(-1)Sigma

[koh]

    Hi,Seqfans

    Once I have studied about (-1)Sigma(m) which is a divisor function.
    I want to verify these old results.
  
    I will submit some sequences related with them.

    Please someone tell me how to write (-1)Sigma(m) in Mathematica.

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    1.Definition. 
    If m=Product p_i^r_i, 
    then (-1)sigma(m)=Product (-1+Sum p_i^s_i, s_i=1 to r_i) 
    Ex. 
    (-1)sigma(24)=(-1+2+4+8)*(-1+3)=1-2-4-8-3+6+12+24 
    all terms are divisors of 24, but it is not the sum of divisors. 
    It is a difference of divisors. 

    See A049060 . Neil described the formula which is a essence of this function.
 

    2. (-1)sigma sociable number.　　A049057, A049058, A049059, A051152 on 
Sloane's table 

    (-1)sigma sequence is defined as follows : 
    a(n)=(-1)sigma(a(n-1)) 
    If a(k+1)=a(1) then these k-tuple numbers are called a (-1)sigma 
sociable number of order k. 

    k=1 
    2^2*5, 2^3*3*13, 2^4*3*7*29, 2^5*3*5*61, 2^10*3*5*17*409, 
    2^9*3*5*17*1021,2^7*3*5*11^2*13*23*131 

    k=2 
    2^2 - 5, 
    2^3*3^3 - 2*13*19, 
    2^5*3*5^2*7 - 2^2*3*29*61 

    k=3 
    2^2*13 - 2^2*3*5 - 2^3*5 
    2^2*3*13 - 2^3*3*5 - 2^3*13 
    2^2*3*5*13 - 2^5*3*5 - 2^3*61 
    2^5*3*13 - 2^3*3*61 - 2^3*3*5*13 
    2^2*3*5*13*17 - 2^9*3*5 - 2^3*1021 
    2^9*3*13 - 2^3*3*1021 - 2^3*3*5*13*17 
    2^3*3*5*17*61 - 2^9*3*5*13 - 2^5*3*1021 
    2^9*3*5*61 - 2^5*3*5*1021 - 2^5*3*5*17*61 

    k=4 
    2^2*3*5*29*61 -　 2^7*3*7*5^2 - 2^2*3*11*23*29 - 2^5*5^2*7*11 

    k=8 
    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 
 
    2^6*3*5^2 - 2*5^3*29 - 2^3*7^2*11 - 2*5^2*11*13 - 2^3*3*5*29 - 
2^5*7*13 - 2^3*3^2*61 - 2^2*3*5*11*13　

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    Yasutoshi

    PS
    Neil seems to think that (-1)Sigma(m) is nice.
    I recommend you to study it.

    The following formulas are well known results.

         Zeta(s)^2=Sum_{m=1 to infinity} d(m)/m^s
         Zeta(s)Zeta(s-1)=Sum_{m=1 to infinity} Sigma(m)/m^s
         Zeta(s)/Zeta(s+1)=Sum_{m=1 to infinity} Phi(m)/m^s
    
    I want to know if any formula which has (-1)Sigma(m) exists.