The hailstone sequencing is the same problem as the Josephus problem in reverse. We want there to be a sequence that doesn't follow the formula but the sequencing eventually hits a binary number or a number made of binary numbers which then make the solution ..4, 2, 1. If you subtract the highest binary number you can see how the base number will react... For example 10 eventually sheds its 8 and becomes a 2 1 which is again the binary sequencing we don't want. What is the issue mostly is that no matter how high the number it still eventually hits a binary number which is antiprime and logs down to 421... So let's take a number like 7;

7=>22 => 11 =>34. 34 is 32 + 2 which are both binary... A few more logs and we will reach 4;2;1. 17=> 52 =>26 =>13 => 40...20...10...=>1. As soon as you hit a decimal multiple of 10 with just the 4..2..1 you are about to finish the series. To see how many moves are left you can count the binary parts; subtract the smaller from the larger and then subtract that product from the whole and multiply the former product to the latter and you have the steps. It works for 10 but probably not for others; just a thought but hopefully I helped in some fashion.

## Hailstone & Binary (Josephus problem correlation)

The hailstone sequencing is the same problem as the Josephus problem in reverse. We want there to be a sequence that doesn't follow the formula but the sequencing eventually hits a binary number or a number made of binary numbers which then make the solution ..4, 2, 1. If you subtract the highest binary number you can see how the base number will react... For example 10 eventually sheds its 8 and becomes a 2 1 which is again the binary sequencing we don't want. What is the issue mostly is that no matter how high the number it still eventually hits a binary number which is antiprime and logs down to 421... So let's take a number like 7;

7=>22 => 11 =>34. 34 is 32 + 2 which are both binary... A few more logs and we will reach 4;2;1. 17=> 52 =>26 =>13 => 40...20...10...=>1. As soon as you hit a decimal multiple of 10 with just the 4..2..1 you are about to finish the series. To see how many moves are left you can count the binary parts; subtract the smaller from the larger and then subtract that product from the whole and multiply the former product to the latter and you have the steps. It works for 10 but probably not for others; just a thought but hopefully I helped in some fashion.