Let $f(0)=2.$ Define for positive integers $n$ :
$f(n+1) = \frac{3}{2} f(n)$ if $f(n)$ is even.
$f(n+1) = \frac{3}{2}(f(n)+1)$ if $f(n)$ is odd.
We now have $\lim_{n->\infty} \dfrac{4* (3/2)^{n} }{f(n)} = C$
Where $C$ is some constant.
Apparantly $C$~$\sqrt[11] {10}$ but I guess that $C$ can not be exactly that value ?
What is the closed form for $C$ ?
I did not find a closed form, but you are correct that $C\ne10^{1/11}$.
If $f(n)$ is even $$ \frac{f(n+1)}{f(n)}=\frac32\tag{1} $$ If $f(n)$ is odd $$ \frac{f(n+1)}{f(n)}=\frac32\left(1+\frac1{f(n)}\right)\tag{2} $$ Thus, $$ \frac{f(n)}{f(0)}=\left(\frac32\right)^n\prod_{\substack{k=0\\f(k)\text{ is odd}}}^{n-1}\left(1+\frac1{f(k)}\right)\tag{3} $$ Since $f(n)\ge\left(\frac32\right)^n$, the product in $(3)$ converges.
Therefore, $$ \begin{align} C &=\lim_{n\to\infty}\frac4{f(n)}\left(\frac32\right)^n\\ &=\lim_{n\to\infty}\frac4{f(0)}\left[\prod_{\substack{k=0\\f(k)\text{ is odd}}}^{n-1}\left(1+\frac1{f(k)}\right)\right]^{-1}\\ &=2\left[\prod_{\substack{k=0\\f(k)\text{ is odd}}}^\infty\left(1+\frac1{f(k)}\right)\right]^{-1}\tag{4} \end{align} $$ Computing $C$ using $n=120$ in $(4)$, I get $$ C=1.2328400204796570012\tag{5} $$ which is accurate to $20$ places. The ISC finds nothing for this number.
Note that $$ 10^{1/11}=1.2328467394420661391\tag{6} $$