## Fibonacci times Euler

Recall the *Fibonacci numbers* given by 1,1,2,3,5,8,13,21… There is no need to define them. You all know. Now take the *Euler numbers (OEIS)* 1,1,1,2,5,16,61,272… This is the number of alternating permutations in with the exponential generating function . Both sequences are incredibly famous. Less known are connection between them.

(1) Define the *Fibonacci polytope* to be a convex hull of 0/1 points in with no two 1 in a row. Then has vertices and vol This is a nice exercise.

(2) (by just a little). For example, . This follows from the fact that

and , where is the g*olden ratio*. Thus, the product . Since and , the inequality is easy to see, but still a bit surprising that the numbers are so close.

Together with Greta Panova and Alejandro Morales we wrote a little note “Why is π < 2φ?” which gives a combinatorial proof of (2) via a direct surjection. Thus we obtain an indirect proof of the inequality in the title. The note is not a research article; rather, it is aimed at a general audience of college students. We will not be posting it on the arXiv, so I figure this blog is a good place to advertise it.

The note also explains that the inequality (2) also follows from Sidorenko’s theorem on complementary posets. Let me briefly mention a connection between (1) and (2) which is not mentioned in the note. I will assume you just spent 5 min and read the note at this point. Following Stanley, the volume of is equal to the volume of the *chain polytope* (=*stable set polytope*), see Two Poset Polytopes. But the latter is exactly the polytope that Bollobás, Brightwell and Sidorenko used in their proof of the upper bound via polar duality.

## The power of negative thinking, part I. Pattern avoidance

In my latest paper with Scott Garrabrant we disprove the *Noonan-Zeilberger Conjecture*. Let me informally explain what we did and why people should try to disprove conjectures more often. This post is the first in a series. Part II will appear shortly.

#### What did we do?

Let *F ⊂ S _{k}* be a finite set of permutations and let

*C*(

_{n}*F*) denote the number of permutations

*σ ∈ S*avoiding the set of patterns

_{n}*F*. The

*Noonan-Zeilbeger conjecture*(1996), states that the sequence {

*C*(

_{n}*F*)} is always

*P-recursive*. We disprove this conjecture. Roughly, we show that every Turing machine T

*can be simulated by a set of patterns*

*F,*so that the number

*a*of paths of length n accepted by by T is equal to

_{n }*C*(

_{n}*F*) mod 2. I am oversimplifying things quite a bit, but that’s the gist.

What is left is to show how to construct a machine T such that {*a _{n}*} is not equal (mod 2) to

**any**P-recursive sequence. We have done this in our previous paper, where give a negative answer to a question by Kontsevich. There, we constructed a set of 19 generators of

*GL(4,Z)*, such that the probability of return sequence is not P-recursive.

When all things are put together, we obtain a set *F* of about 30,000 permutations in *S _{80}* for which {

*C*(

_{n}*F*)} is non-P-recursive. Yes, the construction is huge, but so what? What’s a few thousand permutations between friends? In fact, perhaps a single pattern (1324) is already non-P-recursive. Let me explain the reasoning behind what we did and why our result is much stronger than it might seem.

#### Why we did what we did

First, a very brief history of the NZ-conjecture (see Kirtaev’s book for a comprehensive history of the subject and vast references). Traditionally, pattern avoidance dealt with exact and asymptotic counting of pattern avoiding permutations for small sets of patterns. The subject was initiated by MacMahon (1915) and Knuth (1968) who showed that we get Catalan numbers for patterns of length 3. The resulting combinatorics is often so beautiful or at least plentiful, it’s hard to imagine how can it not be, thus the NZ-conjecture. It was clearly very strong, but resisted all challenges until now. Wilf reports that Richard Stanley disbelieved it (Richard confirmed this to me recently as well), but hundreds of papers seemed to confirm its validity in numerous special cases.

Curiously, the case of the (1324) pattern proved difficult early on. It remains unresolved whether *C _{n}*(1324) is P-recursive or not. This pattern broke Doron Zeilberger’s belief in the conjecture, and he proclaimed that it’s probably non-P-recursive and thus NZ-conjecture is probably false. When I visited Doron last September he told me he no longer has strong belief in either direction and encouraged me to work on the problem. I took a train back to Manhattan looking over New Jersey’s famously scenic Amtrack route. Somewhere near Pulaski Skyway I called Scott to drop everything, that we should start working on this problem.

You see, when it comes to pattern avoidance, things move from best to good to bad to awful. When they are bad, they are so bad, it can be really hard to prove that they are bad. But why bother – we can try to figure out something awful. The set of patterns that we constructed in our paper is so awful, that proving it is awful ain’t so bad.

#### Why is our result much stronger than it seems?

That’s because the proof extends to other results. Essentially, we are saying that everything bad you can do with Turing machines, you can do with pattern avoidance (mod 2). For example, why is (1324) so hard to analyze? That’s because it’s even hard to compute both theoretically and experimentally – the existing algorithms are recursive and exponential in *n*. Until our work, the existing hope for disproving the NZ-conjecture hinged on finding an appropriately bad set of patterns such that computing {*C _{n}*(

*F*)} is easy. Something like this sequence which has a nice recurrence, but is provably non-P-recursive. Maybe. But in our paper, we can do worse, a lot worse…

We can make a finite set of patterns *F,* such that computing {*C _{n}*(

*F*) mod 2} is “provably” non-polynomial (Th 1.4). Well, we use quotes because of the complexity theory assumptions we must have. The conclusion is much stronger than non-P-recursiveness, since every P-recursive sequence has a trivial polynomial in

*n*algorithm computing it. But wait, it gets worse!

We prove that for two sets of patterns *F* and *G*, the problem* “C _{n}*(

*F*) =

*C*(

_{n}*G*) mod 2 for all n” is undecidable (Th 1.3). This is already a disaster, which takes time to sink in. But then it gets even worse! Take a look at our Corollary 8.1. It says that there are two sets of patterns

*F*and

*G*, such that you can never prove nor disprove that

*C*(

_{n}*F*) =

*C*(

_{n}*G*) mod 2. Now that’s what I call truly awful.

#### What gives?

Well, the original intuition behind the NZ-conjecture was clearly wrong. Many nice examples is not a good enough evidence. But the conjecture was so plausible! Where did the intuition fail? Well, I went to re-read Polya’s classic “*Mathematics and Plausible Reasoning*“, and it all seemed reasonable. That is both Polya’s arguments and the NZ-conjecture (if you don’t feel like reading the whole book, at least read Barry Mazur’s interesting and short followup).

Now think about Polya’s arguments from the point of view of complexity and computability theory. Again, it sounds very “plausible” that large enough sets of patterns behave badly. Why wouldn’t they? Well, it’s complicated. Consider this example. If someone asks you if every 3-connected planar cubic graph has a Hamiltonian cycle, this sounds plausible (this is Tait’s conjecture). All small examples confirm this. Planar cubic graphs do have very special structure. But if you think about the fact that even for planar graphs, Hamiltonicity is NP-complete, it doesn’t sound plausible anymore. The fact that Tutte found a counterexample is no longer surprising. In fact, the decision problem was recently proved to be NP-complete in this case. But then again, if you require 4-connectivity, then *every* planar graph has a Hamiltonian cycle. Confused enough?

Back to the patterns. Same story here. When you look at many small cases, everything is P-recursive (or yet to be determined). But compare this with Jacob Fox’s theorem that for a random single pattern π, the sequence {*C _{n}*(π)} grows much faster than originally expected (cf. Arratia’s Conjecture). This suggests that small examples are not representative of complexity of the problem. Time to think about disproving ALL conjectures based on that evidence.

If there is a moral in this story, it’s that what’s “plausible” is really hard to judge. The more you know, the better you get. Pay attention to small crumbs of evidence. And think negative!

#### What’s wrong with being negative?

Well, conjectures tend to be optimistic – they are wishful thinking by definition. Who would want to conjecture that for some large enough *a,b,c* and *n,* there exist a solution of *a*^{n} + *b*^{n} = *c*^{n}? However, being so positive has a drawback – sometimes you get things badly wrong. In fact, even polynomial Diophantine equations can be as complicated as one wishes. Unfortunately, there is a strong bias in Mathematics against counterexamples. For example, only two of the Clay Millennium Problems automatically pay $1 million for a counterexample. That’s a pity. I understand why they do this, just disagree with the reasoning. If anything, we should encourage thinking in the direction where there is not enough research, not in the direction where people are already super motivated to resolve the problem.

In general, it is always a good idea to keep an open mind. Forget all this “power of positive thinking“, it’s not for math. If you think a conjecture might be false, ignore everybody and just go for disproof. Even if it’s one of those famous unsolved conjectures in mathematics. If you don’t end up disproving the conjecture, you might have a bit of trouble publishing computational evidence. There are some journals who do that, but not that many. Hopefully, this will change soon…

#### Happy ending

When we were working on our paper, I wrote to Doron Zeilberger if he ever offered a reward for the NZ-conjecture, and for the disproof or proof only? He replied with an unusual award, for the proof and disproof in equal measure. When we finished the paper I emailed to Doron. And he paid. Nice… 🙂