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 $B_{p_0}(x) = \sum_{n \ge 0} B_{n+p_0} x^n$?

  • $\begingroup$ Curious, what's the point of the subscript $0$? $\endgroup$ – anon Feb 10 '12 at 9:59
  • $\begingroup$ 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. $\endgroup$ – john_leo Feb 10 '12 at 10:03
  • $\begingroup$ 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.) $\endgroup$ – user32970 Jun 5 '12 at 2:46
  • $\begingroup$ 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. $\endgroup$ – john_leo Jun 5 '12 at 8:50

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?

  • $\begingroup$ 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 $P_m(x)$. $\endgroup$ – john_leo Feb 10 '12 at 10:17
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    $\begingroup$ @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... $\endgroup$ – anon Feb 10 '12 at 10:31
  • $\begingroup$ 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? $\endgroup$ – john_leo Feb 10 '12 at 10:43
  • $\begingroup$ Yes. Dunno if it'll help but there's opportunity to be had. $\endgroup$ – anon Feb 10 '12 at 10:44

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