arithmetic sequence $8n+1$ and the collatz conjecture

Question

Is it a known result that if for all $n$ the collatz sequence of $8n+1$ lead to $1$, all natural numbers will?

Answer 1

As Robert Soupe says, why would a result binding on 8n+1 prove the conjecture?

Is there a missing line of background that shows where that expression comes from? Why not 16n+1 or something else?

Checking whether n can coexist in one sequence with 4n+1 (same n) is a bit more interesting because it would be essential for a loop.

Answer 2

A quick sketch to show that 8*n+1 occurs in any trajectory:

Write any positive integer x as m*2^(v+1) + 2^v - 1, or (m,v).

Examples: 1 = (0,1); 6 = (3,0); 27 = (3,2).

For v>=1, succ((m,v)) = (3*m+1,v-1). This is doing the same as (3*x+1)/2.

Example partial trajectory: (3,2) -> (10,1) -> (31,0), which represents 27 -> 41 -> 62.

As now v==0, 62 -> 31 = (0,5), and continue.

At each v==1 the reached value is congruent to 1 (mod 4), so this already means that 4*n+1 is in any trajectory.

For even n that is also congruent to 1 (mod 8).

For odd n, because succ(4*n+1) = (3*(4*n+1)+1)/2/2 = 3*n+1 = succ(n), repeat if feasible, then the final n is not congruent to 1 (mod 4), and can be used as x, from the start. Repeat until an 8*n+1 is reached. As is essential to Collatz, there is no limit to the number of steps needed, as one can always construct a case with more steps.

Answer 3

In https://www.fq.math.ca/Scanned/38-1/andaloro.pdf (Andaloro, 1998), which is mentioned in https://arxiv.org/pdf/math/0608208.pdf (Lagarias, 2012) as "(Andaloro, 2000)", there are proofs for both 1 mod 8 and 1 mod 16 to be "sufficient".