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Is there a easy way to compute the coefficients of the power series which represents

\begin{equation*} \frac{x - x^k + x^{k+1}}{1-2x + x^k - x^{k+1}}. \end{equation*}

I am currently solving this by assuming it has the form

\begin{equation*} a_1x+a_2x^2+a_3x^3+\dots+a_nx^n+\dots \end{equation*}

and solving tedious recurrences obtained from

\begin{equation*} x - x^k + x^{k+1} = (1-2x + x^k - x^{k+1}) \times (a_1x+a_2x^2+a_3x^3+\dots+a_nx^n+\dots). \end{equation*}

Motivation: If my calculations are right, the above is the generating function of the number of compositions of an integer which do not involve a given integer $k$.

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    There is no method faster and simpler than those "tedious recurrences". Maybe you should adjust your expectations! :P2010-08-30
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    The continued fraction form looks interesting: $-\cfrac{1}{1+\cfrac{1}{x^k+\cfrac{1}{1-\frac1{x}}}}$2010-08-30
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    What do you mean by composition? Addition or multiplication?2010-08-30
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    Another interesting thing about your function: if you construct the matrix whose (k,n) entry is the nth derivative of the kth function, the columns in the strictly lower triangular part of the matrix are constant. They seem to be this sequence: research.att.com/~njas/sequences/A002866 . The diagonal of that matrix on the other hand looks like this sequence: research.att.com/~njas/sequences/A0526652010-08-30
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    Have you tried taking the derivative at x = 0?2010-08-30
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    Matt : The compositions of 3 are 1+1+1,1+2,2+1,32010-08-30
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    I thought you meant addition. Are you aware of the exact formula for the number of partitions of a natural number N? http://en.wikipedia.org/wiki/Partition_%28number_theory%29 (Everybody calls these things "partitions", not "compositions")2010-08-30
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    Matt: In partitions order is not important, in compositions, the order is important. So 1+2 and 2+1 will correspond to different compositions but the same partition.2010-08-30

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