I have a seemingly straightforward homogeneous second degree recurrence relationship.
$$Z(a+2) = 3 Z(a+1) - 2 Z(a) $$
If given boundary condition of $a$ at 0 and 1, I could solve it using standard methods. I won't bore people with detail here.
However, if boundary condition is given not at 0 and 1, but rather at 0 and $N$, is this still solvable? $Z(a)$ would be undefined for $a < 0 $ or $a > N$
Essentially I'm given the following. $$Z(a) = 0 \space \space \space if \space a = 0.$$ $$Z(a) = 1 \space \space \space if \space a = N$$ $$Z(a) = 3 Z(a-1) - 2 Z(a-2) \space \space \space if \space 0 < a < N $$
We have $$Z(a+2)-2Z(a+1)=Z(a+1)-2Z(a)$$ so, $$Z(a+1)-2Z(a)=Z(a)-2Z(a-1)=\cdots =Z(1)-2Z(0)=Z(1)\tag1$$
On the other hand, we have $$Z(a+2)-Z(a+1)=2(Z(a+1)-Z(a))$$ so, $$Z(a+1)-Z(a)=2(Z(a)-Z(a-1))=\cdots =2^a(Z(1)-Z(0))=2^aZ(1)\tag2$$
Now $(1)-(2)$ gives $$-Z(a)=Z(1)-2^aZ(1),$$ i.e. $$Z(a)=(2^a-1)Z(1)$$
So, if $Z(N)=1$, then you have $$Z(N)=(2^N-1)Z(1)=1.$$
So, you can set $Z(1)=\frac{1}{2^N-1}$ for a given $N\gt 1$.