I have been playing around with this for awhile. Then I went back to looking at e like this..
Oh thanks for the quick reply. I will delve into it in a bit. But first I think I found a way to count the primes simply using e=(1+1/x)^x.Whether we have [imath] n\, , \,n/2 [/imath] or [imath] (n-2)/2 [/imath] only matters for small values of [imath] n. [/imath] We get [imath] e [/imath] on the right for large [imath] n [/imath] in any of these cases.
If I use [imath] \pi_0(x)=x/ \log(x) [/imath] then I get [imath] e [/imath] on both sides for [imath] n\to \infty [/imath], as before. Your formula
[math] \pi_1(x)=\dfrac{1}{\log(x)-e^{1/\log(x)}} [/math] is more or less identical to [imath] x/\log(x) [/imath] and has the same behavior for very large values of [imath] x. [/imath] This means that there are two possibilities: a) We need a theoretical explanation of where this correction term in the denominator comes from, or b) your formula is a better approximation than the known ones for small values of [imath] x. [/imath]
Let us set [imath] c=e^{\frac{1}{\log(x)}} .[/imath] Then the ratio between your and the known formula is
[math] \dfrac{\pi_1(x)}{\pi_0(x)}=\dfrac{1}{1-\dfrac{c}{\log(x)}} [/math] which converges to one, i.e. the relative error of your formula becomes marginal. However, if we calculate the difference between both formulas, then we get [math] \pi_1(x)-\pi_0(x) =\dfrac{x\cdot c}{\log(x)(\log(x)-c)}\sim \dfrac{x}{\log^2(x)} [/math] for large values of [imath] x. [/imath] Hence, the absolute error becomes larger and larger.
There is a nice list of actual values from [imath] x=10 [/imath] to [imath] x=10^{29} [/imath] on Wikipedia. I will link the German version (scroll to the right) because it also contains a column for a fixed value of [imath] c=1.08366 [/imath] which corresponds to roughly [imath] e^{1/\log(250,000)}. [/imath] The English version is similar, but without [imath] \pi_2(x)=\dfrac{x}{\log(x)-1.08366} [/imath] and without links to https://oeis.org/ where the data were taken from. These data should give you a comparison with your function. Note that the decimal separating signs in German are the other way around:
[math] \text{Engl.:}\;84,367.12345 \quad \text{Germ.:}\;84.367,12345[/math]
Primzahlsatz – Wikipedia
de.wikipedia.org
Yes I apologize. I got ahead of myself and didn’t scrutinize my work hard enough. Sorry about that. Thanks again for everything. We respond when I can.
Finally have some time to go through your msgs. Still going over your math. But as for this yes I’m aware of the wiki page and for awhile I was only doing calculations up to 10^29. But me with my prime density obsession lol needed more. Hence my change to Li on wolfram. As for the sequence site I love it. I use it all the time. My favourite sequence I came up with so far is 2,6,15,28,55…. Great for counting primes. Spent months trying to figure out how to kind of factorialize that sequence. But to no avail.
This looks like legendres constant. I’m not going to google if it is or how to spell his name so hopefully I’m correct on both. If you watched my vids on using the golden ratio for prime density I was so excited thinking I found something so interesting only to find out I just made my own constant. Pretty disappointed but also happy to know I made the same mistake legendre made. Lol
I would love to see my formula like this for comparison…Here is the Wikipedia entry on it:
Legendre's constant - Wikipedia
en.wikipedia.org
I think I fingered out a way to use (1+1/x)^x to count the primes. I have used this technique and it works all they way up to the point you see here….
That was not a very Fermat response. You should have said something like… well I’m kind of busy at the moment. Finishing up my prove for the p vs np problem. But if I have a minute or two later on I will peruse over it.
I have no idea if it goes to infinity. I ran out of time lol. Yes power of tens would be better. No idea how to go about that though.
How (where) did you figure out that there are 141748 primes in [imath] [e^{14},e^{15}] [/imath]?
If I understand your calculations correctly, then your proposal is
... and me stupid asked with [imath] \pi(e^{15}) [/imath] what worked well for small numbers, but resulted in [imath] \pi \cdot e^{15} [/imath] for large numbers. I haven't had the idea to phrase it verbally.
