The Collatz Conjecture viewed from the p-adic world? 2-3 system

Ideas / Approaches
1

Hi, I want to share an interesting view of the Collatz Conjecture hoping anyone can help lol! I’m obsessed and stuck, send help! Anyways… let me know if something is not clear, I’ll be happy to explain.

I use a non-recursive finite reverse Collatz as a vector function (usually, one sees this in the recursive manner f_i(x) = {1 \over 3} \cdot (2^{x_i} \cdot f_{i-1}(x) - 1)) where f_0(x) = 1- this just applies Collatz in reverse):

f(x) = {\displaystyle\prod_{i=1}^{n} 2^{x_i} - \displaystyle\sum_{i=1}^{n}3^{i-1} \displaystyle\prod_{j=i+1}^{n} 2^{x_j}\over 3^n}

where n is the dimension of x, and x is vector of digits- particularly each component is the number of times one needs to divide by 2 in an even Collatz step (note that these make up x in reverse order).

These components of x slowly reconstruct some sequence of odd numbers (here outputs of f(x)), so you can run Collatz on some odd integer, say 17, and you divide 2,3,4 times by 2 in each even step, hence x = (4,3,2), and f( (4,3,2) ) yields 17.

The function will yield an odd integer if the numerator is congruent to 0 mod 3^n, hence x is an admissible vector if this condition is true. Let’s assume S is the set of all admissible vectors, then:

f(S) = \mathbb{Z}_{odd}

is the Collatz Conjecture.

While this is cool, and the function yields some nice bounds on the admissible vectors and their component sums, nothing really popped up for me.

Since any reverse Collatz sequence starting from 1 is infinite, we can think of the congruence condition for a vector to be admissible with vectors of infinite size (in p-adics), which yields the identical condition (due to 2 being primitive root mod 3):

\displaystyle\sum_{i=1}^{\infty}3^{i-1} \displaystyle\prod_{j=1}^{i} 2^{-x_j} = 1

This a mixed 2-3 system, but the important part is that any infinite reverse Collatz sequence is just a p-adic expansion of 1 in this 2-3 system. What about the odd numbers (not divisible by 3) though? Those are just shifts in this expansion:

f(x, n) = \displaystyle\sum_{i=1}^{\infty}3^{i-1} \displaystyle\prod_{j=n+1}^{n+i} 2^{-x_j}

One can obtain this from condition series above, hence any x used in the function will yield an odd integer, but I was hoping to show:

\displaystyle\sum_{i=1}^{\infty}3^{i-1} \displaystyle\prod_{j=1}^{i} 2^{-x_j} = 1 \impliedby f(x,n) = \displaystyle\sum_{i=1}^{\infty}3^{i-1} \displaystyle\prod_{j=n+1}^{n+i} 2^{-x_j} \text{ and } f(x,n) \text{ is an integer}

We can always pick the shifted xs to get any odd integer not divisible by 3, so we have k=f(x,n) (for this go up the reverse tree for this odd integer and pick any admissible digits to continue it →this will converge to the odd integer in the 2-3 p-adic system; even though there is an infinite number of reverse paths, all of them will converge the same way).

While that is nice (but trivial), we lose the information about the first n components of x, so I don’t see any way to show this implication.

Maybe this p-adic view leads somewhere? Lemme know if you’re interested!

1 Like
2

With your formula can you generate an admissible vector for any odd multiple of 3?

3

If the conjecture is true, then any odd integer has an admissible vector and you “generate” it by running Collatz on it, collecting the xs. We can only extend it to be an infinite admissible vector (such that its p-adic 2-3 series converges to 1) if the number is NOT a multiple of 3.

I tried to show that it exists for any odd integer without running Collatz by using \operatorname{f}(x), but nothing really came out of it.

If we ignore even numbers, then you can never get a multiple of 3 as a result after applying a single Collatz step on any number n (by step I mean (3n+1)/2^k which gives an odd number, never a multiple of 3).

This means we can simply ignore all odd integers that are a multiple of 3 in the analysis, because their next Collatz step lands outside all multiples of 3 (and it will never land back in).

Here’s a bunch of admissible vectors for visualization:

I: 3 :: xs: 4, 1, Sum: 5; Count: 2; 
I: 5 :: xs: 4, Sum: 4; Count: 1; 
I: 7 :: xs: 4, 3, 2, 1, 1, Sum: 11; Count: 5; 
I: 9 :: xs: 4, 3, 2, 1, 1, 2, Sum: 13; Count: 6; 
I: 11 :: xs: 4, 3, 2, 1, Sum: 10; Count: 4; 
I: 13 :: xs: 4, 3, Sum: 7; Count: 2; 
I: 15 :: xs: 4, 5, 1, 1, 1, Sum: 12; Count: 5; 
I: 17 :: xs: 4, 3, 2, Sum: 9; Count: 3; 
I: 19 :: xs: 4, 3, 2, 1, 3, 1, Sum: 14; Count: 6; 
I: 21 :: xs: 6, Sum: 6; Count: 1; 
I: 23 :: xs: 4, 5, 1, 1, Sum: 11; Count: 4; 
I: 25 :: xs: 4, 3, 2, 1, 3, 1, 2, Sum: 16; Count: 7; 
I: 27 :: xs: 4, 5, 1, 1, 3, 4, 2, 2, 4, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 2, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 2, 1, Sum: 70; Count: 41; 
I: 29 :: xs: 4, 3, 2, 1, 3, Sum: 13; Count: 5; 
I: 31 :: xs: 4, 5, 1, 1, 3, 4, 2, 2, 4, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 2, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, Sum: 67; Count: 39; 
I: 33 :: xs: 4, 3, 2, 1, 3, 1, 2, 2, Sum: 18; Count: 8; 
I: 35 :: xs: 4, 5, 1, Sum: 10; Count: 3; 
I: 37 :: xs: 4, 3, 2, 1, 1, 4, Sum: 15; Count: 6; 
I: 39 :: xs: 4, 3, 2, 1, 3, 1, 4, 1, 2, 1, 1, Sum: 23; Count: 11; 
I: 41 :: xs: 4, 5, 1, 1, 3, 4, 2, 2, 4, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 2, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 2, Sum: 69; Count: 40; 
I: 43 :: xs: 4, 3, 2, 1, 1, 4, 2, 2, 1, Sum: 20; Count: 9; 
I: 45 :: xs: 4, 3, 2, 3, Sum: 12; Count: 4; 
I: 47 :: xs: 4, 5, 1, 1, 3, 4, 2, 2, 4, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 2, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, Sum: 66; Count: 38; 
I: 49 :: xs: 4, 3, 2, 1, 1, 4, 2, Sum: 17; Count: 7; 
I: 51 :: xs: 4, 3, 2, 1, 3, 3, 1, Sum: 17; Count: 7; 
I: 53 :: xs: 4, 5, Sum: 9; Count: 2; 
I: 55 :: xs: 4, 5, 1, 1, 3, 4, 2, 2, 4, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 2, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 3, 1, 1, Sum: 71; Count: 41; 
I: 57 :: xs: 4, 3, 2, 1, 1, 4, 2, 2, 1, 2, Sum: 22; Count: 10; 
I: 59 :: xs: 4, 3, 2, 1, 3, 1, 4, 1, 2, 1, Sum: 22; Count: 10; 
I: 61 :: xs: 4, 5, 1, 1, 3, Sum: 14; Count: 5; 
I: 63 :: xs: 4, 5, 1, 1, 3, 4, 2, 2, 4, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 2, 1, 1, 1, 2, 1, 1, 2, 1, 4, 1, 1, 1, 1, 1, Sum: 68; Count: 39; 
I: 65 :: xs: 4, 3, 2, 1, 1, 4, 2, 2, Sum: 19; Count: 8; 
I: 67 :: xs: 4, 3, 2, 1, 3, 1, 4, 1, Sum: 19; Count: 8; 
I: 69 :: xs: 4, 3, 4, Sum: 11; Count: 3; 
I: 71 :: xs: 4, 5, 1, 1, 3, 4, 2, 2, 4, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 2, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, Sum: 65; Count: 37; 
I: 73 :: xs: 4, 5, 1, 1, 3, 4, 2, 2, 4, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 2, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 3, 1, 1, 2, Sum: 73; Count: 42; 
I: 75 :: xs: 8, 2, 1, Sum: 11; Count: 3; 
I: 77 :: xs: 4, 3, 2, 1, 3, 3, Sum: 16; Count: 6; 
I: 79 :: xs: 4, 3, 2, 1, 3, 1, 4, 3, 1, 1, 1, Sum: 24; Count: 11; 
I: 81 :: xs: 4, 5, 1, 1, 3, 2, Sum: 16; Count: 6; 
I: 83 :: xs: 4, 5, 1, 1, 3, 4, 2, 2, 4, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 2, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 3, 1, Sum: 70; Count: 40; 
I: 85 :: xs: 8, Sum: 8; Count: 1; 
I: 87 :: xs: 4, 3, 2, 1, 1, 4, 4, 1, 1, Sum: 21; Count: 9; 
I: 89 :: xs: 4, 3, 2, 1, 3, 1, 4, 1, 2, Sum: 21; Count: 9; 
I: 91 :: xs: 4, 5, 1, 1, 3, 4, 2, 2, 4, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 2, 1, 1, 1, 2, 1, 1, 2, 1, Sum: 59; Count: 33; 
I: 93 :: xs: 4, 5, 1, 3, Sum: 13; Count: 4; 
I: 95 :: xs: 4, 5, 1, 1, 3, 4, 2, 2, 4, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 2, 1, 1, 1, 2, 1, 1, 2, 1, 4, 1, 1, 1, 1, Sum: 67; Count: 38; 
I: 97 :: xs: 4, 5, 1, 1, 3, 4, 2, 2, 4, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 2, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 3, 1, 1, 2, 2, Sum: 75; Count: 43; 
I: 99 :: xs: 4, 3, 2, 1, 1, 6, 1, Sum: 18; Count: 7; 
I: 101 :: xs: 4, 3, 2, 1, 3, 1, 4, Sum: 18; Count: 7; 
I: 103 :: xs: 4, 5, 1, 1, 3, 4, 2, 2, 4, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 2, 1, 1, 1, 2, 1, 1, Sum: 56; Count: 31; 
I: 105 :: xs: 4, 3, 2, 1, 3, 1, 4, 3, 1, 1, 1, 2, Sum: 26; Count: 12;

Essentially a vector (x_1, ..., x_n) is admissible \iff \displaystyle\sum_{i=1}^{n}{3^{i-1}\prod_{j=1}^{i}{2^{-x_j}}} \equiv 1 \pmod{3^n}

4

It is critical if you can generate any odd number 1 (mod 12) or (inclusive or not) 5 (mod12) with your formula. Since those previous odd numbers are subsets who can generate by coalescence the whole set of odd number, then you would be proving the Collatz conjecture!

5

Not sure what exactly you mean, but let’s say we want to show it for any
\operatorname{f}(x)=12k+1, k \in \mathbb{N}_1, then you must show that:
12k+1 = {\displaystyle\prod_{i=1}^{n}{2^{x_i}} - \displaystyle\sum_{i=1}^{n}{3^{i-1} \prod_{j=i+1}^{n}{2^{x_j}} } \over 3^n} which is non-trivial (if at all possible).