From https://www.stat.berkeley.edu/~aldous/150/takis_exercises.pdf page 24
Consider a triangle with the verticies 1, 2,3
and transition rules:
$p_{12}=p_{23}=p_{31}= 2/3$, $p_{21}=p_{32}=p_{13}= 1/3$
I am assuming the process begins (n=0) at 1, so hence the p=1
Then it can go from 1 to 2 or from 1 to 3, as denoted below, so the odds of it staying on 1 for (n=1) are zero.
(1,3)
(1,2)
For n=2, we have all the possible paths and probabilities:
(1,2,3) 4/9
(1,2,1) 2/9
(1,3,1) 2/9
(1,3,2) 1/9
So for n=2 (superscript p_2) the odds of returning to the start should be 4/9 instead of 2/9?
$2/9=p^2_{11}=C_1x^2_1+C_2x^2_2+C_3x^2_3$