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Let $B_n$ be the $n$-th Catalan Number. We have $ B(x) = \sum_{n \ge 0} B_n x^n = \frac{1-\sqrt{1-4x}}{2x}$.

Does anyone know a closed form of the generating function of the shifted Catalan Numbers, i.e. for chosen $p_0$, for $B_{p_0}(x) = \sum_{n \ge 0} B_{n+p_0} x^n$?

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    Curious, what's the point of the subscript $0$?2012-02-10
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    Hm, you're right, I could have skipped that. I had the subscript in my notes for a different reason and forgot to take it out for the post.2012-02-10
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    The rational function 1/(1-x/(1-x/(1-x/(1- ... x/1)))) with n x's agrees with B(x) modulo x^{n+1}. Does that help? (Probably not, but maybe.)2012-06-05
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    Well, that's a very interesting result, how do I prove it/where can I find a proof? But I can't yet see how it could help me, because I don't want to take my function mod anything. The real question is linked (asymptotic of shifted Catalan Numbers), I still haven't solved it and I'd be grateful for any help.2012-06-05

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If I understand the question correctly, this isn't too difficult nor is it particular to Catalan numbers in any way. Let $f(x)$ be a generating function for the sequence $\{a_n\}_{n=0}^\infty$. Define the polynomial

$$P_m(x)=\sum_{n=0}^{m-1} a_n x^n.$$

(For $m=0$ we say $P\equiv0$.) This is just the series expansion of $f(x)$ truncated. Then we have

$$\sum_{n\ge0} a_{n+m}x^n=x^{-m}\left(\sum_{n\ge0}a_{n+m}x^{n+m}\right)=\frac{f(x)-P_m(x)}{x^m}.$$

Is this what you're looking for or were you interested in something different?

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    Not quite. I've thought of that, but basically it would mean that I just have $B_p(x) = \sum_{n\ge p} B_{n+p}x^n$. I need to do a singularity analysis close to the singularity $\frac{1}{4}$, and for that I need a different form. I don't know how I could evaluate the truncated polynomial $P_m(x)$.2012-02-10
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    @john_leo: Well, by the work I have above, finding a closed-form for $B_p(x)$ is basically equivalent to a closed form for $P_m(x)$. One issue a lot of people have when approaching MSE with questions is that their *overarching* problem is something others can help with, but they tackle it on their own and get stuck with a subproblem, and only tell others about the subproblem when it's actually a dead end...2012-02-10
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    I understand. I knew before that those two problems are equivalent, I guess I should have noted that. I had also thought about posting the overarching problem, but then decided to start small, because the other question would be rather long...so do you think I should ask the whole question in a new post?2012-02-10
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    Yes. Dunno if it'll help but there's opportunity to be had.2012-02-10