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%I #8 Dec 10 2024 16:51:51
%S 0,4,45,456,4575,45758,457571,4575752,45757486,457574900,4575749050
%N Number of Fibonacci numbers F(k), k <= 10^n, whose initial digit is 9.
%F Limit_{n->oo} a(n)/10^n = log(10/9), where the base is 10. - _Robert Gerbicz_, Sep 05 2002
%e a(2) = 4 because there are 4 Fibonacci numbers up to 10^2 whose initial digit is 9.
%o (PARI) lista(nn) = my(m=log(quadgen(5)), c=0, d=log(10)/m, q, r=log(sqrt(5))/m, s=5-log(9)/m, t=5-d, u=-3); for(n=1, nn, u=10*u+36; until(u<r+=d, q=frac(r); if(q<s && q>t, c++)); print1(c, ", ")); \\ _Jinyuan Wang_, Dec 09 2024
%K nonn,base,more
%O 1,2
%A _Shyam Sunder Gupta_, Aug 15 2002
%E More terms from _Robert Gerbicz_, Sep 05 2002
%E a(9)-a(11) from _Sean A. Irvine_ and _Jinyuan Wang_, Dec 10 2024