The Collatz conjecture asserts that a sequence defined by repeatedly applying the function \begin{align} T_0(n) &= \begin{cases} (3n + 1) / 2, & \text{for odd $n$,}\\ n / 2, & \text{for even $n$,}\\ \end{cases} \end{align} always converges to the cycle passing through the number 1 for arbitrary positive integer $n$.
Theorem. As long as the $T_0$ iterates converge to 1, a sequence defined by repeatedly applying the function \begin{align} T_k(n) &= \begin{cases} (3n + 3^k) / 2, & \text{for odd $n$,}\\ n / 2, & \text{for even $n$,}\\ \end{cases} \end{align} converges to the cycle passing through the number $3^k$ for arbitrary positive integers $n$ and $k$.
Proof. The function $T_{k}(n)$ can be adjusted for multiples of three using \begin{align} \label{eqn:3T0n} 3 \cdot T_{k}(n) = \begin{cases} (3 \cdot 3n + 3^k \cdot 3) / 2, & \text{for odd $n$,}\\ 3n/2, & \text{for even $n$.}\\ \end{cases} \end{align} Now substitute $3n$ for $m$ ($m$ is a multiple of 3), thus $3 \cdot T_{k}(n) = T_{k+1}(3n)$, and therefore \begin{align} \label{eqn:T1m} T_{k+1}(m) = \begin{cases} (3m + 3^{k+1}) / 2, & \text{for odd $m$,}\\ m/2, & \text{for even $m$.}\\ \end{cases} \end{align} Note that $3n$ is odd when $n$ is odd and $3n$ is even when $n$ is even. It means that the sequence of $T_{k+1}$ iterates for $m=3n$ is exactly the sequence of $T_{k}$ iterates for $n$ multiplied by 3.
Now we know what happens to the trajectory of $T_{k+1}$ if $m$ is a multiple of 3. It remains to show what happens to this trajectory if $m$ is not a multiple of 3. If such a number is even, then we can repeatedly pull out all the factors of 2 (the even branch of $T_{k+1}$), and finally get an odd number. Thus we focus on odd $m$. A single iterate of the $T_{k+1}$ function give us $(3m + 3) / 2$ (the odd branch of $T_{k+1}$), where $(3m + 3) / 2$ is a multiple of 3 ($3m+3$ is obviously multiple of 3, the division by $2$ has no effect on divisibility by 3). Note that the iterates of $T_{k+1}$ converges to the cycle passing through the number $3^{k+1}$, which corresponds to $T_0(n) = 1$. $\Box$
There are a few doubts that I have concerning this proof:
- Do the functions $T_k$ and $T_{k+1}$ in the first paragraph make sense? I rewrote them several times.
- Is there any clue missing from the proof? It seems to me that not all the sentences connect to each other.
- Is there any information missing in the proof? For example, I didn't mention what cycles are made of.
- Is there something clearly wrong here?
I would appreciate any feedback.
(this late answer was originally meant for MSE-question About Collatz 3n+3? but possibly fits better here)
Looking at the more general question for $3x+r$ might simplify the understanding as well for the the special case where $r=3$ or even $r=3^k$.
Preliminaries The Syracuse-notation for the Collatz-transformation might be written in this form for odd positive elements $a_k$ and positive integer parameters $A_k$ in the exponents: $$ T_{3,1}:= \qquad a_{k+1} = {3a_k+ 1\over 2^{A_k}} \tag {1.1} $$ Iterating this to $N$ steps can be written as $$ a_{k+N} = { 3^N \over 2^S } a_k + { 3^{N-1} + 3^{N-2} 2^{A_1} + 3^{N-3} 2^{A_1+A_2} + \cdots 2^{A_1 + ...+ A_{N-1} } \over 2^S} \tag {1.2} $$ where here (and always in the following) I denote $S = \sum_{k=1}^N A_k$.
I find it useful to introduce a notation for the long numerator in eq 1.2 : $$ \text{using } \qquad E_{N,S} = [A_1,A_2,...,A_k,...,A_N] \qquad \text{ then } \\ Q(E_{N,S}) = 3^{N-1} + 3^{N-2} 2^{A_1} + 3^{N-3} 2^{A_1+A_2} + \cdots 2^{A_1 + ... +A_{N-1} } \tag {1.3} $$ We have then the expression for an $N$-step iteration from some $a_1$ to $a_{N+1}$ in a much shorter form as $$ T_{3,1}^{\circ N}:= \qquad a_{1+N} = { 3^N \over 2^S } a_1 + { Q(E_{N,S}) \over 2^S} \tag {1.4} $$ and the expression for the first element $a_1$ in a cycle (when $a_1 = a_{N+1}$) of $N$ (odd) steps and $S$ divisions by $2$, with some exponent vector of $A_k$'s (use the symbol $E_{N,S}$ for its representation) by $$ a_1 = {Q(E_{N,S}) \over 2^S - 3^N } \tag {1.5} $$
Derivations Now for the generalization $r$ of the summand in eq 1.1 $$ T_{3,r}:= \qquad a_{k+1} = {3a_k+ r\over 2^{A_k}} \tag {2.1} $$ it comes out that the function $Q()$ is not affected by this generalization and only eq 1.4 changes minimally: $$ a_{1+N} = { 3^N \over 2^S } a_1 + {r \cdot Q(E_{N,S}) \over 2^S} \tag {2.4} $$ and as well for the cycle-formula: $$ a_1 = { r \cdot Q(E_{N,S}) \over 2^S - 3^N } \tag {2.5} $$
From eq 2.4 we see the equivalence of the $T_{3,r}()$-transformation with the original Syracuse if we replace the $a_k$ by $r \cdot b_k$. We get then $$ r \cdot b_{1+N} = { 3^N \over 2^S } r \cdot b_1 + {r \cdot Q(E_{N,S}) \over 2^S} \tag {3.1a} $$ and cancelling the factor $r$ we get $$ b_{1+N} = { 3^N \over 2^S } b_1 + { Q(E_{N,S}) \over 2^S} \tag {3.1b} $$ which is simply the Syracuse-formulation on the variables $b_k$. For instance for $a_1=15 = 3 \cdot 5$ we have with $r=3$ the obvious equivalence $$ T_{3,3}: 3\cdot 1 = {(3 \cdot 5) \cdot 3+3 \over 2^4 } \\ \sim T_{3,1}: 1 \cdot 1 = {(1 \cdot 5) \cdot 3+1 \over 2^4 } \tag {3.2a} $$ as well as after allowance of fractional values for the $T_{3,1}$-Syracuse transformation: $$ T_{3,3}: 3 = { 7\cdot 3+3 \over 2^3 } \\ \sim T_{3,1}: 1=(3/3) = {(7/3) \cdot 3+1 \over 2^3 } \tag {3.2b} $$
Aside of this instructive representation of equivalency between $T_{3,r}$ and $T_{3,1}$ there occurs here first time the problem of dealing with fractional values in the Collatz-problem. It is of course not a big problem to extend the domain for the Collatz-conjecture to fractional values, but once we allow fractional values in the $a_k$ then we shall as well introduce the possibilities of non-integer elements on cycles.
The Collatz-problem is then, more explicitely: "there is no cycle in integral values $a_k>0$ other than $a_k=1$" - but on fractional values cycles might occur!
We start at the cycle-equation eq 2.5: $$ a_1 = { r \cdot Q(E_{N,S}) \over 2^S - 3^N } \tag {4.1} $$ Let us assume, there are no non-trivial cycles for the Collatz-problem, operationalized by the $T_{3,1}$-definition eq 1.1 . Once allowing fractional values we can only say: no integer-valued cycles. It implies, that, given some $N$ and from this $S$, the primefactors in the denominator cannot all be cancelled by the function $Q(E_{N,S})$ in the numerator.
Let's see one example. Assume $N=3$ and $S=5$. Then for $T_{3,1}$ we get $$ a_1 = { 1 \cdot Q(E_{N,S}) \over 32 - 27 } = { Q([A_1,A_2,A_3]) \over 5 } \tag {4.2} $$ and no configuration $\small [A_1,A_2,A_3]$ with $\small A_1+A_2+A_3=S=5$ produces a primefactor $5$ in the numerator and so we stay with a fractional result for $a_1$.
But if we use $r=5$ and thus the transformation $T_{3,5}$ we get $$ a_1 = { 5 \cdot Q(E_{N,S}) \over 32 - 27 } = { 5 \cdot Q([A_1,A_2,A_3]) \over 5 } = Q([A_1,A_2,A_3]) \tag {4.3} $$ and the following solutions are all integral in $a_k$: $$ a_1=Q([1,1,3]) \qquad a_2=Q([1,3,1]) \qquad a_3=Q([3,1,1]) \\ a_1=Q([1,2,2] \qquad a_2=Q([2,2,1]) \qquad a_3=Q([2,1,2]) \tag {4.4} $$ where we have two solutions for $a_1$, each with 3 needed rotations each.
Another example: for the 5-odd-step-cycle $N=5$ from where $S=8$ we have in the denominator $2^8 - 3^5=13$ and thus $r=13$ allows one or more cycles.
By the possible combinations of the exponents $A_k$ (resp. rotations) we find a couple of cycles already mentioned elsewhere in MSE $$\small \begin{array} {llll} a_1=Q([1,1,1,1,4]) & a_2=Q([1,1,1,4,1]) & a_3=Q([1,1,4,1,1]) & \ldots \phantom{\vdots} \\ a_1=Q([1,1,1,2,3]) & a_2=Q([1,1,2,3,1]) & a_3=Q([1,2,3,1,1]) & \ldots \phantom{\vdots} \\ a_1=Q([1,1,2,2,2]) & a_2=Q([1,2,2,2,1]) & a_3=Q([2,2,2,1,1]) & \ldots \phantom{\vdots} \\ a_1 = \; \vdots & a_2 = \; \vdots & a_3 = \; \vdots & \ldots \\ \end{array} \tag {4.5} $$ (giving $7$ cycles (and their rotations) all with $(N,S)=(5,8)$ )
From this we have now a simple recipe to find some $r$ which allows integer cycles for any $N$ (and its according $S$ where $S=\small \lceil N \cdot \log_2(3) \rceil$) : simply use some $r$ which cancels the primefactors in the denominator $\small 2^S-3^N$ by the factors of $r \cdot \small Q(E_{N,S})$.
This last consideration implies as well one more property: since in the denominator there will never occur a primefactor $3$ no $r=3^k$ can have any effect on the cancelling property and any $r=3^k$ shall have no more nontrivial cycles than the original Collatz-transformation $T_{3,1}$ (thus as far as we know: no nontrivial cycles at all).