Even searching for " 'number of trees' leaves " didn't reveal what I am looking for: an approach for calculating the (approximate) number of trees with exactly n nodes and m leaves. Any hints from MO?
3 Answers
The answer to this (very natural) question depends on your notion of "tree" (e.g. free, rooted) and the equivalence relation you employ (e.g. labelled, unlabelled). I haven't gone into the nitty-gritty details of all these results, but here's what I've found so far. There's likely published results I haven't found yet, but hopefully this helps to get you started.
We can compute $T_{m,n}$, the number of non-isomorphic free trees with $m$ leaves and $n$ vertices, for small $m$ and large $m$. For example, (a) $T_{3,n}$ is the number of partitions of $n-1$ into $3$ positive integer parts (Sloane's A001399), (b) $T_{n-2,n}=\lfloor (n-2)/2 \rfloor$ and (c) $T_{n-3,n}=\sum_{j=0}^{n-5} \lfloor (n-3-j)/2 \rfloor$. The first result can be observed by deleting the vertex of degree 3 and the last two can be observed by colouring each non-leaf vertex by the number of adjacent leaves, then deleting the leaves.
Yu (8) seems to have given an algorithm for generating rooted trees with $m$ leaves. Wang (6) and Liu (3,4) considered the number of "structurally different" trees with $m$ leaves (according to MathSciNet). Bergeron, Labelle and Leroux (1) consider the expected number of leaves in trees that admit a certain automorphism. Lam (2) discusses embeddings of trees with $m$ leaves and discusses trees with $(d+1)d^{r+1}$ leaves for integers d and r.
Wilf (7. p. 163) gave a generating function for $\sum_k T_{k,n}^{\text{lab}}$ where $T_{k,n}^{\text{lab}}$ is the number of labelled free trees with $m$ leaves and $n$ vertices. He also gives a formula for the average number of leaves in a labelled tree with $n$ vertices.
There is also this: K. Yamanaka, Y. Otachi, S.-I. Nakano Efficient Enumeration of Ordered Trees with k Leaves, which I haven't looked at yet.
(1) F. Bergeron, G. Labelle, and P. Leroux, Computation of the expected number of leaves in a tree having a given automorphism, and related topics, Discrete Appl. Math., 34 (1991), pp. 49-66.
(2) P. C. B. Lam, On number of leaves and bandwidth of trees, Acta Math. Appl. Sinica (English Ser.), 14 (1998), pp. 193-196.
(3) B. L. Liu, The enumeration of directed trees with a given number of leaves and the enumeration of free trees, Kexue Tongbao, 32 (1987), pp. 244-247. In Chinese.
(4) B. L. Liu, Enumeration of oriented trees and free trees with a given number of leaves, Kexue Tongbao (English Ed.), 33 (1988), pp. 1577-1581.
(5) Q. Q. Nong, The degree sequence and number of leaves in a tree, J. Yunnan Univ. Nat. Sci., 24 (2002), pp. 167-171. In Chinese.
(6) Z. Y. Wang, An enumeration problem on ordered trees, J. Math. (Wuhan), 6 (1986), pp. 201-208.
(7) H. C. Wilf, Generatingfunctionology, Academic Press, 1990.
(8) Q. L. Yu, An algorithm for lexicographically generating ordered rooted trees with constraints on the number of leaves, Chinese J. Oper. Res., 6 (1987), pp. 71-72
I think this is what you want:
OEIS: the triangle of trees with n nodes and k leaves
(You should draw the sequence as a triangle as below to get the 2-dimensional information.)
1 1 0 1 1 0 1 1 1 0 1 2 2 1 0 1 3 4 2 1 0 1 4 8 6 3 1 0 1 5 14 14 9 3 1 0 1 7 23 32 26 12 4 1 0 1 8 36 64 66 39 16 4 1 0 1 10 54 123 158 119 60 20 5 1 0 1 12 78 219 350 325 202 83 25 5 1 0
Edit: I edited to use a different representation of the data. I assume that the n-th row, k-th entry means the number of trees with n nodes and k leaves. See these other displays
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1$\begingroup$ I stumbled over this already, but couldn't make sense of the title: "the TRIANGLE of n nodes and k leaves". And the only reference - Harary's 'Graphical Enumeration' - is almost invisible at books.google.com. Are you sure? $\endgroup$ Commented Apr 16, 2010 at 16:59
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2$\begingroup$ The OEIS is just for (one dimensional) sequences, but this is a two-dimensional function, having two arguments. So they organize it in a triangle. $\endgroup$ Commented Apr 16, 2010 at 17:06
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2$\begingroup$ I edited it to use a clearer representation of the data. $\endgroup$ Commented Apr 16, 2010 at 17:13
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6$\begingroup$ To maybe save others some confusion: The top row is n=2, and the leftmost column is k=2. $\endgroup$ Commented Apr 16, 2010 at 18:19
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$\begingroup$ @Joel: Sorry for not accepting your answer, but I am looking for the "idea" of counting those trees, not for the results only. $\endgroup$ Commented Apr 17, 2010 at 2:22
The number of labelled trees on $n$ vertices with $m$ leaves is $$\binom{n}{m}S(n-2,n-m)(n-m)!$$ where $S(a,b)$ is the Stirling number of the second kind. This can be seen by analysing the multivariate generation which counts trees of any degree sequence given here: https://math.berkeley.edu/~mhaiman/math172-spring10/matrixtree.pdf