Theorem

[y.kohmoto]

    Sorry, I wrote two bytes codes.
    I corrected them.

    -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))

    [ 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))

    Yasutoshi