Let be B(z) the exponential generating function for the number $b_n$ of different rooted unordered binary trees with exactly n leaves labeled only at their leaves (so the internal nodes are unlabeled). Then it's a well know result that $b_n = 1\cdot3\cdot\ldots\cdot(2n-3)=(2n-3)!!$ (see eg. Schröder "Vier combinatorische Probleme"). The EGF satisfies the recursion

$B(z) = z + \frac{1}{2}B(z)^2$.

There is one tree with one leave and every tree is built up from two (smaller) trees where the order doesn't matter (because the trees are unplane).

My question:

So $B(z) = z + z^2 + 3z^3 + 15z^4 + 105z^5 + ...$

And therefore $z + \frac{1}{2}B(z)^2 = z + \frac{1}{2}(z^2 + 2z^3 + 7z^4 + 36z^5 + ...) \neq B(z)$

So the recursion is not satisfied? Where is my mistake?

Thanks a lot!


Since $B(z)$ is an exponential generating function $B(z) = \sum_n b_n \frac{z^n}{n!}$, and not $\sum_n b_n z^n$.

With correct definition the series expansion for $B(z)$ should read:

$$ B(z) \sim z+\frac{z^2}{2}+\frac{z^3}{2}+\frac{5 z^4}{8}+\frac{7 z^5}{8}+\frac{21 z^6}{16}+\frac{33 z^7}{16}+O\left(z^8\right) $$ and this satisfies the recurrence equation:

In[200]:= With[{b = Sum[(2 n - 3)!!/n! z^n, {n, 1, 8}] + O[z]^8}, 
 z + 1/2 b^2 - b]

Out[200]= SeriesData[z, 0, {}, 8, 8, 1]
  • $\begingroup$ Uh yes, thanks a lot! I totally forgot that I've to deal with labeled structures. $\endgroup$
    – lumbric
    Oct 7 '11 at 7:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.