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Jacob Fugal wrote:

> WARNING: Just more math geeking ahead. If you don't care for this
> section of the thread, just skim on past... :)
> 
> On 9/4/06, Michael Ulm <michael.ulm / isis-papyrus.com> wrote:
> 
>> Let the base be  b > 1, and the number x be
>> x   + b * v + b^2 * w,
>> with 0 < , v, w < b, and w > 0.
>> Then
>>
>> x - g(x)   + b * v + b^2 * w - (u^2 + v^2 + w^2)
>>     (1 - u) + v * (b - v) + w * (b^2 - w)
>>   > u (1 - u) + b^2 - 1
>>   > (b - 1) (2 - b) + b^2 - 1   * (b - 1) > 0
> 
> 
> A nitpick, it should be:
> 
>  u (1 - u) + v * (b - v) + w * (b^2 - w) >  (1 - u) + b^2 - 1
> 
> rather than a strict less than. Doesn't affect the outcome of the
> proof, since the following inequality is still correct. In the case
> where v   and w  
--snip more explanations--

You are right, thanks for the correction and explanations.

Some other simple facts:

If x is a two-digit number in base b then  g(x) < 2 * b^2,
so if g(x) is a three-digit number in base b, the most
significant digit is always one. This also means that all
cycles in base b consist of numbers less than 2 * b^2.

No odd number b is a happy base, because the number
x  b + 1)**2 / 2  is a fixed point of g; i.e.
g(x)  .

The only happy bases < 700 are 2 and 4.

To get more on topic for this list, I have attached a ruby
script that computes all cycles for any given base.

Regards,

Michael

-- 
Michael Ulm
R&D Team
ISIS Information Systems Austria
tel: +43 2236 27551-219, fax: +43 2236 21081
e-mail: michael.ulm / isis-papyrus.com
Visit our Website: www.isis-papyrus.com

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