artificial sigma

[y.kohmoto]

    Noe wrote :
    >Except for the first two, these seem very artificial.  For Sigma(n),
the
    >Gaussian primes must be in the first quadrant.


    What is the most artificial function?
    Is the opposite word of "artificial" "natural"? or Is it "wild"?


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    The definition of "First quadrant rule" is as follows.
         Fq(n)=i^k*n , 0<=k<=3 , If  i^k*n=r*e^(i*t) then 0<=t<Pi/2

         ....  it becomes a number in the first quadrant

    Example    Fq(-1+4i)=i^3*(-1+4i)=4+i

    "+" using the first quadrant rule :
         m+n=Fq(m)+Fq(n)

    "-" using the first quadrant rule :
         m-n=Fq(Fq(m)-Fq(n))


    GSigma(n) :
         Fq(Product (Sum(Fq(p_i^s_i) , 0<=s_i<=r_i)))
         Where n=Product Fq(p_i)^ r_i


    +1UnitarySigma(n) :
         Fq(Product (Fq(p_i^r_i) + 1))


    +iUnitarySigma(n) :
         Fq(Product (Fq(p_i^r_i) + i))


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    I like "artificial functions".
    One of "the most artificial " Sigma  which I defined is the following.

    -1Sigma(n) :
         If n=Product p_i^r_i then -1Sigma(n)=Product(-1+{Sum p_i^s_i ,
1<=s_i<r_i})
                                        =Product(-2+(p_i^(r_i+1)-1)/(p_i-1))

         It is a difference of the divisors of n.

    -1Sigma aliquot sequence :
         a(n) = -1Sigma(a(n-1))
    Example
    S0 :    2^3*5*7*29 , 2^5*3*7*13 , 2^4*3^2*61 , 2^2*3*5*11*29 , 2^6*5^2*7
, 2*3*5^3*29 , 2^4*7^2*11 , 2*5^2*11*29
    This is a cyclic sequence whose period is 8, so it is called a "-1Sigma
sociable number of order 8".

    [ Theorem S3 ]　.
　　If 2^(2^(n+1)+2)-3 is prime then
    2^3*13*(Product F_i ,  i=0 to n)　 ,  　 2^3*3*(2^(2^(n+1)+2)-3)　 ,
2^(2^(n+1)+1)*3*13
    are   -1Sigma sociable number of order 3.  F_i  means Fermat Prime.
　　The theorem gives only 2 examples for n=1, 2
　　for n=1
　　a,b,c=　　2^3*13*3*5,
　　　　　　　2^3*3*61,
　　　　　　　2^5*3*13
　　for n=2
　　a,b,c=　　2^3*13*3*5*17
　　　　　　　2^3*3*1021
　　　　　　　2^9*3*13
　　for n=0, it becomes a=b=c
　　for n=3, n=4, 2^18-3 and 2^34-3 are not prime.
　　[ Proof ]
         -1Sigma(2^3*13*(Product F_i , i=0 to n)) = 13*2^2*3*2^(2^(n+1)-1) =
2^(2^(n+1)+1)*3*13
         -1Sigma(2^(2^(n+1)+1)*3*13) = (2^(2^(n+1)+2)-3)*2*2^2*3 =
2^3*3*(2^(2^(n+1)+2)-3)
         -1Sigma(2^3*3*(2^(2^(n+1)+2)-3) = 13*2*2^2*(2^(2^(n+1))-1) =
2^3*13*(2^(2^n)-1)*(2^(2^n)+1) =　2^3*13*(Product F_i , i=0 to n))


    Neil,
    I don't remember if I have posted this sequence S0  to OEIS

    It is periodic. Is it fit to OEIS?
    If not, I will describe it on my OEUAI.

    Yasutoshi