A Collatz-Inspired Puzzle

This is a puzzle.

In prior posts, I used the Collatz Problem, restated here:

Each counting number n past 1 is assigned a successor number, as follows: $successor[n] = \begin{cases}3n+1 & \text{if n is odd} \\ n/2 & \text{if n is even} \end{cases}$
The number “1″ is considered home, and when you’re home, you stop.  If you start at a given number away from home, and cycle through its successors, you may end up home.  Is there any starting number from which you will not eventually reach home?

The Collatz Problem, still unsolved, intrigues me in that it is so simply stated and yet so very deep.  It’s also a very well-balanced problem in the sense that many simple variations of it can be solved at the K-12 level.  Here are some of those variations, offered as a puzzle:

A. Each counting number n past 1 is assigned a successor number, as follows: $successor[n] = \begin{cases}3n & \text{if n is odd} \\ n/2 & \text{if n is even} \end{cases}$
The number “1″ is considered home, and when you’re home, you stop.  If you start at a given number away from home, and cycle through its successors, you may end up home.  For what starting numbers will you eventually reach home?

B. Each counting number n past 1 is assigned a successor number, as follows: $successor[n] = \begin{cases}2n & \text{if n is odd} \\ n/2 & \text{if n is even} \end{cases}$
The number “1″ is considered home, and when you’re home, you stop. If you start at a given number away from home, and cycle through its successors, you may end up home. For what starting numbers will you eventually reach home?

C. Each counting number n past 1 is assigned a successor number, as follows: $successor[n] = \begin{cases}2n-2 & \text{if n is odd} \\ n/2 & \text{if n is even} \end{cases}$
The number “1″ is considered home, and when you’re home, you stop. If you start at a given number away from home, and cycle through its successors, you may end up home. For what starting numbers will you eventually reach home?

D. Each counting number n past 1 is assigned a successor number, as follows: $successor[n] = \begin{cases}2n+2 & \text{if n is odd} \\ n/2 & \text{if n is even} \end{cases}$
The number “1″ is considered home, and when you’re home, you stop. If you start at a given number away from home, and cycle through its successors, you may end up home. For what starting numbers will you eventually reach home?

Note that in each of these we’ve left the treatment of even numbers alone, and yet get very different behaviors. The treatment of odd numbers in the original Collatz Problem is one where you can neither find an obvious counter example nor a straightforward demonstration of why you always reach home.
However, the 3n+1 formula is not unique in this. There is at least one other treatment of the odd numbers that is essentially equivalent to the Collatz problem (in the sense that if you solve one, you’ve also solved the other one.)

E. Can you find a treatment for the odd numbers (leaving the n/2 treatment of the even numbers alone) that creates a problem equivalent to the Collatz problem?

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