Perfect number on Z[i]

[y.kohmoto]

    Hello, Seqfans


    I considered Gaussian Perfect number.
    The following condition should be  necessary for defining "Sum" of
Gaussian integer.

    If a Gaussian integer is of the form : r*e^(i*t), then 0<=t<Pi/2.
[C]

    Definition of GSigma(n) :
    If n=Product p_i^r_i then GSigma(n)=Product (Sum p_i^s_i, 0<=s_i<=r_i)
[GS]

    The formula for ordinal Sigma(n) : Product{(p_j^(r_i+1)-1)/(p_i-1)} is
useless.
    Example :
    GSigma((1+i)^7)=1+1+i+2+2+2*i+4+4+4*i+8+8+8*i=30+15*i
    ((1+i)^8-1)/(1+i-1)=-16*i=16

    G-Perfect number is defined as follows.

         GSigma(n)=m*n, for some Gaussian integer m.
[GPN]

    My algorithm for Ordinal Perfect number gives some solutions.

    GPN                           m
    1+i * 2+i                     1+2i ,
    1+i * 2+i * 1+2i              2+2i ,
    1+i ^2 * 4+i * 2+3i           3 ,
    1+i ^2 * 4+i * 2+3i * 3       4 ,
    1+i ^3 * 2+i * 1+2i           3+3i ,
    1+i ^3 * 2+i * 1+2i * 3       4+4i ,
    1+i ^4 * 10+3i * 2+i * 1+2i    1+5i

    S1 : 1, 5, 10, 10, 30, 30, 140, 200
         Real part of GPN
    S2 : 3, 5, 10, 28, 30, 84, 140, 600
         Imaginary part of GPN
    S3 : 1, 2, 3, 3, 4, 4, 4, 1,
         Real part of m
    S4 : 2, 2, 3, 0, 4, 0, 4, 5,
         Imaginary part of m

    I have no example for m=2. Is it possible?

    These numbers are solutions of [GPN] and indeed we are able to verify
sum of all divisors of the first example is (1+2i)*n as follows.
       1+1+i+2+i+1+3i = 5+5i =(1+i)*(2+i)*(1+2i)

    But after having calculated the solutions, I became to think that the
definition of GSigma is not fit to [C] because "Product" doesn't support the
condition [C].
    What is a good formula for "Sum of divisors of a Gaussian integer"?
    Even though [GS] is a formal sum of divisors, [GPN] is an interesting
equation.

    I proved the following theorem :
    If
    n=(1+i)^(2*m+1)*(2+i)*(1+2*i)*M(m+1)
     =2^m*5*(1+i)*M(m+1)
     =5*(1+i)*PN
                                  M(m+1) means  Mersenne Prime : (2^(m+1)-1)
                                  PN means Perfect number
    Then
    GSigma(n)=(1+i)^5*n

    I will soon post S1, S2, S3 and S4  to OEIS.
    I am not sure if I don't lack any term.

    Do Mathematicians know the idea of Gaussian Perfect number?
    Is there any study about GSigma(n) function?

    Yasutoshi