Divisor

koh zbi74583 at boat.zero.ad.jp
Sat Jan 6 06:40:01 CET 2007

```    Hi, Seqfans

I defined four divisor functions which are "difference" of divisors.

1.

SENSiguma(m) = (-1)^((Sum_i  r_i)+Omega(m))*Sum_{d|m} (-1)^((Sum_j r_j)+Omega(d))*d
=Product_i (Sum_{1<=s_i<=r_i} p_i^s_i)+(-1)^(r_i+1)

Where m=Product_i p_i^r_i , d=Product_j p_j^r_j

ex.    SENSigma(240)=(-1+2+4+8+16)*(1+3)*(1+5)
SEN for Signed by Exponents of prime factors and Number of prime factors.

2.

SEPSigma(m) = (-1)^(Sum_i  r_i)*Sum_{d|m} (-1)^(Sum_j r_j)*d

=Product_i (Sum_{1<=s_i<=r_i} p_i^s_i)+(-1)^r_i

Where m=Product_i p_i^r_i , d=Product_j p_j^r_j

ex.    SEPSigma(240)=(1+2+4+8+16)*(-1+3)*(-1+5)
SEP for Signed by Exponents of Prime factors .

3.

SENUnitarySigma

4.

SEPUnitarySigma

[SENSigma Multiply Perfect Number]

SENSigma(m)=k*m , for some integer k.

2^11*3^3*5^4*7^3*13*19*41     k=5
2^9*3*11*31                   k=3
2^9*3^3*5*11*31               k=4
2^9*3^3*5^2*11*29*31          k=4
2^9*3^4*7*11*17*31            k=4
2^9*3*11^2*31*131             k=3
2^9*3^3*5*11^2*31*131         k=4
2^9*3^3*5^2*11^2*29*31*131    k=4
2^9*3^4*7*11^2*17*31*131      k=4
2^6*3*5^4*7*19*41             k=4
2^6*3^2*5^4*7*11*19*41        k=4
2^6*3^4*5^4*7^2*11*17*19*41   k=5
2^6*3^2*5^4*7*11^2*19*41*131  k=4
2^6*3^4*5^4*7^2*11^2*17*19*41*131      k=5
2^5*3*7                                k=3
2^5*3^2*7*11                           k=3
2^5*3^3*5*7                            k=4
2^5*3^3*5^2*7*29                       k=4
2^4*3*5*29                             k=3
2^4*3^2*5*11*29                        k=3
2^4*3^2*5*11^2*29*131                  k=3
2^4*3^4*5*7*17*29                      k=4
2^3*3*5*29                             k=3
2^3*3^2*5*11*29                        k=3
2^3*3^2*5*11^2*29*131                  k=3
2^3*3^4*5*7*17*29                      k=4
2^2*3*5                                 k=2
2^2*3*5^2*29                            k=2
2^2*3^2*5*11                            k=2
2^2*3^2*5^2*11*29                       k=2
2^2*3^2*5*11^2*131                       k=2
2^2*3^2*5^2*11^2*29*131                  k=2
2*3                                       k=2
2*3^2*11                                  k=2
2*3^2*11^2*131                            k=2

[SEPSigma Multiply Perfect Number]

SEPSigma(m)=k*m , for some integer k.

2^10*3^5*5*7*11^2*19*23*89       k=2
2^9*3*5*17*1021                  k=1
2^8*3^5*5*7*73*181               k=2
2^8*3^6*7^2*13*19*1093           k=3
2^7*3^5*5*7*11^2*19*23*181       k=2
2^6*3^5*5*7*127*181              k=2
2^6*3^6*7^2*13*19*127*1093       k=3
2^5*3*5*61                       k=1
2^4*3*5*31                       k=1
2^3*3*13                          k=1
2^2*3*7                           k=1
2*3                                k=1

Is the following conjecture correct?

"All SEPSigma Perfect Numbers are of the form 2^n*3^5*5*181*k, GCD(2*3*5*181,k)=1"

Yasutoshi

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