Is$$\pi(N+2)\sim \sum_{p_{n+2}\in\Bbb P}^{N+2} p_n^{\frac{1}{\log(p_{n+2})}}=3^{\frac{1}{\log(5)}}+5^{\frac{1}{\log(7)}}+11^{\frac{1}{\log(13)}}+17^{\frac{1}{\log(19)}}+...+N^{\frac{1}{\log(N+2)}}$$ where $N+2\in \Bbb P?$
When $N+2=181$
$\pi(181)=42$ and the summation is approximately equal to $33.7.$ The difference is about $8.$
What is the difference when $N+2=8011 ? $
I think the difference between the sum and the prime counting function will continue to increase slowly but they will be asymptotic.
First of all, note that the individual terms you have are of the form $e^{\log(p_n)/\log(2+p_n)}$ (I'm writing it this way since it seems like you mean $p_n+2\in\mathbb{P}$, i.e., that you're interested in the twin primes, but you've written $p_{n+2}$; please clarify if I'm wrong!). Now, $\log(x+2)$ $=\log(x)+\log(1+2/x)$ $\approx\log(x)+2/x$, so $\log(x)/\log(x+2)\approx \log(x)/(\log(x)+2/x)$ $\approx 1-\frac2{x\log x}$. Similarly, near $x=1$, $e^x\approx ex$, so each of the individual terms is roughly $e\cdot(1-\frac2{x\log x})$. Since there are $\pi_2(N)$ terms in your sum, you should expect that the value of your sum is roughly $e\pi_2(N)$; best estimates are that this is $\approx e\frac{N}{\log^2(N)}$, whereas $\pi(N)\approx\frac{N}{\log N}$, so the two are off by a logarithmic factor.