Are my calculations of a recursive prime-generating function based on logarithms correct?

So say that for some $C\in\Bbb{R}$, we have: $$p_n\sim C^{(\log2)(n+3)}$$ Now: $$L_n\sim\dfrac{\log(C^{(\log 2)(n+3)}+L_{n-1})}{\log 2^{n+3}}$$ We already know that $L_{n-1} < 1$, while $C^{(\log 2)(n+3)}$ tends to infinity: $$L_n\sim \dfrac{\log(C^{(\log 2)(n+3)})}{\log 2^{n+3}} = \dfrac{1}{\log 2^{n+3}}\cdot \log(C^{(\log 2)(n+3)})=\\ \log(C^{\dfrac{(\log 2)(n+3)}{\log 2^{n+3}}})=\log(C^{\dfrac{(\log 2)(n+3)}{(\log 2)(n+3)}})= \log C$$ We can now say: $$L_n \sim C \Longleftrightarrow p_n \sim e^{C^{(\log 2)(n+3)}}$$ So you'll still have some Mills-like constant involved.