I realized that if the hailstone sequences of two consecutive numbers took the same amount of steps to reach 4, then they must start coinciding (by this I mean that the sequences become the same) at some point. For any two consecutive numbers that do this, lets call them n and n+1, their sequences coincide after m steps. Therefore every consecutive two numbers of the form n,n+1 (mod 2^m) are a double. Also, if you remove every consecutive two numbers of the form 4n+1, 4n+2, the pattern might become clearer, because if 4n+1, 4n+2 is a double, then 3n+1, 3n+2 are.
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