Hanoi tower with forbidden direct move from source to destination
Notation
Let's denote least number of moves needed to move $n$ disks from column $i$ to column $j$ with $T_{ij}$. Let $A$ be the source and $C$ be the destination column.

Recurrence relations
We can say the followings about each $T$:
\begin{equation}
T_{AC}(n)=T_{AC}(n-1)+1+T_{CA}(n)+1+T_{AC}(n-1)
\end{equation}
To move $n$ disks from source to destination, we first have to move the top $n-1$ disks to column $C$ and then move the last one to Column $B$, Then return the $n-1$ disks to $A$ so as to free up destination column, then move the biggest disk to $C$ and once again move the $n-1$ disks to $C$. It's easy to see that this is the optimal set of operations. This equation can be written as:
\begin{equation}
T_{AC}(n)=2T_{AC}(n-1)+2+T_{CA}(n-1)
\end{equation}

For other $T$s we have a different situation:
\begin{equation}
T_{CA}(n)=T_{CB}(n-1)+1+T_{BA}(n-1)\\
T_{CB}(n)=T_{CA}(n-1)+1+T_{AB}(n-1)\\
T_{BA}(n)=T_{BC}(n-1)+1+T_{CA}(n-1)\\
T_{AB}(n)=T_{AC}(n-1)+1+T_{CB}(n-1)\\
T_{BC}(n)=T_{BA}(n-1)+1+T_{AC}(n-1)
\end{equation}

Solving the system of recurrence relations
Let's start with $T_{AB}$ or $T_{BC}$ . We already know:
\begin{equation}
T_{AB}(n)=T_{AC}(n-1)+1+T_{CB}(n-1)
\end{equation}
We try to rewrite $T_{CB}$'s equation by replacing $T_{AB}$ from the above equation.
\begin{align}
T_{CB}(n)&amp;=T_{CA}(n-1)+1+(T_{AC}(n-2)+1+T_{CB}(n-2))\\
&amp;=T_{CA}(n-1)+T_{AC}(n-2)+T_{CB}(n-2)+2
\end{align}
\begin{equation}
T_{CB}(n)-T_{CB}(n-2)=T_{CA}(n-1)+T_{AC}(n-2)+2
\end{equation}

By the same method we get:
\begin{equation}
T_{BA}(n)=T_{CA}(n-1)+T_{AC}(n-2)+T_{BA}(n-2)+2
\end{equation}

It can immediately be seen that $T_{BA}(n)=T_{CB}(n)$. Then we rewrite $T_{CA}$ as:
\begin{equation}
T_{CA}(n)=2T_{CB}(n-1)+1 \Rightarrow
\end{equation}
\begin{align}
T_{CA}(n)-T_{CA}(n-2)&amp;=2(T_{CB}(n-1)-T_{CB}(n-3))\\
&amp;=2(T_{CA}(n-2)+T_{AC}(n-3)+2)\\
&amp;=2T_{CA}(n-2)+2T_{AC}(n-3)+4
\end{align}

Now let's go back to the main relation, to $T_{AC}$:
\begin{equation}
T_{AC}(n)=2T_{AC}(n-1)+2+T_{AC}(n-1) \Rightarrow\\
\end{equation}
\begin{align}
&amp;T_{AC}(n)-3T_{AC}(n-2)\\
=&amp;2(T_{AC}(n-1)-3T_{AC}(n-3))+(T_{CA}(n-1)-3T_{CA}(n-3))=\\
=&amp;2(T_{AC}(n-1)-3T_{AC}(n-3))+(2T_{AC}(n-4)+4)\\
=&amp;2T_{AC}(n-1)-6T_{AC}(n-3)+2T_{AC}(n-4)+4 \Rightarrow
\end{align}
\begin{equation}
T_{AC}(n)=2T_{AC}(n-1)+3T_{AC}(n-2)-6T_{AC}(n-3)+2T_{AC}(n-4)+4
\end{equation}

There! We have reduced the system of recurrence relations to a single relation, involving only the target series. Ok, the rest is easy from here on. I leave it as an exercise to the reader! $\ddot\smile$