Saw-mon & Natalie

@AcerFur Tell us more about your new friends and your workflow 🦔

damn I should have gotten hooked onto Codex way sooner it's so cool to see a whole bunch of agents edit different parts of a paper

@maruroido2 Pretty cool 💍: the infinity germ of sequences of prime residues …

環Aの超越数論の論文です！ arxiv.org/abs/2604.25566

🤖 also came up with an adjacent idea of instead of phrasing the solution using markov chains, one can instead look at divergence of flows on the related poset graph. It does again require tail estimates of some sums but the idea is pretty intuitive …

Created some exercises to prove Erdös problem 1196 on my own (for those of us who don’t want to spoil the proof). At the heart of it are two tail estimations and the basic identity regarding von Mangoldt function plays the key role: log = Λ ∗ 𝟙 🪡

@litecoin List of fixes matches with the git diff (including the safe fee accounting discovered by additional audits) litecointalk.io/t/litecoin-mwe…

Postmortem is out regarding the April 25th Litecoin MWEB incident. Please use the link in the reply to access the report. 👇

@ChatGPTapp gist.github.com/Saw-mon-and-Na…

Never talk about goblins, gremlins, raccoons, trolls, ogres, pigeons, or other animals or creatures unless it is absolutely and unambiguously relevant to the user’s query.

🦝🧌👹🐦

@arb8020 github.com/openai/codex/c…

gpt-5.5 prompt for codex seems to have a duplicated line trying to get it to not talk about creatures? Never talk about goblins, gremlins, raccoons, trolls, ogres, pigeons, or other animals or creatures unless it is absolutely and unambiguously relevant to the user's query. [...] Never talk about goblins, gremlins, raccoons, trolls, ogres, pigeons, or other animals or creatures unless it is absolutely and unambiguously relevant to the user's query gh link: #L55" target="_blank" rel="nofollow noopener">github.com/openai/codex/b…

@Czar102 You should write down your findings for n² + 1 (maybe a tiny article). Note in general the unit equilateral triangles can be rotated with any angle.

@sw0nt answering the first question (with a proof sketch) took me 3x more time than the second one very cool problem!

⟁ can you cover an equilateral triangle of side length n + ɛ with: - n² + 2 unit equilateral triangles ? - n² + 1 unit equilateral triangles? fun puzzle to think about. I came up with Alexander Soifer's construction.

@Czar102 You should read for other related problems ...

@wiz_io @github The better thread to follow: x.com/sagitz_/status…

🚨 BREAKING: Wiz Research discovered Remote Code Execution on GitHub.com with a single git push The flaw in @github allowed unauthorized access to millions of repositories belonging to other users and organizations 🤯

Pretty cool through unsanitised X-Stat header key=value injection …

@Bayesprof Are the ChatGPT threads public?

Just posted a paper (with Aaron Smith) that solves the mixing time of Kac's walk on the rotation group; this walk was introduced by Hastings in his seminal paper in 1970 that started the MCMC revolution and is used in statistics, statistical physics, and cryptography, to name a few. Its mixing time has been open for decades. Even just a few months ago, I thought this problem was impossible! This problem has been studied by many people over the last 30 years. The main difficulty was that, as a community, we don't have the tools to study mixing time in continuous state spaces. AI influence was key (GPT 5.4 pro) in our solution: it acted as a Rosetta Stone, translating ideas from another part of probability (Malliavin calculus) to Markov chains, and also pointed out a specific lemma from the literature that makes this translation work! Of course, the hard work was in verification. This also gives a genuinely new tool to study the mixing times of such constrained systems. Very satisfying and grateful to be able to prove this result and be a part of this journey. I am more optimistic than ever about what's possible as the tools continue to get better. arxiv.org/abs/2604.23828

@mgrczyk @Liam06972452 I was wondering if Liam had shared it. Very cool to see the prompt.

@sw0nt @Liam06972452 Liam, I just thought it was cool

This is one of the coolest things I've ever seen (erdos 1196)

@mgrczyk Is this convo by you or @Liam06972452 ?

@antimath3 if you consider this field as a 1-form ω, then one can show ω generates the first de Rham cohomology of the punctured plane since ∫ω ≠ 0 and dω = 0. H¹(ℝ² \ {0}) ≅ ℝ given a curve C on the plane the integral or pairing <ω, C> = 2π x the winding number of C around the 0

Choose your favorite 2d vector field, and explain why it is <-y/(x^2+y^2), x/(x^2+y^2)>

@7homaslin @EricaKlarreich @QuantaMagazine Lots of conjectures also would be interesting to know how they came up with those knot invariants … arxiv.org/pdf/2509.18456

Math news: A powerful new invariant offers "a combination of strength and speed… that... can probe knots that were previously far out of reach." Reported by @EricaKlarreich in @quantamagazine: quantamagazine.org/a-powerful-new…

some refs: - wfnmc.org/mc20101.pdf - seewoo5.github.io/jekyll/update/… (many related refs ...) funny coming across @antimath3 again after the @axiommathai 's (@KenOno691) announcement from a few months ago ...

@mathemetica youtu.be/OgbJApTCTPE?si…

λ · μ = lk(∂λ, ∂μ) + ∑ handles intersections Watching Lisa Piccirillo own the blackboard like this; intersection forms on homology modules, handle cancellations in 4-manifolds; hits different. The precision, the chalk dust, the way she just sees the topology. This is pure 4D magic: where smooth vs topological distinctions go wild, and one clever trace or form can kill a 50-year knot problem. Her Conway knot proof lives here; showing it's not smoothly slice, unlocking exotic phenomena that change how we understand spacetime surgery and manifold classification.

@__SimonCoste__ wait till you permute sets with negative or complex cardinalities … ps: the statement should be the other way around

Fact : there are more permutations of a set with 0.4 elements than permutations of a set with 0 elements (0.4! = Gamma(1.4) ≈ 0.9 while 0! = Gamma(1) = 1)