Let the vertecies of the cube be $\{O\}\cup \{A_1,A_2,A_3\}\cup \{B_1,B_2,B_3\} \cup \{C\}$. The $A_i$ are the three vertecies of distance 1 from $O$. $B_i$ are distance 2 and $C$ is distance 3 from $O$. Let $o_i,a_i,b_i,c_i$ be the number of paths of length $i$ that start at $O$ and end in sets $O,A_i,B_i,C_i$ respectively. Then we have $o_1=b_1=c_1=0, a_1=3$ and \begin{align} o_i &= a_{i-1}\\ a_i &= 3o_{i-1} + 2b_{i-1}\\ b_i &= 3c_{i-1} + 2a_{i-1}\\ c_i &= b_{i-1} \end{align} Let $x_i = a_i + b_i$ and solve the $4$ equations above to find $x_i = 2x_{i-1} + 3x_{i-2}$, which can be solved in the standard way to yield $x_i = \frac{3}{4}(3^n - (-1)^n)$. Now note that $a_i$ is always $0$ when $i$ is even because only paths of odd length can start at $O$ and end in $A$. Hence $b_{1386} = x_{1386} = \frac{3}{4} (3^{1386}-1)$, and $o_{1386} = \frac{3}{4}(3^{1385}+1)$ By symetry the number of paths from from $O$ to any specific $B_1,B_2,B_3$ is $\frac{1}{3}b_{1386}$. Therefore have $o_{1386} > \frac{1}{3}b_{1386}$, hence there are more from from $O$ to itself then $O$ to $A$ EDIT: my notation is a bit unfortunate because in the question A is acutally an element of $\{B_1,B_2,B_3\}$. I can't be bothered changing it sorry

Thanks! Can you prove that the 1386-runs from $O$ to itself is $1$ number more than 1386-runs from $O$ to $A$?

sororak wrote: Thanks! Can you prove that the 1386-runs from $O$ to itself is $1$ number more than 1386-runs from $O$ to $A$? $\frac{3}{4}(3^{1385}+1) - \frac{1}{4}(3^{1386}-1) = 1$ QED

Assume the rectangle passing through 4 vertices of cube which A,O are not on it. if we reflect all the ways from O to A about it ( parts from the rectangle to A, not O to the rectangle ) then we will have 1386-run from O to O. but clearly there's at least 1 way from O to O that is not a reflection so 1386-runs from O to O is at least 1 more than 1386-runs from O to A.