A105519 Number of times 9 is the leading digit of the first n+1 Fibonacci numbers in decimal representation.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5

Offset: 0

Author

Reinhard Zumkeller, Apr 11 2005

Keywords

• nonn
• base

Links

• Winston de Greef, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A000030, A000045, A008963, A105509.

Cf. A105511, A105512, A105513, A105514, A105515, A105516, A105517, A105518.

Programs

Mathematica

Table[If[First[IntegerDigits[Fibonacci[n]]]==9,1,0],{n,0,110}]// Accumulate (* Harvey P. Dale, Nov 27 2018 *)

PARI

(leadingdigit(n, b=10) = n \ 10^logint(n, b));
(isok(n) = leadingdigit(fibonacci(n))==9);
(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 18 2023

Formula

a(n) = #{k: A008963(k) = 9 and 0<=k<=n};

a(A105509(n)) = a(A105509(n) - 1) + 1;

n = A105511(n) + A105512(n) + A105513(n) + A105514(n) + A105515(n) + A105516(n) + A105517(n) + A105518(n) + a(n).

a(n) ~ (1 - log_10(9)) * n. - Amiram Eldar, Jan 12 2023