Is this Algorithm's Result concerning Collatz Sequences provable?

Question

I have searched the www for this little algorithm and its result thoroughly, but it was nowhere to be found, do you know if my observation is provable?

The Algorithm:

Step 1: Select an arbitrary prime number p, and calculate its Collatz Sequence $C(p)$.

Step 2: Now select the largest prime number m in this sequence. $ m := Max.Prim \{C(p)\}$

Step 3: Select the next higher "neighbour" general prime number $M \in \mathbb{P}$, $M \gt m$ and calculate the Collatz Sequence $C(M)$. (Go to Step 2 and repeat)

Observation:

The sequence of the next higher prime number continues the previous number sequences.

What I'd like to proove: $C(m) \subset C(M), \forall p \in \mathbb{P}$

Is this even provable as long as the Conjecture is not proven?

I'm reallly struggling to find a starting point here...

Do you have any ideas?

Answer 1

I believe that your question is this:

1. For any $p$, compute $s = collatz(p)$, the collatz sequence for $p$.

2. let $m$ be the largest prime in $s$.

3. Let $M$ be the next-larger prime than $m$.

4. Let $t = collatz(M)$.

5. It is always true that $m$ is in the sequence $t$, regardless of the starting prime number $p$?

What you've asked is whether, if we repeat the process, it'll continue to be true, but since the process starts with a prime, if we can prove my shorter conjecture true for all primes, we'll have proven it for all iterations of the process. And if we can prove that the one-iteration thing fails for some prime, we'll have shown that the conjecture is false. (In between is the possibility that we cannot prove it for all primes, but the overall conjecture is true because certain primes never occur as the largest prime in a collatz sequence...but that third possibility need not concern us, as we'll see.)

Unfortunately, my smaller conjecture is false, hence your larger one is false as well. For $p = 19$, the Collatz sequence's largest prime is $m = 29$; the next prime after that is $M = 31$. The Collatz sequence for $M = 31$ is

      31          94          47         142          71         214
     107         322         161         484         242         121
     364         182          91         274         137         412
     206         103         310         155         466         233
     700         350         175         526         263         790
     395        1186         593        1780         890         445
    1336         668         334         167         502         251
     754         377        1132         566         283         850
     425        1276         638         319         958         479
    1438         719        2158        1079        3238        1619
    4858        2429        7288        3644        1822         911
    2734        1367        4102        2051        6154        3077
    9232        4616        2308        1154         577        1732
     866         433        1300         650         325         976
     488         244         122          61         184          92
      46          23          70          35         106          53
     160          80          40          20          10           5
      16           8           4           2           1

which does not contain the number $m = 29$.