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Let $X = 2\Bbb{Z} + 1$ or $2 \Bbb{N} + 1$ where $0 \in \Bbb{N}$, this approach will probably play well with both forms. See Extending the Collatz function to larger domains.

Define the shortcut Collatz function to be:

$$ f : X \to X \\ f(x) = \dfrac{3x + 1}{2^{\nu_2(3x + 1)}} $$

where $\nu_2(z)$ is the common notation meaning the maximum exponent $e$ of $2$ such that $2^e \mid z$.

Clearly then, the Collatz conjecture is equivalent to this function always taking its input (either in $\Bbb{Z}$ or $\Bbb{N}$) to one of the well-known finite loops. For the $X = 2\Bbb{N} + 1$ form, this is the original Collatz conjecture.

But $f$ satisfies this equation:

$$ f(3xy + x +y) = f(x)f(y) $$

Proof. Simply multiply the RHS out and notice that $2^{\nu_2(3x + 1)} 2^{\nu_2(3y+1)} = 2^{\nu_2((3x + 1)(3y+1))}$. So the numerator is $(3x + 1)(3y+1) = 9xy + 3x + 3y + 1 = 3(3xy + x + y) + 1$, and we're done, but the reader should write this out fully on paper. QED

So for instance $3(1)(1) + 1 + 1 = 5$ so that we know without a doubt that $f(5) = f(1)f(1) = 1$.

So if no reference exists that starts out with this simple observation, my question is this; in either form ($2\Bbb{N} + 1$ or $2\Bbb{Z} + 1 = X$), does the set:

$$ \{3xy + x + y: x,y \in X\} $$

cover almost all of $X$? By that I mean all but a finite number of values in $X$ are taken on by values of that polynomial expression where inputs are all pairs $(x,y) \in X^2$.

No idea how to solve this!

Notice that the law $x\star y := 3xy + x + y$ is a semigroup law on the set $X$. I didn't prove this recently, but I recall that this is true from other posts I've made in the past. You can prove it directly. I don't know if this matters to Collatz, but it's just a note. This implies that $f(x\star y) = f(x)f(y)$ or that the standard shortcut Collatz function, is also a semigroup homomorphism, although the $\star$ law is less natural to comprehend than the usual $\cdot$ on the RHS.

I want to say this kind of seems related to Erdős covering systems, since:

$$ \vdots \\ 3(-3)x - 1 + x = -8x - 1 \\ 3(-1)x - 1 + x = -2x - 1 \\ 3(1)x + 1 + x = 4x + 1 \\ 3(3)x + 3 + x = 10x + 3 \\ 3(5)x + 5 + x = 16 x + 5 \\ \vdots $$

But we merely want to cover odd numbers. So to align this with the usual Erdos covers, we would take each residue class $a\Bbb{Z} + b$ and do: $\frac{a \Bbb{Z} + b - 1}{2}$, which will work because each modulus is even and each residue representative is odd. We also must only pass in an odd number for $x$, so that would change the above list of residue classes completely. It's doable though!

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    $\begingroup$ Your numbers are all $1\pmod 4$ Indeed, writing $x=2n+1, y=2m+1$ tells us that $3xy+x+y=3(2n+1)(2m+1)+2n+1+2m+1\equiv 1\pmod 4$. $\endgroup$ Commented Jul 22, 2023 at 1:15
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    $\begingroup$ Oh, but not all numbers of the form $4n+1$ can be obtained by your method. Barring computational error, I get the list of your numbers to be $\{5,13,21,29,33,37,45,53,\cdots\}$ so you are missing quite a few (I am taking $X$ to be the odd natural numbers here). $\endgroup$ Commented Jul 22, 2023 at 1:28
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    $\begingroup$ In any case, I do not see how this helps with Collatz. It's not much of an observation to note that we can take out all the factors of $2$ in one step. $\endgroup$ Commented Jul 22, 2023 at 1:29
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    $\begingroup$ Of course one can do that, there is no problem doing that. If $k$ is even, then $2^k\equiv 1 \pmod 3$ so we can solve $3n+1=2^k$, and Collatz is trivial for all such $n$. And so on. $\endgroup$ Commented Jul 22, 2023 at 1:33
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    $\begingroup$ No offense intended. I just don't see the idea. Good luck, I will not comment further on this. $\endgroup$ Commented Jul 22, 2023 at 1:33

1 Answer 1

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Code

from sympy import * N = 50 def v2(n: int) -> int: k = 1 while n % (2 ** k) == 0: k += 1 k -= 1 return k def f(x: int) -> int: """Shortcut Collatz function""" assert x % 2 == 1 return (3 * x + 1) // v2(3*x +1) def semigroup_law(x:int, y:int): assert x % 2 == 1 and y % 2 == 1 return 3 * x * y + x + y values = set() for m in range(-N, N): m = 2 * m + 1 for n in range(-N, N): n = 2 * n + 1 s = semigroup_law(m, n) values.add(s) values = list(values) values.sort(key=lambda v: abs(v)) print(values)

Prints:

 [1, -3, 5, -7, 9, -11, 13, -15, 17, -19, 21, -23, 25, -27, 29, -31, 33, -35, 37, -39, 41, -43, 45, -47, 49, -51, 53, -55, 57, -59, 61, ...]

Result

Probably easy to prove, is that while $3xy + x + y$ doesn't cover all $(4\Bbb{Z} + 1) \cup (4\Bbb{Z} + 3)$ for $x, y \in\Bbb{Z}$, we have that the absolute value $|3xy + x + y|$ covers all $2\Bbb{N} + 1, (0 \in \Bbb{N})$. And so if we consider all the $X = 2\Bbb{Z} + 1$ form of Collatz, then $4\Bbb{N} + 1$ gets covered and so does $-(4\Bbb{N} + 3)$.

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  • $\begingroup$ Donnely : I heard that collatz is equivalent to proving collatz for imputs 3n. $\endgroup$ Commented Feb 19, 2024 at 12:21

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