Convergence of Collatz sequences I may be using the wrong terminology here, but please bear with me, I'm not a mathematician, just a hobbyist... ;-)
I noticed that for some pairs of numbers n and n+1, when you put them through the Collatz rule, sometimes they fall into step with each other. The rule - as I assume will be known - is: when n is even, divide by two, when n is odd, multiply by three and add 1. Repeat this until you get 1 as a result.
As an example: when you put the numbers 350 and 351 through this algorithm, they will have the same sequence after step 12. Like so:
350,  175, 526,  263, 790,  395, 1186,  593, 1780,  890,  445, 1336, 668, 334, etc
351, 1054, 527, 1582, 791, 2374, 1187, 3562, 1781, 5344, 2672, 1336, 668, 334, etc
Other examples are:
242 and 243 (after only 5 steps), 1346 and 1347 (same), 237 and 238 (8 steps)
If the sequences converge it seems to happen after only a small number of steps. 
Has any study been done on this phenomenom? Does anyone have a hint on where to start looking?
 A: I think the simplest answer for this is that this doesn't just happen with number's n and n+1, it happens with any two numbers you choose (assuming the conjecture is true). You can see this easily if you look at a Collatz Tree. The sequence for every number is obtained by just following the number down towards the root of the tree. If the conjecture is true, then every 2 numbers has at least some of their paths in common.
A: I've just seen something interesting that Zander has more or less explained.
For every odd number m between 17 and 99, excluding 31, if:
m $\equiv 13 \pmod{8}$ , then  $f^3(m)= f^3(m-1)$
m $\equiv 19 \pmod{16}$ , then $ f^5(m)= f^5(m-1)$
m $\equiv 23 \pmod{32}$ , then  $f^7(m)= f^7(m-1)$
m $\equiv 15 \pmod{64}$ , then $ f^9(m)= f^9(m-1)$
m $\equiv 95 \pmod{128}$ , then  $f^{11}(m)= f^{11}(m-1)$
m $\equiv 63 \pmod{256}$ , then $f^{13}(m)= f^{13}(m-1)$
m  $\equiv 27 \pmod{16}$ or $\equiv 39 \pmod{32}$ or $\equiv 47 \pmod{64} $ then $f^2(m)= f^2(m-1)+1$
else $ m  \equiv 17 \pmod{8}$ , and $f^2(m) +2 = f^2(m+1)$
