When I was looking into the Collatz conjecture, I ran into this sequence:

0, 1, 2, 5, 4, 7, 10, 19, 8, 11, 14, 23, 20, 29, 38, 65, 16, 19, 22, 31, 28, 37, 46, 73, 40, 49, 58, 85, 76, 103, 130, 211, 32, 35, 38, 47, 44, 53, 62, 89, 56, 65, 74, 101, 92, 119, 146, 227, 80, 89, 98, 125, 116, 143, 170, 251, 152, 179, 206, 287, 260, 341, 422, 665

Please let me know if you can find a general way to predict numbers in that sequence.

Update: This is now published on the Online Encyclopedia of Integer SequencesÂ http://oeis.org/A119733

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## Comments

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@fulldecent

If n is a power of 2 then f(n) = n, making a nice binary tree pattern. If n mod 4 = 0, then f(n) mod 4 = 0. The number of embedded binary trees increases for each leaf by twofold. Somehow 3^n is involved, powers of three are rampant, more so for the even elements. Because of the Collatz relationship, f(2x), f(2x-1) is probably the layout. Don't leave me hanging, explain math ASAP, ima PTFO after looking at this shit so long.

Anonymous

Please discuss this topic anywhere and let me know any great comments or media coverage I should link here.