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Probability that Fib(n) contains no 3's decreases to zero as n goes to infinity. I suppose that the maximum number is Fib(223) having 47 digits, none of them being a "3".
Table of n, a(n) for n=1..31.
a(6)=21 since 21 is the 6th Fibonacci having no 3's
Cf. A000045, A177194, A177195, A177231, A176253
Sequence in context: A121568 A001005 A009735 * A137095 A092097 A195295
Adjacent sequences: A177242 A177243 A177244 * A177246 A177247 A177248
Carmine Suriano, May 06 2010