The above chart consists entirely of counting numbers that are not divisible by 2 or 3. (For reasons already discussed, these are the only numbers with which we need concern ourselves.)

The chart extends infinitely to the right and down.

Every counting number not divisible by 2 or 3 is represented in this chart exactly twice: once above the line, and once below the line.

Above the line, all counting numbers not divisible by 2 or 3 are simply listed in numerical order.

Below the line, each column consists of all counting numbers not divisible by 2 or 3 that go to their above-the-line number in

Within any one below-the-line column, the numbers are arranged in numerical order.

It so happens that the numbers in any one below-the-line column progress by the procedure: multiply by 4; then add 1. (But numbers divisible by 3 are omitted.)

To perform

(Note: 7 is not part of this sequence; the last up-arrow skips directly from 29 to 11; it does not visit 7.)

Forward hailstone is strictly deterministic; you have no choice about what to do next.

The conjecture claims that forward hailstone will always reach 1.

To perform

(Note: 13 is not part of this sequence; the second down-arrow skips directly from 5 to 53; it does not visit 13.)

Reverse hailstone provides you with an

The conjecture claims that using reverse hailstone from 1, you can reach any number in this chart.

(Note: A formal proof of the conjecture that utilizes the above chart would have to include proofs that all the above statements about the chart are correct and/or consistent. I believe that such would be trivially easy to do.)

Suppose we start from any number N that is greater than 1 (and, of course, not divisible by 2 or 3).

Obviously, if you could prove that forward hailstone will, in one or more steps, reach a number less than N, then that would prove the conjecture, because the same rule would apply to

Less obviously, if you could prove that

Similarly, if you could prove that from N, it must be possible to get to a smaller number via either forward or reverse hailstone (without specifying which for any particular N), then the conjecture is proven.

Further still: If you could prove that from N, it must be possible to get to a smaller number via either forward hailstone, reverse hailstone, or

That’s it for now!