### Collatz research

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 Sequences http://oeis.org/A119733

Anonymous said…
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 imbedded 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.

Dont leave me hanging, explain math ASAP, ima PTFO after looking at this shit so long.
-D
I got on the integer sequences with this: http://oeis.org/A119733