Written: Aug. 11, 2020.
As mentioned in this opinion , co-EIC Akihiro MUNEMASA, after an "initial review" decided that this article is "unlikely to meet the standards of depth and originality that ALCO is seeking". This erroneous decision was upheld and condoned by Sara BILLEY, Head of the journal's Steering Committee.
I would like to explain why this decision was erroneous, and why the Kauers-Z submission not only meets the "standards of depth and originality" that ALCO is seeking, but far exceeds them.
The rejected paper beautifully intertwines several central themes of algebraic combinatorics.
Enumeration: The mother of algebraic combinatorics.
I invite you to read the paper and find out for yourself. I offer donations to the OEIS in honor of the first provers for each of the four fascinating conjectures at the end. Let me just comment on Conjecture 2a. The total number of walks from the origin [0,0,0] to [n,n,n] with unit positive steps in the cubic lattice is trivially (3n)!/n!3 whose asymptotics (use Stirling or otherwise) is Ω(27n/n). From the Young-Frobenius formula, it follows that the analogous quantity for walks that stay in x ≥ y ≥ z is Ω(27n/n4).
Now add the restriction that each "run" in each of the three directions must be of length at least two. As we proved (beautifully) in the paper (a deep result!) for the lattice walk case, the enumerative sequence is P-recursive, and in fact the order 13, degree 10, linear recurrence can be found in this output file that implies (rigorously!) using standard methods, that the asymptotics is Ω(8n/n). Using our efficient enumeration algorithm for the Young tableaux case, we conjecture that the analogous asymptotics is Ω(8n/n4), in exact analogy with the unrestricted case. I have no clue how to prove it, and would love to donate to the OEIS in honor of the first prover.
Even more interesting is Conjecture 2b, where the growth constant is irrational (7+5*sqrt(2)). Once again the lattice-path version can be proved rigorously to be Ω((7+5*sqrt(2))n/n) but I have no clue how to prove the Young tableaux analog, that it equals Ω((7+5*sqrt(2))n/n4).
To be honest, I don't like the words "depth" and "originality". They are so pompous. What is more important is interesting and significant, and in these respects my paper with Manuel Kauers far exceeds even the "better" papers of ALCO.
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