Unitary sigma - Unitary phi perfect number

[y.kohmoto]

    To Niel :
    I post the sequence of USUP perfect number.

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%I A000001
%S A000001 6, 20, 72, 272, 5280, 12480, 65792, 251719680, 4295032832,
375297105592320,
%T A000001 4238621367336960, 20203489717239782783648394117120
%U A000001 84353101158454670682576150304666023245622804480,
276593818698572865168167196848999890222397175889920
%N A000001 Unitary sigma-Unitary phi perfecct number.
                     USUP(n)=1/k*n , for some  integer k.
                    If n=Product p_i^r_i then USUP(n)=
UnitarySigma(2^r_1)*UnitaryPhi(n/2^r_1)
                        =(2^r_1+1)*Product(p_i^r_i-1) , 2<p_i
%e A000001 USUP(2^4*7^2)=UnitarySigma(2^4)*UnitaryPhi(7^2)=17*48= 816
                     So,  USUP(n) = UnitarySigma(n)        if n=2^r
                                        = UnitaryPhi(n)            if
GCD(2,n)=1

%C A000001 Numbers of form 2^(2^m)*F_m appear on the sequence.
                 Where F_m means Fermat prime 2^(2^m)+1.
                 Because,

USUP(2^(2^m)*F_m)=UnitarySigma(2^(2^m))*UnitaryPhi(F_m)=(2^(2^m)+1)*(F_m-1)=
F_m*2^(2^m))

                 Numbers of the following form exist on the sequence.
                 For j=0 to 4 , k*product F_i , i=0 to j , F_i means Fermat
prime 2^(2^n)+1 , k is an integer
                 examples :
                a(1)=2*F_0 , a(5)=2^5*11*F_0*F_1 , .... ,
a(12)=2^20*4278255361*F_0*F_1*F_2*F_3*F_4

    Factorizations : 2*3; 2^2*5; 2^3*3^2; 2^4*17; 2^5*3*11*5; 2^6*5*13*3;
2^8*257; 2^12*3*5*17*241; 2^16*65537; 2^17*3*5*17*257*43691;
2^20*3*5*17*257*61681; 2^40*3*5*17*257*65537*4278255361;
2^48*3^6*5*7*11*13*17*23*47*137*193*65537*115903*22253377;
2^48*3^7*5*7*11*13*17*23*47*137*193*1093*65537*115903*22253377


    %K A000001 nonn
    %O A000001 0, 1
    %A A000001 Yasutoshi Kohmoto (zbi74583 at boat.zero.ad.jp)
    %Y A000001 A000002 , A091321 , A092356


    %I A000002
    %S A000002 1, 1, 1, 1, 2, 2, 1, 2, 1, 2, 2, 2,
    %N A000002 USUP perfecct number.
                     USUP(n)=1/k*n
                     If n=Product p_i^r_i then USUP(n)=
UnitarySigma(2^r_1)*UnitaryPhi(n/2^r_1)
                         =(2^(r_1)+1)*Product(p_i^r_i-1) , 2<p_i
                     Sequence gives k numbers.
    %K A000002 nonn
    %O A000002 0, 1
    %A A000002 Yasutoshi Kohmoto (zbi74583 at boat.zero.ad.jp)
    %Y A000002 A000001

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    To seqfans :

    The fact that Mersenne prime is related with perfect number is well
known.
    I found a relationship between Fermat prime and Unitary sigma - Unitary
phi perfect number.
         USUP(n)=1/k*n
          2^40*F_0*F_1*F_2*F_3*F_4*4278255361  is a solution.
    It is a magic, isn't it?
    Fermat primes exist finitely. Does it mean the number of USUP perfect
number is finite?

    Yasutoshi