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

1,1

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

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

Mathematica

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

Crossrefs

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

Sequence in context: A368913 A354820 A188086 * A188291 A287372 A188221

Adjacent sequences: A105560 A105561 A105562 * A105564 A105565 A105566

Keyword

nonn,base,changed

Author

Reinhard Zumkeller, Apr 14 2005

Status

approved