Talk:Collatz conjecture

Another fractal

https://yozh.org/2012/01/12/the_collatz_fractal/ Rudxain (talk) 02:20, 11 October 2025 (UTC)

Function for sequence length

Apologies if this has been covered earlier.

Is there a function that will convert an input integer into a Collatz sequence length? For example, F(12) = 10, F(19) = 21, F(27) = 111, etc.

I suspect the answer is no, because if such a function did exist, it would be a simple step to then prove or disprove the conjecture. but I'd like an explanation if possible. Thanks. -- Jack of Oz ^{[pleasantries]} 18:40, 3 November 2025 (UTC)

The answer is yes, if and only if the Collatz conjecture is true. If the conjecture is false, and some $n$ diverges, then $f(n)$ does not exist. But if the conjecture is true, then the function exists and is computable: just run the Collatz iteration starting from $n$, count how many steps it takes, and return that number. Whether it is a primitive recursive function or has a closed-form expression are different questions. The same non-rigorous probabilistic reasoning that suggests the conjecture to be true also suggests that the number of steps is typically only logarithmic in $n$ but that is not even close to being proven. —David Eppstein (talk) 19:01, 25 November 2025 (UTC)

I do nor know of a function, but I do know a method that makes it easier to determine the number of steps. It is best used when the number of steps will be large, since it would be quicker to just do the iteration if the number is small. Method: You create a spreadsheet with combinations of 3^n/2^m (2^m is always larger than 3^n). The size of the spreadsheet depends on the size of the selected number for which the number of steps is to be calculated. Then multiple the selected number by the various fractions in the spreadsheet. Then look for values between 0.95 and 0.80. There may be a few within this range. The number of steps will be the sum of the exponent of 3 with the exponent of 2. For example 19 has 20 steps (3^6/2^14) x 19 [0.845398] and 27 has 111 steps (3^41/2^70) x 27 [0.834133]. There probably will be several fractions that will give a result in the range, so you have to manually check each one to make sure you get the correct fraction. It is a lot of work at the beginning, setting up the spreadsheet and learning how to check the various fractions, but it gets easier after a few times. I am not allowed to give the citation of the paper, since this is consider PROMOTING. Just search for a paper publish last year around late summer / early fall. ~2025-36800-42 (talk) 06:58, 28 November 2025 (UTC)

Some of the math text isn't properly formatted :(

It looks like under the "Statement of the Problem" section that there is unformatted text. I don't know how to fix it, but does anyone here know how? ✧𝒎𝒂𝒕𝒄𝒉𝒂✧【💬】 17:07, 25 November 2025 (UTC)

It looks fine to me. Can you be more specific, or upload a picture of what the problem is? --JBL (talk) 18:45, 25 November 2025 (UTC)

might be an issue on my end, it says this: $ {\displaystyle f(n)={\begin{cases}n/2&{\text{if }}n\equiv 0{\pmod {2}},\\3n+1&{\text{if }}n\equiv 1{\pmod {2}}.\end{cases}}} $ ✧𝒎𝒂𝒕𝒄𝒉𝒂✧【💬】 19:30, 25 November 2025 (UTC)

Check your preferences, under appearance, at the bottom: do you have math rendering set to "LaTeX source (for text browsers)"? Use one of the other three choices. —David Eppstein (talk) 19:37, 25 November 2025 (UTC)

okay! i'll try that! thank you :D ✧𝒎𝒂𝒕𝒄𝒉𝒂✧【💬】 20:40, 25 November 2025 (UTC)

Adding a visualization for the mapping (rather than convergence)

https://kentslaney.github.io/ideal-bassoon/

I'm interested in adding a visualization from the above URL to the article (probably the base 3 version), but I'm new to contributing here and wanted to double-check the original research requirements and whatnot before writing the JavaScript to export it properly to GIF. Is there anything I should know before posting? What would be an appropriate sequence length? Here's a watermarked screen recording for 27. Kslaney (talk) 18:54, 1 December 2025 (UTC)

Desent of 4n+3 numbers under the collatiz transform

Theorem For any starting value x of the form 4n+3 that follows a Collatz sequence for t=d+m steps until it first reaches a smaller value y (where d is the number of divisions by 2 and m is the number of 3n+1 steps), we can define a family of starting values X_n=2^d(n)+x. Applying the exact same sequence of t steps to X_n results in Y_n=3^m(n)+y, where the inequality Y_n<X_n holds true for all n>=0. Proof The proof relies on establishing the consistent parity sequence and comparing the arithmetic progressions formed by the starting and ending values. Step 1: Parity Pattern Consistency The sequence of operations (multiply by 3 then add 1, or divide by 2) is determined entirely by the parity of the current number. By constructing the expression X_n=2^d(n)+x, where 2^d is the common difference, all terms in this new arithmetic progression share the exact same parity sequence for the first t steps as the original number x. This ensures the sequence of operations for X_n mirrors that of x. Step 2: Derivation of the Resulting Expression Following the t steps (composed of m multiplications by 3/2 implicitly and d divisions by 2 explicitly), the transformation on the generalized term X_n yields the expression Y_n=3^m(n)+y, where y is the result of applying the same sequence to the original x. Step 3: Proving the Inequality The inequality to prove is 3^m(n)+y<2^d(n)+x. Rearranging the terms gives (2^d-3^m)n>x-y. The starting values form an arithmetic progression (AP) with a common difference of C_X=2^d. The resulting values form an AP with a common difference of C_Y=3^m. As demonstrated by the example (d=4, m=2, 2^4=16, 3^2=9), the common difference of the starting AP (2^d) is always greater than the common difference of the resulting AP (3^m) because the process definition ensures the sequence eventually decreases (y<x). Step 4: Conclusion of Inequality Since the initial values (x) are larger than the resulting values (y), and the rate of increase of the starting terms (2^d) is greater than the rate of increase of the resulting terms (3^m), the difference between the corresponding terms, (2^d-3^m)n-(x-y), grows positively with n. Therefore, the inequality 3^m(n)+y<2^d(n)+x holds true for all n>=0. Answer: The provided theorem and proof, rigorously demonstrate that for any number of the form x=4k+3, a related family of numbers X_n=2^d(n)+x always produces a smaller result Y_n=3^m(n)+y after following the same parity-driven Collatz sequence path, because the common difference of the initial arithmetic progression is always greater than that of the resulting progression. ~2025-39107-45 (talk) 13:33, 8 December 2025 (UTC)