# Collatz Conjecture exclusivity

I have been wondering if there are any numbers that exist only in their own string of the 3n+1 problem.

I need to explain that better. Basically, when you follow the rules of the conjecture, you end up with a string of numbers, starting with whatever you chose and ending with 1. Example with 995: 995 -> 2986 -> 1493 -> 4480 -> 2240 -> 1120 -> 560 -> 280 -> 140 -> 70 -> 35 -> 106 -> 53 -> 160 -> 80 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1

Basically, I'm wondering if there are numbers that don't appear in strings of any other numbers. So with the example given, we know that 2986, 1493, 4480, 2240, 1120, 560, 280, 140, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 all don't meet that criteria.

So if you took every single number, then wrote the string of numbers that comes from each number, would any numbers only appear once?

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Would you please rephrase the question to make it more clear? Especially from this part: "So with the example given, we know that.."? – user2468 Aug 1 '11 at 22:22

Every number is half of an even number so $n$ appears in the string starting $2n$ (or $4n$, $8n$, etc.).

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See this Wikipedia image. Collatz sequences fit into a directed graph. Sometimes sequences run together, which is where the branching of the graph occurs. But every number has its own unique place in the graph. The Collatz conjecture says that this directed graph is connected.

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To close off the rest of the analysis from Henry's answer: if $n$ is a multiple of 3, then so will be all of its 'even' predecessors $p_t=2^tn$, so none of these could be of the form $3q+1$; they are the only predecessors of $n$ in this case. Otherwise, $n$ has some odd predecessor: if $n$ is of the form $3t+1$ for odd t, then obviously $t$ is a predecessor of $n$. If $n$ is $3t+1$ for even $t$, then $4n=12t+4=3(4t+1)+1$ and clearly $4t+1$ is odd, and thus it's an odd predecessor of $n$. If $n$ is $3t+2$ for some $t$, then $2n=6t+4=3(2t+1)+1$ and similarly $2t+1$ is an odd predecessor of $n$ here.

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Others answered this question already sufficiently, I'd say. But one more different view might be enlightening here.

First assume the Collatz-transformation in the version as "Syracuse-transformation", where one 3x+1-step and all subsequent A x/2 -steps are collected into one step $\small c_{k+1} = (3 c_k + 1) / 2^A$ . Then it suffices to look only at positive odd numbers whose trajectories eventually fall down to the "circuit" at 1.

Next turn the view around: look at the 1 as the beginning of the tree of inverse operation $\small d_{k+1} = (2^A d_k - 1)/3 \quad ; \quad d_0=1$ (The Collatz-conjecture is then equivalent to the conjecture that each odd natural number can be reached by iteration of this transformation.)
Then obviously at each $d_k \ne 0 \pmod3$ we have the choice for $A$ from infinitely many values: it may either be one of $1,3,5,7,\ldots$ or $2,4,6,8,\ldots$ - and this choice is at every $d_k$ except at that $d_k$ divisible by 3, which are endpoints.

So there cannot be other "isolated" numbers than multiples of 3 in your original question. Well - it may be, there are cycles, but even then the involved numbers are not isolated ( contrarily: they are even infinitely repeating connected...)

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Any number $n \equiv {\left(3\; mod \;6\right)}$ will only be reached from starting numbers beginning with $n(2^a)$

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see: math.stackexchange.com/questions/1571469/… and observe that all odd numbers lead to a succeeding odd number in a subset of (5 mod 6) or (1 mod 6) – frogfanitw Jan 16 at 2:27