Prove by induction that $(Y_n)_{n\ge0}$ is:
$\frac1 2$$\begin{bmatrix}\frac{2^{n-1}+1} {2^n} & 1 & \frac{2^{n-1}-1} {2^n}\\\frac1 2 & 1 & \frac 1 2 \\ \frac{2^{n-1}-1} {2^n} & 1 & \frac{2^{n-1}+1} {2^n}\end{bmatrix}$
is the n-step matrix for
$Y_0$:
$\begin{bmatrix}0.5 & 0.5 & 0\\ 0.25 & 0.5 & 0.25 \\ 0 & 0.5 & 0.5\end{bmatrix}$
I think $(Y_n)_{n\ge0}$ is wrong because if you set $n=0$ you get negative values for the top right and bottom left index, which you can't have a stochastic matrix.
Let $$ Y_1 = \left( \begin{array}{ccc} \frac{1}{2} & \frac{1}{2} & 0 \\ \frac{1}{4} & \frac{1}{2} & \frac{1}{4} \\ 0 & \frac{1}{2} & \frac{1}{2} \\ \end{array} \right).$$ Assume for some $n\geqslant 1$ that $$ Y_n = \frac{1}{2} \left( \begin{array}{ccc} 2^{-n}+\frac{1}{2} & 1 & 2^{-n}-\frac{1}{2} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ 2^{-n}-\frac{1}{2} & 1 & 2^{-(n+1)}+\frac{1}{2} \\ \end{array} \right). $$ Then we have \begin{align} Y_{n+1} &= Y_1Y_n\\ &= \frac{1}{2}\left( \begin{array}{ccc} \frac{1}{2} & \frac{1}{2} & 0 \\ \frac{1}{4} & \frac{1}{2} & \frac{1}{4} \\ 0 & \frac{1}{2} & \frac{1}{2} \\ \end{array} \right) \left( \begin{array}{ccc} 2^{-n}+\frac{1}{2} & 1 & 2^{-n}-\frac{1}{2} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ 2^{-n}-\frac{1}{2} & 1 & 2^{-(n+1)}+\frac{1}{2} \\ \end{array} \right)\\ &= \frac{1}{2} \left( \begin{array}{ccc} 2^{-(n+1)}+\frac{1}{2} & 1 & 2^{-(n+1)}-\frac{1}{2} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ 2^{-(n+1)}-\frac{1}{2} & 1 & 2^{-(n+1)}+\frac{1}{2} \\ \end{array} \right), \end{align} so by induction the claim holds for all $n\geqslant 1$.