Ben Grant

Vishal Bhatoy, Colin Ingalls
Discrete Invariants of Koszul Artin-Schelter Regular Algebras of Dimension four
https://arxiv.org/abs/2602.13178

'measles outbreak at the child prison' seems entirely avoidable, it's really the kind of thing that only happens if you do several unthinkably evil things on purpose all at once

Breaking: ICE is now reportedly following white people on grocery runs in Minnesota, suspecting they're delivering food to neighbors too afraid to leave their homes.

Neighborhood instructions:

Don't put the address in your phone.

Don't use GPS.

Write it on paper.

And if you get stopped, eat it.

Benjamin Grant, Zhongyang Li: Self-avoiding walk, connective constant, cubic graph, Fisher transformation, quasi-transitive graph https://arxiv.org/abs/2601.12571 https://arxiv.org/pdf/2601.12571 https://arxiv.org/html/2601.12571

If you’re wondering why your friends in academia are a little on edge right now, it’s because an eighteen-year-old who hasn’t done the reading, doesn’t look at the assignment, and has does no critical thinking skills more complex than “because I think it’s in the Bible” can literally end your career

Jonah Berggren, Khrystyna Serhiyenko: Classical tilting and $\tau$-tilting theory via duplicated algebras https://arxiv.org/abs/2512.13893 https://arxiv.org/pdf/2512.13893 https://arxiv.org/html/2512.13893

For the longest time, I tried to convince myself that I didn’t find combinatorics all that interesting, and that if I found anything interesting about it, it was probably just because it was similar to something I liked about abstract algebra. Turns out it was the other way around this whole time

Scott Carter, Benjamin Cooper, Mikhail Khovanov, Vyacheslav Krushkal: An Extension of Khovanov Homology to Immersed Surface Cobordisms https://arxiv.org/abs/2510.14760 https://arxiv.org/pdf/2510.14760 https://arxiv.org/html/2510.14760

let alone one of the form we are interested in, x_n=f^n(x). So f cannot exist.

Thanks for reading!

/end/

contradiction with the assumption that (x_n) is an integer sequence, since L=1+1/k requires our (integer!!) differences between consecutive terms to eventually be arbitrarily close to 1+1/k, which no integer is. So there doesn’t even exist an integer sequence (x_n) with x_{n+k}-x_n=k+1 for all n,

We can rewrite the LHS as a telescoping sum with k terms:

x_{n+k}-x_{n+k-1}+…+x_{n+1}-x_n=k+1.

Taking the limit on both sides as n—>infty (and letting L be the same limit as in the previous problem), we see that kL=k+1.

But then L=1+1/k, which is not an integer (since k is at least 2). This is a

use this technique for: suppose we are told to show that for any fixed natural number k at least 2, there does not exist a function f:Z—>Z such that f^k(x)=x+(k+1) for all x in Z. Run the same “replace f^i(x) with x_{n+i}” bit we did before and rearrange to get x_{n+k}-x_n=k+1 for all n \geq 0.

f(x)=x-4 does indeed satisfy this functional equation, but this is easy to verify).

I think this is a pretty neat method. Like I said, I wasn’t familiar with it until yesterday, so forgive me if this is a common trick you have seen before, since I certainly hadn’t.

Another quick application we can

In this case, x_0=x is some fixed x in Z and x_1=f(x_0)=f(x). We now know that x_1=x_0-4=x-4 by the above, so f(x)=x-4. But this holds for arbitrary x in Z, so we have our answer: the only function f:Z—>Z such that f(f(x))-4f(x)+3x=8 for all x in Z is f(x)=x-4 (one should technically also check that

n+1 on, then it also is from term n on. This allows us to conclude that (x_n) is genuinely an arithmetic progression.

Therefore, all of (x_n) just depends on x_0: we see that x_n=x_0-4n for any n \geq 0. In particular, x_1=x_0-4.

Let’s go back to the case we were interested in to begin with.

arithmetic progression, but that is an arithmetic progression in its entirety. Indeed, if n is such that x_{n+2}-x_{n+1}=-4, then by massaging our two-step recurrence a tiny bit, we can see that

x_{n+1}-x_n=-1/3*(8-(x_{n+2}-x_{n+1}))
=-1/3*(8-(-4))
=-1/3*12
=-4.

So if (x_n) is arithmetic from term

means that if the differences between consecutive terms *approach* -4, then at some point they actually *are* -4 and they stay that way (since these differences are integers!). Thus, (x_n) is eventually arithmetic with common difference -4.

Next, we can show that (x_n) is not just eventually an

L=lim_{n—>infty} (x_{n+1}-x_n). Then (after some mild regrouping on the LHS), we may take limits on either side of our recurrence to get L-3L=8, which gives L=-4. Therefore, the difference between consecutive terms in our sequence approaches -4. But recall that (x_n) is an INTEGER sequence. This

when applied to x_n for any n. The idea is that if we can determine properties of *any* integer sequence (x_n) satisfying this two-step linear recurrence, then we can do it for our specific sequence x_n=f^n(x).

We can first show that (x_n) is *eventually* arithmetic with common difference -4. Let

about integer sequences (x_n : n \geq 0). Fix x, and let x_n=f^n(x) be the nth composite power of f applied to x (i.e., we’re taking f^0=id, f^1=f, f^2=f°f, etc.). Then the functional equation above becomes x_2-4x_1+3x_0=8 when applied to our specific x, or more generally, x_{n+2}-4x_{n+1}+3x_n=8

Learned a really neat trick yesterday for a select family of competition-style problems with certain kinds of functional equations over the integers. As an example, suppose we’re tasked with finding all functions f:Z—>Z such that f(f(x))-4f(x)+3x=8 for all x. The trick is to turn this into a problem

A drawing of the connected sum of a (2,7) torus knot and its mirror image (much modified).

An image of a crazy knot I claim is the unknot.

New blog post! Through a sequence of images, I verify that the unknotting number of the connected sum of a (2,7) torus knot and its mirror is less than 6: I show that this first image is the connected sum, and after changing those crossings, it produces the unknot! divisbyzero.com/2025/10/08/t...

from now on I'm just going to truncate the taylor series for sin at the first term. sin(x) = 0 for all x. this approximation

- is efficient to compute
- has bounded error
- has excellent analytic properties
- has very high accuracy for small values, which occur frequently in applications

the award or an HM, but because it feels like this reviewer didn’t take my application seriously at all, and I worry that I’m not the only person this happened to. End of rant, thanks for reading.

algebraic combinatorics— this reviewer was leaving comments about my desire to do research in “set theory,” about “supercompactness,” and about “the combinatorics of the first singular cardinal.” I did not mention any of these whatsoever.

I am mostly just disappointed, not because I didn’t receive

were almost directly copied but with additional typos that I didn’t make (???), but by far the most annoying part to me is that other bullet points were about things that I just *did not talk about at all* in my application. For context, the focus of my proposal was in representation theory and

detailed reviews that demonstrated they, at a minimum, read and thought about my statement and proposal. The third reviewer, on the other hand, left as their comments a relatively incoherent mix of bullet-point-style remarks. Some of these points were directly copied pieces of my statements, some

Very mild rant. GRFP reviews from last year’s application cycle came out a couple days ago. I applied last year but did not receive the award or an honorable mention (oh well, this happens, not the part that’s frustrating). Two of the three reviewers assigned to my application left very positive,

question, but it seems (to me) like a very natural point of view nonetheless.

the “regular functions” on A^n into an object R_n (i.e., taking R_n=Hom(A^n,A)) should give you the “algebraic” dual of the “geometric” A^n. Then duals of mappings f:A^n—>A^m are just their images f^*:R_m—>R_n under Hom(-,A).

I am not sure if this gives a reasonable answer to the original duality