Solve pair of recurrence relations

Question

[Corrected equations and added simple example]

Can you solve a system of (loosely) coupled recurrence relations like this in Mathematica somehow?

{A[k] ==1+((n-k-2)/n) A[k+1] +(2/n) B[k+1] + (k/n) (A[0]), 
 B[k]==1+((n-k-1)/n) B[k+1] + (k/n) A[0]}

We want to solve it for $A[0]$ as a closed form function of $n$.

$A[k]$ is defined for $0 \leq k \leq n-2$ and $B[k]$ is defined for $1 \leq k \leq n-1$.

Version 8 doesn't do anything useful but I suspect I am not asking it in the right way.

As a simple example, if you set $n=3$ then you get four simulataneous equations.

A[0] == 1 + (2/3)*B[1] + (1/3)*A[1]
B[1] == 1 + (1/3)*A[0] + (1/3)*B[2]
B[2] == 1 + (2/3)*A[0]
A[1] == 1 + (2/3)*B[2] + (1/3)*A[0]

Solving for $A[0]$ gives you $33/5$ I believe. To start things off, how do you get Mathematica to do this?

Update. If you just take the second recurrence alone.

B[k]==1+((n-k-1)/n) B[k+1] + (k/n) A[0]

How can you get Mathematica to give a sensible solution for $B[k]$ in terms of $n$ and $A[0]$? It seems Rsolve ought to be able to do this. I even tried

RSolve[{B[k] == 1 + ((n - k - 1)/n) B[k + 1] + ((k)/n) (A[0]), 
  B[n - 1] == k A[0]/n}, B[k], k]

which should be identical. However this now gives an empty solution with the following warning.

RSolve::bvnul: For some branches of the general solution, the given boundary conditions lead to an empty solution.

I would like to tell Mathematica to only try to solve it for the defined range of $k$. Is that possible?

Answer 1

There may be a better way, but this solves it the same way you did. It generates all the equations and then solves them. If you only want $A_0$, use Sol[n][[1,1]].

Sol[n_] := 
Solve[Take[
Flatten[Table[{Subscript[B, k] == 
   1 + ((n - k - 1) Subscript[B, k + 1])/n + (k Subscript[A, 0])/
    n, Subscript[A, k] == 
   1 + ((n - k - 2) Subscript[A, k + 1])/n + (
    2 Subscript[B, k + 1])/n + (k Subscript[A, 0])/n}, {k, 0, 
  n - 1}]], {2, -2}]]

Edit:

Here's a graph of the first 80 values:

And the first set of differences:

That looks somewhere between $\sqrt{x}$ and $\log{x}$.