Matthew England is a great collaborator. I am glad to be in the same project as him. Coventry University is also a great host. I enjoy my visits to this group a lot. After working for 3 days non-stop, we had a short time to have this chat. I am so happy that we get to talk about important questions like what motivates him to do research. I learned a lot, I think you can too.

Matthew also has great humor and such a quick wit. It is quite fun to chat with him. That is why, this time, I am not editing the early minutes of our conversation out of this video. I think you’d like it. I am so dissapointed I don’t seem like I am enjoying the jokes but I love how he “threatened me with a pen”. 😛

I was just too preoccupied, thinking how I am supposed to start the video, what to ask, and in which order, etc. These early jokes deserve much more laughter that I didn’t bring to the table… I hope you will enjoy them more than I couldn’t back then.

Hebron & Medlock Professor of Information Technology James H. Davenport needs no introduction. I have been meaning to have this interview for a while now, but 15 minutes is a long time to sit and chat when you are as busy as James. He is so courteous to spare me this time, so that I can share some of his knowledge, and some of his thought on academia with you.

I hope you will enjoy this conversation as much as I did. I should schedule the next one with James really soon.

The thumbnail is from the Oberwolfach photo collection and it is used under the creative commons license.

## Some Recent Advances in Cylindric Partitions

Studying theory of partitions, I come across many natural and exotic definitions. Cylindric partitions is one of those that is both. I don’t know how but it somehow is. Informally cylindric partitions encodes information about non-intersecting lattice walks and that is clearly natural. However, it is also a finite vector of partitions that can be put together in such a way that you can wrap this shifted arrangement around a cylinder. I don’t think that as something I would be able to come up, so I find it quite exotic in that sense. Thankfully we have great visionaries around such as Ira Gessel and Christian Krattenthaler in this case.

Cylindric partitions are actually really nice. An integer partition is a finite list of non-increasing natural numbers. So given a partition $\lambda = (\lambda_1,\lambda_2,\dots)$, we have an order relation that the parts satisfy: $\lambda_i \geq \lambda_{i+1}$. In the cylindric partitions, we start with a finite list of partitions $(\lambda^{(1)}, \dots, \lambda^{(k)})$ and a list of non-negative integers called a profile $c = (c_1,c_2,\dots,c_k)$ where the parts $\lambda^{(i)}_j$ of the partitions $\lambda^{(i)}$ satisfy an interlacing order relation $\lambda^{(i)}_{j} \geq \lambda^{(i+1)}_{j+c_i},$ where $\lambda^{(k+1)} = \lambda^{(1)}$.

Why do I/we care? Most classical partition identities gets realized as analytic identities where one side is an infinite sum and the other side is an infinite product. Number theoretically, to me, this is a bone chilling structural finding. We see that an infinite series can be completely factored. In the classical results, the sums are easier to find and easier to associate with the partition theory. However, it is quite hard to prove (and unclear to do so) that this sum is equal to a product even when the numerics provides indubitable evidence.

In cylindric partitions, all thanks to Alexei Borodin, we already know that the generating function for the number of such objects is an infinite product. To be precise, generating function for the number of cylindric partitions with profile $c = (c_1,c_2,\dots,c_k)$ (and $t=c_1+c_2+\dots+c_k +k$) is

$\displaystyle\frac{1}{(q^t;q^t)_\infty} \prod_{i=1}^k \prod_{j=i}^k \prod_{m=1}^{c_i} \frac{1}{(q^{m+j-i+s(i+1,j)};q^t)_\infty} \prod_{i=2}^k \prod_{j=2}^i \prod_{m=1}^{c_i} \frac{1}{(q^{t-m+j-i-s(j,i-1)};q^t)_\infty}$

*This product of products is written using q-Pochhammer symbols. Later we will also use q-Binomial coefficients.

This goes against the grain of classical results. This time we have the product representations for these generating functions for free but we have no idea what the infinite series are supposed to look like. For the profile sizes $k=2$, we know the cylindric partitions’ related products coincide with the Andrews-Gordon and Bressoud products, so the sum sides are known through these powerful theorems. However, the $k=3$ cases are still filled with unknowns and conjectures for the sum representation of the generating functions, most advancements are at this level now. $k\geq 4$ is a mystery. That is the product generating functions are there, but we don’t know the closed formula sum representations.

The recent $k=3$ journey starts with a paper by Sylvie Corteel and Trevor Welsh. Their idea is simple and that is what makes it extra brilliant. In simple words, we know the interlacing order relations the cylindric partitions have to satisfy. Therefore, we know at what location or locations the largest part of a cylindric partition can be; that is once given the profile $c$. If those largest parts are removed from the object, a new cylindric partition with a new profile $c^*$ emerges. So we can derive recurrences for the generating functions and associate them with generating functions for different profiles. The number of generating functions involved in such a construction is finite so we are not churning water.

Sylvie and Trevor used this recurrence relations to find the sum generating functions for $k=3$ cylindric partitions with $c_1+c_2+c_3=4$. They coincided with Andrews-Schilling-Warnaar‘s $A2$ Rogers-Ranaujan identities plus found an identity that was not identified before. Later, Sylvie, Jehanne Dousse and I come together and identified the $k=3$ case with $c_1+c_2+c_3 = 5$ in this paper.

As a side note, it is not easy to guess these sum representations at all. They are infinite multisums with 3 and 4 variables in the cases mentioned above, respectively. Once guessed/identified, it is not easy to prove that these objects satisfy the recurrences the generating functions for the number of cylindric partitions with certain products is not easy either due to the high number of variables.

Sylvie and Jehanne were the ones who were able to identify the 4-fold sum generating functions. I cannot take any credit in that miraculous find. I was working on the qFunctions package and as a part of that I was implementing the (coupled) recurrence system of Corteel-Welsh. My objective was to carry this system to a computer algebra system so that we can use other this in conjunction with other symbolic computation tools to prove (or disprove) guessed identities in the spirit of Corteel-Welsh paper.

This was a perfect match. Once we combined our powers, we proved 7 sum-product formulas such as the following.

$\displaystyle\sum_{n_1,n_2,n_3,n_4\geq 0 } \frac{q^{n_1^2+n_2^2+n_3^2+n_4^2+n_1+n_2+n_3+n_4-n_1n_2 + n_2n_4}}{(q;q)_{n_1}} {n_1\brack n_2}_q{n_1\brack n_4}_q{n_2 \brack n_3}_q=\frac{1}{(q^2,q^3,q^3,q^4,q^4,q^5,q^5,q^6;q^8)_\infty}$

Not long after we submitted our paper, Ole Warnaar was able to guess the generic sum representiton for the $k=3$ profile cylindric partitions with $c_1+c_2+c_3\not\equiv 0$ mod $3$. For example, the claim is that for every $j$ the following sum is the sum representation of the generating function for some particular cylindric partitions.

$\displaystyle\sum_{\substack{n_1,\dots,n_j\geq 0\\m_1,\dots,m_{j-1}\geq 0}} \frac{q^{n_j^2+\sum_{i=1}^j}}{(q;q)_{n_1}} \prod_{i=1}^{j-1} q^{n_i^2 - n_im_i + m_i^2 + m_i} {n_i\brack n_{i+1}}_q {n_i-n_{i+1}+m_{i+1}\brack m_i}_q$

If we can prove this, then sum (due to Borodin) factors completely and has a product represenatiton that we wrote above. More conjectures and many interesting results can be found in Ole’s paper.

The way Ole come up with these results is also commendable. He looked at the $k=3$ results that we mentioned above as well as two results from $k=2$ that can be viewed as initial cases of $k=3$ using some symmetries of cylindric partitions. There two steps were enough for him to guess the whole chain. This is an amazing advancement. I wonder if I could do this induction even if I had 20 steps on this chain… Ole only needed 2. Who knows maybe he would guess parts of the $k=4$, etc. chains in the future too.

Technically, we should be able to use the code in qFunctions with Christoph Koutschan‘s HolonomicFunctions package and prove these cases one by one. However, we do not have the computational power to prove that the multi-fold sums satisfy a given recurrence. This is not about physical limiting factor of a computer hardware either. We know that for these objects the theory says that Doron Zeilberger‘s creative telescoping algorithm would terminate, but we do not know when. My personal experience is that for the conjecture above with $j=3$, the code did not terminate in a month on a RISC server even thought the server was not having any memory shortage or anything of sorts.

There you have it. We have all these identities and conjectures, and that is just the beginning of it. We still haven’t touched anything with $k\geq 4$. So be my guest and/or be my collaborator. If you have a good idea on how one can guess these sums?, how we can prove Ole’s guessed sums?, etc. I’m all ears.

But wait there is more! Walter Bridges was looking at a paper about the asymptotics related to the number of cylindric partitions by Han-Xiong. He saw that they have presented Borodin like product representations for symmetric cylindric partitions and double skew shifted plane partitions. Not only that, he also observed that we can adapt Corteel-Welsh idea for all these objects. I was happy to jump in this project right away, and this led to our joint paper. We discovered many nice new sum-product identities such as

$\displaystyle\sum_{n,m \geq 0} (-1)^m q^{3{n+1 \choose 2}-3m(m+1)} \frac{(-q,-q^5;q^6)_m}{(q^6;q^6)_m(q^3;q^3)_{n-2m}}=\frac{(q^4,q^8;q^{12})_{\infty}}{(q^6;q^{12})_{\infty}}.\\$

Furthermore, the more that we looked at the Borodin and Han-Xiong product generating functions and how they are constructed, the more clear it became that we can weight the counts of different diagonals (partitions) in the cylindric partitions and the product generating functions would reflect this change without the underlying structure changing. This opened us up to infinitely many combinatorial connections that we don’t know what to do with (yet). There are really cute ones, and there are the ones that were discovered before outside of the scope of cylindric partitions/skew shifted plane partitions.

For example, in our paper, we prove some Schmidt-type partition results, my favorite being the Schmidt’s theorem itself. (A great read on this with many new results by George Andrews and Peter Paule here.) The theorem says that counting the partitions into distinct parts ($\lambda = (\lambda_1,\lambda_2,\dots)\in\mathcal{D}$ i.e. $\lambda_i>\lambda_{i+1}>\dots$) with the “every other” part counting statistic takes you to the count of all the partitions:

$\displaystyle\sum_{\lambda\in\mathcal{D}} q^{\lambda_1 + \lambda_3+\lambda_5+\dots} = \frac{1}{(q;q)_\infty}$

*For any confused audience members: $1/(q;q)_\infty$ is the generating function for the number of all partitions.

In one of my previous papers, I proved this result without knowing that it already known. I was using a completely different technique, I am happy that it was not a full rediscovery. With Walter, I also proved this theorem using the weighted skew shifted plane partitions (see the line above Prop 5.1 here). Basically, we put distinct parts in a staircase of size two and then weight the diagonal that has the even indexed parts with 0 so that their contribution vanishes and the weighted version of Han-Xiong’s products did the work for us.

Anyway, I will leave it here. There is a lot more to discover, and next time when you have an infinite product generating function in hand check if it is a weighted cylindric partition or a skew shifted plane partition, you might just see a new combinatorial connection to your problem. We have these functionalities build in qFunctions now…

## Prove One, Get Infinitely Many Free!

This is a better deal than any deal you can find on the market.

Bailey’s lemma is an amazing tool. At its core it is a really simple observation but in practice it is a beast. I can (and will) state the actual lemma and you will think to yourself “is this it?”. I am serious. It is as easy as teaching a kid that they can roll a snowball on snow to get it larger to make a snowman. Yet, in practice the same action of snowball rolling down a hill can lead to huge avalanches.

Here is the lemma: Wilfrid N. Bailey noticed that if you start two identities of the form

$\displaystyle \beta_L = \sum_{r=0}^L \alpha_r u_{L-r} v_{L+r}$ and $\displaystyle \gamma_L = \sum_{r=L}^\infty \delta_r u_{L-r} v_{L+r}$

for some sequences $\alpha_j,\ \beta_j,\ \gamma_j, \delta_j,\ u_j,$ and $v_j$ , then we have

$\displaystyle \sum_{L=0}^\infty \alpha_L \gamma_L = \sum_{L=0}^\infty \beta_L \delta_L.$

OK, so I can hear you say that “Ali, you lied, this doesn’t look all that easy.” Well, it is not so easy to roll a snowball for a kid at their first try. Yes, Bailey’s observation is sophisticated but let me give you the proof. It will make you agree that this implication is actually easy. Follow from left to right:

$\displaystyle \sum_{L=0}^\infty \alpha_L \gamma_L = \sum_{L=0}^\infty \alpha_L \sum_{r=L}^\infty \delta_r u_{L-r} v_{L+r} = \sum_{r=0}^\infty \delta_r \sum_{L=0}^r \alpha_L u_{r-L} v_{r+L} = \sum_{r=0}^\infty \beta_r \delta_r.$

Of course, the most rigorous among us must be telling me that we need many convergence conditions to justify the magic trick I did above. You would be right, but this is supposed to be a general writing that I am failing to keep light. Please excuse me. I will not justify them here.

Now back to the main point. There are standard choices for $u_j,\ v_j,\ \delta_j,$ and $\gamma_j$ which makes Bailey’s lemma work. With these standard choices, we are left with free choices for $\alpha_j$ and $\beta_j$. A pair $(\alpha_j,\beta_j)$ that satisfies Bailey’s lemma is conveniently called a Bailey pair.

Bailey and later his student Lucy Slater did amazing things with this result. I am talking about hundreds of novel sum-product identities by merely applying Bailey’s lemma. Skipping ahead, two big visionaries, Peter Paule and George E. Andrews independently observed something Bailey and Slater missed completely. They observed that if you start with a Bailey Pair $(\alpha_j,\beta_j)$ then you can find a new pair $(\alpha'_j,\beta'_j)$ that is also a Bailey pair. So you can apply the lemma again and generate a new-new Bailey pair $(\alpha''_j,\beta''_j)$… That is the avalanche. You start with a snowball -the initial seed identity that defines your Bailey pair- then you apply Bailey’s lemma as many times as you like. This creates an infinite hierarchy on top of your seed identity.

Andrews-Gordon identities is one of the most famous q-series identities that builds on the Rogers-Ramanujan identities with this very lemma. These identities are the main content of Andrews’ American Mathematical Society Presidential Signature:

I am sure you also noticed Srinivasa Ramanujan‘s silhouette watching over this infinite hierarchy. He sow the case $k=2$ seed that turned into this infinite family of sum-product identities by Bailey’s lemma in the hands of Andrews.

One remark I need to make is that at each step of the application of the lemma you are summing both sides of your seed identity in a special way. It is the reason that you see an implicit $\underline{n} = (n_1,n_2,\dots,n_{k-1})$ vector on the left-hand side of the above identity. That is actually a $k-1$-fold sum.

To sum it up, Bailey’s lemma is an incredible tool. Every day there are more and more being discovered using this technique. In my recent paper with Alexander Berkovich, we reaped the seeds we sowed. For example, with a special form of Bailey’s lemma, we saw that

$\displaystyle \sum_{m,n\geq 0} \frac{q^{2m^2+6m n+6n^2}(q^3;q^3)_M}{(q;q)_m (q^3;q^3)_n(q^3;q^3)_{M-2n-m}} = \sum_{j=-M}^M q^{3j^2+j} {2M\brack M-j}_{q^3}$

(which we proved in an earlier joint paper) implies

\begin{aligned} \sum_{m,n,n_1,n_2,\dots,n_f\geq 0 } &\frac{q^{2m^2+6mn+6n^2+3(N_1^2 + N_2^2+\dots+N_f^2)} (q^3;q^3)_{n_f}}{(q;q)_m (q^3;q^3)_n (q^3;q^3)_{n_f-2n-m }(q^3;q^3)_{n_1}(q^3;q^3)_{n_2}\dots(q^3;q^3)_{n_{f-1}}(q^3;q^3)_{2n_f}}\\&\hspace{9cm} = \frac{(q^{6f+6}, -q^{3f+2}, -q^{3f+4}; q^{6f+6})_\infty}{(q^3;q^3)_\infty},\end{aligned}

for any $f=1,2,3,\dots$, where $N_i:=n_i + n_{i+1}+\dots + n_f$ with the standard definitions of q-Pochhammer symbols and q-Binomial coefficients.

## Reflecting (on) the Kanade-Russell conjectures

I want to write a short note on the Kanade-Russell conjectures especially concerning their brief history and my involvement with them. This comes right after my new manuscript titled “Reflecting (on) the modulo 9 Kanade-Russell (conjectural) identities” with Wadim Zudilin. Long story way too short, I find these conjectures to be marvelous and equally frustrating. I hope the day will come that I will be able to either produce or follow a proof of them. Moreover, instead of solving the original conjectures, Wadim and I produced more sum-product type conjectures directly related to the modulo 9 Kanade-Russell conjectures in our work.

Creating clear targets in partition theory research is usually a hard task for me. I sometimes find myself like a car mechanic with no cars in his shop. It is a lazy existence. I have tools but no target to aim them at. It is boring, and even worrisome at times. You just don’t know when or where a new problem will arise or if the last paper you wrote is indeed your last paper (modulo all the projects you left behind for a future date). Kanade-Russell modulo 9 conjectures, as they stand, completely eliminates this problem for me. They are clear targets, I just don’t have the tools/don’t know how to use the tools to crack them open.

I should maybe start with a small recap. In their Ph.D. studies, Shashank Kanade and Matthew Russell put together a Maple program called IdentityFinder which does exact enumeration of partitions with some gap conditions in distance rules. They computed many truncated q-series generating functions for various gap conditions, and then checked to see if any one of these series’ conjecturally have a product representation (indicating the existence of equinumerous sets of partitions with some congruence conditions). In their first paper, published in 2015, they presented 4 modulo 9 and 2 modulo 12 conjectures. The 5th modulo 9 conjecture was noted in Matthew’s thesis the next year. In 2018, they wrote down another 9 modulo 12 conjectures. Around this time both Kanade-Russell, and Kağan Kurşungöz independently wrote down the analytic versions of these conjectures too. The one I stared at the most must be the first modulo 9 conjecture:

$\displaystyle \sum_{m,n\ge0} \frac{q^{m^2+3mn+3n^2}}{(q;q)_m(q^3;q^3)_n} \overset?= \frac{1}{(q,q^3,q^6,q^8;q^9)_\infty},$

where we use the standard definitions (and abbreviations) of the q-Pochhammer symbols $(a;q)_n = (1-a)(1-aq)\dots(1-aq^{n-1})$ for $n=0,1,2,\dots$, $(a;q)_\infty = \lim_{n\rightarrow\infty} (a;q)_n,$ and $(a_1,a_2,\dots,a_k;q)_n = (a_1;q)_n(a_2;q)_n\dots(a_k;q)_n.$

For anyone not-so-familiar with q-series, this is a good time to stop and ask yourself:

Isn’t it beautiful to see the full factorization of an infinite series?

If this conjecture is true, it will mean that this double series on the left has a reciprocal product representation. Not only that, this product representation has a clear structure where we know all the exponents of the variable q to appear. Looking on the right-hand side these exponents are coming from the residue classes 1, 3, 6, 8 modulo 9. Hence, the naming of this being a “modulo 9 conjecture“. Also we call this product symmetric since the residue classes can be grouped into additive inverse pairs, such as $\pm 1$ and $\pm 3$ modulo 9 in this example.

In 2018, Kathrin Bringmann, Chris Jennings-Shaffer, and Karl Mahlburg together were able to crack 7 of the 11 modulo 12 conjectures. They reduced some of the other modulo 12 conjectures. Their proofs were not out of this world either. In fact, from start to finish, it was just known techniques applied brilliantly. Their hard work payed off. This breakthrough, of course, raised my hopes towards all the other conjectures. In 2019, Hjalmar Rosengren re-proved some of the identities of Bringmann et al. and he also proved the remaining 4 modulo 12 conjectures. This part needs confirmation from Rosengren, but I believe he became aware of these conjectures for the first time in Chris’ talk in my OPSFA 2019 special session*. So I feel that by organizing a special session and inviting Chris to Linz, I played an infinitesimal (and clearly a non-mathematical) role in the proofs of the last 4 modulo 12 conjectures.

*Chris confirmed this on June 15, 2021.

We reached 2020 and there were only the modulo 9 conjectures left. I talked about them with many great researchers; Alexander Berkovich, Jehanne Dousse, Chris Jennings-Shaffer, Carsten Schneider to name just a few. I also wrote many papers about Capparelli’s identities (joint with Alexander Berkovich) which seems to be the closest cousin to these conjectures. In fact, take a look at the first Capparelli’s identity in its analytic form

$\displaystyle \sum_{m,n\ge0} \frac{q^{2(m^2+3mn+3n^2)}}{(q;q)_m(q^3;q^3)_n} = (-q^2,-q^4;q^6)_\infty(-q^3;q^3)_\infty.$

You can mistake the sum side of this identity with the first modulo 9 Kanade-Russell conjecture above any day. The difference between the sum sides is only a factor of two in the exponent of q. On the other hand, the difficulty level is completely opposite. Capparelli’s identities are well studied. There are many proofs of Capparelli’s identities starting from George E. Andrews‘ 1994 proof. Not only q-series proofs either. For example, Stefano Capparelli himself gave a representation theoretic proof to this identity in 1995. With Alexander Berkovich, I proved this Capparelli’s identity 3 times already, each time by discovering a new finite analogue!

In the first proof we gave, we found a finite version of the left-side sum using the combinatorial techniques Kurşungöz laid out. I used this technique on Kanade-Russell conjectures too. I found combinatorially meaningful polynomial analogues of the sum sides for the modulo 9 Kanade-Russell conjectures. Not to drown you in details but the left-hand side sum of the first Kanade-Russell conjecture we presented above is counting some integer partitions with some gap conditions. I was able to show that if you want to count those partitions with the extra condition that the parts are all $\leq N$, the left-hand side sum turns into

$\displaystyle \sum_{m,n\geq 0} q^{m^2+3mn+3n^2}{N-m-3n+1 \brack m}_q {\lfloor\frac{2}{3}N\rfloor-m-n+1\brack n}_{q^3},$

where we use the standard q-Binomial coefficients defined by

$\displaystyle {m+n \brack m}_q := \left\lbrace \begin{array}{ll}\frac{(q;q)_{m+n}}{(q;q)_m(q;q)_{n}},&\text{for }m, n \geq 0,\\0,&\text{otherwise.}\end{array}\right.$

Now coming back to my recent paper with Wadim. He visited the Research Institute for Symbolic Computation in February 2020 and we had some good time to talk about various possible projects that we can work on. These conjectures were already in both of our radars so we started sharing ideas and get to experimenting with things. Nothing really solved the conjecture (and I am not going to announce that we prove them at the end or anything) but we started to have a mathematical discussion together. We started sending TeX files back and forth. Unfortunately, and with sincere regrets on my part, with the events of 2020 plus me getting more and more occupied with the search of the next academic job, I started to slack and let our conversation to fizzle out.

Later came an unforeseen and golden opportunity. Wadim contacted me out of the blue and let me know, one month in advance, that he will be attending the 85th Séminaire Lotharingien de Combinatoire that was going to take place in September. This was supposed to take place in Bundesinstitut für Erwachsenenbildung, Strobl; only 2 hours away from my apartment in Linz. It was incredibly unexpected to hear about an in-person conference. It was the 40th anniversary of Séminaire Lotharingien de Combinatoire and Christian Krattenthaler was not letting it be an online event. I wanted to go but I was too anxious to join. There were 14 participants on the list but that was still a high number for me. I watched the conference participants list maybe every other day and kept on seeing it shrink; 12 people, 11 people, 9… Exactly 10 days before the conference’s start date the list shrunk to only 7 people. That was my cue to join. I send Christian an email and asked if I can join…

It was an enjoyable conference. It was nice to be back at Strobl. It was pleasant to be able to meet new people again. And most importantly, it was delightful to catch up with people that I already know. The change of place was a nice refresher and I was able to bounce some ideas. My mind was at the right place again. Wadim and I picked up where we left off. He had new ideas, we started to try them one by one. The modulo 9 conjectures were not unraveling but at least we were learning a little more as each attempt failed.

My TeX file traffic with Wadim picked itself up again after the conference. We kept on chatting on the conjectures. After a while, Wadim reminded me of some observations he shared with me, which I completely failed to address months prior. Ole Warnaar in a private conversation shared his observations about the generating function for the partitions with the gap conditions of the 4th modulo 9 conjecture where the parts are $\leq N$. He observed that if one reflects $q\mapsto 1/q$ and clears denominators, depending on the residue class of $N$ mod 3, these sub-sequences converge to some modulo 45 products or to a sum of two mod 45 products. Ole observed these new conjectures using the defining recurrences of these generating functions. I already had the exact expression for these generating functions. So the formulae for the reflected sum sides were already there for us.

The 4th and the, later found, 5th modulo 9 conjectures involve asymmetric modulo 9 products. After the reflections we look at sub-sequences with respect to $N$ modulo 3. We conjecturally see that the limit of two sub-sequences are modulo 45 products and for the last residue class modulo 3 the limit of the reflection is the sum of the two modulo 45 products that appear as the limit of the other cases. On the other hand, the reflection of the first three symmetric conjectures are not as easy to identify. None of the sub-sequences seemingly converge to a single product. In those cases, Wadim’s intuition was what took us to the finish line. He reduced the problem down to a finite search space for me. Only after that, we saw that although the first 3 mod 9 conjectures involve symmetric products, their reflections are nowhere symmetric. In fact, the limits of the reflected images are not even single products. For each case, these limits seem to be sums of three modulo 45 products.

Here is a small sample set of how these new conjectures look like:

\begin{aligned}\sum_{a,b\ge 0} \frac{q^{a^2 - 3ab + 3b^2 + b}}{(q^3;q^3)_b} {3b - a \brack a}_q&\overset?= \frac{1 }{(q^2;q^3)_\infty(q^3,q^9,q^{12},q^{21},q^{30},q^{36},q^{39};q^{45})_\infty},\\\sum_{a,b\ge 0} \frac{q^{a^2-3ab+3b^2+a-2}}{(q^3;q^3)_b}{3b-a-1 \brack a}_q & \overset?= \langle 2,10,16,19\rangle+q\langle 4,10,16,22\rangle-q^2\langle 8,10,14,19\rangle ,\end{aligned}

where

$\displaystyle \langle c_1,c_2,c_3,c_4\rangle =\frac{(q^{45};q^{45})_\infty}{(q^3;q^3)_\infty\prod_{j=1}^4(q^{c_j},q^{45-c_j};q^{45})_\infty}.$

These are only 2 out of the 10 main conjectures proposed in the paper. We also reflected Capparelli’s theorem in the same spirit and proved that its reflection yields mod 48 products. So, I highly suggest you look into that paper.

We dedicated this paper to our dear friend and legendary mathematician Doron Zeilberger on the occasion of his 20th prime birthday.

I do not know when the modulo 9 Kanade-Russell conjectures will be proven. I also don’t know when their reflections will be proven. I was trying to prove some conjectures and be a part of the solution. We ended up with more conjectures. I became a part of the problem so to say. Thankfully, it is the kind of the problem mathematicians like. I hope people will enjoy attempting these new conjectures as much as I enjoy trying to solve Kanade-Russell conjectures.

As my final word, I want to thank my new co-author Wadim wholeheartedly for his collaboration, patience, and leadership.

## 10 days of quarantine

I don’t believe anyone wants me to recap what is up these days (the numbers, the numbers are up) or why there would be a need for quarantining. So I’ll just jump into my mandatory quarantine days.

### March 1, 2021: Quarantine day 0

I started this day fairly early. I actually did the Vienna to Bath part of my trip on this very day. So when I finally came to the University of Bath campus, there wasn’t much more of the day left. The evening was closing in when I finally got my keys and burrowed in my room at the Marlborough Court. I was the only occupant of a flat with 6 ensuite rooms and a shared kitchen. Perfect for quarantining.

The University’s quarantine team was already on top of everything. There was a microwave, a kettle, and a crate of food to last me the next 10 days. It was satisfactory, to say the least. Except, there wasn’t any crockery or cutleries. That was just an email to the quarantine away though. They responded really fast.

I had a meeting with the HR representative to prove my identity and that I entered the UK. That took no more than 15 minutes. Right then and there, I became a new member of the University of Bath.

### March 2: Just settling in

This was an eventless day really. The weather is not the greatest either. It was more about settling in and trying to list the things with respect to their urgency. I had my first meeting with, my new mentors, James H. Davenport and Russell Bradford. We had meetings before but this was the first time we were all in the same country. We mostly talked about the university and its structure.

James sent out some emails to jump-start things for me. Next thing, I was getting bombarded by different offices of the University to get things done fast. I set up my university account, email, sent a request for a library card… and the day ended before you know.

### March 3: First test day

Another foggy day with all grey skies. At least that’s what I can see from my dorm window. Nevertheless, this was an exciting day. I was going to self-administer my first test and send it back to NHS for testing. Of course, a negative result was desirable but this test will also show how the preventative measures I used held up against this virus.

I opened a test kit. Nothing surprising was in there. A booklet, a swab, a tube with some saline solution, a bag, and a return box was all there is. I read the documentation, did the registration, and watched a video in which a doctor demonstrated the test. Let me tell you, he made it look so easy. My test experience was completely different. Swabbing my tonsils? Almost impossible. I didn’t know that I could show so much gag reflex. Swabbing my sinuses? I started sneezing uncontrollably. Anyway, the experience wasn’t the best but it made me appreciate how proficient the tester in Linz was once more. The young soldier that administered the test on me just made it look so easy.

The university’s security picked the test from me later in the day. I kept on catching up with emails.

### March 4: I am negative

Once again the day started all foggy, but there was some excitement going around on the campus. I saw more students going up and down the street and I think some students were moving into the building across from my dorm.

The quarantine food pack came with some quite similar breakfast options. At least that is how I perceived it, maybe someone would have considered the frozen dinners as possible breakfasts too. I was given instant oatmeal or bar cereal with fruit yogurt. For the curious ones, I had milk. I would be okay with these options any day of the week but I noticed that I wanted something else, anything else today. I think it was quarantine creeping in.

My test results came back negative later in the day. Another 5 days to the second test and this will be over.

### March 5: University training courses

I got contacted by the Computer Science department admin about the department/university orientation among other things. I started with the departmental induction module. It was a fair amount of reading, especially if you are following the links and getting into the crevices of the site. It needs to be said that I am really impressed with how well put-together and organized this university is. The documentation is impeccable and there seems to be some reference document for every possible scenario.

By pure chance, I saw that a friend was giving a talk on Facebook. I directly jumped on it and asked if I can join too. Next thing, I was in a math talk with all the top people in the field. I was so surprised that I didn’t hear about this seminar series before.

At night the students threw what seems like a block party on the street in front of my window. I think they had a great time. I didn’t pay too much attention to them but I doubt there was much social distancing.

### March 6: A fire alarm chirp

Bright day, chilly but sunny. A boring breakfast and a cup of coffee later, I am ready for the day.

I continued going through the University’s induction information. Among many other things, I found a concrete set of mandatory training courses expected of me. I started with basically the easiest one, fire safety. I liked, the Windows 98 type, old interface this course had. This was actually true for all the online courses. They must have been prepared many years back but the information was still relevant. Why fix something that is not broken, right?

In the afternoon, while I was chatting with my girlfriend, the fire alarm went off for a short time. The information was fresh in my mind, if only the alarm went off for a tad bit longer, I would have darted all the way out. It would have been a great excuse to stretch my legs a little.

I figured that I have been eating rice one way or another for some number of days. Cup-noodle-like rice dishes, microwave rice, rice vermicelli… Suddenly the frozen dinners looked boring too.

Another night of partying on the streets. These kids know how to have fun.

### March 7: Frustrations build

Today I saw two new quarantine signs on doors in the apartment. Soon there will be some people here. One even got his commissary food box ready for their arrival on their doorstep. So far I was being careful but relaxed in the apartment but it became clear that more caution will be required around the kitchen from now on.

I was completely bored with the food though. That was the real issue. I didn’t want to prepare something from the quarantine rations. It might have also been that I got really bored of the statical discharge shocking me every time I touched the microwave. I wanted something delivered. This plan failed horribly due to the food delivery service not liking my credit card’s zip code. I got a little frustrated with that but at least I still had food.

To bribe my own morale, I decided to order some items from the online shop of the on-campus grocery store. I didn’t see this shop before, and I couldn’t wait to see it for myself. Their proportions of listings were really bizarre. They had only 3 types of cereal but at least 5 types of Asian sliced flavored tofu snacks. Sifting through the items was still some good fun. I collected things in my shopping cart and paid for them to be delivered in a day or two.

### March 8: Fire drill

I was calling realtors and property managers to find a long-term place for me and my girlfriend for the last week. Today, I extended my radius and called even more realtors to no avail. So things were getting a bit frustrating. The quarantine food was boring, staying indoors was boring,. my ear was at the door to see if I was going to get my grocery order (I didn’t). You know, there was some restlessness. I wish I could go out and maybe walk around the campus a little.

James gave an overview talk on our project, the literature, and possible future questions to tackle. Then I was added to the Math Foundations research group’s Discord server. It was so pleasant to see a group or colleagues socializing (as much as the internet allows) and joking around in a friendly setting. Although, I haven’t met them in person yet, I also received a warm welcome. That was a great interlude to a gloomy day.

Once again I was reminded to be careful of what to wish for. At 11:30 pm (23:30) all the fire alarms went off. Red lights and siren sounds, I followed the route with everyone to the outside meeting point. The fire marshal was not nice (nor they need to be) in any shape or form. I stayed away from the others with a mask on. I knew I just refreshed my fire safety information but that was quite enough for me to be honest. I wanted to be outside but not under those circumstances.

### March 9: Final test day

This was one of the most exciting and full days. I decided to rip the band-aid off and directly do the test. It was just as bad as the first time. Voluntarily swabbing your own tonsils and/or sinuses is not easy. I sneezed uncontrollably for minutes on end once again. They should give a medal to the guy that tested me in Linz.

More stimulating or shall I say more exciting was the arrival of my grocery order! Normally, such things are so mundane. So what if I got some milk? or a frozen pizza? This wouldn’t even be a topic of conversation. That is on what we deem a normal day. Today, after days of weak-incarceration, no matter how socially connected it was, the knock on my door and the two bags of goodies felt like the outside came to me.

The rest of the day was in the frame of the new normal. Sent some emails, did some work/typing, joined the Alcyon Lab‘s weekly seminar…

### March 10: Good ol’ hard work

Finally, a day where I wake up with a clear mind and willingness to focus heavily on work. Maybe I finally get used to the quarantine environment. And the progress showed. I was able to finish some tasks I was hoping to finish in my quarantine period. This meant now I needed to start all the tasks I needed to do.

Nevertheless, this was a good day.

### March 11: Negative again. Freedom.

I woke up to NHS emails telling me that my test results show no signs of the virus. Perfect start of a day. This was followed by the University quarantine team paying me a quick visit and taking the quarantine sign off my door. I was officially free.

Sadly, it was a rainy day. Normally I wouldn’t mind being home for 10 days and I would definitely go out when it’s raining. On this occasion, I made an exception and stretched my legs around the campus a little. It is a nice university and a nice campus for sure. I hope to walk this campus up and down for at least another 3 years.

## How to get your money back from Emma Mattress?

Disclaimer: This is my story (originally posted in mid August) and what I experienced. There is no intention of giving unsolicited legal advice. Experiences are to be shared so that people can judge and use any and all information they acquire. That being said…

# I Almost Sued a Company!

Internet shopping always comes with the possibility of fraudulent businesses and unfulfilled promises. Long story short, I encountered one of these companies. Emma Matratzen GmbH is a mattress company all around Europe with lots of internet presence. They advertise their products in all the social media platforms as well as on classical media. They are proud of their product and flash their awards on every advertisement. On the other hand, you meet hoards of upset customers if you even pay a little bit of attention to the comments of this company’s advertisements online.

Well… I should have paid some attention to those internet testimonials. I recently moved apartments. I was in the need of a bed and ordered one from Emma Matratzen GmbH. They failed to send my bed and made me sleep on my couch for a month. When asked, they said they will not cancel my order either! The full version of the story is below. Let me move onto how I got my money back from Emma right away.

## Here Is The Algorithm of How I Got My Money Back From Emma:

My business with Emma ended in a record breaking 45 days. I see people are struggling to get a refund for months/years with this company. What I wrote below is what worked for me:

1. I sent Emma regular customer support requests and asked them for updates on my unfulfilled order/cancellation/refund.
2. I found and contacted the nearest Customer Rights Office and asked for their advice.
3. In a formal language, I sent Emma Matratzen GmbH an email asking for the cancellation and gave them a formal 14 days grace period to refund my account. Under the guidance of the Upper Austria’s Customer rights office, I cited the following law + reason:

> “European Union Trade Law: [CELEX-No .: 32011L0083 ] Chapter 33, Items 37, 40, 42: The present contract is a business without the physical presence of the contracting parties, outside of the business premises of the entrepreneur. With this type of business the customer has a right of withdrawal in accordance with [CELEX-No .: 32011L0083]. If the withdrawal has been declared, the company is obliged to repay it.”

4. No matter what they said and did after the point 3. I reminded them that I have given them the 14 days formal grace period and my decision is final.
5. I made it clear that if they fail to refund me in the grace period, contacting the nearest Customer Rights Office again would be the first thing I do to start a legal action against them.

I feel like this is a solid algorithm. It worked quite well for me. I hope you will not let them keep your honestly earned money.

## Here is the Longer Version of What Happened to Me:

This is a story as old as time. A company does some customer wrong, that customer has lots of quarantine time on their hands, and next thing you know, they decide to procrastinate on everything they are actually supposed to do and fight back. Does it make this a Don Quixote or an actual standing for your rights story? You decide.

I moved apartments recently and in that I needed to purchase a mattress. I decided to purchase a mattress from Emma Matratzen over the internet due to their 100 days no-questions-asked return option and the one award they keep flashing on their advertisements. It looked like a safe bet. I ordered a mattress on June 29th, 2020, to be delivered between 7th to 11th of July. This was supposed to be my new mattress in my new apartment. I found having a mattress in an apartment to be quite important. Silly me. This order got unfulfilled. I contacted the Emma Matratze GmbH through their customer support system many times and finally I got a new delivery date of 22nd of July. I have not received the order on that date either. I told them that I want to cancel this order on July 22nd for the first time to no avail. I tried to contact them many times through their online customer support but they did not respond to my inquiries.

After the second failed delivery date and unanswered customer support emails, I called the company directly and asked the representative what was happening. They told me that the mattress was in their warehouse, it was supposed to be sent soon. I was told that I was going to receive a tracking number on July 24th and receive the mattress exactly on July 27th, 2020 (jokes on me, this did not happen). I clearly stated that their business conduct was not appropriate and asked them to cancel my order. They told me that they will not cancel my order. Then I asked for a return and refund my order, if they will not cancel it. This company offers 100 days of no-questions-asked refunds. They said they will not refund my order either. To add insult to injury, the representative hung up the phone on me when I asked “What would you like me to do? You are not sending the bed. You are not cancelling the order. You are not giving my money back. What would you do, if you were in my shoes?”.

Right after my phone conversation with the Emma Matratzen GmbH, I contacted my credit card company and told them about this issue. I was hoping that we can dispute this fraudulent charge at the credit card level. I explained the situation. They suggested that I contact their Risk Management Department and also to contact the Consumer Rights Office. Sadly, the Risk Management Department were late to respond and I did pay for a bed that I will never see. Hence, Emma Matratzen GmbH really got my money without providing any service and left me in an apartment with no mattress to sleep on.

At this point I was already sleeping on my couch for weeks. Now think about this, you have a new apartment, new life, new hopes, a new bed frame, a new box spring, new sheets but no mattress… for weeks.

I decided enough was enough and contacted the Upper Austria’s Customer Rights Office on July 29th. Exactly one month after my initial order. They told me that I was in the right and that I should cite the law and ask for my money. I wrote a formal email to Emma and told that I have the Customer Rights Office involved, and they have guided me to give them a formal 14 days to cancel my order and fully refund me. Trust me, you’ll start getting responses from Emma Matratzen GmbH once you tell them that you contacted the Customer Rights.

Emma replied with the same shebang, how they have the mattress advanced in the shipment process, whatever and ever. But you have to understand that they are so sneaky and will do anything to keep the money. Here is the BS Emma Matraten suggested I do to get my refund (even after I told them that I have the law on my side):

1. I wait for the shipment to arrive,
2. Reject the delivery,
3. Wait for the shipment I rejected to go back to Emma,
4. Then wait for them to start the refund process.

This is for an item they failed to ship after a month. Oh! The idiocy of this company’s practices! How dumb do they think I am? I flat out rejected this and reminded that they have been given a 14 days grace period to refund me and that was final. And in each email from that point on I reminded them that they have been given this grace period to refund me and I expect my funds reinstated by the end of August 12th.

I interacted with couple of Emma Matratzen victims on the company’s Instagram page. I told them that I had enough of this company and I already contacted Customer Rights Office on August 2nd or 3rd. I advised that people should consider that as an option.

Surprise! Surprise! My comments disappeared on Instagram and furthermore, on August 4th, I got an automated email from Emma saying my mattress is on its way with a DPD tracking number. This is 15 days after I first asked them to cancel the order and a week after I got Customer Rights Office involved. This was easy though, I told Emma that they sent a mattress into the void and I would reject their shipment even if it comes and their grace period was still in effect. I contacted DPD and told them that I will reject the item and this was Emma Matratzen GmbH’s pitiful attempt to delay my cancellation and refund. DPD cancelled the shipment for me. After I told Emma I even got their shipment cancelled, they finally (and unwillingly) told me that they will refund me.

They still waited till the literal end of the grace period. I got an automated email from Emma stating that they started their refund process and that I will get my money back on August 12th at 20:20 (8:20 pm). I got my money reinstated couple of days later. In my mind they still failed to finalize the refund in time. I was callused by now though.  After all this was Emma Matratzen, a good-for-nothing business. No surprises there. They seem to suck at everything business related. They are late, but at least they have done one thing that they were legally obligated to do. All in all, I finally got my money, in a record breaking 45 days, from the claws of this fraudulent company and I am happy now.

## Postlude and Reflection

I kept on thinking if my expectations were too high in the times of Covid19. Then again, all the other companies I purchased stuff from delivered their shares to my apartment on time.  Moreover, I was not the one that put the delivery estimates on Emma Matratzen’s website. If only they wrote:

> “If you order this item on June 29th 2020, we will lie to you constantly, force you to the brink of a legal battle with us, and maybe ship your order on August 4th after you asked us to cancel it many times.”

I would have decided not to order it. Instead they wrote, guaranteed delivery dates to be in 7 to 10 days.

To make it even more clear that shipping a mattress is/was possible even at this time: I actually ordered a mattress from another company on 27th of July and got it delivered on 29th of July successfully, while I was still trying to get a refund from Emma.

I also thought about sharing this online or not. I am not known for being an activist. I do not know myself as an activist. This time though, I found a huge database of people that fell victim to this company on social media platforms (especially on Instagram and Facebook, but Emma Matratzen seems to actively suppress bad publicity especially on Instagram, so this group of people keeps on changing), on https://at.trustpilot.com/review/emma-matratze.de, https://de.trustpilot.com/review/emma-matratze.ch, and the list goes on.  I felt the need to share my experience at least with them.

I hope this story would entertain someone, or maybe even give the courage to take a step against wrongdoing.

## Where do the Maximum Absolute q-series Coefficients of (q;q)_n occur?

##### Dedicated to the MACH2 group, in particular to Prof. Wolfgang Schreiner and Johann Messner.

In mathematics research, some questions are too easy to ask that one cannot help but feel amazed when the answers to these questions are nowhere to be found. The title is an example of such a question, I and Alexander Berkovich (my former advisor from the University of Florida) asked in 2017. Our account on this question recently got published in Experimental Mathematics. I suggest checking the ArXiv version of this paper or the actual published article, if my explanation of the problem sparks your interest.

I want to motivate the problem by introducing a well-known q-series identity first. The Euler Pentagonal Number Theorem (EPNT)

$\displaystyle \sum_{i=-\infty}^{\infty} (-1)^i q^{\frac{i(3i-1)}{2}} = (q;q)_\infty,$

where for any non-negative integer $n$

$\displaystyle (q;q)_n := \prod_{i=1}^{n} (1-q^i)$ and $\displaystyle (q;q)_\infty = \lim_{n\rightarrow\infty} (q;q)_n$.

When one looks at the sum side of the EPNT, it is clear that all the coefficients that appear in this formal power series are merely one of the three options: $-1$, $0$, or $1$. This property is not visible from the product representation. Moreover, it is highly surprising to see such cancellations to appear in the series expansion of an infinite product!

But wait, there is more! The products $(q;q)_n$ will always have a coefficient that is

• $> 1$ in size if $n\not= 0,\ 1,\ 2,\ 3,\ 5$ or $\infty$.
• $> 2$ in size if $n\not= 0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,\ 11$ or $\infty$.
• $> 3$ in size if $n\not= 0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,\ 10,\ 11,\ 13,\ 14$ or $\infty$.
• $> 4$ in size if $n\not= 0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,\ 10,\ 11,\ 12, 13,\ 14, 15$ or $\infty$.
• $...$

This is what Berkovich and I proved in 2017* using elementary observations and simple calculations.

For me, the above list highlights how special the infinite product $(q;q)_\infty$ that appears in EPNT once again. There is no way of seeing these unbelievable cancellations that lead to small coefficients of the infinite product when we look at the finite products. Furthermore, the size of the coefficients of the $(q;q)_n$ polynomials grow, and they grow exponentially. In 1964, Sudler studied the size of the maximal absolute coefficients of these products and proved their exponential growth*.

Let’s get a little more technical and write out the polynomial

$\displaystyle (q;q)_n = \sum_{i=0}^{n(n+1)/2 } a_{i,n} q^n$.

We can observe how fast these coefficients grow with the following animation where we plot $(i, a_{i,n})$ from $n=0$ to $n=100$ in steps of $5$. On the top of the image, you can also see the Maximum Absolute Coefficient (denoted as $M$) and the first location at which this coefficient appears (denoted by $L$).

At $n=100$, this polynomial has $5051$ terms, about $1/3$ of which are huge and all around the place! The maximum absolute coefficient at $n=10$ is $3$. This becomes $11,493,312$ when we reach $n=100$. They grow really fast; just as Sudler discovered.

Whenever we have a list of integers, it is a good idea to check the On-line Encyclopedia of Integer Sequences. We already had the absolute maximum coefficients ($M$s) of $(q;q)_n$ for $n\leq 50$ for our 2017 paper, so we made a quick search of this sequence only to find that this sequence was already listed* and our main question-to-be was right in front of us in the comments.

This is how the website looked in 2019:

The comment is clear. People believe that the absolute maximum coefficient of $(q;q)_n$ appears at the midpoint if $n$ is even (I haven’t seen a proof of this, please contact me if you find one) and it is a mystery if $n$ is odd. Berkovich and I studied these locations by calculating all the $(q;q)_n$ for $n\leq 75,000$. With that, now, we have confirmed the even index belief all the way up to 75,000 and we actually have a conjectural answer to the location question for the odd cases too. There are so many interesting properties of this elusive location that we supposedly found, but this is not the place to discuss those. I suggest checking the article.

The calculation of these polynomials one by one, finding the maximum absolute coefficient and its location was a huge task. We started calculating $(q;q)_n$s on Maple, which gave up around $n=10,000$. We moved to Mathematica and a larger server at Research Institute for Symbolic Computation (RISC) and we got stuck around $n=15,000$. So I abandoned all the computer algebra systems and started programming in Python and later in C++. Carrying out these calculations correctly is/was not an easy task. I doubt it will ever be in my lifetime. After using all the symmetries and simplifications, one still needs at least $\lceil n(n+1)/4 \rceil$ operations to calculate $(q;q)_{n+1}$ from $(q;q;)_n$. There is no way of splitting the data. The numbers are growing exponentially and each calculation needed to be carried with arbitrary precision. RISC IT Ralf Wahner dealt with all my incompetence on the RISC side. I need to thank him a lot.

I attended a presentation by Wolfgang Schreiner on the MACH2 massively parallel shared-memory supercomputer project while I was struggling to make progress in my calculations. Using MACH2 was clearly what I needed. My calculations were already crashing computers with 1.5 Tb Ram at RISC. It would have been a great challenge to crash a computer with 20 Tb of shared Ram (I did crash a 10 Tb section of MACH2 once, so I can rest happily).

I contacted the MACH2 team, got an account, and gave a brief explanation of what I wanted to calculate. After a couple of emails, Johann Messner and I met in person and went through my code, line by line. I took his feedback and suggestions and started working on those. In the meantime, he started running tests with then-code. It took so much back and forth. Johann kept on asking me to parallelize my code but I didn’t want to do that in order to save used memory. I resisted as much as I can but he won. He knew the MACH2 beast the most after all. Also, he made the great point that MACH2 has slow processors but it has tons of memory at our disposal.

I didn’t know how to parallelize anything in C++ so I put it on the backburner. It was already more than a year at this point since we first started. The project could have easily sat for another couple of months… and it did. Later in the year, at a conference, I accidentally learned how to parallelize my code from a couple of numerical analysis postdocs. All of a sudden the missing link was there. After Johann’s okay, we cranked the last portion of the calculations (with lots of intermediate checks) from $n=40,000$ to $n =75,000$. It still took 3 months even though we were using a little more than 200 cores and 10 Tb of memory.

The mathematical claims of the article came after all these crazy calculations. I came together with Prof. Berkovich at the University of Florida in October 2019. We compiled the data, looked at it left and right. In a short time, we made the best claims we can make about the elusive location. While we were at it, we refined Sudler’s growth constants as well. This ended my first 2 year-long experimental math project.

Long story short, the mathematical claims I made stand on the shoulders of many computers, so much wasted electricity, many sleepless nights, and the input of many people many of which haven’t even been mentioned here.

If you’d like to confirm or disprove that of the maximum absolute coefficient when $n=5,050,029$ appears as the coefficient of the $q^{6,375,697,892,084}$ term, be my guest. 🙂

##### I shared this work with some audiences:
1. Austrian High-Performance Computing Meeting 2020, IST Austria, Klagenfurt AUT. February 19, 2020
Where do the maximum absolute q-series coefficients of $(1-q)(1-q^2)...(1-q^{n-1})(1-q^n)$ occur? (Slides)
2. The 85th Seminaire Lotharingien de Combinatoire, Strobl AUT. September 6, 2020
Where do the maximum absolute q-series coefficients of $(1-q)(1-q^2)...(1-q^{n-1})(1-q^n)$ occur?
3. Hudson Colloquium, Georgia Southern University, Savannah GA. September 23, 2020 (Online Meeting)
Where do the maximum absolute q-series coefficients of $(1-q)(1-q^2)...(1-q^{n-1})(1-q^n)$ occur? (Pdf)
##### Of course there are some simple typos here and there in the article, let me address the ones I know about here:
1. There are two 68,324s written by accident. Those should be 62,624s as in everywhere else in the paper.