Perfect number on Z[i]
y.kohmoto
zbi74583 at boat.zero.ad.jp
Fri Dec 24 12:41:12 CET 2004
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
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