In fact, weird cycles among the rationals are quite easy to find, by a sort of engineering process. Suppose we write H for a halving step, and T for a tripling step. Next we pick a sequence of step types, obeying the rule that two T's cannot appear sequentially, say, THTHH. Then if the number we start with is K, after performing those steps we have (3[(3K+1)/2] + 1)/4 = (9K + 5)/8 (if I did my algebra right). We would like to find a starting value of K for which this step sequence makes a cycle. So we solve the equation K = (9K + 5)/8, which gives K = -5. And sure enough, -5, -14, -7, -20, -10, -5 is a cycle of the desired shape. If the desired cycle shape were THTHHH, then the equation to solve would be K = (9K + 5)/16, which gives K = 5/7. And checking again: 5/7, 22/7, 11/7, 40/7, 20/7, 10/7, 5/7 is a cycle of the desired shape. (Note that "even" generalizes here to "even numerator".) We can engineer a cycle of any valid shape. Several of the references in Lagarias's comprehensive bibliographies on the problem have nosed down this byway. A couple of impossibility results for various categories of cycle-shapes among the integers have fallen out of this approach. On Sun, Mar 25, 2018 at 8:07 PM, Victor Miller <victorsmiller@gmail.com> wrote:
Jeff Lagarias has been obsessed with the Collatz conjecture for years. Here’s one annotated bibliography https://arxiv.org/pdf/math/0608208.pdf
There’s also his book https://bookstore.ams.org/mbk-78
On Sun, Mar 25, 2018 at 19:15 Dan Asimov <dasimov@earthlink.net> wrote:
Three observations:
1) The Collatz mapping can be thought of as just going from odd integers to odd integers, via
f(x) = 2K+1
for any odd integer x, where
3x+1 = 2^n * (2K+1) for integers n >= 0 and K >= 0.
This seems much more natural than dividing by 2 only once per "time step".
Then the Collatz conjecture can be phrased as saying that for odd x > 0, iterating f to get f(x), f(f(x)), ... must end up in the (1 ) cycle.
2) No reason we can't look at this for odd x < 0. In that case the cycles I know of are (-1 ), (-5 -7 ), and (-17 -25 -37 -55 -41 -61 -91 ).
3) It seems even more natural to look, more generally, at Collatz on rational numbers of the form
x = odd/odd.
Let x be such a rational. Then much as in 1) set
f(x) = p/q
in lowest terms where
3x+1 = 2^n * p/q
for n, p, q integers with p and q odd. There seem to be lots of cycles, but they don't seem easy to find. And it's not clear if studying this more general situation will shed any light on the original conjecture.
—Dan
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