Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I)
#7
I added one more I on top, so now iteration becomes:

z_0=I^I

z=c^z

I got a symmetric picture of fish in large scale.
   

Ivars
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RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/20/2008, 09:08 PM

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