A105565 a(n) = if (exactly 5 Fibonacci numbers exist with exactly n digits) then 1, otherwise 0.

0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1

Offset: 1

Author

Reinhard Zumkeller, Apr 14 2005

Keywords

• nonn
• base

Comments

The sequence is almost periodic, see also A105566;

a(n) = 1 - A105563(n), for n>1.

a(n) = A050815(n) - 4, for n>1. - Hans J. H. Tuenter, Jul 29 2025

Examples

			The Fibonacci numbers with two decimal digits are 13, 21, 34, 55, 89; a total of five, so that a(2)=1.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Igor Szczyrba, Rafał Szczyrba, and Martin Burtscher, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
• Jürgen Spilker, Die Ziffern der Fibonacci-Zahlen, Elemente der Mathematik 58 (Birkhäuser 2003).
• Eric Weisstein's World of Mathematics, Fibonacci Number
• Eric Weisstein's World of Mathematics, Almost Periodic Function

Crossrefs

Cf. A050815, A060384, A000045.

Programs

Maple

n:= 1: count:= 2: a:= 0: b:= 1:
for m from 2 while n < 101 do
  c:= b; b:= a+b; a:= c;
  s:= ilog10(b)+1;
  if s = n then count:= count+1
  else
    if count = 5 then A[n]:= 1 else A[n]:= 0 fi;
    count:= 1; n:= s
  fi
od:
seq(A[i],i=1..100); # Robert Israel, Dec 17 2018

Formula

For n>1, a(n) = [{n*alpha+beta}<{alpha}], where alpha=log(10)/log(phi), beta=log(5)/(2*log(phi)), [X] is the Iverson bracket, {x}=x-floor(x), denotes the fractional part of x, and phi=(1+sqrt(5))/2. - Hans J. H. Tuenter, Jul 29 2025