r 

Nan de bagay ki ekskli yo youn lòt, nou gen youn fòseman.

m se yon nonm pè ou m se yon nonm enpè 

1)
 m pè   \Leftrightarrow \exists n\in \mathbb{N} {/}m = 2n 

\{_{m={2n}}^{2^r {

ò pa ipotèz rekirans, inegalite a vrè pou tout n ki nan entèval presedan an. Inegalite a vrè pou \theta\left(n\right) 
Kidonk selon premyè dediksyon an li vrè pou  \theta\left(2n\right). Li vre donk pou tout m pè ki nan entèval  ]{2^r} , {2^{r + 1}}] 

2)

m enpè  \Leftrightarrow \exists n\in \mathbb{N}{/}m = {2n - 1} 

\{_{m={2n-1}}^{2^r {

selon ipotèz rekirans

\theta\left( n\right) {

 inegalite a vrè pou tout n ki nan entèval presedan an. Inegalite a vrè pou \theta\left(n\right) 
Kidonk selon premyè dediksyon an li vrè pou  \theta\left(2n - 1\right). Li vrè donk pou tout m enpè ki nan entèval  ]{2^r} , {2^{r + 1}}] 

An fendkont pwopriyete a vrè pou tout {n}\ge{2}

 \forall n\ge 2, \theta\left( n\right) {

### Teyorèm Legendre

Si nou gen yon nonm premye fiks, pi gwo ekspozan k tèl ke {p^k}\mid n! alò 

 k = \sum_{i=1}^{\infty}{\left[\frac{n}{p^i}\right]} 

p^k \|{n!} 

An nou demontre sa pa rekirans

fòmil la vrè pou  n = 1

 {\sum_{i=1}^{\infty}{\left[\frac{1}{p^i}\right]}} = 0 

 p^i > 1 

 {\frac{1}{p^i}} 

an sipoze fòmil la bon pou  n epi an nou montre li bon pou n + 1

swa 
p^\alpha\|{n+1} 

{\sum_{i=1}^{\infty}{\left[\frac{n+1}{p^i}\right]}} = {\sum_{i = 1}^{\alpha}{\left[\frac{n + 1}{p^i}\right]} + \sum_{i=\alpha + 1}^{\infty}{\left[\frac{n + 1}{p^i}\right]}}

{\sum_{i=1}^{\infty}{\left[\frac{n+1}{p^i}\right]}} = {\sum_{i = 1}^{\alpha}{\left(\left[\frac{n }{p^i}\right] + 1\right)} + \sum_{i=\alpha + 1}^{\infty}{\left[\frac{n}{p^i}\right]}}

{\sum_{i=1}^{\infty}{\left[\frac{n+1}{p^i}\right]}} = {\sum_{i = 1}^{\alpha}{\left[\frac{n }{p^i}\right]} + \sum_{i = 1}^{\alpha}{1} + \sum_{i=\alpha + 1}^{\infty}{\left[\frac{n}{p^i}\right]}}

{\sum_{i=1}^{\infty}{\left[\frac{n+1}{p^i}\right]}} ={ \sum_{i = 1}^{\alpha}{\left[\frac{n }{p^i}\right]} + \alpha + \sum_{i=\alpha + 1}^{\infty}{\left[\frac{n}{p^i}\right]}}

{\sum_{i=1}^{\infty}{\left[\frac{n+1}{p^i}\right]}} =  

{\sum_{i=1}^{\infty}{\left[\frac{n+1}{p^i}\right]}} = {\sum_{i=1}^{\infty}{\left[\frac{n}{p^i}\right]}} + {\alpha} 

{\sum_{i=1}^{\infty}{\left[\frac{n+1}{p^i}\right]}} = {\alpha + k} 

selon ipotèz rekirans lan

ò    {\left(n + 1\right)!} = {\left(n + 1 \right)\times{n!}} 

 {p^k} \| {n!}  

ak  

 {p^{\alpha}} \| { n + 1}

enplike 

{ {p^{\alpha}}\times {p^k}}\| 

Sa ki enplike

 {p^{\alpha + k}} \|  \Rightarrow {n 

pou tout  i \ge 1 

 \frac{n}{p^i} 
kidonk  \left[\frac{n}{p^i}\right]= 0  
pou tout  i \ge 1 

\prod_{n

Nou ka ekri

 N = \frac\right)}}\right)}}} 

 N = \frac\right)}}\right)}}}\times{\prod_{n 

 N = \frac\right)}}{\left(p^{2\times\sum_{i = 1}^{\infty}{\left[\frac{n}{p^i}\right]}}\right)}}} 

 N = \frac{\left(p^{\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\right)} 

An nou pran logarithm Neperyen de manm egalite ya.

 Ln N = Ln\left( \prod_{p\le {2n}}{\left(p^{\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\right)}\right) 

 Ln N =\sum_{p\le{2n}}{\left(Ln\left({\left(p^{\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\right)}\right)\right)}

 Ln N =\sum_{p\le{2n}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}

An nou diseke sòm sa an 4 tèrm  .

 Ln N =\sum_{n 

pou premye tèm lan nou genyen :

 \sum_{n 

....

kidonk

 \sum_{n 

 n 

 \forall i > 0 , \frac{n}{p^i}

 \left[\frac{n}{p^i}\right] = 0 

sou yon lòt ang

 n 

 \forall p , p\le 2 

 p\le 2 \Rightarrow \frac{1}{p} \le \frac{1}{2} 

nou genyen sou baz ipotèz la

 \frac{1}{p} \le \frac{2n}{p^2} 

 \frac{2n}{p^2} 

pou tout  i > 2 , \left[ \frac{2n}{p^i}\right]=0 

An nou konsidere dezyèm sòm lan

\sum_ 

 \frac{2n}{3} 

 \frac{2n}{3} 

 \left[\frac{2n}{p}\right] - 2\left[\frac{n}{p}\right] = 2 - 2\times 1 = 0 

sou yon lòt ang , 

 \frac{1}{p} \le \frac{n}{p^2} 

  \left[\frac{n}{p^2}\right] = 0 

 \frac{2}{p}\le\frac{2n}{p^2} 

 \frac{2}{p}\le\frac{2n}{p^2} 

si p = 2

 1\le\frac{2n}{2^2} 

nap  genyen lè sa a :

\left[\frac{2n}{p^2}\right] = 1 

 1\le\frac{n}{2} 

kidonk tou nou tap genyen

 n 

si  n \le 3 

An nou gade ki sa nou ap genyen

 2n \ge 3\times 2

 2n \ge 6 
 \frac{2n}{3}\ge 2

 2\le \frac{2n}{3}

kidonk  p> 2 

 p > 3 
 \frac{2n}{p^2} 

 \frac{1}{p} 

 \frac{3}{p} 

 \frac{2n}{p^2} 

\frac{2n}{p^2} 

 \left[\frac{2n}{p^2}\right] = 0 

pou  n\le 3  e  i \ge 2

 \left[\frac{2n}{p^i}\right] = 0 

 \left[\frac{n}{p^i}\right] = 0 

 \left[\frac{2n}{p^i}\right] - 2\left[\frac{n}{p^i}\right] = 0 

kidonk pou tout  n\ge 3

\sum_ 

An nou konsidere twazyèm sòm lan 

\sum_                

An konsidere nan ki kondisyon inegalite ki anba somasyon ak gen sans :

 {\sqrt{2n}} \le \frac{2n}{3} 

2n\le \frac{4n^2}{9}

18n\le4n^2

18n - 4n^2 \le 0

2n\left(9 -2n\right) \le 0

 9 - 2n \le 0

 9\le 2n

 \frac{9}{2}\le n

n\ge\frac{9}{2}

n\ge 5

An nou endike dabò ke  \left[2\times\frac{n}{p^r}\right] - 2\left[\frac{n}{p^r}\right]   toujou egal oubyen ak zero oubyen ak 1 .

anfèt pou tout reyèl x nou genyen 

 \left[2x\right] - 2\left[x\right]  egal oubyen a 0 oubyen ak 1 .

 x = \left[x\right]+\{x\} 

 2x = 2\left[x\right]+ 2\{x\} 

 \left[2x\right] = \left[2\left[x\right]+ 2\{x\}\right] 

\left[2x\right] = 2\left[x\right] + \left[2\{x\}\right] 

\left[2x\right] - 2\left[x\right] = \left[2\{x\}\right] 

 0\le\{x\}

0\le 2\{x\}

\left[2x\right] - 2\left[x\right] = \left[2\{x\}\right]  egal swa zero soit 1 .

\left[\frac{2n}{p^i}\right] - 2\left[\frac{n}{p^i}\right] vo 0 oubyen 1 si  i = 1

An nou montre ke li vo 0 nan tout lòt ka .

\sqrt{2n}

\sqrt{2n} 

 \forall i\ge 2 , \frac{2n}{p^i} 

 \forall i\ge 2 , \left[\frac{2n}{p^i}\right] = 0 

 
 \forall n\ge 5

\sum_  

ò 

 \left[  \frac{2n}{p} \right] - 2\left[\frac{n}{p}  \right]  oubyen egal ak 1 oubyen egal ak 0

Kidonk  \left[  \frac{2n}{p} \right] - 2\left[\frac{n}{p}  \right] \le 1

{\left(\left( \left[  \frac{2n}{p} \right] - 2\left[\frac{n}{p}  \right] \right)\times Ln P\right)}\le Ln p

{\sum_

 {\sum_

{\sum_

\sum_ 

 \theta\left(\sqrt{2n}\right) = Ln2 + Ln3+ ... + Ln p_{\left(\pi\left(\sqrt{2n}\right)\right)}\ge{\pi\left(\sqrt{2n}\right)\times{Ln2}}  

 - \theta\left(\sqrt{2n}\right) \le {- \pi\left(\sqrt{2n}\right)\times{Ln2}}  

\theta\left( \frac{2n}{3}\right) - \theta\left(\sqrt{2n}\right) \le {\theta\left( \frac{2n}{3}\right) - \pi\left(\sqrt{2n}\right)\times{Ln2}}  

Pou tout  n \ge 5 

\sum_ 

An nou konsidere katryèm sòm lan

\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}   

An nou raple ke  \left[\frac{2n}{p^i}\right] - 2\left[\frac{n}{p^i}\right] swa egal ak 0 swa egal ak 1 .

yon lòt kote  \left[\frac{2n}{p^i}\right] \ge {  2\left[\frac{n}{p^i}\right]} 

An nou gade apati de kilè   \left[ \frac{2n}{p^i}\right]  toujou vo zero pou yon p fikse

 
  \left[ \frac{2n}{p^i}\right] = 0 \Leftrightarrow  \frac{2n}{p^i}  {2n}\Leftrightarrow Ln{p^i} > Ln{2n} \Leftrightarrow i Ln{p} > Ln{2n} \Leftrightarrow i > \frac{Ln2n}{Ln p} \Leftrightarrow i > \left[ \frac{Ln2n}{Ln p}\right] \Leftrightarrow i \ge \left[ \frac{Ln2n}{Ln p}\right] + 1  , etandone ke i  se yon antye natirèl ki pa egal ak 0 .

 i \ge \left[ \frac{Ln2n}{Ln p}\right] + 1 \Rightarrow \left[ \frac{2n}{p^i}\right] -2\left[ \frac{n}{p^i}\right] = 0 

sou ipotèz  n\ge 5

\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} =    \sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\left[ \frac{Ln2n}{Ln p}\right]}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}   +   \sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{\left[ \frac{Ln2n}{Ln p}\right] + 1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}

\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} =    \sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\left[ \frac{Ln2n}{Ln p}\right]}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}  + 0 

\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} =    \sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\left[ \frac{Ln2n}{Ln p}\right]}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} \le{\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\left[ \frac{Ln2n}{Ln p}\right]}{1}\right)}\times Ln p\right)}} 

\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} \le \sum_{p\le{\sqrt{2n}}}{\left(\left[ \frac{Ln2n}{Ln p}\right]\times Ln P\right)} 

\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} \le \sum_{p\le{\sqrt{2n}}}{\left(\left[\frac{Ln 2n}{Ln p}\right]\times Ln p\right)} \le \sum_{p\le{\sqrt{2n}}}{\left(\frac{Ln 2n}{Ln p}\times Ln p\right)}

\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} \le \sum_{p\le{\sqrt{2n}}}{\left(\left[\frac{Ln 2n}{Ln p}\right]\times Ln p\right)} \le \sum_{p\le{\sqrt{2n}}}{Ln 2n}

\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} \le \sum_{p\le{\sqrt{2n}}}{\left(\left[\frac{Ln 2n}{Ln p}\right]\times Ln p\right)} \le Ln 2n\sum_{p\le{\sqrt{2n}}}{1}

\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} \le \pi\left(\sqrt{2n}\right)Ln 2n 

An nou raple ke  Ln N =\sum_{n 

lè nou anvizaje tout estimasyon yo oswa tout estime yo nou kapab ekri :

pou tout  n\le 5

 Ln N \le \left(\theta\left(2n\right) - \theta\left(n\right)\right)  + 0 + \left( \theta\left(\frac{2n}{3}\right)-\pi\left(\sqrt{2n}\right)\times  Ln 2\right) + \left( \pi\left(\sqrt{2n}\right)\times  Ln 2n\right) 

\theta\left(2n\right) -\theta\left(n\right) \ge Ln N - \theta\left(\frac{2n}{3}\right) + \pi\left(\sqrt{2n}\right)\left(Ln2 - Ln 2n\right) 

\theta\left(2n\right) -\theta\left(n\right) \ge Ln N - \theta\left(\frac{2n}{3}\right) - \pi\left(\sqrt{2n}\right)\left(Ln\left(\frac{2n}{2}\right)\right) 

\theta\left(2n\right) -\theta\left(n\right) \ge Ln N - \theta\left(\frac{2n}{3}\right) - \pi\left(\sqrt{2n}\right)Ln2 

Li rete pou nou pwouve ke Ln N - \theta\left(\frac{2n}{3}\right) - \pi\left(\sqrt{2n}\right)Ln  estriteman pozitif .

Nou te genyen  N > \frac{2^n}{2\sqrt{n}} 

 Ln N > Ln \left( \frac{2^n}{2\sqrt{n}}\right)

Ln N > Ln 2^{2n} - ln{2\sqrt{n}} 

 Ln N > 2n ln2 - ln\left(2\sqrt{n}\right)

Ln N > 2nLn2 - Ln2 -Ln{\sqrt{n}} 

 Ln N > 2n Ln2 -Ln2 - Ln{n^{\frac{1}{2}}}   

  Ln N >2n Ln2 -Ln2 - \frac{1}{2}Ln n 

Anplis lè  n \le 2  , nou genyen 

 \theta\left(\frac{2n}{3}\right) = \theta\left(\left[\frac{2n}{3}\right]\right) 

 - \theta\left(\frac{2n}{3}\right) > - \frac{4n}{3}Ln2 

yon lòt kote,  \pi\left(n\right)\le\frac{n}{2}, \forall n\ge 8 , etandone ke tout nonm pè pi gran ke 2 pa premye ou di mwens pa konpoze. 

Nou vin genyen :

\theta\left(2n\right) -  \theta\left(n\right) > 2nLn2 - Ln\left(2\sqrt{n}\right)-\frac{4n}{3}Ln2 - \frac{\sqrt{2n}}{2}Ln n 

\theta\left(2n\right) -  \theta\left(n\right) > 2nLn2 -Ln2 -Ln{\sqrt{n}} -\frac{4n}{3}Ln2 - \frac{\sqrt{2n}}{2}Ln n 

\theta\left(2n\right) -  \theta\left(n\right) > 2nLn2 -Ln2 -\frac{1}{2}Lnn -\frac{4n}{3}Ln2 - \frac{\sqrt{2n}}{2}Ln n 

 \theta\left(2n\right) -  \theta\left(n\right) > \left(2n - \frac{4n}{3} - 1\right)Ln2 - \left(\frac{1}{2} + \frac{\sqrt{2n}}{2}\right)Lnn 

 \theta\left(2n\right) -  \theta\left(n\right) > \left(\frac{2n}{3} - 1 \right)Ln2  - \left( \frac{1 + \sqrt{2n}}{2}\right)Ln n

Li rete pou nou pwouve ke manm dwat la pi gran ke zero e konsa ant n ak 2n nap va asire ke gen o mwen yon nonm premye .

 \left(\frac{2n}{3} - 1 \right)Ln2  - \left( \frac{1 + \sqrt{2n}}{2}\right)Ln n  > 0 \Leftrightarrow \sqrt{2n} -\frac{3}{2}\frac{Ln n}{Ln 2} -\frac{3\sqrt{2}}{Ln2}\frac{Ln{\sqrt{4n}}}{\sqrt{4n}} > 0     

An nou pwouve ekivalans lan 
 

\sqrt{2n} -\frac{3}{2}\frac{Ln n}{Ln 2} -\frac{3\sqrt{2}}{Ln2}\frac{Ln{\sqrt{4n}}}{\sqrt{4n}}>0 \Leftrightarrow \frac{2\sqrt{4n}Ln2\times{\sqrt{2n}}- 3\sqrt{4n}Ln n - 6\sqrt{2}Ln{\sqrt{4n}} }{2\sqrt{4n}Ln 2} > 0 \Leftrightarrow \sqrt{4n}Ln2\times{\sqrt{2n}}- 3\sqrt{4n}Ln n - 6\sqrt{2}Ln{\sqrt{4n}} > 0 

\sqrt{4n}Ln2\times{\sqrt{2n}}- 3\sqrt{4n}Ln n - 6\sqrt{2}Ln{\sqrt{4n}} > 0 \Leftrightarrow 2\sqrt{8n^2}Ln2 -6\sqrt{n}Ln n - 6\sqrt{2}Ln{\sqrt{4n}} \Leftrightarrow 4n\sqrt{2}Ln2 - 6\sqrt{n}Ln n - 6\sqrt{2}Ln{\sqrt{4n}} > 0 \Leftrightarrow 4n\sqrt{2}Ln2 - 6\sqrt{n}Ln n -6\sqrt{2}\left(Ln{\sqrt{4}}+Ln{\sqrt{n}}\right)        \Leftrightarrow 4n\sqrt{2}Ln2 - 6\sqrt{n}Ln n -6\sqrt{2}\left(Ln2 + \frac{1}{2}Ln2\right)>0 

   \Leftrightarrow 4n\sqrt{2}Ln2 - 6\sqrt{n}Ln n -6\sqrt{2}\left(Ln2 + \frac{1}{2}Ln2\right)>0 \Leftrightarrow \frac{4n}{6}\sqrt{2}Ln2 -\sqrt{n}Ln n - \sqrt{2}Ln2 - \frac{\sqrt{2}}{2}Ln n > 0  \Leftrightarrow   \frac{2n}{3}\sqrt{2}Ln2 -\sqrt{n}Ln n - \sqrt{2}Ln2 - \frac{\sqrt{2}}{2}Ln n > 0  \Leftrightarrow \sqrt{2}\times \left( \frac{2n}{3}Ln 2 -\frac{\sqrt{2}}{2}\sqrt{n}Ln n -Ln 2 -Ln \sqrt{n}\right) > 0 

 \Leftrightarrow \sqrt{2}\times \left( \frac{2n}{3}Ln 2 -\frac{\sqrt{2}}{2}\sqrt{n}Ln n -Ln 2 -Ln \sqrt{n}\right) > 0 \Leftrightarrow \left(\frac{2n}{3} - 1\right)Ln2 -\frac{\sqrt{2}}{2}\sqrt{n}Ln n - Ln {\sqrt{n}} > 0  \Leftrightarrow \left(\frac{2n}{3} - 1\right)Ln2 -\frac{\sqrt{2}}{2}\sqrt{n}Ln n - \frac{1}{2}Ln n > 0 \Leftrightarrow \left(\frac{2n}{3} - 1 \right)Ln2  - \left( \frac{1 + \sqrt{2n}}{2}\right)Ln n  > 0 

An nou deziye pa  f\left(n\right)  

 \sqrt{2n} -\frac{3}{2}\frac{Ln n}{Ln 2} -\frac{3\sqrt{2}}{Ln2}\frac{Ln{\sqrt{4n}}}{\sqrt{4n}} 

An nou montre ke li pozitif pou tout  n \ge 2^6 

An etann fonksyon an sou tout  \mathbb{R}

 f\left(x\right) =  \sqrt{2x} -\frac{3}{2}\frac{Ln x}{Ln 2} -\frac{3\sqrt{2}}{Ln2}\frac{Ln{\sqrt{4x}}}{\sqrt{4x}}  

an nou pwouve ke   f\left(2^6\right) > 0 

 f\left(2^6\right) = 2^3\times\sqrt{2} - \frac{3}{2}\times\frac{6Ln2}{Ln2} - \frac{3\sqrt{2}}{Ln2}\times\frac{Ln\sqrt{4\times2^6}}{\sqrt{4\times2^6}} = 8\sqrt{2} - 9 - \frac{3\sqrt{2}}{Ln2}\times\frac{4Ln2}{2^4}  =  8\sqrt{2} - 9 - \frac{12\sqrt{2}}{2^4} =   8\sqrt{2} - 9 - \frac{3\sqrt{2}}{4} = \frac{32\sqrt{2} - 36 - 3 \sqrt{2}}{4} = \frac{29\sqrt{2} -36}{4} 

 f\left(2^6\right) = \frac{29\sqrt{2} -36}{4} = \frac{\left({29\sqrt{2} -36}\right)\times\left({29\sqrt{2} +36}\right)}{4}\times\frac{1} = \frac{\left(29\times29\times2 - 36\times36\right)}{4}\times\frac{1} = \frac{386}{4}\times\frac{1} > 0 

an nou pwouve tou ke  f'\left(x\right) > 0  pou tout  n \ge 2^6 

kidonk fonksyon f la estriteman kwasant

   \left(\sqrt{2x}\right)' = \left( \left(2x\right)^{\frac{1}{2}}\right) = \frac{1}{2}\left(2x\right)^{-\frac{1}{2}}\times2 = \left(2x\right)^{-\frac{1}{2}} = \frac{1}{\sqrt{2x}} = \frac{\sqrt{2x}}{\sqrt{2x}\times \sqrt{2x}} = \frac{\sqrt{2x}}{2x}

 \left(-\frac{3}{2}\frac{Lnx}{Ln2}\right)' = - \frac{3}{2Ln2}\times\frac{1}{x} = \frac{-3}{2xLn2}

 \left(\sqrt{4x}\right)'  = \left( \left(4x\right)^{\frac{1}{2}}\right)' = \frac{1}{2}\times\left(4x\right)^{\frac{1}{2}}\times4 = 2\times\left(4x\right)^{\frac{1}{2}} = \frac{2}{\sqrt{4x}}= \frac{2\sqrt{4x}}{4x} = \frac{\sqrt{4x}}{2x} = \frac{\sqrt{x}}{x}

 \left(Ln\sqrt{4x}\right)' = \frac{1}{\sqrt{4x}}\times\frac{\sqrt{4x}}{2x} = \frac{1}{2x} 

oubyen 

 \left(Ln\sqrt{4x}\right)' = \left(Ln\left( 4x\right)^{\frac{1}{2}}\right)' = \left( \frac{1}{2}Ln{4x}\right)' = \frac{1}{2}\times\frac{1}{4x}\times 4 = \frac{4}{8x} = \frac{1}{2x}

 \left(\frac{Ln\sqrt{4x}}{\sqrt{4x}}\right)' = \frac{\frac{1}{2x}\times\sqrt{4x} - \frac{\sqrt{4x}}{2x}\times Ln{\sqrt{4x}}}{\left({\sqrt{4x}}\right)^2}  =  \frac{\frac{\sqrt{4x}-\sqrt{4x}\times Ln{\sqrt{4x}}}{2x}}{4x}  = \frac{\sqrt{4x}\left(1 - Ln{\sqrt{4x}}\right)}{8x^2}

 \left(\frac{-3\sqrt{2}}{Ln2}\times\frac{Ln\sqrt{4x}}{\sqrt{4x}}\right)' = \frac{\frac{\sqrt{4x}-\sqrt{4x}\times Ln{\sqrt{4x}}}{2x}}{4x}  = \frac{-3\sqrt{2}}{Ln2}\times\frac{\sqrt{4x}\left(1 - Ln{\sqrt{4x}}\right)}{8x^2}

 f'\left(x\right) = \frac{\sqrt{2x}}{2x} - \frac{3}{2xln2}  - \frac{3\sqrt{2}}{Ln2}\times \frac{\sqrt{4x}\times\left( 1- Ln{\sqrt{4x}} \right)}{8x^2}

 f'\left(x\right) =  \frac{4x\times Ln2\times\sqrt{2x} - 12x - 3\sqrt{8x}\left(1 - Ln{\sqrt{4x}}\right)}{8x^2Ln2}

 f'\left(x\right) =  \frac{4x\times Ln2\times\sqrt{2x} - 12x - 3\sqrt{8x} + 3\sqrt{8x}ln{\sqrt{4x}}}{8x^2Ln2}

f'\left(x\right)>0\Leftrightarrow \frac{4x\times Ln2\times\sqrt{2x} - 12x - 3\sqrt{8x} + 3\sqrt{8x}Ln{\sqrt{4x}}}{8x^2Ln2}> 0 \Leftrightarrow 4x\times Ln2\times\sqrt{2x} + 3\sqrt{8x}\times Ln{\sqrt{4x}} > 12x +3\sqrt{8x} 

si nou miltiplye pa  \sqrt{2x} \ne 0

f'\left(x\right)>0\Leftrightarrow \frac{4x\times Ln2\times\sqrt{2x} - 12x - 3\sqrt{8x} + 3\sqrt{8x}Ln{\sqrt{4x}}}{8x^2Ln2}> 0 \Leftrightarrow 4x\times Ln2\times\sqrt{2x} + 3\sqrt{8x}\times Ln{\sqrt{4x}} > 12x +3\sqrt{8x} \Leftrightarrow 4xLn2\times2x + 3\times4x\times Ln{\sqrt{4x}} > 12x \times\sqrt{2x} + 3\times 4x \Leftrightarrow 8xLn2 + 12Ln{\sqrt{4x}} > 12\sqrt{2x} + 12 

sou ipotèz  x > 2^6  an nou pwouve ke    8xLn2 + 12Ln{\sqrt{4x}} > 12\sqrt{2x} + 12 

 x > 2^6 \Rightarrow x> 18 \Rightarrow 16x > 288 \Rightarrow 16x \times Ln4 \times Ln4 > 288\times Lne \times Lne \Rightarrow 2\times 2\times Ln2 \times Ln2 \times 16x \Rightarrow 64x\times \left(ln2\right)^2>288 \Rightarrow 64x^2\times \left(ln2\right)^2>288x \Rightarrow 64x^2\times \left(ln2\right)^2>144\times 2x \Rightarrow 8xln2 > 12\sqrt{2x}

yon lòt kote nou genyen 

 12 Ln\sqrt{4x} > 12 \Leftrightarrow Ln\sqrt{4x} > 1  \Leftrightarrow sqrt{4x} > e \Leftrightarrow 4x > e^2 \Leftrightarrow x > \frac{e^2}{4} 

non genyen tou :

 e 

 0 

 2^6 > \frac{9}{4} > \frac{e^2}{4}  
 
 x > 2^6 \Rightarrow x > \frac{e^2}{4} \Rightarrow 4x > e^2 \Rightarrow \sqrt{4x} > e \Rightarrow Ln\sqrt{4x} > 1 \Rightarrow 12 Ln\sqrt{4x} > 12 

donk 

 x > 2^6 \Rightarrow 8xLn2 + 12Ln{\sqrt{4x}} > 12\sqrt{2x} + 12 

### Lis nonm premye superyè ou egal ak 3 epi ki enferyè ak 64
3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61

## Dezyèm demonstrasyon

## Majorasyon apati postila Bètran

P_n \le {2^n} 

 P_1 = 2 = 2^1 
 2^2 =4 

 P_2 = 3 

 P_3 = 5 

 3^2 
P_3 

an sipoze majorasyon an vrè pou  n > 3  e an montre ke li vrè pou n+1

 P_n \le 2^n \Rightarrow \pi\left(P_n\right) \le \pi\left(2^n\right) \Rightarrow n\le \pi\left(2^n\right)\ge n

kidonk nan entèval \left[1, 2^n\right] gen o mwen n nonm premye selon ipotèz rekirans lan .

selon ipotèz  Bètran  an egziste pou pi piti yon nonm premye nan entèval \left]2^n , 2^{n+1}\right[ 

nou kapab di tou ke nan entèval  \left]2^n , 2^{n+1}\right] 
gen pou piti yon nonm premye

 \pi\left(2^{n+1}\right) - \pi\left(2^n\right)  \ge 1

 \pi\left(2^{n+1}\right) \ge 1+\pi\left(2^n\right) 

 \pi\left(2^{n+1}\right) \ge 1+ n 

gen pou pi piti n+1  nonm premye nan entèval

 \left[1, 2^{n+1}\right]    

 P_{n+1} 

Anfendkont si majorasyon an vrè pou n, li vrè pou n + 1 .

 \forall n \in \mathbb{N} , P_{n}\le 2^n 

.....

P_k\left(n\right) \le {n\times{2^k}}

sa  nou rele 
P_k\left(n\right) se k yèm nonm premye ki siperyè oubyen egal ak n .

Avèk , sipoze fikse an nou montre ke majorasyon an vre pou  k+1  .

P_k\left(n\right) \le n\times 2^k

Anvan nou fè sipozisyon an , an nou verifye ke majorasyon vre pou k = 1 

 
  P_1\left( n\right) \le 2n  paske nan entèval \left] n, 2n \right[   li egziste pou pi piti yon nonm premye selon postila Bètran  .

Anfèt si n premye,  P_1\left(n\right) = n 

si n pa premye  P_1\left(n\right) \in \left]n, 2n\right] 

kidonk  P_1{\left(n\right)} \in \left[n, 2n\right] 

 P_1\left(n\right) \le 2n 

## Postila Bètran avèk de twa fòmil matematisyen Ayisyen Lainé Jean Lhermite Junior konsènan nonm premye

### Ekspresyon nonm premye ran n selon modèl flèch Jonatan  anakò avèk postila Bètran an

  P_n = \sum_{m=1}^{2^n}{\left(    \left[  \frac  {1+\sum_{m=1}^{i}{\left( -\left[ \frac{\left(m!\right)^2}{m^3} \right] -\left[ -  \frac{\left(m!\right)^2}{m^3}   \right]     \right)}  }       {n+1} \right]   \times      \left[ \frac{n+1}  {1+\sum_{m=1}^{i}{\left( -\left[ \frac{\left(m!\right)^2}{m^3} \right] -\left[ -  \frac{\left(m!\right)^2}{m^3}   \right]     \right)}  }    \right] \times i \times \left(  -\left[\frac{\left(i!\right)^2}{i^3}\right] -   \left[-\frac{\left(i!\right)^2}{i^3}\right] \right)       \right)}

### K yèm nonm premye ki siperyè oubyen egal ak n
#### Premye apwòch

P_K\left(n\right) = \left(\sum_{i=1}^{2^k\times n}{\left(\left[\frac{1+\sum_{m=1}^{i}{\left(1 - \left[\frac{\left[\frac{k + 1 + \sum_{m=1}^n{\left(1 - \left[\frac{\left[\frac \right]\times \left[\frac{k + 1 + \sum_{m=1}^{n}{\left(1 - \left[\frac{\left[\frac{ 1 + \sum_{m=1}^i{\left(1 - \left[\frac{\left[\frac \right]\times i \times \left(1 - \left[\frac{\left[\frac\right]\right)\right)}\right)\times\left[\frac{\left[\frac{k+1}\right]\times\left[\frac{k+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)}  

## Gade tou
* Fonksyon ki konte kantite divizè n
* Fonksyon ki konte nonm premye ki sou fòm an+b lè a ak b premye entre yo
* Fonksyon lanmbda Liouville
* Fonksyon pi
* Fonksyon Omega ki konte kantite faktè premye ki rantre nan dekonpozisyon yon nonm antye n
* Fonksyon omega ki konte kantite nonm premye ki diferan ki divize yon nonm n
* Fonksyon phi Euler
* Fonksyon pi, ki konte kantite nonm premye
* Fonksyon psi Tchebycheff
* Fonksyon tèta Tchebycheff
* Fonksyon Von Mangolt
* Fòmil pou nonm konpoze
* Modèl boul wouj ak boul ble
* Modèl flèch Jonatan
* Modèl Lhermite yo
* Modèl mouton lòt patiraj
* Modèl pitit prodig ou modèl ti mouton pèdi
* Modèl predesesè
* Modèl siksesè
* Modèl ti triyang ak gwo triyang
* Modèl triyang ak kwadrilatè
* Nonm pafè
* Nonm premye
* Nonm premye ak modèl boul rouj ak boul ble
* Nonm premye Ezau
* Nonm premye Fermat
* Nonm premye Izaak
* Nonm premye Jakòb
* Nonm premye Mersenne
* Nonm zanmi
* Primoryel
* Pwisansyèl
* Fonksyon ki konte nonm premye ki sou fòm an+b lè a ak b premye entre yo
* Rezolisyon ekasyon premye degre ak sentèz

## Referans

## Lyen deyò