OU-Sigma function

[y.kohmoto]

    Hello, Seqfans
    I considered an artificial divisor function as follows.

    Definition of ordinary unitary sigma :
         If n=Product p_i^r_i then
OU-Sigma(n)=Sigma(2^r_1)*UnitarySigma(n/2^r_1)
         =(2^(r_1+1)-1)*Product(p_i^r_i+1) , p_i is not 2

    e.i.
    OU-Sigma(2^4*7^2)=Sigma(2^4)*UnitarySigma(7^2)=31*50=1550
    So,  OU-Sigma(n) = Sigma(n)               if n=2^r
                            = UnitarySigma(n)      if GCD(2,n)=1

    I calculated OU-Sigma perfect number.
         OU-Sigma(n) = k*n

    sequence of OU-Sigma perfect number :
    2*3, 2*3^2*5, 2^2*7, 2^2*5^2*7^2*13, 2^3*3*5, 2^4*31, 2^5*3^2*5*7,
2^6*127, 2^7*3^3*5*7*17*, 2^7*3^4*5*7*17*41, 2^8*5*7*19*37*73,
2^8*3*5*7*19*37*73, 2^9*3*11*31, 2^9*3^2*5*11*31, 2^10*3^3*5*7*23*89,
2^10*3^4*5*7*23*41*89, 2^11*3^3*5*7^3*11*13*43, 2^11*3^4*5*7^3*11*13*41*43,
2^12*8191,
    sequence of k :
    2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 4, 3, 3, 3, 3, 3, 3, 2,

    The sequence contains perfect numbers.
    Because,
         OE-Sigma(2^(m-1)*M_m) =  Sigma(2^(m-1))*UnitarySigma(M_m)
                                 = Sigma(2^(m-1))*Sigma(M_m)
                                 = 2^m*M_m

    To Neil :
    Do you need this sequence?
    If Yes  then I will format it, and if No then I will throw it in my junk
box.

    Yasutoshi