Counting binary strings of length $n$ that contain no two adjacent blocks of 1s of the same length? Is it possible to count exactly the number of binary strings of length $n$ that contain no two adjacent blocks of 1s of the same length? More precisely, if we represent the string as $0^{x_1}1^{y_1}0^{x_2}1^{y_2}\cdots 0^{x_{k-1}}1^{y_{k-1}}0^{x_k}$ where all $x_i,y_i \geq 1$ (except perhaps $x_1$ and $x_k$ which might be zero if the string starts or ends with a block of 1's), we should count a string as valid if $y_i\neq y_{i+1}$ for every $1\leq i \leq k-2$.
Positive examples : 1101011 (block sizes are 2-1-2), 00011001011 (block sizes are 2-1-2), 1001100011101 (block sizes are 1-2-3-1)
Negative examples : 1100011 (block sizes are 2-2), 0001010011 (block sizes are 1-1-2), 1101011011 (block sizes are 2-1-2-2)
The sequence for the first $16$ integers $n$ is:  2, 4, 7, 13, 24, 45, 83, 154, 285, 528, 979, 1815, 3364, 6235, 11555, 21414. For $n=3$, only the string 101 is invalid, whereas for $n=4$, the invalid strings are 1010, 0101 and 1001.

Here I am going to use generating functions like in this answer to a related problem to compute columns of @RobPratt table for $k \ge 3$.
We can define:
$$S_y(k,i) = \left\{\text{n. of solutions for} \sum_{j=1}^{k-1} y_j = i \text{ with } y_j \neq y_{j+1}\right\} \tag{1}\label{1}$$
and then scompose the problem as follows:
$$\left\{\text{n. of solutions for} \sum_{j=1}^k x_j + \sum_{j=1}^{k-1} y_j = n-2k+3 \right\}=\\ \sum_{i=0}^{n-2k+3}\left\{\text{n. of solutions for} \sum_{j=1}^k x_j = n-2k+3-i \right\}S_y(k) =\\ \sum_{i=0}^{n-2k+3}{n-k+2-i \choose k-1}S_y(k,i) \tag{2}\label{2}$$
When $k=3$, the problem of determining $S_y(k,i)=S_y(3,i)$ is all the same as in the above linked problem, only with $2$ variables instead of $4$. Instead of repeating all calculations we can reuse the above answer, removing all terms with an exponent for $y$ greater than $2$, to get the generating function:
$$f(x)=\left[\frac{y^2}{2!}\right]\prod_{n\ge0}(1+yx^n) = \left[\frac{y^2}{2!}\right]\left( 1+\frac y{1-x}+ \frac12\frac{y^2}{(1-x)^2}\right)\left( 1-\frac12\,\frac{y^2}{1-x^2}\right)=\\ \frac{1}{(1-x)^2}-\frac{1}{1-x^2}=\sum_{n=0}^{\infty}\left\{\frac 12 \left[1+(-1)^{n+1}\right]+n\right\}x^n$$
where in the last step I have used WolframAlpha because I am lazy, and then:
$$S_y(3,i) = [x^i]f(x) = \frac 12 \left[1+(-1)^{i+1}\right]+i \tag{3}\label{3}$$
OK, yes, using generating functions for $k = 3$ and $y_1+y_2=i$ is a little overkill, because the $\eqref{3}$ result is obvious (once we choose a value for $y_1$, and this can be done in $i+1$ ways, then $y_2$ is determined; after that the first addendum is needed to discard the $y_1=y_2=i/2$ solution when $i$ is even).
Anyway, replacing in $\eqref{2}$ we obtain the formula for the third column of @RobPratt table:
$$\sum_{i=0}^{n-3}{n-1-i \choose 2}\left\{\frac 12 \left[1+(-1)^{i+1}\right]+i\right\}=\\ \frac 1{48} (2 n^4 - 8 n^3 + 4 n^2 + 8 n + 3 (-1)^n - 3)\tag{4}\label{4}$$
where again I have used WolframAlpha for the last step (verified against @RobPratt table here).
Still thinking how to extend this to $k \gt 3$...
