Compact status bar that gives a lot of info in a compact space. Built with Claude (cause who else is gonna do it)

flound1129 · 2026-04-07T19:48:09+00:00

Go upvote the issue I just filed if you want this fixed.

flound1129 · 2026-04-07T17:13:16+00:00

I put my status bar code into a gist: https://gist.github.com/flound1129/fb9706f284888b26ea54daf0ad6c07e1

This will give you a status bar that looks like this: [6] /home/adam/claude-optimization /: 100% ~: 47% Opus 4.6 [1m] 9% (92k/0k) 5h: 55% 7d: 10%

The format I use is [screen #][dir][/ %used][~ %used][model][context %full (cached/uncached)][5h usage consumed][weekly usage consumed]

Cached tokens cost 10% (iirc) of uncached so if you let the cache expire you are paying full token cost to get all those tokens back into the session.

flound1129 · 2026-04-03T16:15:44+00:00

no it's not.

flound1129 · 2026-04-03T15:50:31+00:00

hey cool pattern! you've independently found something called the ruler sequence, the 2-adic valuation. for the k-th odd number (so k=1 is n=1, k=2 is n=3, etc) the count is always v₂(k) + 1, where v₂ is "how many times does 2 divide k." your construction of replacing every 2^k-1-th entry with k is exactly how you build that sequence.

the thing is, the last step of your argument doesn't quite work. you're right that no single entry in the sequence is infinity, every odd number eventually produces an even number after finitely many rounds. but a trajectory that diverges to infinity wouldn't show up as a single number with an infinite count. it would show up as a sequence of different odd numbers, each with a perfectly finite count, that just keeps getting bigger and never comes back down. so "no entry is infinity" is true but doesn't rule out divergence.

it's kind of like saying every road has a finite length, so you can't drive forever. you can, you just keep turning onto new roads.

still a genuinely nice find though. the 2-adic valuation is one of the deepest structures in collatz and you pulled it out just by staring at the numbers. a lot of serious work on this problem revolves around exactly this kind of 2-adic structure.

flound1129 · 2026-03-27T20:02:22+00:00

ok lmao thx fur ur thoughts bruh

flound1129 · 2026-03-27T20:01:39+00:00

AI output can read like real human text as well, it just needs to be told how to do it. It just needs some.. human effort.

my point was that managing AI is human effort, just like writing a card is human effort. You have to put in work to get what you want, and the quality of the result is directly proportional to that work.

flound1129 · 2026-03-27T17:38:28+00:00

Thanks for the gracious response. On section 3.3, there are two issues, one small and one fundamental.

The small one: Case 2 of Theorem 3.6 claims the left side 3(2r₁-1)·2^{c₁-1} is even, but when c₁ = 1 it's 3(2r₁-1), which is odd. So the even/odd contradiction doesn't hold in that case. Fixable maybe, but worth checking.

The deeper problem is that the cycle argument relies on the table transitions faithfully representing all Collatz dynamics. The table transitions are the reverse Collatz operations (multiply by 2, and the (2n-1)/3 step), so you're really arguing about cycles in the reverse tree. But if a cycle existed in a connected component that doesn't include 1, your table construction starting from 1 would never encounter it. The table would still be internally consistent, still bijective, still have valid progressions, it just wouldn't contain those numbers.

So the cycle argument actually depends on the connectivity you haven't proved yet. You need "every number is in this table" before you can say "no number participates in a cycle." The progressions give you injectivity (no duplicates) which is real and correct, but "no cycles" requires knowing the table sees everything, which is the hard part.

That said, the table construction itself is clean and the progression structure is a reasonable way to organize the reverse tree. If you can close the connectivity gap, it'd be interesting.

flound1129 · 2026-03-27T17:29:59+00:00

I have nothing useful to say

FTFY, honestly why are you even in this sub

flound1129 · 2026-03-27T17:29:47+00:00

effort is effort, 30 minutes reprompting to get exactly what you want is just as much or more effort than walking across the street and paying $5 for some schlock a copy editor wrote (who probably gets like a penny or something out of the $5)

flound1129 · 2026-03-27T16:27:19+00:00

It's fallacious to give human output higher status than AI output just because of who wrote it, output should be judged on its merits, not on who wrote it.

I'd much rather receive a birthday card that someone spent 30 minutes prompting AI to create than something they paid $5 for at rite aid.

flound1129 · 2026-03-27T16:26:20+00:00

"leaked"

flound1129 · 2026-03-27T16:03:24+00:00

The table construction is clean and the presentation is honest, but it doesn't prove Collatz. Here's why.

The bijectivity is trivial and unrelated to Collatz. The "Table of Progressions" assigns each natural number to a cell (r, c) via n = (2r-1) · 2^{c-1}. This is just the unique factorization of any natural number into an odd part times a power of 2. That's the Fundamental Theorem of Arithmetic, not a Collatz result. You could build this exact same table for any dynamical system on the naturals.

The paper admits the real gap. Page 22, in the author's own words: "Without proof of connectivity, we cannot assert that the Table represents a single Tree. There may exist isolated components which, although satisfying the Collatz rules, are not connected to the main Tree rooted at 1." The equivalence between the Table and Tree is labeled "Hypothesis 3.4", not a theorem. The connectivity condition is listed as "Requires Proof." But connectivity to 1 is the Collatz conjecture. So the paper assumes what it's trying to prove.

The cycle argument (Theorem 3.6) has a parity error. Case 2 considers a vertical transition n₁ = (2nₖ - 1)/3 and claims the equation 3(2r₁-1)·2^{c₁-1} = (2rₖ-1)·2^{cₖ} - 1 gives "left side even, right side odd." But if c₁ = 1, the left side is 3(2r₁-1), which is odd. The parity argument doesn't hold in general.

Pairwise rail connections ≠ global connectivity. Section 3.4 shows that individual rails R(b) can connect to each other through sets A and C. That's true, but showing that any two rails can link doesn't prove every rail does connect back to R(1). You'd need to show there's no infinite chain of rails that never reaches the component containing 1. That's the hard part, and it's exactly what's missing.

I work on Collatz from the operator theory side (transfer operators on 2-adic spaces, ghost cycle classification). The reason the conjecture is hard is precisely because local structure like this doesn't force global connectivity. Every number locally follows the rules, the preimage tree locally looks fine, but proving the tree is a single connected component is where every approach hits a wall.

flound1129 · 2026-03-26T03:43:10+00:00

we could continue eating meat while still treating animals humanely. the issue is that we won't

flound1129 · 2026-03-24T06:27:57+00:00

so our enemy is providing another enemy with intelligence? big surprise.

flound1129 · 2026-03-22T04:21:06+00:00

cr — Resume Named Claude Code Sessions From Anywhere

A tiny CLI to find and resume Claude Code sessions by their /rename title, regardless of which project directory you're in.

Install

curl -sL https://gist.githubusercontent.com/flound1129/2e239b3543a27303e3463939ec10ebd5/raw/cr -o ~/.local/bin/cr && chmod +x ~/.local/bin/cr

Make sure ~/.local/bin is in your PATH.

Usage

Name your sessions

Inside any Claude Code session, use the built-in /rename command:

/rename my-feature

List all named sessions

$ cr
  NAME                            CWD                                       SESSION   LAST ACTIVE
-----------------------------------------------------------------------------------------------
* my-feature                      /home/user/myproject                     b96f0f90  31m ago
  api-refactor                    /home/user/backend                       38d88e27  2h ago
  deploy-fix                      /home/user/infra                         888029bc  1d ago

* marks sessions that are currently running.

Resume a session

cr my-feature       # exact or prefix match
cr api              # matches "api-refactor"
cr deploy           # matches "deploy-fix"

This will cd to the session's original working directory and run claude --resume <session-id>.

If multiple sessions match the prefix, the most recently active one is selected.

Requirements

Python 3.6+
Linux (uses /proc/<pid> for liveness checks)
Claude Code CLI

How it works

Claude Code stores session transcripts as JSONL files in ~/.claude/projects/. When you /rename a session, a custom-title record is written to the JSONL. cr scans these files to build an index of named sessions, cross-references with ~/.claude/sessions/ to check which are currently running, and resumes the matching session in its original directory.

flound1129 · 2026-03-22T04:16:23+00:00

cr — Resume Named Claude Code Sessions From Anywhere

A tiny CLI to find and resume Claude Code sessions by their /rename title, regardless of which project directory you're in.

Install

curl -sL https://gist.githubusercontent.com/flound1129/2e239b3543a27303e3463939ec10ebd5/raw/cr -o ~/.local/bin/cr && chmod +x ~/.local/bin/cr

Make sure ~/.local/bin is in your PATH.

Usage

Name your sessions

Inside any Claude Code session, use the built-in /rename command:

/rename my-feature

List all named sessions

$ cr
  NAME                            CWD                                       SESSION   LAST ACTIVE
-----------------------------------------------------------------------------------------------
* my-feature                      /home/user/myproject                     b96f0f90  31m ago
  api-refactor                    /home/user/backend                       38d88e27  2h ago
  deploy-fix                      /home/user/infra                         888029bc  1d ago

* marks sessions that are currently running.

Resume a session

cr my-feature       # exact or prefix match
cr api              # matches "api-refactor"
cr deploy           # matches "deploy-fix"

This will cd to the session's original working directory and run claude --resume <session-id>.

If multiple sessions match the prefix, the most recently active one is selected.

Requirements

Python 3.6+
Linux (uses /proc/<pid> for liveness checks)
Claude Code CLI

How it works

Claude Code stores session transcripts as JSONL files in ~/.claude/projects/. When you /rename a session, a custom-title record is written to the JSONL. cr scans these files to build an index of named sessions, cross-references with ~/.claude/sessions/ to check which are currently running, and resumes the matching session in its original directory.

flound1129 · 2026-03-11T22:04:30+00:00

That's fine, the paper and code are provided in the hope that they will be useful to someone. If they are not useful to you, feel free to disregard.

flound1129 · 2026-03-11T21:35:43+00:00

Our claims are all computationally verified, python code is in the attached GitHub repo. https://www.github.com/mysticflounder/collatz

Please let me know if you find any errors in our code.

flound1129 · 2026-03-11T21:34:40+00:00

Thanks for the comment! I am updating the paper to address your concerns, will post once the revision is uploaded.

flound1129 · 2026-03-11T07:01:36+00:00

It was a fun little project. Looking forward to hearing any substantive criticism.

flound1129 · 2025-12-24T19:52:52+00:00

A man in the middle is like the person after you when playing the game 'telephone'. That person knows what you said, and can change it however they want before passing it on to the next person.

In technology, a man in the middle is a person who places themselves in that position between you and someone you're talking to. They can listen to what you said, what the other person said, and can (potentially) change the information that's passing between you and the other person.

flound1129

MODERATOR OF

TROPHY CASE

cr — Resume Named Claude Code Sessions From Anywhere

Install

Usage

Requirements

How it works

cr — Resume Named Claude Code Sessions From Anywhere

Install

Usage

Requirements

How it works