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Progress:

I observed this result:

$\pi(n)\simeq\Bigg\lfloor\dfrac{n}{10}+\dfrac{5}{\sqrt{n}}+\dfrac{\log{n}+n}{\log_2{n}}\Bigg\rfloor$

while studying several inequalities and relations concerning prime gaps and PNT and it tends to be extremely accurate as I tested so far. But how can I prove this one? I would appreciate any help or insight you can provide. Thanks.

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    You know this contradicts the Prime Number Theorem? – 2017-01-04
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    Where did you find this result? – 2017-01-04
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    *Hint*: You can put your term into wellkown inequalities for $\pi(n)$ and check if it holds. See e.g. https://en.wikipedia.org/wiki/Prime-counting_function . – 2017-01-04
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    No, I don't. Can you show me how? @dezdichado – 2017-01-04
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    I accidently found this result by myself while studying about prime numbers. @JanEerland – 2017-01-04
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    PNT says $\pi(n)\sim\dfrac{n}{\log n}$ but yours says $\pi(n)\sim\dfrac{n}{10}$ – 2017-01-04
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    I'm not sure what you are trying to say but that is clearly not what I wrote. My statement says this: $\pi(n)\simeq\Bigg\lfloor\dfrac{n}{10}+\dfrac{5}{\sqrt{n}}+\dfrac{\log{n}+n}{\log_2{n}}\Bigg\rfloor$ @dezdichado – 2017-01-04
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    Can you please show me some results in heuristics and numbers regrading the function I mentioned. @user90369 – 2017-01-04
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    @user18724 Can you show us how you found it?! – 2017-01-04
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    @user18724 : Sorry it's not clear for me what you mean. I think you want to check how far your assumption is correct and you can check this, if you substitute $\pi(n)$ by your term in any inequality for $\pi(n)$, e.g. $\frac{x}{\ln x}<\pi(x)<1.25506 \frac{x}{\ln x}$ for $x\geq 17$ . – 2017-01-04
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    Others have noted how asymptotically it's wrong, but I'm wondering what tests you ran where "it tends to be extremely accurate". Running over some increasing numbers shows it getting worse and worse. For example, $\pi(10^{11}) = 4118054813$. n/log(n) is 0.96 times the real value. Your formula is 2.47 times the real value -- that's a horrible estimate. li(n) is 1.0000028 times the real value, which is quite good. – 2017-01-05

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Divide both sides by $n$: $$\frac{\pi(n)}n\simeq \frac1{10}+\frac5{n\sqrt{n}}+\frac{\log n + n}{n\log_2n}>\frac1{10}$$ So you say that $\frac{\pi(n)}{n}$ goes to $\frac{1}{10}$ as $n$ goes to infinity. But the PNT says that $\pi(n)\sim\frac{n}{\log n}$, so: $$\lim_{n\to\infty}\frac{\pi(n)}{n}=\lim_{n\to\infty}\frac1{\log n}=0<\frac1{10}$$ This means that, sadly, your result is incorrect.