While researching the Collatz Conjecture, I came across some formulas written by Ben Crossley in attempt to write a proof for the Conjecture. Despite my lacking background in mathematics, I decided to try and understand how his formulas work and their connection to the Conjecture. I am certain I am missing some obvious theme or idea due to my inexperience.

I am having trouble understanding how $T_d$ works in Ben Crossley's formulas. Is $T_d$ supposed to be a random set of integers, a certain sequence of even numbers, or something else entirely?

I plugged in some values for $x_1$ with the hope of better understanding his formulas. I started using some positive integers for $x_1$ and plugging them into his first formula $T_1(x_1) = \frac{2^{x_1}}{3^1} - \frac{3^0}{3^1}$ , finding the sequence 5, 21, 85, 341... with the even values {2, 4, 8, ...} for $x_1$. I tried experimenting with a couple of variables, however I am concerned my attempt to understand his formulas only lead me astray and further from the main idea I did not understand to begin with.


I've looked into the linked article, but have only a slight impression about what is written. I think it is the following.
Consider the Collatz- transformation $a \to b$ in the form $b = {3a+1 \over 2^A}$ which is thought to work only on odd positive numbers $a$ and $b$. By this the value $A$ is uniquely determined.

Now consider the backwards transformation (again on odd positive numbers $a$ and $b$) $$ a = {b2^A - 1 \over 3} $$ Here $A$ is no more uniquely determined; instead it can assume each second positive integer. Let for instance $b=1$ then $$a_A = {1 \cdot 2^A - 1 \over 3} \in \{1,5,21,85,... \}_{A=2,4,6,8,...} $$ After that, a second step of that backwards transformation, transforming $a_A \to z_{A,Z}$ by $$ z_{A,Z_A} = {a_A \cdot 2^{Z_A} - 1 \over 3} = { {1 \cdot 2^A - 1 \over 3} \cdot 2^{Z_A} - 1 \over 3} = { (1 \cdot 2^A - 1 ) \cdot 2^{Z_A} - 3 \over 3^2} = { 1\cdot2^{A+Z_A} - (3 + 2^{Z_A} ) \over 3^2} $$ implements actually a two-dimensional array of results $z_{A,Z_A}$ Which $Z_A$ can be chosen depends on $a_A \pmod 3$ and can either be element of the odd or of the even natural numbers .

I think the $T_d()$ with $\lim_{d \to \infty}$ in the article is nothing else than this ansatz, only a different notation, and also assuming the obvious (?) generalization to three and more backwards transformations.

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  • $\begingroup$ Thank you for your response, that helped clarify a lot. I did not know anything about sequece transformations before you mentioned it. Do you have any suggestions for readings that may cover sequence transformations in more detail? Thank you! $\endgroup$ – Griffon Theorist697 Jul 15 '17 at 12:35

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