Infinite sets that are proven to be true for Collatz Conjecture? I know that the set of powers of 2 and the set of 1,5,21,85,341,... are proven to be true for Collatz conjecture. Are there other sets with infinite number of numbers that are also proven to satisfy Collatz conjecture?
 A: $\qquad \{ 1,5,21,85,\cdots,{1 \cdot 4^k-1 \over 3}, \cdots \}$ go to $1$,
$\qquad \{ 3,13,53,253,\cdots,{10 \cdot 4^k-1 \over 3}, \cdots \}$ go to $5$,
$\qquad \{ 113,453,\cdots,{340 \cdot 4^k-1 \over 3}, \cdots \}$ go to $85$,
$\qquad \{ 17,69, \cdots \}$ go to $13$ ...
$\qquad \qquad$ and so on: infinitely many subsets!     
You might like to see some examples of various forms of trees (of course which are recursive) to see a lot more subsets of odd numbers being proved to converge by these schemes. Obviously they are infinitely many such infinite subsets. But unfortunately that does not mean, that it would have been proven that all odd positive natural numbers are in that single tree.
See mainpage and then go to subpage "about numerical and graphical trees". My favourite is that "bottle-brush" like tree at the end ...  
A: One example is any set containing the odd integer $(2^n-1)/3$, for $n>1$. 
Indeed, applying the "$3x+1$" rule to this will yield a power of $2$ and therefore the chain ends in the trivial cycle $\{4,2,1\}$.
A: One can simplify the Collatz iteration by bypassing the even operation using $g_n(x)=(2^{e(x)}x-1)/3$, where $x \not \equiv 0\pmod{3}$ is any odd natural number, $e(x)$ is any positive integer exponent for which $3$ divides $2^{e(x)}x-1$. This leads to $g_n(x)=z_n+2^{e(x)}m_r$, where $e(x)=2n$ if $x \equiv 1\pmod{3}$ or $e(x)=2n-1$ if $x \equiv 2\pmod{3}$ for $n \in \mathbb{N}=\{1,2,3,\ldots\}$, $z_n$ is any element of $Z=\{1,5,21,85,\ldots\}$, and $m_r$ is the multiple of any admissible $x\pmod{3}$, i.e. $x=3m_r+r$ for $r=1,2$. With this, one finds an infinite set $\{g_n(x)\}_{n\in\mathbb{N}}$ satisfying the Collatz conjecture and repeatedly iterating $g_n$ leads to convergent odd Collatz sequences. See further details here.
