1,3

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: A276464 A001005 A009735 * A283511 A137095 A092097

Adjacent sequences: A177242 A177243 A177244 * A177246 A177247 A177248

nonn,base

Carmine Suriano, May 06 2010

approved