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Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I)
#3
Oops, thanks

I am thinking something else and doing paralel calculations in my head, stupid mistake.

Anyway, thanks for the link, I will try the programm You suggested as i of course wanted to know distribution over values of z ( or real a) , but did not know how to get there.

Ivars
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RE: Calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/20/2008, 06:43 AM

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