Is my estimate of the number of primes less than N = n/LOGn + n/(LOGn*LOG(n/10)) always better than n/LOGn or even n/(LOGn-1)

Question

My estimates are always better up to 10^13 which is the highest number I can calculate on my calculator.
For 10^13 the actual number of primes is 346065536839

using n/LOGn (LOG = the natural Log ) gives 334039269656, a difference of 12026267183

Using n/(LOGn-1) gives 345618934387 a difference of 446602452

my equation n/LOGn + n/(LOGn*LOG(n/10)) yields 346128563190 a difference of 63026351
but will this improvement always be the case?

Answer 1

The PNT gives that $|\pi(n)-li(n)| \leq \frac{n}{\ln^4 n}$ where $li(n) = \int \limits_{2}^{n} \frac{dt}{\ln t}$ so $ \pi(n) = \frac{n}{\ln n} +\frac{n}{\ln^2 n}+\frac{2n}{\ln^3 n} + O(\frac{n}{\ln^4 n})$

And

1) $|\pi(n)-\frac{n}{\ln n}| = \frac{n}{\ln^2 n}+\frac{2n}{\ln^3 n}+O(\frac{n}{\ln^4 n})$

2) $|\pi(n)-\frac{n}{\ln n-1}| = \pi(n) - \frac{n}{\ln n} \frac{1}{1-\frac{1}{\ln n}} = \pi(n) - \frac{n}{\ln n} \sum \limits_{j=0}^{\infty} (\frac{1}{\ln n})^j = \frac{n}{\ln^3 n} + O(\frac{n}{\ln^4 n})$

3) $|\pi(n)-\frac{n}{\ln n}-\frac{n}{\ln n(\ln n-\ln 10)}| = -\frac{n}{\ln^2 n}-\frac{2n}{\ln^3 n}+O(\frac{n}{\ln^4 n}) + \frac{n}{\ln^2 n} \frac{1}{1-\frac{\ln 10}{\ln n}} =-\frac{n}{\ln^2 n}-\frac{2n}{\ln^3 n}+O(\frac{n}{\ln^4 n}) + \frac{n}{\ln^2 n} \sum \limits_{j=0}^{\infty} (\frac{\ln 10}{\ln n})^j = -\frac{n}{\ln^2 n}-\frac{2n}{\ln^3 n}+O(\frac{n}{\ln^4 n})+ \frac{n}{\ln^2 n} ( 1+\frac{\ln 10}{\ln n} +O(\frac{1}{\ln ^2 n})) = n\frac{\ln 10-2}{\ln^3 n} +O(\frac{n}{\ln^4 n}) \approx \frac{0.3n}{\ln^3 n} + O(\frac{n}{\ln^4 n})$

So in conclusion your estimate is the best of the three, yet studying ($li,\pi$) functions we can give better estimate which is just $\frac{n}{\ln n}+\frac{n}{\ln^2 n}+\frac{2n}{\ln^3 n}$ which is better than your estimate.