Chapter 2: The nth term of primes

Chapter 2 The nth term of primes

there have lots of formulas for nth term of prime in mathematics but i think what i am describing is very simple and based on simple concept.

1.1 The Euclidean form of prime

P(n) = h*n+b

How the idea came into my mind. I was thinking about the Number of chromosomes of human and in the development of the baby.  Each parent contributes 23 chromosomes and time taken for development of baby is 9 months.   p(9) = 23

                         P(n)=h*n+b                               h= [p(n)/n]                    pn%n=b

[] represent greatest integer function.% sign stands for remainder. H is group number , b is random integer, and n is nth term of prime. Such that

         (-1)^b=------                              when h is even

When h is odd then

             (-1)^b=±±±±                          or         (-1)^b=∓∓∓∓∓

1.2 The Euler’s form of prime

Let highest random integer in a group h is denoted byb∞h . And highest difference in a group h is d∞h .then it has been found that         

              ln(b∞h)>h                               and                                 d∞h<2e^(h/2)

  P(n)=hn+b   now dividing both side by n we get   p(n)n=h+bn    .   Now when h approaches to h+1 then b becomes highest random integer also b→n .

Then putting b=n and b∞h=eh  we get

P(n)=h*n+e^h  also pn= ln(e^n*n^n)

Now this equation will give the precise value of prime so to make it accurate I introduce o (omicron).since we need to change something  it can’t be n so I change e with o(omicron).so the given equation becomes. pn=m+n*ln⁡(n)/ln(o)⁡(o)  . o is  called prime constant.

              ln(o)= n*ln(n)/(p(n)-n)     range of ln(o) is found to be

                         Logπ8.3<lno<0

Also ln(o) is maximum at p(4)=7. If ln(o)→ 0 then p(n)→∞ which simply means that no.of primes are infinite.