Enumeration of Multi-rooted Plane Trees

Abstract

We give closed form expressions for the numbers of multi-rooted plane trees with specified degrees of root vertices. This results in an infinite number of integer sequences, some of which are known to have an alternative interpretation. We also propose recursion relations for numbers of such trees as well as for the corresponding generating functions. Explicit expressions for the generating functions corresponding to plane trees having two and three roots are derived. As a by-product, we obtain a new binomial identity and a conjecture relating hypergeometric functions.

Appendix A

Consider the following sums:

$$\begin{aligned} S_1 (n,r,s):= & {} \sum _{k=0}^n \frac{1}{k+r+1} \left( {\begin{array}{c}2n-2k+s\\ n-k\end{array}}\right) \, \left( {\begin{array}{c}2k+r\\ k\end{array}}\right) ,\\ S_2 (n,r,s):= & {} \sum _{k=0}^{n+s} \frac{1}{k+r+1} \left( {\begin{array}{c}2n-2k+s\\ n-k\end{array}}\right) \, \left( {\begin{array}{c}2k+r\\ k\end{array}}\right) ,\\ S_3 (n,r):= & {} \sum _{k=0}^{n} \frac{1}{n-k+1} \left( {\begin{array}{c}2n-2k\\ n-k\end{array}}\right) \, \left( {\begin{array}{c}2k+r\\ k\end{array}}\right) ,\\ S_4(n,r):= & {} \sum _{k=0}^n \frac{1}{n-k+1} \left( {\begin{array}{c}2n-2k+2\\ n-k\end{array}}\right) \, \left( {\begin{array}{c}2k+r\\ k\end{array}}\right) , \\ S_5 (n,r,s):= & {} \sum _{k=0}^{n} \left( {\begin{array}{c}2n-2k+s\\ n-k\end{array}}\right) \, \left( {\begin{array}{c}2k+r\\ k\end{array}}\right) ,\\ S_6(n,r,s,t):= & {} \sum _{k=0}^n \frac{r}{tk + r } \left( {\begin{array}{c}tk +r\\ k\end{array}}\right) \, \left( {\begin{array}{c}tn-tk+s\\ n-k\end{array}}\right) . \end{aligned}$$

Not all these sums appear in Sect. 7, but all are closely related to sums we needed and we include them for completeness.

The sum $S_6$ is given in Equation (5.62) in Ref. [34]. It is understood that if the parameters considered are such that the factor $tk+r$ is equal to zero for some value of k, then in that term the binomial factor $\left( {\begin{array}{c}tk+r\\ k\end{array}}\right)$ is taken to cancel the factor $\frac{1}{tk+r}$. With this convention, $S_6$ is well defined for $n \ge 0$ and $r,s,t \in {\mathbb {Z}}$ and is equal to

$$\begin{aligned} S_6(n,r,s,t) = \left( {\begin{array}{c}tn+r+s\\ n\end{array}}\right) . \end{aligned}$$

The sums $S_1$ to $S_4$ can be obtained from $S_6$ after appropriate changes of variables, with the following results:

$$\begin{aligned} S_1 (n,r,s)= & {} \frac{1}{r+1} \left( {\begin{array}{c}2n+r+s+1\\ n\end{array}}\right) , \\ S_2(n,r,s)= & {} \frac{1}{r+1} \left( {\begin{array}{c}2n+r+s+1\\ n+s\end{array}}\right) , \\ S_3(n,r)= & {} \left( {\begin{array}{c}2n+r+1\\ n\end{array}}\right) , \\ S_4(n,r)= & {} \left( {\begin{array}{c}2n+r+2\\ n\end{array}}\right) , \end{aligned}$$

where, in $S_1$ and $S_2$, n and r are integers satisfying $n,r \ge 0$ and $s \in {\mathbb {Z}}$ (the results are actually valid for a wider range of values, but the general results are not needed for this work). The conditions for $S_3$ and $S_4$ are $n\ge 0$ and $r \in {\mathbb {Z}}$.

We could not find the sum $S_5$ in closed form in the literature, as mentioned in the main text, see (34).