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Scott Neville

@drScott@mathstodon.xyz
mastodon 4.7.2
  • Open on mathstodon.xyz

Mathematics PhD at ICERM Fall 2025. I will be at LACIM in Winter 2026.
I like thinking about combinatorics, theoretical computer science, and number theory.

58 Followers
91 Following
15 Posts
Joined August 22, 2024
Webpage:
https://sites.google.com/umich.edu/nevilles/about
Open post
Scott Neville @drScott@mathstodon.xyz
· 4mo ago
Replying to
spoiler (solution)
@gjm@mathstodon.xyz @benleis@mathstodon.xyz Whew; I stared at that bit for a long time. That factorization of 3^5 + 1 is really nice in balanced ternary. 11 * 1m1m1; I had the urge to write it that way, but was too lazy.
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Open post
Scott Neville @drScott@mathstodon.xyz
· 4mo ago
Replying to
a step
@benleis@mathstodon.xyz We should probably rewrite that as $3^15 + 3 \cdot 3^10 + 3 \cdot 3^5 + 1$, and perhaps look at a triangle.
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Open post
Scott Neville @drScott@mathstodon.xyz
· 7mo ago
Replying to
@2something this is an excellent observation! I get a similar effect where when I'm reading a paper on a laptop I'll take fewer notes and do less of the examples vs when it's printed. I always blamed it on the pain of media/context switching, but I've no data...
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Open post
Scott Neville @drScott@mathstodon.xyz
· 5mo ago

So pleased to have posted a preprint with Amanda Burcroff this week, on eventual sign coherence for quiver mutation. http://arxiv.org/abs/2605.12865

I'm especially happy to now be a scholar of brogs https://www.xenoserieswiki.org/wiki/Brog
#combinatorics #xenoblade #quivers #clusteralgebras

Eventual sign coherence
arXiv.org

Eventual sign coherence

The sign coherence of $c$-vectors is one of the fundamental theorems of cluster algebras with principal coefficients. In 2019, Gekhtman and Nakanishi posed the asymptotic sign coherence conjecture for arbitrary cluster algebras of geometric type, which says sign coherence should eventually hold in any sufficiently generic infinite mutation sequence. We prove that their conjecture holds almost always for skew-symmetric cluster algebras of arbitrary rank. That is, we prove that with probability $1

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Open post
Scott Neville @drScott@mathstodon.xyz
· 5mo ago

Aristotle, famous for work on naming Boops Boops ( https://en.wikipedia.org/wiki/Boops_boops )

Boops boops - Wikipedia
en.wikipedia.org

Boops boops - Wikipedia

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Open post
Scott Neville @drScott@mathstodon.xyz
· 5mo ago
Replying to
AI
@2something@transfem.social the analogy of DRM seems very insightful. Do you know of any longer/extended/further analysis? The company finance history could be interesting.
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Open post
Scott Neville @drScott@mathstodon.xyz
· 5mo ago
Replying to
The devs are discriminating against the bug type.
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Open post
Scott Neville @drScott@mathstodon.xyz
· 5mo ago
Replying to
@2something Don't stop at teens! Talk with the Toddlers! Let siblings strike in solidarity. They have to live with our choices much longer than we do.
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Open post
Scott Neville @drScott@mathstodon.xyz
· 6mo ago
Replying to
@yoriyuki I know one teaching model that still requires code experience and gets students to prep: each class has a random student coding on the projector in front of the class, working on an exercise with help from the peanut gallery + prof. However, unless you are really careful this will slow down the class a lot. I learned a lot from the programming languages course that worked this way, but it didn't cover all that much material.
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Open post
Scott Neville @drScott@mathstodon.xyz
· 5mo ago
Replying to
fortunately, I think most repeated tokens will suffice? Hopefully?
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Open post
Scott Neville @drScott@mathstodon.xyz
· 5mo ago
Replying to
@HM @EUCommission confused by this reply. Seems pretty small scale marketing effort, roughly in line with what a public organization needs to do to avoid being labeled as useless, without demanding space and being annoying. The right amount of marketing is not 0.
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Open post
Scott Neville @drScott@mathstodon.xyz
· 5mo ago
Replying to
@tomkalei@machteburch.social @tao@mathstodon.xyz this makes a lot of sense to me - similar to how 'run it through spell check' has become.
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Open post
Scott Neville @drScott@mathstodon.xyz
· 4mo ago
Replying to
spoiler (solution)
@benleis@mathstodon.xyz @gjm@mathstodon.xyz I didn't do more; let's see if there's fun ahead. It seemed to me that factors of 3 digit numbers can't be that bad? (3⁵+1)³, so we're looking for the largest prime factor of 3⁵+1 . This is 3*81+1=244=4*61. So our answer is 61. And now to reflect on if there's a cleaner way to get that... 61 is 2⋅3³+2⋅3+1 , which doesn't jump out at me from the initial trinary. Famously 2^3 = 3^2-1, so our other factor is 2⁶=3⁴−23²+1, which seems not that great. Spoiler please!
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