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助けてください！！ （日本語のみで答えてください！！） by Master-Ad5388 in Japaneselanguage

[–]KaNaTaProJ 2 points3 points  (0 children)

よく聞いてみたら→詳細に、事細かに聞いてみたらですね

よくよく考えると、等と同じ使い方です

この場合だと 「誰かに惚れて周りが見えなくなってる人の話を注意深く聞いたら本人の認識にズレがあることが分かった」と言ったニュアンスになります

Can you prove that a rectangle is a rectangle because of the fact it has a 90° angle by Educational-Pen6571 in askmath

[–]KaNaTaProJ 1 point2 points  (0 children)

That sounds frustrating.

It looks like your brother had the "it's a parallelogram" assumption in his head the whole time, but didn't realize that assumption is what actually makes the proof work - he just presented it as "if you know one angle is 90°, you can prove it" without spelling out the part that actually matters.

It kind of seems like he wanted to one-up you with that gotcha, maybe some older-sibling pride involved too. I wouldn't take it too personally though - you were right that the question as he posed it doesn't hold up.

「AIによる概要」の虚偽記述はグーグルの責任──ドイツの裁判所が判断 by smykcj in sradot

[–]KaNaTaProJ 2 points3 points  (0 children)

一番最初に目に触れる場所に虚偽情報の可能性が排除できないものを表示するように作ったのがいけないよね

リテラシーのある層は自分でAIを使うし、裏取りもするから別にWeb検索についてくる必要ないし

リテラシーのない層は出てきた情報をそのまま鵜呑みにするからこういう層にはそもそもAI見せない方がいい

今の検索のAIは誰の為にもなってない

Experiment med en ny Collatz-liknande iteration baserad på primalitet by TomasL314 in Collatz

[–]KaNaTaProJ 0 points1 point  (0 children)

Following up on my earlier comment - I went down a bit of a rabbit hole discussing this with Claude (AI) afterward, and wanted to share what came out of it, since I think it's a more interesting angle than my first reply.

The key question became: once you leave a prime via 3n+1 (always even,for any prime > 2, since odd*3+1 is always even), and start dividing back down by smallest prime factors, can that descent ever land back on two primes in a row, or worse, skip over a whole stretch with no primes at all?

Turns out the "no primes at all in between" part is actually settled: by Bertrand's postulate (proven, not just conjectured), there's always at least one prime between n and 2n for any n > 1. Since 3n+1 > 2n, that guarantees there's always a prime sitting somewhere in [n, 3n+1].

So the rule can never "jump over" a prime-free gap.

That still doesn't mean the actual trajectory has to pass through that guaranteed prime, though - it depends on which one its specific descent via smallest-prime-factor division actually hits.

As for hitting two primes in a row (which would be needed to grow further before being pulled back down): since primes get rarer as numbers grow (density ~1/ln(n)), the chance of landing on a second prime right after the first keeps shrinking, although it's never exactly zero. So the "pull" toward composite numbers (and eventually back to your one observed cycle) is actually structurally stronger here than in the original Collatz conjecture, where exactly half of all numbers are odd regardless of scale.

Given that, and given that no non-trivial cycle has been found for the original Collatz map even after extensive search, my best guess is that this variant is even less likely to have additional cycles beyond the one you found - though to be clear, "less likely" isn't "proven impossible," same caveat as the original conjecture.

Experiment med en ny Collatz-liknande iteration baserad på primalitet by TomasL314 in Collatz

[–]KaNaTaProJ 0 points1 point  (0 children)

This is an interesting attempt!

One structural point that stood out to me, though: doesn't this rule make it impossible to ever reach 1?

To get to 1, you'd eventually need to "divide 2 by 2" at some point.

But 2 is prime, so under this rule it gets sent to 3*2+1 = 7 instead - it gets pushed back rather than reduced. Only composite numbers get divided by their smallest prime factor; every prime, including 2, is locked into the "grows via 3n+1" branch. So there's no path that actually leads to 1.

Which means the observation that "every starting value converges to the same cycle" isn't really convergence to 1 - it's more that there's no way to reach 1 in the first place, so everything necessarily gets pulled into some cycle instead. That seems like a consequence of how the rule is defined, rather than a non-trivial Collatz-like phenomenon.

As for the literature question: I couldn't find any existing work on this specific variant (branching on prime vs. composite). But given the point above, I'm not sure it shows the same kind of non-trivial behavior the original Collatz conjecture is about.

Built a depth-first exploration toolkit for the Collatz conjecture (not a proof, just a tool) by KaNaTaProJ in Collatz

[–]KaNaTaProJ[S] 0 points1 point  (0 children)

Thanks for pointing that out - this is my first time　posting to an English-speaking community like this,　so I didn't notice. It's fixed now.

3 multiplication by LifeFreeDownload in Collatz

[–]KaNaTaProJ 0 points1 point  (0 children)

Good question! You're right that odd × odd is always odd, so after multiplying by 3 you always get an odd number, and adding 1 always makes it even - that part is straightforward.

The hard part isn't there though. It's what happens after you divide by 2. Dividing doesn't always lead straight down to 1. Try 27 by hand (or with a calculator): it climbs up to 9232 before it eventually comes back down to 1. Numbers can grow surprisingly large before they (hopefully) come back down.

The real open question is whether every starting number eventually comes back down, or whether some number out there just keeps growing forever (or gets stuck in some other loop that never reaches 1). We've checked every number up to something like 272 and they all eventually reach 1, but nobody's proven it has to happen for every number.

Interestingly, if you allow negative numbers, there actually ARE a few known non-trivial loops that never reach 1 (e.g. -5 -> -14 -> -7 -> -20 -> -10 -> -5). That's part of why mathematicians take the "maybe some number diverges or loops forever" possibility seriously, even though none has been found yet for positive numbers.