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I am trying to solve the following recurrence relation:

c[1] = 0,

c[2] = a[1]*b[1],

c[n] = a[n]*b[n] + a[n]*c[n-1]+b[n]*c[n-1],

Modulus -> 2, a[n]*a[n]-a[n] = 0, b[n]*b[n]-b[n] = 0

c[n] is the recurrence that grows out of control that I am trying to solve/simplify

Everything I try in RSolve is producing

"For some branches of the general solution, the given boundary conditions lead to an empty solution"

Can anyone help me understand the right way to approach this recurrence problem?

Here is what I tried in the worksheet:

RSolve[ {c[1] == 0, c[2] == a[1]*b[1], 
         c[n] == a[n]*b[n] + a[n]*c[n - 1] + b[n]*c[n - 1]}, 
         c[n], n
]

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

Now if I try to include the equations specifying that values for a and b must be 0 or 1 (a*a-a==0), I get a different error:

RSolve[ {c[1] == 0, c[2] == a[1]*b[1], 
        c[n] == a[n]*b[n] + a[n]*c[n - 1] + b[n]*c[n - 1], 
        a[n]*a[n] - a[n] == 0, b[n]*b[n] - b[n] == 0}, 
        c[n], n
]

RSolve::deqx: Supplied equations are not difference equations of the given functions.

Lastly, I have no idea how to specify Modulus -> 2 or if that matters for the purposes of recurrence relation solving.

share|improve this question
    
Can you please write the equations as they are in the notebook. –  Sektor Aug 17 '13 at 8:17
    
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Note that Modulus -> 2 is not a valid option for RSolve[] –  Sektor Aug 17 '13 at 11:42
    
It seems you are seeking a solution to a recurrence in which the coefficients a[n], b[n] are arbitrary (0s and 1s). On the face of it, I would think the solution would be a function of the (infinitely many) coefficients and thus wouldn't be possible, at least in this form for a computer program. Is there a reason for thinking this problem has a solution? –  Michael E2 Aug 17 '13 at 13:52

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