a) Y cannot appear two or more times
Then
- if Y does not appear, we are left with a binary (X,Z) string of length $n=N$;
- if Y appears once, by removing it, we are left with two binary (X,Z) string of length $n$ and $N-n-1$, with $0 \le n \le N-1$.
b) The string does not contain one (or more) runs of three (or more) consecutive X
Consider a binary string with $s$ $X\; \leftrightarrow \,1$'s and $m$ $Z\; \leftrightarrow \,0$'s in total.
The number of these strings in which the runs of consecutive ones have a length not greater than $r$, is given by
$$N_{\,b} (s,r,m+1) = \text{No}\text{. of solutions to}\;\left\{ \begin{gathered}
0 \leqslant \text{integer }x_{\,j} \leqslant r \hfill \\
x_{\,1} + x_{\,2} + \cdots + x_{\,m+1} = s \hfill \\
\end{gathered} \right.$$
which is equal to
$$
N_b (s,r,m + 1)\quad \left| {\;0 \leqslant \text{integers }s,m,r} \right.\quad
= \sum\limits_{\left( {0\, \leqslant } \right)\,\,k\,\,\left( { \leqslant \,\frac{s}{r}\, \leqslant \,m + 1} \right)}
{\left( { - 1} \right)^k \left( \begin{gathered}
m + 1 \\
k \\
\end{gathered} \right)\left( \begin{gathered}
s + m - k\left( {r + 1} \right) \\
s - k\left( {r + 1} \right) \\
\end{gathered} \right)}
$$
as thoroughly explained in this and this other posts.
In our case $r=2$, and for a string of length $n$ we shall put $m=n-s$ and sum for $0 \le s \le n$
$$
S(n)\quad = \sum\limits_{\left( {0\, \le } \right)\,s\,\left( { \le \,n} \right)} {\sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{s \over 2}\,} \right)}
{\left( { - 1} \right)^k \binom{n-s+1}{k} \binom{n-3k}{s-3k}
} }
$$
For $n=0,1,2,\cdots ,6$ we obtain that $S(n)$ equals
$$1, 2, 4, 7, 13, 24, 44, \cdots$$
c) Conclusion
Standing what said in point a) we can conclude that the sought number $T(N)$ is given by
$$
T(n) = S(N) + \sum\limits_{0\, \le \,n\, \le \,N - 1} {S(n)\,S(N - 1 - n)}
$$
For $n=0,1,2,\cdots ,8$ $T(n)$ results to be
$$1, 3, 8, 19, 43, 94, 200, 418, 861, \cdots$$
which checks correctly with a direct count.
