u/Honests-Anonymous

edit: doesn't work :( ill tray pascsls and fractals mod n (2, 3, 6)

collatz is genuinely keeping me up at night.

i really don't know how to solve this (skill) issue.

if u saw, i wrote an embarrassing program of an "equivalent, collatz-like conjecture" that was genuinely the most atrocious absolute SHIT you could see from an amateur/noob mathematician.

i will state again, i am an amateur mathematician.

again, amateurish.

let's starting with what we DO know what we can genuinely prove with 0% mistake.

all integers 2^n where n is a natural number.

1, 2, 4 (well 4 2 1 loop)

i had a thought. instead of "proving" the collatz conjecture with the collatz conjecture, what if we created a "reverse" collatz conjecture that's like the "evil" collatz conjecture?

2^n starting at the natural numbers {1, 2, 3, 4, ... infinity}

{2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, etc. etc.}

now let's create a reverse collatz:

if n is even, subtract one.

{1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, 4095, 8191, 16383, etc. etc.}

if n-1 is divisible by three, (n-1)/3.

now it is quite easy to check divisibility by 3 with "elementary" number theory.

2 == -1 (mod 3) --> 2^n == (-1)^n (mod 3) --> 2^n - 1 == (-1)^n - 1 (mod 3), so when n is even, it becomes 0, and when n is odd, then n is 1 (mod 3), so yeah.

That means we have proven collatz rigorously for, starting from 3, 15, 63, 255, 1023, etc.

{1, 5, 21, 85, 341, 1365, 5461...}

why? well, when we apply 3x+1 to all of these odds, we know we will get to an even power, which is already trivially proven.

3( 3(n-1)/3 ) + 1 = (n-1) + 1 = n.

n is an even power.

if n-1 is not divisible by three, well, i was stuck.

if n-1 is not divisible by three, quite the kenundrum.

then i realized, wait, i can just multiple by 2 and make it even anyway! That's the EVIL collatz conjecture lmao.

evil collatz reverse conjecture:

if n is even:

(3n-1)/3 if n-1 is divisible by 3, go back to loop

if not divisible by 3 (n-1), go back to loop

if n is odd:

(2n), go back to loop

if it lands on 2^n, exit the loop.

the conjecture is this: it will always exit the loop.

the thing we start with is a rigorous ALL even powers (2^n), including 2^0 technically!

now this struck to me an awfull like the odd perfect square problem, or the mersenne numbers. 2^n - 1... huh. interesting

{1, 7, 31, 127, 511,...}

multiply by 2

{2, 14, 62, 254, 522,...}

subtract 1

{1, 13, 61, 253, 521,...}

but then i realized something massive and sad and depressing: none of these will EVER be divisible by 3. this is the limiting point of elementary number theory.

multiply by 2

{2, 26, 122, 506, 1042...}

subtract 1

{1, 25, 121, 505, 1041...}

hmmmm, this is weird, every FIFTH term is now eliminated from our nice list.

again, this is by elementary number theory.

hmmmm, could the collatz be undecidable because it encodes the primes?

lemme know what you think!

Formal proof of (undecidability of) collatz i wrote in one day while i was in high school with elementary number theory. i'm not taking a math class lol i'm just interested in math. needs ZFC well-ordering to order all of the "collatz primes."

Amateur mathematician here, doesn't know how to code (well): New method for tackling the conjecture? "Reverse" or "inverse" collatz conjecture. Only elementary number theory needed to understand.