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 ($k$). First, I hoped that the even terms would cancel out, but that does not seem to be the case.

Any idea where is the problem?

[Edit: Starting from the expected result, by the same elementary binomial manipulations, we have $$ B_{n,h} = A_{nn} + \sum_{k \in \mathbb{Z}}\left[\binom{2n-2}{n-k(h+1)} - \binom{2n-2}{n-1+k(h+1)}\right]. $$ But if we try to find a combinatorial interpretation to this sum in terms of bounded lattice paths, we fail: the destination point has coordinates $(n,n-2)$, which is below the inferior boundary $y = x - 1$, so the number of paths is $0$, but Mohanty's formula gives negative numbers in this kind of situation (although he does not mention this). Therefore, if we find a combinatorial interpretation for the absolute value of these numbers, we can understand the result in the same terms.]

[Edit: In response to a comment below, here are all the details in slow motion. \begin{align} B_{n,h} &= \sum_{k \in \mathbb{Z}}\left[\binom{2n-2}{n-(2k+1)(h+1)} - \binom{2n-2}{n-1+(2k+1)(h+1)}\right]\cr 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]\cr &= \sum_{k \in \mathbb{Z}}\left[\binom{2n}{n+1-(2k+1)h} - \binom{2n}{n-(2k+1)h}\right]\cr &= \sum_{k \geqslant 0}\left[\binom{2n}{n+1-(2k+1)h} - \binom{2n}{n-(2k+1)h}\right]\cr &+ \sum_{k > 0}\left[\binom{2n}{n+1+(2k-1)h} - \binom{2n}{n+(2k-1)h)}\right]\cr &= \sum_{k \geqslant 0}\left[\binom{2n}{n+1-(2k+1)h} - \binom{2n}{n-(2k+1)h}\right]\cr &+ \sum_{k \geqslant 0}\left[\binom{2n}{n+1+(2k+1)h} -\binom{2n}{n+(2k+1)h}\right]\cr &= \sum_{k \geqslant 0}\left[\binom{2n}{n+1-(2k+1)h} - \binom{2n}{n-(2k+1)h}\right]\cr &+ \sum_{k \geqslant 0}\left[\binom{2n}{n-1-(2k+1)h} -\binom{2n}{n-(2k+1)h}\right]\cr &= \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]. \end{align} ]

  • $\begingroup$ After "This can be simplified in" did you perform some simplification beyond replacing $n$ with $n+1$ and replacing $h$ with $h-1$? Because performing those two substitutions gives something that looks slightly different from what you've displayed; namely, $\sum_{k \in \mathbb{Z}} \binom{2n}{n-1-(2k-1)h} - \binom{2n}{n + (2k+1)h} = \sum_{k \in \mathbb{Z}} \binom{2n}{n+1+(2k-1)h} - \binom{2n}{n - (2k+1)h}$. Note that the former binomial in the final sum looks slightly different from what you have (by a single minus sign before the $2k$), with its $(2k-1)h$ instead of your $-(2k+1)h = (-2k-1)h$. $\endgroup$ – Benjamin Dickman Aug 30 '12 at 23:59
  • $\begingroup$ But perhaps this makes no difference since you're summing over all $k \in \mathbb{Z}$... $\endgroup$ – Benjamin Dickman Aug 31 '12 at 0:03
  • $\begingroup$ [See above for the details.] The $2k+1$ appears right at the start, from $m$ in the first binomial and from $s$ in the second (notation of Mohanty's formula). $\endgroup$ – Christian Aug 31 '12 at 2:58

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