Skip to main content

All Questions

Filter by
Sorted by
Tagged with
18 votes
2 answers
1k views

A combinatorial interpretation for $n$-ary trees for negative $n$

The ordinary generating function $T_n=T_n(x)$ for the $n$-ary trees satisfies the functional equation $$ T_n=1+xT_n^n. $$ This is usually defined for $n\ge 0$, but the functional equation can be ...
Alexander Burstein's user avatar
6 votes
0 answers
276 views

Power law correction factor in tree enumeration via naïve division

It is a theorem of Otter, building on fundamental work of Pólya, that the number of unlabeled trees on $n$ vertices is $\approx C \alpha^{n} n^{-5/2}$, where $C = 0.534\ldots$ and $\alpha = 2.955\...
Sam Hopkins's user avatar
  • 24.2k
14 votes
1 answer
647 views

Bijective proof of recurrence for rooted unlabeled trees

Would've been a better question for Christmas than Thanksgiving, but alas... Let $t_n$ denote the number of rooted, unlabeled trees on $n$ vertices (OEIS A000081). These are the isomorphism classes of ...
Sam Hopkins's user avatar
  • 24.2k
1 vote
0 answers
85 views

Varieties of trees with logarithmic degree function

I am interested in varieties of trees with a logarithmic degree function. I am currently looking at Bergeron, Flajolet, and Salvy's work "Varieties of increasing trees." They discuss exactly ...
Samuel Crew's user avatar
7 votes
1 answer
2k views

How does the number of trees on $n$ vertices *up to isomorphism* grow as $n \to \infty$?

It is well known that the number of labelled trees on $n$ vertices is equal to $n^{n-2}$. We do not expect any such exact formula for the number of isomorphism types of trees on $n$ vertices. But ...
Matthew Kahle's user avatar
1 vote
1 answer
248 views

mapping integers to k-ary trees

Is there an algorithmic way to map the natural numbers to unique k-ary trees? I am familiar with the work of Tychonievich who created a mapping from integers to binary trees. https://www.cs.virginia....
Sohrab T's user avatar
  • 113
7 votes
1 answer
571 views

Are there Prüfer sequences for rooted forests?

One well-known, extremely slick proof of Cayley's tree enumeration theorem is the use of Prüfer sequences. Cayley also proved a version for forests, namely that the number of forests with $n$ ...
Chuck Newton's user avatar
7 votes
1 answer
705 views

Is there a natural relationship between OEIS A127670 and Cayley's tree formula?

I apologize in advance that this question must sound highly amateurish, but I am wondering if there is any connection between the formula https://oeis.org/A127670 , which counts the number of fixed $n$...
Tom Solberg's user avatar
  • 4,049
5 votes
2 answers
351 views

Asymptotics of unrooted labeled forests

It is well known that the number of unrooted labeled trees on vertex set $[n]={1,2,...,n}$ is $n^{n-2}$. Let $U(z)$ be the exponential generating function of the sequence of these numbers. Then $F(z)=\...
Miklos Bona's user avatar
12 votes
0 answers
330 views

The number of labeled pairs of edge disjoint trees and related questions

I wonder what is known on the following: 1) What is the number $T_k(n)$ of $k$-tuples of (pairwise) edge-disjoint trees $(T_1,T_2,\dots, T_k)$ with $n$ labelled vertices? 2) (harder, it seems) What ...
Gil Kalai's user avatar
  • 24.7k
8 votes
1 answer
344 views

Bijective proof of formula for rooted binary forests

For $n\ge 1$, let $f(n)$ be the number of rooted complete (unordered) binary trees with $n$ leaves labeled from $1$ to $n$ ("complete binary" means that every vertex has either $0$ or $2$ children and ...
Timothy Chow's user avatar
  • 82.6k
7 votes
1 answer
455 views

More asymptotics for trees

This is a follow up to my recent question on the asymptotics of A003238. Lucia gave a fine answer to that question, but as I hinted the 'real' problem I have in mind is slightly different, and I've ...
Michael Albert's user avatar
14 votes
1 answer
696 views

Are the asymptotics of A003238 known?

Sequence A003238 of the OEIS counts ``rooted trees with $n$ vertices in which vertices at the same level have the same degree.'' The sequence, $a$, begins 1, 1, 2, 3, 5, 6, 10, 11, 16, ... and it is ...
Michael Albert's user avatar
13 votes
5 answers
1k views

Is the following invariant of rooted trees a complete invariant?

Recall that rooted trees may be generated by starting with a trivial rooted tree (just a vertex), along with the operations of grafting a number of trees (identify their roots) and adding a new vertex ...
Spice the Bird's user avatar