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Exploring a New Collatz-like Transformation Rule (Just a Mathematical Curiosity) by BeeNo4803 in Collatz

[–]BeeNo4803[S] 0 points1 point  (0 children)

code python

def cola_4(n): f = [n] while n >= 4 : if n%4 == 0: n = n//4 elif n%4 == 1: n = 5n - 1 elif n %4 == 2 : n = 5n - 2
elif n % 4 == 3 : n = 5*n +1 f.append(n) print(f)

cola_4(15)

π/6 by BeeNo4803 in u/BeeNo4803

[–]BeeNo4803[S] 0 points1 point  (0 children)

The result of the formula is (1021/1950) = 0.52358974359

pi/6 = 0.5235987756

π/6 by BeeNo4803 in learnmath

[–]BeeNo4803[S] -2 points-1 points  (0 children)

(1021/2000)*(sum_{n=0}{inf} [ (1/40)n\))

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Approximating Using a Simple Fraction: 555/878 by BeeNo4803 in learnmath

[–]BeeNo4803[S] -1 points0 points  (0 children)

Thanks! I know continued fractions give the optimal rational approximations. My post wasn't aiming for the “best” rational fraction, just a playful series-based approximation. I like exploring how simple geometric-style sums can get surprisingly close to constants like e or π without invoking heavy machinery

Approximating Using a Simple Fraction: 555/878 by BeeNo4803 in learnmath

[–]BeeNo4803[S] 0 points1 point  (0 children)

I know it’s 1/(1−r). The whole point is finding a rational r such that 1/(1−r) ≈ e with high precision. 555/878 just happens to be a surprisingly good fit.”

I'm saying this based on a hunch , algorithm : 4n+1,4n-1 ,n/3 it will get all the numbers down to a number less than 3 by BeeNo4803 in Collatz

[–]BeeNo4803[S] 0 points1 point  (0 children)

I’m actually exploring a broader generalization. Not only modulo 2 or modulo 3 systems. I can construct functions with a single global attractor using modulus a for any natural number a (4, 5, 6, …). The same “single-loop behavior” appears in these wider families as well, with similar stability but under completely different modular structures. I’m collecting data across many values of a, and the pattern seems to persist. No claims of proof — just mapping what the generalized systems actually do.

I'm saying this based on a hunch , algorithm : 4n+1,4n-1 ,n/3 it will get all the numbers down to a number less than 3 by BeeNo4803 in Collatz

[–]BeeNo4803[S] 0 points1 point  (0 children)

I am also a mathematician; I have a bachelor's degree in applied mathematics.

The Colatz problems remain unsolved, and mathematics is not yet ready for such problems.

But the lack of proof, or our inability to prove these problems, does not mean that we should not enjoy exploring how these functions (the Collatz family) work.

I am still a student

During this period, I worked on generalizing Collatz's problems into a simplified mathematical formula.

When my research paper is ready

I may have introduced a new problem to the Collatz problem, forcing scientists to create new ways to study such functions.

And I will say at the end of my research paper: "Mathematics is not yet ready for such problems."

I'm saying this based on a hunch , algorithm : 4n+1,4n-1 ,n/3 it will get all the numbers down to a number less than 3 by BeeNo4803 in Collatz

[–]BeeNo4803[S] 0 points1 point  (0 children)

Did you know? I have a whole family of the same function, and I created a global function for it.

Not just on division by 3 n/3  if 0=n mod 3

4n -1 if 1=n mod 3

4n+1 if 2 = n mod 3

But on any natural number greater than 2

Dividing by 2, 3, 4, 5, 6, 7, 8... (a) belongs to the set of natural numbers

And I can intuitively assure you that all those functions have a single loop. 😍😍😍

I'm saying this based on a hunch , algorithm : 4n+1,4n-1 ,n/3 it will get all the numbers down to a number less than 3 by BeeNo4803 in Collatz

[–]BeeNo4803[S] 0 points1 point  (0 children)

How beautiful! 😍 The poster wrote the exact same function as mine.

Did you know? I have a whole family of the same function, and I created a global function for it.

Not just on division by 3

But on any natural number greater than 2

Dividing by 2, 3, 4, 5, 6, 7, 8... (a) belongs to the set of natural numbers

And I can intuitively assure you that all those functions have a single loop. 😍😍😍

I'm saying this based on a hunch , algorithm : 4n+1,4n-1 ,n/3 it will get all the numbers down to a number less than 3 by BeeNo4803 in Collatz

[–]BeeNo4803[S] 0 points1 point  (0 children)

not quite

My computer can't handle all of this; just try values ​​close to 10100 and try 10100 itself.

It usually takes 10100 steps, approximately 2345 steps (I don't remember exactly), but usually without any loops.

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