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OPN's don't exist

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  • Bill Bouris
    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
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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 (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
      already proven EPN formula.

      Bill



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    • elevensmooth
      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
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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 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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