For N, perfect number (PN), we have:
(1) h(N)=S(N)/N=2,
where S(N) is a sum off all divisors of N. According to Euler for n, odd PN
(OPN), we have:
(2) n=p1^k*m^2,
where p1 is a prime, p1=1(mod4), k=1(mod4).
May be there are no OPNs, but exist n*s,
approximately OPNs (AOPNs), complying with (2) and with h(n*), which very
slightly differs from 2.
Carl Pomerance proved that there are infinitely many AOPNs such, that with
increasing r
lim h(n*)=2, where r is number of different prime divisors of AOPN (NT
archives, 2006, Sept, #5Re1).
But the proof is a pure theorem of existence and will be more interest to
propose the convergence procedure for searching for sequence of such AOPNs.
May be useable the most simple procedure. Let us consider the example.
For r=8, p1=5 we have 29-zero AOPN:
(3) h(3^2*5^1*13^26*17^24*419^12*22717^6*169853617^4*1603344813625411^2)=
2.0000000000000000000000000000079…
For r=9 on the next step we have 57-zero AOPN:
(4) h(3^2*5^1*13^50*17^46*419^22*22717^14*169853617^6*1603344813625447^4*
114717727802748788483526000271^2)=
2.00000000000000000000000000000000000000000000000000000000023…
It is clear that sufficiently to consider constant or infinite divisor
exponents. Note qi the i-th prime divisor of AOPN, then in the common
shape we have:
(5) h1=h(3^2*5^1*13^inf*…*qr^inf)=h0*h(qr^inf)>2,
where for some m qr=tm is m-th prime number. We use qr such that take place
inequality:
(6) h2=h0*h(tm+1^inf)<2
On the next step we use tm+j (j>1) such that
(7) h3=h0*h(tm+1^inf*tm+j^inf)>2 and
(8) h4=h0*h(tm+1^inf*tm+j+1^inf)<2.
Then, using (8) and (5) we have
(9) h3/h1= h(tm+1^inf*tm+j^inf)/h(tm^inf)< h(tm+1^inf*tm+j+1^inf)/
h(tm^inf)= 2/h0*h(tm^inf)<1.
If h3=h0*h(tm+1^inf*tm+2^inf)<2, then instead tm+2 we can use
tm+2*tm+3 or tm+2*tm+3*tm+4,… and so on until h3>2. It is
possible because with increasing m the sum 1/pm+1/pm+2+…
infinitely grow and we’ll have:
(10) h(tm+1^inf*tm+2^inf*tm+3^inf…)=[1+1/(tm+1-1)]*[1+1/(tm+2-1)]*[1+1/
(tm+3-1)]…
But whether exist or not the case h3=h0*h(tm+1^inf*tm+2^inf)<2?
