On Mon, November 19, 2012 5:51 pm, Bill Allombert wrote:
> On Mon, Nov 19, 2012 at 05:05:42PM -0000, Stefan Kohl wrote:
>> Proving finiteness of all cycles may be pretty hard, however.
>
> Indeed, this problem is too close to the Collatz ((3n+1)/2) conjecture
> for comfort.
Most likely yes. -- But I wouldn't claim right away that it is really
of similar difficulty. There are some regularities, e.g. except for
the cycle containing 1, all cycles seem to have either length 2 or length
congruent to 8 mod 16 etc.
> The Collatz sequence can be restated in your notation by
> asking whether the orbits of Z under M are finite, where M is the submonoid
> of Z^Z generated by a = (1(2)->2(3)) and b = (0(2)->0(1))
> (a and b are not one-to-one).
You can actually do the same with a group. -- The group
G := <(1(2),4(6)),(1(3),2(6)),(2(3),4(6))> < Sym(Z)
acts transitively on the set of positive integers which are not
divisible by 6 if and only if the 3n+1 conjecture holds.
The group G can be entered by pasting the following into GAP:
a := ClassTransposition(1,2,4,6);
b := ClassTransposition(1,3,2,6);
c := ClassTransposition(2,3,4,6);
G := Group(a,b,c);
Best wishes,
Stefan
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