Problem: An ant makes a sequence of moves on a cube where a move consists of walking from one vertex to an adjacent vertex along an edge of the cube. Initially the ant is at a vertex of the bottom face of the cube and chooses one of the three adjacent vertices to move to as its first move. For all moves after the first move, the ant does not return to its previous vertex, but chooses to move to one of the other two adjacent vertices. All choices are selected at random so that each of the possible moves is equally likely. The probability that after exactly $8$ moves that ant is at a vertex of the top face on the cube is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution: For all positive integers $k,$ let
The base case occurs at $k=1,$ from which $\left(N(1,\mathrm{BB}),N(1,\mathrm{BT}),N(1,\mathrm{TB}),N(1,\mathrm{TT})\right)=(2,1,0,0).$
Suppose the ant makes exactly $k$ moves for some $k\geq2.$ We perform casework on its last move:
Alternatively, this recursion argument is illustrated below, where each dashed arrow indicates $1$ way, and each solid arrow indicates $2$ ways: [asy] /* Made by MRENTHUSIASM */  size(9cm);  pair A, B, C, D, E, F, G, H, X, Y;  A=(0,6);  B=(0,4);  C=(0,2);  D=(0,0);  E=(10,6);  F=(10,4);  G=(10,2);  H=(10,0);  X=(-1,8);  Y=(11,8);  label("BB", A, (-2,0));  label("BT", B, (-2,0));  label("TB", C, (-2,0));  label("TT", D, (-2,0));  label("BB", E, (2,0));  label("BT", F, (2,0));  label("TB", G, (2,0));  label("TT", H, (2,0));  label("\textbf{The \boldmath{$k$}th Move}", shift(0.3,0)*X);  label("\textbf{The \boldmath{$(k+1)$}th Move}", shift(-0.3,-0.085)*Y);  draw(A--E,0.8+black+dashed,EndArrow);  draw(A--F,0.8+black+dashed,EndArrow);  draw(B--H,0.8+black,EndArrow);  draw(C--E,0.8+black,EndArrow);  draw(D--G,0.8+black+dashed,EndArrow);  draw(D--H,0.8+black+dashed,EndArrow);  dot(A^^B^^C^^D^^E^^F^^G^^H, 5+black); [/asy] Therefore, we have the following relationships: \begin{align*} N(1,\mathrm{BB})&=2, \\  N(1,\mathrm{BT})&=1, \\ N(1,\mathrm{TB})&=0, \\ N(1,\mathrm{TT})&=0, \\ N(k+1,\mathrm{BB})&=N(k,\mathrm{BB})+2\cdot N(k,\mathrm{TB}), \\ N(k+1,\mathrm{BT})&=N(k,\mathrm{BB}), \\ N(k+1,\mathrm{TB})&=N(k,\mathrm{TT}), \\ N(k+1,\mathrm{TT})&=N(k,\mathrm{TT})+2\cdot N(k,\mathrm{BT}). \end{align*} Using these equations, we recursively fill out the table below: \[\begin{array}{c||c|c|c|c|c|c|c|c}    \hspace{7mm}&\hspace{6.5mm}&\hspace{6.5mm}&\hspace{6.75mm}&\hspace{6.75mm}&\hspace{6.75mm}&\hspace{6.75mm}&& \\ [-2.5ex] \boldsymbol{k} & \boldsymbol{1} & \boldsymbol{2} & \boldsymbol{3} & \boldsymbol{4} & \boldsymbol{5} & \boldsymbol{6} & \boldsymbol{7} & \boldsymbol{8} \\  \hline \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{N(k,\mathrm{BB})} &2&2&2&6&18&38&66&118 \\ \hline   &&&&&&&& \\ [-2.25ex] \boldsymbol{N(k,\mathrm{BT})} &1&2&2&2&6&18&38&66 \\ \hline  &&&&&&&& \\ [-2.25ex] \boldsymbol{N(k,\mathrm{TB})} &0&0&2&6&10&14&26&62 \\ \hline    &&&&&&&& \\ [-2.25ex] \boldsymbol{N(k,\mathrm{TT})} &0&2&6&10&14&26&62&138 \\ \hline \hline &&&&&&&& \\ [-2.25ex] \textbf{Total}&\boldsymbol{3}&\boldsymbol{6}&\boldsymbol{12}&\boldsymbol{24}&\boldsymbol{48}&\boldsymbol{96}&\boldsymbol{192}&\boldsymbol{384} \end{array}\] By the Multiplication Principle, there are $3\cdot2^{k-1}$ ways to make exactly $k$ moves. So, we must get \[N(k,\mathrm{BB})+N(k,\mathrm{BT})+N(k,\mathrm{TB})+N(k,\mathrm{TT})=3\cdot2^{k-1}\] for all values of $k.$
Finally, the requested probability is \[\frac{N(8,\mathrm{BT})+N(8,\mathrm{TT})}{N(8,\mathrm{BB})+N(8,\mathrm{BT})+N(8,\mathrm{TB})+N(8,\mathrm{TT})}=\frac{66+138}{118+66+62+138}=\frac{204}{384}=\frac{17}{32},\] from which the answer is $17+32=\boxed{049}.$

Answer:  049