Collected wisdom of Rahul k well , you are trying to count exact no. of prime but actually it is not possible accurately, but rather possible precisely. let's invent a function i called it god function. if y=p^a+q^b,then d(y)=ap+qb with the help of it you can calculate the number of primes.ex d(112)=15 ,113=p(30),also 15*2=30,precise formula of nth term of prime number is n=z*d(p1) where z=1,2,3,... 
Collected wisdom of Rahul k i am not a mathematician but a lover, i think in near by future there will be lot's of importance of prime number will be (especially in atomic physics). 
Collected wisdom of Rahul k definition of God: The thing which is always YOUNG. mathematics could be consider as always young,and in mathematics pi could be consider as always young. do you agree? e^π= p(9)/〖10〗^0 +(r^(1) (9))/〖10〗^2 +d(9)/〖10〗^4 +(h^(1) (9))/〖10〗^6 +(h^(1) (d(9))/〖10〗^8 +(r^(1) (9))/〖10〗^10 +(h^(1) (9))/〖10〗^12 +(h^(1) (d(9) ))/〖10〗^14 +9/〖10〗^15 
PI inventing formula freinds i need your help to develop the formula for pi,if k is developed this formula will produce aroud 20 digit per term(highest till now). 1/π=(√3√2)√(2^9&log_5〖((27√24√3)/〖(0.999)〗^2 〗 √(2^22&(cos(k^0))/130)) could you help me to find the series of k(k is in degrees) 
Two suggestions:

let root z= x^2 then apply it you will get the answer. 
Prime Numbers: 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)=hn+b h=[P(n)/n] p(n)%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=∓∓∓∓∓ 
Prime Numbers: 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+b/n . Now when h approaches to h+1 then b becomes highest random integer also b→n. Then putting b=n and b_∞^h=e^h we get P(n)=hn+e^h also p(n)= 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. p(n)=m+(n*ln(n))/(ln(o)) . o is called prime constant. ln(o)= (nln(n))/(p(n)n) range of ln(o) is found to be Log_π8.3<ln(o)<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. 
Prime Numbers: Fibonacci form of prime I had observed that highest random integer in a group h is equal to f_(h+2)^2.f is fibonnacci sequence.then p(n) could be writtern as p( n)=f_(h+2)^2 (h+1)+y where y is fibonnacci coefficient of prime. This formula could be used to calculate last prime of the group precisely. 
Prime Numbers: Ramunujan's form of prime Reciprocal of prime is define as R(n,h)= 〖10〗^h/(p(n)) , it has been found that Ln(R(n,h))=h+x(n) where x(n) is ramunujan’s factor for prime. p(n)= 〖10〗^h e^(hx(n)) Also if p_i^h be the first prime of group h then it has been found that p_i^h≥e^h √h 
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