# To what extent is it possible to generalise a natural bijection between trees and $7$-tuples of trees, suggested by divergent series?

In the paper "Seven Trees In One" by Andreas Blass, a "very explicit" bijection is found between trees and 7-tuples of such trees.

The idea to construct such a bijection stems from looking at some particular divergent series, as mentioned in Chapter 2 of the paper:

To see why seven is special, we ﬁrst give an argument to establish Theorem 1 in the style of eighteenth-century analysis, where meaningless computations (e.g., manipulating divergent series as though they converged absolutely and uniformly) somehow gave correct results. This argument begins with the observation that a tree either is 0 or splits naturally into two subtrees (by removing the root). Thus the set T of trees satisﬁes $T = 1 + T^ 2$ . (Of course equality here actually means an obvious isomorphism. Note that the same equation holds also for the variant notions of tree mentioned at the end of Section 1.) Solving this quadratic equation for T, we ﬁnd $T = 1/2 ± i \sqrt{3}/2$ . (The reader who objects that this is nonsense has not truly entered into the eighteenth-century spirit.) These complex numbers are primitive sixth roots of unity, so we have $T^ 6 = 1$ and $T^ 7 = T$. And this is why seven-tuples of trees can be coded as single trees.

A more elaborate explanation can be found here.

Mr. Blass does not elaborate a lot on the divergent series he mentioned. This is explained in a bit more detail in a blog post on the "Everything Seminar". The divergent sum of the Catalan numbers (where the $n$'th Catalan number is defined as $C_n = \frac{1}{n+1} \binom{2n}{n}$) is computed by looking at the corresponding generating function of these numbers: $$T(z) = 1 + z + 2z^2 + 5z^3 + 14z^4 + 42z^5 + \dots$$

It is mentioned in that blog post that the coefficient of $x^n$ is the number of binary rooted trees with $n+1$ leaves. Combining these pieces of information, it can be computed that $$T(z) = \frac{1- \sqrt{1-4z}}{2z} .$$

By plugging in $x=1$, we obtain that $$1 + 1 + 2 + 5 + 14 + 42 + \dots "=" \frac{1}{2} - i \frac{\sqrt{3}}{2} = e^{-\pi i/3} .$$

Therefore, we "know" that $$1 + 1 + 2 + 5 + 14 + 42 + \dots "=" (1 + z + 2 + 5 + 14 + 42 + \dots)^7$$

This suggested the bijection in the paper by Blass. I am curious, though, whether there are more instances of divergent series that suggest bijections between certain combinatorial objects. For instance, we also have the Fibonacci generating function $$F(z) = 1 + z + 2z^2 + 3z^3 + 5z^4 + 8z^5 + \dots = \frac{z}{1-z-z^2} .$$ (Here, the $n$'th coefficient is defined as $f_{n} = f_{n-1} + f_{n-2}$, where $f_0 = f_1 = 1$.)

If we let $z$ go to $1$, we find that $$1 + 1 + 2 + 3 + 5 + 8 + \dots "=" -1 .$$

Therefore, we find that $$1 + 1 + 2 + 3 + 5 + 8 + \dots "=" (1 + 1 + 2 + 3 + 5 + 8 + \dots)^{2k+1} \qquad (1),$$ is true as well, for integer $k > 0$ .

Question 1: Does this "equality" (1) involving the Fibonacci divergent series admit a combinatorial interpretation as well? Can a bijection between combinatorial structures similar to the one mentioned in Blass' paper be constructed based on this interpretation as well?

Furthermore, we can compute that $$1 + 2 + 4 + 8 + 16 + \dots = -1$$ by means of a generating function as well. (Here, the $n$'th term is of the form $2^n$.) So we have $$1 + 2 + 4 + 8 + 16 + \dots = (1 + 2 + 4 + 8 + 16 + \dots)^{2k+1} \qquad (2)$$ for integer $k>0$ as well.

Question 2: Does equality (2) admit a combinatorial interpretation and a corresponding "very explicit" bijection as well?

Question 3: If there are combinatorial interpretations of equalities (1) and (2), could there exist "very explicit" bijections between these different combinatorial structures too?

• fr.arxiv.org/abs/math/0004133 it doesn't answer your question or come close, but maybe it's related. Also en.wikipedia.org/wiki/Combinatorial_species Jun 9, 2014 at 21:06
• @johnmangual thank you! Jun 10, 2014 at 11:56
• @Alex J Best: thanks a lot for the bounty! Jun 11, 2014 at 13:51
• @MaxMuller No problem, shame it doesn't seem to have worked so well! Jun 11, 2014 at 13:57
• I've asked the same question on MO some years ago. It has an interesting answer: mathoverflow.net/questions/168913/… Sep 10, 2022 at 0:24