Basically how many times we should roll the dice to get $n$ even numbers back to back? But the catch is we have already started the trial and we got $m$ even numbers, where $m \ge 0$ and $m < n$.
That means I have rolled already $m$ times and I got even number every time. Now what is the expected number of additional rolls that require to get $n$ consecutive even numbers?
Call this number $e(m,n)$, and also set $e(n,n)=0$ (if you've just had a run of $n$ consecutive evens you need no more rolls).
Conditioning on the next roll gives, for $0\le m<n$, $$e(m,n)=1+\frac{e(m+1,n)+e(0,n)}{2}.$$ Subtracting two consecutive instances gives $$e(m+1,n)-e(m,n)=\frac{e(m+2,n)-e(m+1,n)}{2}$$ for $0\le m\le n-2$. Therefore $$e(m+1,n)-e(m,n)= 2^mA$$ with $A=e(1,n)-e(0,n)$. Adding these up gives $$e(n,n)-e(0,n)=1-e(0,n)=A(1+2+4+\cdots+2^{n-1}).$$ This provides a relation between $e(1,n)$ and $e(0,n)$ and from the initial relation we get another: $$e(0,n)=1+\frac{e(1,n)+e(0,n)}{2}.$$ These should be enough to find $e(0,n)$, $e(1,n)$ and so then $A$ and all the $e(m,n)$.