Solution:
Let $m$ and $n$ be positive integers and $n=m q+r, 0 \leq r<m$. Then $\left[\frac{n+m-1}{m}\right]=q+1+\left[\frac{r-1}{m}\right]$ and $\frac{n}{m}=q+\frac{r}{m}$. Hence $\left[\frac{n+m-1}{m}\right] \geq \frac{n}{m}$ with equality if and only if $m$ divides $n$.

Applying the above inequality we obtain
$$2.3^{n}=\prod_{i=1}^{k}\left[\frac{a_{i}+a_{i-1}-1}{a_{i-1}}\right] \geq \prod_{i=1}^{k} \frac{a_{i}}{a_{i-1}}=\frac{a_{k}}{a_{0}}=2.3^{n}$$
Therefore $a_{i-1}$ divides $a_{i}$ for $i=1,2, \ldots, k$. We have to find the number of the sequences $1=a_{0}<a_{1}<\cdots<a_{k}=2.3^{n}$ such that $a_{i-1}$ divides $a_{i}$ for $i=1,2, \ldots, k$. Starting by an arbitrary sequence we construct a sequence of $k$ symbols $\star$, $n$ digits 3 and one digit 2 in the following way: if $\frac{a_{i+1}}{a_{i}}=2^{p} 3^{q}$, $i=1,2, \ldots, k-1$ then we put $p$ digits 2 and $q$ digits 3 between $\star$ number $i$ and $\star$ number $i+1$. Since $a_{i}<a_{i+1}$ there are no two consecutive symbols $\star$.

It is clear that any sequence of $k$ symbols $\star$, $n$ digits 3 and one digit 2 with no two consecutive symbols $\star$ corresponds to a sequence $1=a_{0}<a_{1}<\cdots<a_{k}=2.3^{n}$.