A prime counting function I have never seen

That was a beautiful start. I have seen before the prime number theorem which is \(\displaystyle \sim \frac{x}{\ln x}\) as well as Ramanujan's formula \(\displaystyle \frac{x}{\ln x - 1}\).

Also the \(\displaystyle \text{Li}(x)\) function is very famous in the integral world.

\(\displaystyle \text{Li}\left(10^{300}\right) = \int_{2}^{10^{300}} \frac{1}{\ln x} \ dx \approx 1.449750053 \times 10^{297}\)

So, the main idea of this thread is that your discovery is closer to the \(\displaystyle \text{Li}(x)\) function than other approximation methods?

binbots said:
It gets more accurate the higher you go.
Click to expand...
Now I understand what you mean here because I know that the logarithmic integral \(\displaystyle [ \ \text{Li}(x) \ ]\) yields better approximation for the number of prime numbers as \(\displaystyle x\) gets bigger.

Let me apply your formula to my example.

\(\displaystyle \large \pi(100) = \frac{100}{\ln 100 - e^{\frac{1}{\ln 100}}} \approx 30\)

It is still not a bad approximation compared to the actual value which is \(\displaystyle 25\) while we used here a very small number.
 
logistic_guy said:
That was a beautiful start. I have seen before the prime number theorem which is \(\displaystyle \sim \frac{x}{\ln x}\) as well as Ramanujan's formula \(\displaystyle \frac{x}{\ln x - 1}\).

Also the \(\displaystyle \text{Li}(x)\) function is very famous in the integral world.

\(\displaystyle \text{Li}\left(10^{300}\right) = \int_{2}^{10^{300}} \frac{1}{\ln x} \ dx \approx 1.449750053 \times 10^{297}\)

So, the main idea of this thread is that your discovery is closer to the \(\displaystyle \text{Li}(x)\) function than other approximation methods?


Now I understand what you mean here because I know that the logarithmic integral \(\displaystyle [ \ \text{Li}(x) \ ]\) yields better approximation for the number of prime numbers as \(\displaystyle x\) gets bigger.

Let me apply your formula to my example.

\(\displaystyle \large \pi(100) = \frac{100}{\ln 100 - e^{\frac{1}{\ln 100}}} \approx 30\)

It is still not a bad approximation compared to the actual value which is \(\displaystyle 25\) while we used here a very small number.
Click to expand...
Yes the goal here is to find the best approximation only using arithmetic in order to get a more intuitive understanding of the primes.
 
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