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For my math internal assessment, I'm looking at the Collatz Conjecture and different ways to try and solve it.

I wanted to see if there were any possible partial solutions for it by which I mean if we take a specific type of number the conjecture would be true for all of those numbers.

For example, the collatz conjecture is true for all m where m=2^n since the powers of two will keep being even until it reaches one. However, I wanted a more complex proof for another type of numbers(Say, for all powers of 3 or for all prime numbers if it were possible) so I wanted to know if there are any such partial proofs you could think of

Samik Maini
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  • Please specify which sets are allowed! If it is the set of odd numbers $n$, such that $3n+1$ is a power of $2$, then the Collatz-conjecture is trivial to prove. – Peter Jun 04 '18 at 10:43
  • There really isn't a specific type of sets are allowed which means the one you've given is also viable I just wanted to see the different kind of sets that did exist for which the conjecture could be proven but ideally I would like non-trivial proofs with some mathematics that is worthy of investigation and is of some high level since the one you've given is quite self explanatory. THank you – Samik Maini Jun 04 '18 at 11:44
  • see my comment here https://math.stackexchange.com/questions/2716155/are-there-specific-numbers-for-which-the-collatz-conjecture-is-proven/2716668#2716668 – Collag3n Jun 04 '18 at 17:23

1 Answers1

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The Collatz conjecture is proven for:

$2^n$

$2^n\cdot x$ for all integers $x$ for which it is proven.

Every positive integer up to some really large number like say $2^{50}$

Every number of the form $\frac {4^{n}-1}3$; i.e. $0,1,5,21,85,341,\ldots$

Every number of the form $4^n\cdot x+\frac{4^n-1}3$ for all odd $x$ for which it is proven, i.e. if $3$ converges then $3,13,53,213,\ldots$ converges.

Every $4x+1$, for all odd $x$ that converge.

Every linear combination of the Lucas sequences $U_n(5,4)$ and $V_n(5,4)$, for which the first element of the sequence is a convergent odd integer.

$\frac{y-1}{3}$ converges for every $y=64^n\cdot x+\frac{64^n-1}{3}$, for every $\frac{x-1}{3}$ known to converge.

A number of these are largely equivalent and proving any of these is a relatively simple exercise and should keep you going for a while (the $\{x<2^{50}\}\to1$ exercise requires a computer and some time).

it's a hire car baby
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