cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A072353 a(n) is the index of the largest Fibonacci number containing n digits.

Original entry on oeis.org

6, 11, 16, 20, 25, 30, 35, 39, 44, 49, 54, 59, 63, 68, 73, 78, 83, 87, 92, 97, 102, 106, 111, 116, 121, 126, 130, 135, 140, 145, 150, 154, 159, 164, 169, 173, 178, 183, 188, 193, 197, 202, 207, 212, 216, 221, 226, 231, 236, 240, 245, 250, 255, 260, 264, 269, 274
Offset: 1

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Author

Shyam Sunder Gupta, Jul 17 2002

Keywords

Comments

Partial sums of A050815: a(n) = Sum_{k=1..n} A050815(k). - Reinhard Zumkeller, Apr 14 2005
Equivalently, a(n) is the number of Fibonacci numbers < 10^n including F(0) = 0 and F(1) = F(2) = 1 once. - Derek Orr, Jun 01 2014

Examples

			a(3)=16, as the 16th Fibonacci number is the largest Fibonacci number with 3 digits.
		

Crossrefs

Programs

  • Mathematica
    With[{fibs=Fibonacci[Range[300]]},Flatten[Position[fibs,#]&/@ Table[ Max[ Select[fibs,IntegerLength[#]==n&]],{n,60}]]] (* Harvey P. Dale, Nov 09 2011 *)
  • Python
    def A072353_list(n):
        list = []
        x, y, index = 1, 1, 1
        while len(list) < n:
            if len(str(x)) < len(str(y)):
                list.append(index)
            x, y = y, x + y
            index += 1
        return list
    print(A072353_list(57)) # M. Eren Kesim, Jul 19 2021

Formula

Limit_{n->oo} a(n)/n = 1/log_10((1+sqrt(5))/2) = 1/A097348 = 4.784... . - Reinhard Zumkeller, Apr 14 2005.
a(n) = floor(n*log(10)/log(phi)+log(5)/(2*log(phi))), where phi=(1+sqrt(5))/2, the golden ratio. - Hans J. H. Tuenter, Jul 08 2025

Extensions

More terms from Reinhard Zumkeller, Apr 14 2005
Name edited by Michel Marcus, Jul 19 2021

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

Original entry on oeis.org

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

Views

Author

Reinhard Zumkeller, Apr 14 2005

Keywords

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

Crossrefs

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

A105566 Number of blocks of exactly 5 Fibonacci numbers having equal length <= n.

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 5, 5, 6, 7, 8, 9, 9, 10, 11, 12, 13, 13, 14, 15, 16, 16, 17, 18, 19, 20, 20, 21, 22, 23, 24, 24, 25, 26, 27, 27, 28, 29, 30, 31, 31, 32, 33, 34, 34, 35, 36, 37, 38, 38, 39, 40, 41, 42, 42, 43, 44, 45, 45, 46, 47, 48, 49, 49, 50, 51, 52, 53, 53, 54, 55, 56, 56, 57
Offset: 1

Views

Author

Reinhard Zumkeller, Apr 14 2005

Keywords

Comments

a(n) = Sum_{k=1..n} A105565(k); a(n) = n - A105564(n);
lim_{n->inf} a(n)/n = 1/log_10((1+sqrt(5))/2) - 4 = 0.784....

References

  • Juergen Spilker, Die Ziffern der Fibonacci-Zahlen, Elemente der Mathematik 58 (Birkhäuser 2003).

Crossrefs

Showing 1-3 of 3 results.