A105563 a(n) = if (exactly 4 Fibonacci numbers exist with exactly n digits) then 1, otherwise 0.

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

Offset: 1

Author

Reinhard Zumkeller, Apr 14 2005

Keywords

• nonn
• base
• changed

Comments

The sequence is almost periodic, see also A105564;

Asymptotically, a fraction of 1-alpha=0.215028... of the terms are 1. For the partial sums S(n) = Sum_{k=1..n} a(k), this implies S(n)~(1-alpha)*n. Conjecture: -beta < S(n)-(1-alpha)*n < 1-beta. The constants alpha and beta are as defined in the formula section. - Hans J. H. Tuenter, Aug 28 2025

Links

• Hans J. H. Tuenter, Table of n, a(n) for n = 1..10000
• Jürgen Spilker, Die Ziffern der Fibonacci-Zahlen, Elemente der Mathematik, 58(1):26-33, 2003.
• Eric Weisstein's World of Mathematics, Fibonacci Number
• Eric Weisstein's World of Mathematics, Almost Periodic Function

Crossrefs

Cf. A000045, A050815, A060384, A105565, A001622.

Programs

Mathematica

If[#==4,1,0]&/@Tally[IntegerLength/@Fibonacci[Range[500]]][[;;,2]] (* Harvey P. Dale, Nov 15 2023 *)

Formula

a(n) = 1 - A105565(n), for n>1.

a(n) = 5 - A050815(n), for n>1. - Hans J. H. Tuenter, Aug 28 2025

For n>1, a(n) = [{n*alpha+beta}>alpha], where alpha=log(10)/log(phi)-4, beta=log(5)/(2*log(phi))-1, [] is the Iverson bracket, {x}=x-floor(x), denotes the fractional part of x, and phi = (1+sqrt(5))/2 = A001622. - Hans J. H. Tuenter, Aug 28 2025