I am trying to express the following in term of n, $$\underbrace{\hbox{$ \left \lceil \frac{3}{2}.\left \lceil { \frac{3}{2}.\left \lceil { \ldots \left \lceil \frac{3}{2}.\left \lceil { \frac{3}{2}.2 } \right \rceil \right \rceil \ldots } \right \rceil } \right \rceil \right \rceil$}}_{\hbox{n times}}$$ We can easily see that, if there exits no ceil in the expression, then it is equal to $2.(\frac{3}{2})^{n}$. But in this case, is it might equal to $ 2.(\frac{3}{2})^{n} + f(n)$ ?
Also, when we think as each $\frac{3}{2}=1+\frac{1}{2}$, this problem is equal to following recurrence equation $$a_n=a_{n-1}+ \left \lceil \frac{a_{n-1}}{2} \right \rceil $$ where $a_0=2$, $a_1=3$, $a_2=5$. If this recurrence can be expressed in term of n, this will correspond to the problem.
I don't know whether it works.
Do you have any idea about that ?
Thanks in advance.