The Fifth Conjecture

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Author : Spanu Dumitru Viorel Address: Street Marcu Mihaela Ruxandra no. 5 , 061524 , flat 47 , Bucharest , Romania E-mail : [email protected] [email protected] [email protected] Phones : +4021413107 0731522216 The English version will be soon ready .

Conjectura a 5 – a O conjectura scrisa de Spanu Dumitru Viorel Fie n {0} . Fie

un numar natural

par , adica

n€N\

i un numar natural si i

n = ------2

.

Atunci 1

∑ p

------p

=

p + n prim

unde Ki

sunt toate

constante .

Ki

,

Secventa

1 ----pα

se repeta doua ori in fiecare

serie , atunci cind gapul egal cu n apare prima oara si apoi nu se mai repeta in cadrul seriei .

Remarca : Cind n este egal cu 2 , atunci K1 = B , unde B este Constanta lui Brun si are valoarea B = 1,902160583104 … . Altfel

spus ( Conjectura a 5 – a ) :

Seriile formate cu reciprocele tuturor numerelor prime care alcatuiesc perechi de numere prime distantate printr-un gap de 2 , sau un gap de 4 , … , sau un gap de n , sunt toate serii convergente , fiind egale , fiecare dintre aceste serii convergente , cu o constanta ;

Aceste constante care reprezinta limitele spre care tind seriile convergente amintite mai sus , sunt diferite intre ele . Aceste serii convergente formate in modul precizat mai sus sunt in numar infinit . Secventa

1 ----pα

se repeta doua ori in fiecare

serie , atunci cind gapul egal cu n apare prima oara si apoi nu se mai repeta in cadrul seriei .

Exemple : 1/3 + 1/5 + 1/5 + 1/7 + 1/11 + 1/13 + 1/17 + 1/19 + 1/29 + 1/31 + 1/41 + 1/43 + 1/59 + 1/61 + 1/71 +1/73 … = B = Constanta lui Brun = 1,902160583104 … 1/3 + 1/7 + 1/7 + 1/11 + 1/13 + 1/17 + 1/19 + 1/23 + 1/37 + 1/41 + 1/43 + 1/47 + 1/67 + 1/71 + … = K2 .

1/5 + 1/11 + 1/ 11 + 1/17 + 1/23 + 1/29 + 1/31 + 1/37 + 1/41 + 1/47 + 1/53 + 1/59 + 1/61 + 1/67 + … = K3 1/3 + 1/11 + 1/11 + 1/19 + 1/23 + 1/31 + 1/29 + 1/37 + 1/53 + 1/61 + 1/59 + 1/67 + 1/71 + 1/79 + … = K4 1/3 + 1/13 + 1/13 + 1/23 + 1/7 + 1/17 + 1/19 + 1/29 + 1/31 + 1/41 + 1/37 + 1/47 + 1/43 + 1/53 + 1/61 + 1/71 + 1/73 + 1/83 + 1/79 + 1/89 + … = K5 ………………………………………………………………

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