Let $\pi$ be the prime counting function, and let $p_x$ be a prime number.
Is it possible to show that $\pi(p_x^2) \geq 2p_x -2$?
Alternatively;
Is there an upper bound to $p_x$ in terms of $x$?
Asked
Active
Viewed 138 times
1
Brad Graham
- 1,171
-
1Not to be a swearword about this, but $\pi$ (lowercase) is the usual symbol for this function; the uppercase letter can be confused with the product operator. – Mr. Brooks Mar 12 '16 at 22:42
2 Answers
1
By Prime number theorem,
$$\pi(x)>\frac x{logx}$$
$$\pi(p^2)>\frac{p^2}{2logp}$$
(1):$$\frac{p^2}{2logp}>2p-2$$ Is true if $$\frac{p^2}{2logp}>2p$$
$$p>4logp$$
Using graph( http://m.wolframalpha.com/input/?i=y%3Dx%2Cy%3D4logx&x=0&y=0 ) , the inequality hold for $p>8.61...$
So your original postulate holds at least for prime numbers starting from 11. However, for $p=7$, there are 13 prime numbers under $49$ (it is due to our approximation).
But for $p=5$, there are 9 prime numbers under 25. So it doesnt satisfy the inequality.
For your second question, there are a lot of bounds for the nth prime number, but the results should be hard to prove. Bounds for $n$-th prime
-
Thank you very much. $9\geq 2 \times 5 -2 = 8$ so it does hold doesn't it? – Brad Graham Mar 12 '16 at 18:48
-
@BradGraham yes that was what I mean, it holds for all prime larger than 5 – lEm Mar 13 '16 at 04:43
0
I find here (you have to scroll down a little) that a bound of $p_x < (1.2)^x$ for $x > 25$ is obtained from a result of J. Nagura, cited as "On the interval containing at least one prime number, Proc. Japan Acad. 28 (1952) 177–181". The result used is similar to Bertrand's postulate, which itself gives $p_x < 2^x$, only it is tighter.
There is also this answer here on MSE, though it answers a slightly different problem. The bound stated is $n \ln n + n(\ln\ln n - 1) < p_n < n \ln n + n \ln \ln n$ for $n \geq 6$.
