I have obtained an equation which needs to be approximated in order to be true. The equation is -
$ 2^{ \epsilon_1 } - 2^{\frac{- \log(n!)}{2}+ \frac{\epsilon_1}{2} +1}=1$
$\epsilon_1$ is a decreasing function of $n$ and $0< \epsilon_1 < 1/2$. We need to take floor of the left hand side, if we consider the equation as true.
To be precise- $\epsilon_1= -\log(2^{n-s_2(n)}-2^{\frac{-1}{2}(( \log(n!)+ \epsilon_1)- 2( n- s_2(n)-1))}) +\log(2^{n-s_2(n)}) $
Where $s_{2}(n) $ is the number of $1$'s in the binary representation of $n$.
But I have not used any kind of approximation in the whole derivation.
How Can it be explained?
Denote $c=2^{-\ln(n!)/2}\ll 1$, $x=2^{ϵ_1/2}>1$ so you get the quadratic equation $$ x^2-2cx=1\iff(x-c)^2=1+c^2\\\iff x=c+\sqrt{1+c^2} =1+\frac{2c}{\sqrt{1+c^2}+1-c} $$ so essentially $x=1+c$ or $$ ϵ_1=2\ln(1+2^{-\ln(n!)/2})\approx 2^{1-\ln(n!)/2} $$ which is small and falling with $n$.