%I #24 Mar 18 2023 11:21:54
%S 0,0,0,0,1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,5,
%T 5,5,5,5,6,6,6,6,6,6,6,6,6,6,6,6,6,6,7,7,7,7,7,8,8,8,8,8,8,8,8,8,8,8,
%U 8,8,8,9,9,9,9,9,10,10,10,10,10,11,11,11,11,11,11,11,11,11,11,11,11,11,11
%N Number of times 3 is the leading digit of the first n+1 Fibonacci numbers in decimal representation.
%H Winston de Greef, <a href="/A105513/b105513.txt">Table of n, a(n) for n = 0..10000</a>
%F a(n) = #{k: A008963(k) = 3 and 0<=k<=n};
%F a(A105503(n)) = a(A105503(n) - 1) + 1;
%F n = A105511(n) + A105512(n) + a(n) + A105514(n) + A105515(n) + A105516(n) + A105517(n) + A105518(n) + A105519(n).
%F a(n) ~ log_10(4/3) * n. - _Amiram Eldar_, Jan 12 2023
%t Accumulate[Table[If[IntegerDigits[Fibonacci[n]][[1]] == 3, 1, 0], {n, 0, 100}]] (* _Amiram Eldar_, Jan 12 2023 *)
%o (PARI)
%o (leadingdigit(n, b=10) = n \ 10^logint(n, b));
%o (isok(n) = leadingdigit(fibonacci(n))==3);
%o (lista(n)=my(a=vector(1+n), r=0); for (i=1, n, r+=isok(i); a[1+i]=r); a) \\ _Winston de Greef_, Mar 17 2023
%Y Cf. A000030, A000045, A008963, A105503.
%Y Cf. A105511, A105512, A105514, A105515, A105516, A105517, A105518, A105519.
%K nonn,base
%O 0,10
%A _Reinhard Zumkeller_, Apr 11 2005