(1+2i)x+1 sequence

[kohmoto]

----- Original Message ----- 
From: "Marc LeBrun" <mlb at fxpt.com>
To: <seqfan at ext.jussieu.fr>
Sent: Saturday, April 30, 2005 1:27 AM
Subject: Re: (1+2i)x+1 sequence


> >=Kohmoto
> >  I considered about an analog of 3x+1 sequence in Gaussian Integer.
>
> Interesting idea.
>
    Thank you.

> > [Definition of 3x+1 sequence]
> > a(n)=(3*a(n-1)+1)/2^k , where 2^k is the highest power of two dividing
> 3*a(n-1)+1.
>
> I believe the traditional "Collatz" definition is simpler: "x --> x/2 if x
> even, 3x+1 if x odd".  These are essentially similar, but  I'd recommend
> the simpler form to encourage easier communication.
>
    I agree with you, but I think my definition represents the essence of 
the sequence.
    The definition is from the furthest generalization of Collatz's 3x+1 
sequence as follows.

     a(n)=Floor[A*a(n-1)+B]/p^k , where p^k is the highest power of p 
dividing  Floor[A*a(n-1)+B] . A,B are real numbers. p is prime.

     For example, if we set a(0)=1, A=2.0001-,  B=3.0, p=2, and display a 
graph of (x,y)=(n, log a(n)), then a figure such that a linear sequence - a 
random sequence - a linear sequence - a random sequence - a linear 
sequence - .... is observed.
     See a graph on my home page.
         http://boat.zero.ad.jp/~zbi74583/another02.htm
     If we use the traditional definition then the phenomenon never appears.

> Then, generalizing with respect to some divisor (2, (1+i), etc), "even"
> just becomes "evenly divisible" while "odd" becomes "not evenly 
> divisible".
>
> > [A translation to Gaussian integer]
> >
> >  (2+i)x+1 sequence :
> >   S_1  1, 1+2i, 1, 1+2i, ....
> >   S_2  3, 2+5i, 3, 2+5i, ....
> >   S_3  7,  4+11i, 6+7i, 3+10i, ....
> >
> >   (1+2i)x+1 sequence :
> >   T_1  1, 1, 1, 1, ....
> >   T_2  3, 2+3i, 2+5i,  1+8i, 6+i, ....
> >   T_3  5, 3+10i, 1, 1, ....
> >
> > I  am not sure if they are correct, because the factorization is
> difficult without a computer.
>
> This illustrates why simpler definitions might be better.  To simply 
> divide
> x by (1+i) we need only to multiply it by (1-i)/2.  We can then do this
> until x becomes "not evenly divisible".  This is easy to test, since if
> x=(a+ib) then x/(1+i) is (a+b)/2+i(a-b)/2 so we can quickly tell by 
> looking
> at a and b, we don't need to perform difficult factorizations (although a
> computer is still handy).
>
> We thus just need to specify when x/(1+i) doesn't divide out "evenly". 
> Now
> obviously when a and b have opposite parity the ratio gives half-integers,
> so we at least know that all such numbers are "odd" (although there may be
> others, see next comment).
>
> > Numbers are calculated in the first quadrant of Z[i] plane.
>
> And this is true just when a>=b.
>
> However I don't think factorization of Gaussian integers is usually 
> defined
> this way.  Instead the "irreducible" factors are taken to lie in the
> "tilted" quadrant between x=y and x=-y containing the positive
> x-axis.  This produces the simple traditional conjugate factorizations 
> such
> as 2=(1+i)(1-i).
>
> Both definitions might be interesting, but we're proliferating 
> generalizations.
>
> > Do  S_3 , T_2   become periodic?
>
> I don't know, since it depends on what choices you want make in defining
> these things.  However I will say that my quick computer-aided 
> calculations
> seemed to get quite different values than the ones you sent, no matter
> which variations I chose.  They all looked like they were diverging, but I
> didn't have time to dig deeper.
>
    I think your definitions are different from mine.

    Let us start with a(0)=7 .  f(n)=((2+i)*a(n-1)+1)
    f(7)=(2+i)*7+1=15+7i=(1+i)*(4+11i)
                       because (1+i)*(4+11i)=-7+15i=i*(15+7i) , "i" is 
ignorable. So, it is the factorization of 15+7i.
    (1+i) is deleted.
     The result is 4+11i.

    f(4+11i)=(2+i)*(4+11i)+1=-2+2-i=(1+i)^3*(6+7i)
                              because 
(1+i)^3*(6+7i)=2i*(-1+1-i)=i*(-2+2-i), "i" is ignorable. So, it is the 
semi-factorization of -2+2-i. 6+7i is still factorized.
    (1+i)^3 is deleted.
    It becomes 6+7i.

    f(6+7i)=6+20i=(1+i)^2*(3+10i)
    The result is 3+10i .

    The sequence becomes, 7, 4+11i, 6+7i, 3+10i .... .



> I'd suggest re-analyzing this starting with the simpler definitions,
> eliminating as many cases as possible a priori, to see if there's a "core"
> that are worth exploring via more extensive calculations.
>
    I have no time to do it.
    I wish some one will study about them.

> An interesting diagram might be produced by drawing the vectors connecting
> each Gaussian integer with its successor.
>
>
    To calculate the analogs of  ideas "Glide","Delay","Path", Records
 whose definitions is described on Eric Roosentaar's home page is also 
interesting.
   http://personal.computrain.nl/eric/wondrous/

   The sequences which are related with these ideas may be fit to OEIS.

    Yasutoshi