## OPN's don't exist

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• Group, I believe that (O)dd (P)erfect (N)umbers do NOT exist in the same manner as (E)ven (P)erfect (N)umbers do! If we allow the generic formula of
Message 1 of 2 , Sep 26, 2007
Group,

I believe that (O)dd (P)erfect (N)umbers do NOT exist
in the same manner as (E)ven (P)erfect (N)umbers do!

If we allow the generic formula of (k^(m-1))*(k^m -
(k-1)) to define the situ-ation, then we will see that
it encompasses both the proven EPNn == (2^(n-1))*(2^n
-1) and a closely-related OPNn == (3^(n-1))*(3^n -2).

However, we can't be satisfied with our find, since
these two instances of a preferred formula don't have
comparable 'pole' positions when 'n' is set equal to
zero.

If we let n= 0, then the 'pole' for EPN0 == (1/2) *(0)
is located at the origin, and the 'pole' of OPN0 ==
(1/3) *(-1) is shifted by a (-1). Comparing results
with respect to perfection between these two instances
is already suspect from the onset.

After a closer examination, though, the formula
(3^(n-1))*(3^(n+1) -2) turns out to be a more
appropriate function but with a little twist involved;
if we let n= 0, the 'pole' of OPN0 == (1/3) *(1) is
shifted by a (+1). We will use this fact later to
offset our closest OPN formula by a (-1) to bring the
'poles' into agreement.

Now, a 'proper-incipient' EPN is generated when the
exponent 'n' is less than the value inside the portion
of the (second)parenthesis of its generic formula; n=
1 implies that EPN1 == (1)*(1), but it's improper
since n= 1 is not less than the second (1); n= 2
implies that EPN2 == (2)*(3) and since n= 2 < (3), we
quickly discover that 6 = 1 *2 *3 = 1 +2 +3 = 6 (an
unshifted 'proper-incipient' characteristic that could
be shared by the latter OPN formula).

If n= 1, then the preferred OPN formula gives an OPN1
== (1)*(7) and 1 < (7). So, the OPN1 could share the
proper 'incipient' quality of its cousin EPN, but not
without an adjustment to its value with an offset of
(-1), allowing for a little twist on the OPN's
existance-- they're not only odd, they're quirky!

If we subtract '1' from the summand and allow the use
of an 'improper' divisor, then we have OPN1 == (1)*(7)
= (-1) + (1) +7 = 7 (offsetting the summand), and the
incipient OPN presents itself; but it wasn't created
using the expected 'EPN' axiom. It's the only OPN
attached to k= 3 from the preferred OPN formula, but
not the only OPN... if k= 5 and n= 1, then 21 isn't
the next OPN, and...

Finally, if all OPN's are == (2^(n-1))*(2^(n+1) -1),
or (3^(n-1))*(3^(n+1) -2), (5^(n-1)) *(5^(n+1) -4),or
(7^(n-1))*(7^(n+1) -6),or (11^(n-1))*(11^(n+1) -10),
or (13^(n-1))* (13^(n+1) -12), etc., then we can see
that the sequence of OPN's emerge as 3, 7, (not 21),
43, (not 111), 157, etc.

They are not governed by the same rules as their
cousin EPN and are not the ex-pected 'gigantic'
numbers that contain a multitude of factors, but
actually a set of individual (must be)prime numbers
that are generated by a continually changing formula
and by modifying the calculations of both the product
and sum-mations.

Thus, OPN's could barely exist but not in the usual
EPN sense... they're quirky!
At best, the term OPN is a fictional misnomer, because
all possible formulas for an OPN lack the offset that
is necessary to coincide with the zero 'pole' of an

Bill

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• Bill, Your analysis appears to be restricted to consideration of odd perfect numbers with only two distinct prime divisors. It s known that any odd perfect
Message 2 of 2 , Sep 26, 2007
Bill,

Your analysis appears to be restricted to consideration of odd perfect
numbers with only two distinct prime divisors. It's known that any odd
perfect must have are least nine distinct prime factors and at least
75 prime factors counting multiplicity. Wolfram and Wikipedia are
good starting points for learning what is known about the possible
existence of odd perfect numbers.

William Lipp
Poohbah of OddPerfect.org
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