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Article #121

Is 2^(2^13466918-4)*(2^13466917-1) a big number?

#121
From: mountain man
Date: Tue, 18 Sep 2007 10:25
47 lines
3047 bytes
It has been conjectured (April 2004) that the largest known balanced*
number is the number: 2^(2^13466918-4)*(2^13466917-1).

This is a number with about 10^4053946 digits,
(not 4053946 digits but 10^4053946 digits!!!),
but perfectly balanced, having precisely
2^(2^13466918-4)*(2^13466917-1) ordered factorizations.



* BALANCED Number?

The ancient notion of aliquot type historically (Euclid) defined by 
summation of the component aliquots, and notes the three separate threads of 
number so formalised. Abundant numbers have the sum of their divisors 
greater than themselves, deficient numbers have the sum of their divisors 
less than themselves. In the middle, with the sum of their divisors exactly 
equal to the number are the "Perfect Numbers".

A redefinition in terms of the number of ordered factorisations (divisors) 
for any given number (rather than the summation of the divisors) is 
investigated.
That is, we define the following categories in regard to the number of 
ordered factorizations.
Let the number of ordered factors of n be H(n), then:


  a.. Deficient Ordered Factorizations ... H(n) < n
  Note that this set includes the primes for which by definition H(n) = 1.

  b.. Balanced Ordered Factorizations ... H(n) = n
  The term balanced was used in preference to "perfect".
  The first six balanced numbers are: 1, 48, 1280, 2496, 28672, 29808.
  These numbers are scant and reclusive like their cousins the Perfect 
Numbers. In the following analyses these are classed as abundant.

  c.. Abundant Ordered Factorizations ... H(n) > n
  The abundant numbers are those numbers for which
  the number of ordered factors is greater than the number itself.

For further information, and graphs, see this page:
http://www.mountainman.com.au/harmonics_01.htm

Best wishes,


P.R.F. Brown
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Subject: Is 2^(2^13466918-4)*(2^13466917-1) a big number?
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