Solve the N steps on a cube problem

 

 First we see that given a fixed corner as apex, there are 4 classes of corners:

 

1)A(1): the apex.

2)B(3): next to apex.

3)C(3): 2 steps from apex.

4)D(1): the opposite corner.

 

In general we have PAA(N)=PAB(N-1)=1/3 PAA(N-2)+ 2/3 PAC(N-2)

 

PAC(N)=1/3 PAD(N-1) + 2/3 PAB(N-1) = 1/3 PAC(N-2) + 2/3(1/3 PAA(N-2)+ 2/3 PAC(N-2))

      = 7/9 PAC(N-2)+ 2/9 PAA(N-2)

 

Given PAA(1)=PAC(1)=0, all PAA(N)=0 for N= odd.

 

 

Now let's work out the N=even number case

 

 

Therefore working with this pair:

 

PAA(2N)=1/3 PAA(2N-2)+ 2/3 PAC(2N-2)

PAC(2N)= 2/9 PAA(2N-2) + 7/9 PAC(2N-2)

 

Solve the eigenvalue problem of 

 

 1/3, 2/3

 2/9,  7/9

 

or

 

lam^2-(10/9) lam +17/81=0... eigen value = (5+/-sqrt(8))/9

 

Someone finish this for me?