I am trying to derive the classic paper in the title only by elementary means (no generating functions, no complex analysis, no Fourier analysis) although with much less precision. In short, I "only" want to prove that the average height $h_n$ of a tree with $n$ nodes (that is, the maximum number of nodes from the root to a leaf) satisfies $h_n \sim \sqrt{\pi n}$.

The outline is as follows. Let $A_{nh}$ be the number of trees with height less than or equal to $h$ (with the convention $A_{nh} = A_{nn}$ for all $h \geqslant n$) and $B_{nh}$ the number of trees of $n$ nodes with height greater than or equal to $h+1$ (that is, $B_{nh} = A_{nn} - A_{nh}$). Then $h_n = S_n/A_{nn}$, where $S_n$ is the finite sum $$ S_n = \sum_{h \geqslant 1} h(A_{nh} - A_{n,h-1}) = \sum_{h \geqslant 1} h(B_{n,h-1} - B_{nh}) = \sum_{h \geqslant 0} B_{nh}. $$ It is well known that $A_{nn} = \frac{1}{n}\binom{2n-2}{n-1}$, for the set of general trees with $n$ nodes is in bijection with the set of binary trees with $n-1$ nodes, counted by the Catalan numbers.

Therefore, the first step is to find $B_{nh}$ and then the main term in the asymptotic expansion of $S_n$.

At this point the authors use analytical combinatorics (three pages) to derive $$ B_{n+1,h-1} = \sum_{k \geqslant 1} \left[\binom{2n}{n+1-kh} - 2\binom{2n}{n-kh} + \binom{2n}{n-1-kh}\right]. $$

My own attempt is as follows. I consider the bijection between trees with $n$ nodes and monotonic paths on a square grid $(n-1) \times (n-1)$ from $(0,0)$ to $(n-1,n-1)$ which do not cross the diagonal (and are made of two kinds of steps: $\uparrow$ and $\rightarrow$). These paths are sometimes called Dyck paths or excursions. I can express now $B_{nh}$ in terms of lattice paths: it is the number of Dyck paths of length 2(n-1) and height greater than or equal to $h$. (Note: a tree of height $h$ is in bijection with a Dyck path of height $h-1$.)

Without loss of generality, I assume that they start with $\uparrow$ (hence stay above the diagonal). For each path, I consider the first step crossing the line $y = x + h - 1$, if any. From the point above, all the way back to the origin, I change $\uparrow$ into $\rightarrow$ and vice versa (this is a reflection wrt the line $y=x+h$). It becomes apparent that the paths I want to count ($B_{nh}$) are in bijection with the monotonic paths from $(-h,h)$ to $(n-1,n-1)$ which avoid the boundaries $y=x+2h+1$ and $y=x-1$. (See figure.)

In the classic book Lattice Path Counting and Applications by Mohanty (1979, page 6) the formula $$ \sum_{k \in \mathbb{Z}} \left[\binom{m+n}{m-k(t+s)} - \binom{m+n}{n+k(t+s)+t}\right], $$ counts the number of monotonic paths in a lattice from $(0,0)$ to $(m,n)$, which avoid the boundaries $y = x - t$ and $y = x + s$, with $t > 0$ and $s > 0$. (This result was first established by Russian statisticians in the 50s.) Therefore, by considering a new origin at $(-h,h)$, we satisfy the conditions of the formula: $s=1$, $t=2h+1$ and the destination point (the upper right corner) is now $(n+h-1,n-h-1)$. Then $$ B_{nh} = \sum_{k \in \mathbb{Z}} \left[\binom{2n-2}{n+h-1-k(2h+2)} - \binom{2n-2}{n-h-1+k(2h+2) + 2h+1}\right]. $$ This can be simplified in $$ B_{n+1,h-1} = \sum_{k \in \mathbb{Z}} \left[\binom{2n}{n+1-(2k+1)h} - \binom{2n}{n-(2k+1)h}\right], $$ which, in turn, is equivalent to $$ B_{n+1,h-1} = \sum_{k \geqslant 0} \left[\binom{2n}{n+1-(2k+1)h} - 2\binom{2n}{n-(2k+1)h} + \binom{2n}{n-1-(2k+1)h}\right]. $$ The difference with the expected formula is that I sum over the odd numbers ($2k+1$), instead of all positive integers ($k$).

Any idea where the problem is?

  • $\begingroup$ You say you want to use only elementary things, yet you use a result from a book. How does Mohanty derive the identity you use? $\endgroup$
    – Raphael
    Sep 29, 2012 at 10:01
  • $\begingroup$ I define in the first sentence what I mean by "elementary": no generating functions, no complex analysis, no Fourier analysis. In his book, Mohanty uses elementary means to derive that formula, more precisely, the principles of reflection and inclusion-exclusion on lattice paths. (I use the former above.) If you insist, I will add his proof at the end of the question. $\endgroup$ Sep 29, 2012 at 14:04
  • $\begingroup$ Not at all, just wanted to make sure you were not breaking your rule yourself. $\endgroup$
    – Raphael
    Sep 29, 2012 at 18:02
  • $\begingroup$ It's very weird to me to see 'generating functions' listed as a non-elementary technique when analytical combinatorics is apparently considered elementary. $\sqrt{\pi}$ seems like an almost inherently non-elementary value; do you have e.g. a comparable proof of the asymptotics of the central binomial coefficient to give a better sense of what you're looking for? I suspect the two are closely related... $\endgroup$ May 7, 2014 at 19:26

1 Answer 1


The monotonic paths from $(−h,h)$ to $(n−1,n−1)$ that you construct only avoid the boundary $y=x+2h+1$ before they cross $y=x+h$ for the first time. Thus the formula you use is not applicable.


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