The answer is obviously \(1\) for \(N = 2, 3\). Below, we assume \(N>3\).

Let us consider the case \(c_{\mathrm{AB}} =\) B. (The case \(c_{\mathrm{AB}} =\) A is almost the same.)

If \(c_{\mathrm{BB}} =\) B, there is only one string that can be made. Below, we assume \(c_{\mathrm{BB}} =\) A.

If \(c_{\mathrm{BA}} =\) A here, we can construct every string that begins with AB and ends with B, so the answer is \(2^{N-3}\).

If \(c_{\mathrm{BA}} =\) B, among the strings that begin with AB and end with B, we can construct everything that does not contain AA as a substring. In this case, the answer is \(F_{N-2}\), where \(F\) is defined as follows:

\( F_{n} = \begin{cases} 1 & \text{if } n \leq 1 \cr F_{n-1} + F_{n-2} & \text{otherwise} \end{cases}\)