A Bijective Proof of Richard Stanley’s Observation that the sum of the cubes of the n-th row of Stern’s Diatomic Array equals 3 times 7 to the power n-1

Shalosh B. EKHAD and Doron ZEILBERGER

Abstract: In a delightful article, Richard Stanley derived, algebraically, the “surprisingly simple” formula, 3 · 7^{n−1}, for the sum of the cubes of the n-th row of Stern’s diatomic array. In this note, we find an elegant bijective proof of this surprising fact, that explains it and gives insight. The novelty is that this gorgeous bijection was discovered by a computer (SBE), with minimal guidance by a human (DZ). This debunks the conventional wisdom, held by some human supremacists, that computers can only compute, but they can’t give insight.

In [S], the following double sequence a(n, k) is defined

∑_{k=0}^{2(2^{n}−1)} a(n, k)x^{k} = ∏_{i=0}^{n−1} ( 1 + x^{2^{i}} + x^{2·2^{i}} ) .

For any vector in a ∈ {0, 1, 2}^{n}, define the valuation, v(a), by

v(a) = ∑_{i=1}^{n} a_{i} 2^{i−1} ,

then, obviusly, a(n, k) counts the number of vectors in {0, 1, 2}^{n} with valuation k, and hence a(n, k)^{r} counts the number of r × n matrices with entries in {0, 1, 2} such that the valuation of each of its r rows equals k. Hence

A_{r}(n) := ∑_{k=0}^{2(2^{n}−1)} a(n, k)^{r} ,

counts the set, let’s call it A_{r}(n), of all r × n matrices with entries in {0, 1, 2} such that all rows have identical valuations.

Stanley ([S], p.4) proved (among many other results), the

Surprisingly Simple Fact: A_{3}(n) = 3 · 7^{n−1} .

This cries for a bijective proof! The right side obviously counts the set of words in the alphabet {1, 2, 3, 4, 5, 6, 7} of length n, whose last letter is in {1, 2, 3}. Let’s call this set S(n).

Theorem: There exists an elegant bijection, implemented in procedure StanBij(M) in the Maple package

http://www.math.rutgers.edu/~zeilberg/tokhniot/BijectionStern.txt ,

between the set A_{3}(n) and S(n), whose inverse is given by InvStanBij(w), also implemented there.

The actual definition of the bijection can be easily read directly from the Maple source code, let’s just describe how we constructed it. The same methodology should work to provide bijective proofs to any of the recurrences listed in [S], but the details will not be as elegant.

The set A_{3}(3) has 3·7^{2} = 147 members, but the set obtained by deleting the third column happens to only have 63 elements. In fact this is true if you take A_{3}(n) and remove the last n − 2 columns, you always get the same set. Let’s call this set of 3 × 2 matrices, B_{3}.

On the other hand the set of 3 × 1 matrices obtained by only retaining the first column of the members of A_{3}(n) (for n ≥ 2) has exactly 9 elements, let’s call it C_{3}.

C_{3} = \{( \begin{matrix}0 \\ 0 \\ 0 \end{matrix} ) , ( \begin{matrix}0 \\ 0 \\ 2 \end{matrix} ) , ( \begin{matrix}0 \\ 2 \\ 0 \end{matrix} ) , ( \begin{matrix}0 \\ 2 \\ 2 \end{matrix} ) , ( \begin{matrix}1 \\ 1 \\ 1 \end{matrix} ) , ( \begin{matrix}2 \\ 0 \\ 0 \end{matrix} ) , ( \begin{matrix}2 \\ 0 \\ 2 \end{matrix} ) , ( \begin{matrix}2 \\ 2 \\ 0 \end{matrix} ) , ( \begin{matrix}2 \\ 2 \\ 2 \end{matrix} )\} .

Given a matrix M in A_{r}(n), we define v(M) to be the vector of length r consisting of the valuations of its rows. If the smallest entry is larger than zero, we subtract it from all the entries, getting a reduced valuation vector, so one of its entries must be 0.

It so happens that the 63 members of B_{3} neatly fall into 7 classes, each with the same (after adjusting by dividing by 2) valuations as the members of C_{3}. So we can ‘extract’ a letter in {1, 2, 3, 4, 5, 6, 7} as well as a member of C_{3} in such a way that if you replace the first two columns by that new column, it is still true that all the rows have the same valuation, in other words, you get a member of A_{3}(n − 1). Now keep going, recursively, until we have a member of A_{3}(1), and map its three members to {1, 2, 3}.

Of course, there are many choices for these mappings, so we showed how to construct many bijective proofs to this surprisingly simple result. For a table of the bijection for n up to 5, see the output

file http://www.math.rutgers.edu/~zeilberg/tokhniot/oBijectionStern1.txt .

Reference

[S] Richard P. Stanley, Some linear recurrences motivated by Stern’s Diatomic Array, arXiv:1901.04647v1

[math.CO], 15 January 2019. https://arxiv.org/abs/1901.04647 .

Also in: Amer. Math. Monthly 127 (2020), 99-111.

Shalosh B. Ekhad and Doron Zeilberger, Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen Rd., Piscataway, NJ 08854-8019, USA.

Email: [ShaloshBEkhad, DoronZeil] at gmail dot com .

Exclusively published in the Personal Journal of Shalosh B. Ekhad and Doron Zeilberger and arxiv.org

Written: March 3, 2021.