Assume we had an exact formula for $\pi (n)$, how could we get from that formula an exact expression for the $n$th prime?
I tried looking at approximations we have of $\pi (n)$ like $\frac {n}{\ln (n)}$, and from that I was able to state an approximation for the $n$th prime as $n \ln (n)$, which, with the previous definition, could be expressed as $\frac{n^2}{\pi (n)}$ but this does not seem correct.
If we had a correct expression for $\pi (n)$, how could we calculate the $n$th prime as a function of $\pi (n)$?
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GuPe
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Related: http://math.stackexchange.com/questions/507178/most-efficient-algorithm-for-nth-prime-deterministic-and-probabilistic – DanaJ Sep 04 '15 at 17:04
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1Use $n=\pi(p_n)$ and write this with your formula of $\pi(n)$. – Dietrich Burde Sep 05 '15 at 18:32
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Maybe it's too late, But I think I should answer this.
Now if you're trying to compute the nth prime, it has been proven that the nth prime must be greater than
nlog(n)+n(loglog(n))−1)
and less than
nlog(n)+nlog(log(n))
When n≥6. So if you're searching for the nth prime, you can use these ranges to search the nth prime number. Here you can use binary search to get the nth Prime number.
l = n * log(n) + n*(log(log(n))-1);
u = n * log(n) + n*log(log(n));
while(l != u){
m = l + (u - l) /2;
if(pi(m) >= n) u = m;
else l = m + 1;
}
so m will me the prime number you are looking for.
where pi(m) is prime count function (number of primes less than m).
I use above concept to solve on SPOJ problem. http://www.spoj.com/problems/NTHPRIME/
Lakshman
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2The downside is that as n becomes larger, it becomes much cheaper to use a better starting estimate and sieve the small difference, versus repeatedly calling prime count. SPOJ uses small values for n, but even at 10^9 there would be a pretty big performance difference. The faster way is described in the math.stackexchange link. – DanaJ Aug 01 '17 at 16:00