[FOM] Continuous tower function

Joe Shipman JoeShipman at aol.com
Tue Jan 15 10:19:59 EST 2013

I asked the following question here a few years ago but never got an answer. Has anyone ever addressed this question with enough specificity that a numerical answer (precise to the nearest integer) can be obtained for f(3.5)?
Let f(0)=1, f(n+1)=2^f(n). Thus f(0)=1, f(1)=2, f(2)=4, f(3)=16, f(4)=65536, etc. 
What is the "natural" value for f(3.5)? (to the nearest integer if this is easier, but an algorithm for calculating arbitrarily close approximations would be preferable).
-- JS

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