The number of $3$-valent rooted plane trees with $n$ internal nodes is (I think) the Catalan numbers $C_n$. What is the formula for $4$-valent planar trees? $n$-valent? Apologies for the simple question, but I can't seem to find the right keywords to google.
Asked
Active
Viewed 382 times
0
-
That would mean that there is only one $3$-valent rooted plane tree with $1$ internal node -- but a root with $1$, $2$ or $3$ children already makes three different $3$-valent rooted plane trees with $1$ internal node? Or am I misunderstanding some of the terms you used? – joriki Mar 06 '16 at 18:21
-
@joriki: I’m pretty sure that the OP means rooted plane trees such that each node has degree $1$ or $3$, and the root has degree $1$. Assuming that plane means that we can distinguish left from right subtrees, there are indeed $C_n$ of these with $n$ internal nodes. – Brian M. Scott Mar 07 '16 at 07:17
-
@joriki: Oops: I meant that the root has degree $3$. (otherwise the root is just a pointless appendage.) – Brian M. Scott Mar 07 '16 at 07:23
1 Answers
3
If I’ve understood correctly the way you’re using the term $n$-valent tree, the number of $4$-valent rooted plane trees with $n$ internal nodes is
$$\frac1{2n+1}\binom{3n}n\;;$$
this is OEIS A001764.
More generally, let $a_n^{(m)}$ be the number of $(m+1)$-valent rooted plane trees with $n$ internal vertices; these numbers satisfy the recurrence
$$a_n^{(m)}=[n=0]+\sum_{k_1+\ldots+k_m}a_{k_1}^{(m)}a_{k_2}^{(m)}\ldots a_{k_m}^{(m)}\;,$$
where the first term is an Iverson bracket. Example $\mathbf5$ in Section $\mathbf{7.5}$ of Graham, Knuth, and Patashnik, Concrete Mathematics, shows that the solution to this recurrence is
$$a_n^{(m)}=\frac1{mn+1}\binom{mn+1}n=\frac1{(m-1)n+1}\binom{mn}n\;.\tag{1}$$
For the $3$-valent case we have $m=2$, so that $(1)$ does indeed yield the Catalan numbers.
The OEIS entry has many more applications/interpretations of this sequence.
Brian M. Scott
- 631,399