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%I #3 Oct 02 2013 16:02:48
%S 2,8,80,792,7918,79182,791817,7918132
%N Number of Fibonacci numbers F(k), k <= 10^n, whose initial digit is 5.
%F Lim as n -> inf. a(n)/10^n=log(6/5), where the base is 10. - _Robert Gerbicz_, Sep 05 2002
%e a(2)=8 because there are 8 Fibonacci numbers up to 10^2 whose initial digit is 5.
%K base,nonn
%O 1,1
%A _Shyam Sunder Gupta_, Aug 15 2002
%E More terms from _Robert Gerbicz_, Sep 05 2002
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