How Can a Closed Form for This Complex Difference Equation Be Determined?

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phoenixthoth
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While working on another problem, the following difference equation came up.

[tex]s_{n}=\frac{1}{r_{1}-r_{1}^{n}}}\sum_{k=1}^{n-1}s_{k}B_{n,k}\left(<br /> r_{1},r_{2},...,r_{n-k+1}\right)[/tex]

The B_{n,k} are (nonlinear) polynomials in the variables r_{1}, ..., r_{n,k-n+1} that don't involve the s's. (In fact they are Bell polynomials though I'm not sure it is necessary to know B's to get a formula for s_{n}.)

Initial condition: s_{1}=1.

You can see that s_{n} is equal to some function of the previous s_{k} for k running from 1 through n-1.

I want a closed form formula for s_{n} that does not involve any other s_{k}. I've taken it up to 5 to look for a pattern and it seeeeems like there is some function of the B's that's being iterated to get the result of simplifying and substituting previous values of s_{k}.

Actually, I'm not sure this is a difference equation because the sum is not fixed but increases in complexity as n increases.

Any feedback or inquiries as to how this problem arose are quite welcome and appreciated!
 
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phoenixthoth said:
though I'm not sure it is necessary to know B's to get a formula for s_{n}

You'll need to use some kind of relationship between the B's to simplify the formula, otherwise s_n will have 2^(n-2) terms (this can be shown by induction).
 
Here is a recurrence I found in the literature which might help.

[tex]B_{n,k}\left( r_{1},r_{2},...,r_{n-k+1}\right) =\frac{1}{k}\sum_{j=k-1}^{n-1}\binom{n}{j}r_{n-j}B_{j,k-1}\left(<br /> r_{1},...,r_{j-k+2}\right)[/tex]