correction

[kohmoto]

    Sorry, I found typo mistakes in my early mail about (1+2i)x+1.

    The corrected explanation is as follows.

    --------------------

    (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.



    [Calculation of S_3]

    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+26i=(1+i)^3*(6+7i)
                              because (1+i)^3*(6+7i)=2i*(-1+13i)=i*(-2+26i), 
"i" is ignorable. So, it is the semi-factorization of -2+26i. 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 .... .

    Yasutoshi