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%I #12 Mar 01 2023 14:08:48
%S 3,30,301,3011,30103,301031,3010300,30103001,301029995,3010299957
%N Number of Fibonacci numbers F(k), k <= 10^n, whose initial digit is 1.
%F Limit_{n->infinity} a(n)/10^n = log(2), where the base is 10. - _Robert Gerbicz_, Sep 05 2002
%e a(2) = 30 because there are 30 Fibonacci numbers up to 10^2 whose initial digit is 1.
%o (PARI) default(realprecision, 10^4); m=log((1+sqrt(5))/2);
%o lista(nn) = {my(d=log(10)/m, r=log(sqrt(5))/m, s=log(5-sqrt(5))/m, t=0, u=1); for(n=1, nn, u=10*u; while(s<u, if(floor(r+=d)==floor(s+=d), t++, t+=2)); print1(t+(r<u||floor(r)==floor(s)), ", ")); } \\ _Jinyuan Wang_, Feb 21 2020
%Y Cf. A000045, A047855 (numbers of integers <= 10^n, whose initial digit is 1).
%K nonn,base,more
%O 1,1
%A _Shyam Sunder Gupta_, Aug 15 2002
%E More terms from _Robert Gerbicz_, Sep 05 2002
%E a(9)-a(10) from _Jinyuan Wang_, Feb 21 2020
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