A073560 - OEIS

%I #10 Dec 10 2024 10:02:11

%S 0,9,96,968,9690,96908,969101,9690998,96910014,969100129,9691001297

%N Number of Fibonacci numbers F(k), k <= 10^n, whose initial digit is 4.

%F Limit_{n->oo} a(n)/10^n = log(5/4), where the base is 10. - _Robert Gerbicz_, Sep 05 2002

%e a(2) = 9 because there are 9 Fibonacci numbers up to 10^2 whose initial digit is 4.

%o (PARI) lista(nn) = my(m=log(quadgen(5)), c=-1, d=log(10)/m, q, r=log(sqrt(5))/m, s=3-log(4)/m, t=4-log(5)/m, u=1); for(n=1, nn, u=10*u; until(u<r+=d, q=frac(r); if(q<s || q>t, c++)); print1(c - (q<s&&r-d+2>u || q>t&&r-d+3>u), ", ")); \\ _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