Haskell, 62 chars 63 76 83, 86, 97, 137
c 1=[1]
c n=n:c(div(n`mod`2*(5*n+2)+n)2)
main=readLn>>=print.c
User input, printed output, uses constant memory and stack, works with arbitrarily big integers.
A sample run of this code, given an 80 digit number of all '1's (!) as input, is pretty fun to look at.
Original, function only version:
Haskell 51 chars
f n=n:[[],f([n`div`2,3*n+1]!!(n`mod`2))]!!(1`mod`n)
Who the @&^# needs conditionals, anyway?
(edit: I was being "clever" and used fix. Without it, the code dropped to 54 chars.
edit2: dropped to 51 by factoring out f())
