(1+2i)x+1 sequence

[kohmoto]

    Hello, Seqfans.
    I considered about an analog of 3x+1 sequence in Gaussian Integer.


    [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 .


    [A translation to Gaussian integer]

         3 -> 2+i or 1+2i       ....        Secondly small prime
         1 -> 1                    ....        Unit
         2 -> 1+i                  ....        The smallest prime


    (2+i)x+1 sequence is defined as follows


    a(n)=((2+i)*a(n-1)+1)/(1+i)^k , where (1+i)^k is the highest power of 
(1+i) dividing (2+i)*a(n-1)+1 .
　

    (1+2i)x+1 sequence is defined as follows



    a(n)=((1+2i)*a(n-1)+1)/(1+i)^k , where (1+i)^k is the highest power of 
(1+i) dividing (1+2i)*a(n-1)+1 .


    [Examples]

    (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, ....


    Numbers are calculated in the first quadrant of Z[i] plane.

    I calculated only  six  examples on real number line by  hand.
    I  am not sure if   they are collect, because the factorization is 
difficult without a computer.             .
    Do  S_3 , T_2   become periodic?

    Yasutoshi
 
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