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 1-10 of 10 results.

A105501 Numbers n such that 1 is the leading digit of the n-th Fibonacci number in decimal representation.

Original entry on oeis.org

1, 2, 7, 12, 17, 21, 22, 26, 27, 31, 36, 40, 41, 45, 46, 50, 55, 60, 64, 65, 69, 70, 74, 79, 84, 88, 89, 93, 94, 98, 103, 107, 108, 112, 113, 117, 122, 127, 131, 132, 136, 137, 141, 146, 151, 155, 156, 160, 161, 165, 170, 174, 175, 179, 180, 184, 189, 194, 198, 199
Offset: 1

Views

Author

Reinhard Zumkeller, Apr 11 2005

Keywords

Comments

A008963(a(n)) = 1; A105511(a(n)) = A105511(a(n) - 1) + 1.

Examples

			a(10)=31: A008963(31) = A000030(A000045(31)) =
A000030(1346269) = 1.
		

Crossrefs

Programs

  • Maple
    filter:= proc(n) local t;
      t:= combinat:-fibonacci(n);
      t < 2*10^ilog10(t)
    end proc:
    select(filter, [$1..200]); # Robert Israel, May 02 2018
  • Mathematica
    fQ[n_] := IntegerDigits[Fibonacci[n]][[1]] == 1; Select[Range@200, fQ] (* Robert G. Wilson v, May 02 2018 *)
  • PARI
    is(n)=digits(fibonacci(n))[1]==1 \\ Charles R Greathouse IV, Oct 07 2016

Formula

a(n) ~ kn by the equidistribution theorem, where k = log(10)/log(2) = 3.321928.... - Charles R Greathouse IV, Oct 07 2016

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

Original entry on oeis.org

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

Views

Author

Reinhard Zumkeller, Apr 11 2005

Keywords

Crossrefs

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

A105512 Number of times 2 is the leading digit of the first n+1 Fibonacci numbers in decimal representation.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 16
Offset: 0

Views

Author

Reinhard Zumkeller, Apr 11 2005

Keywords

Crossrefs

Programs

  • Mathematica
    Accumulate[Table[If[IntegerDigits[Fibonacci[n]][[1]] == 2, 1, 0], {n, 0, 100}]] (* Amiram Eldar, Jan 12 2023 *)
  • PARI
    (leadingdigit(n, b=10) = n \ 10^logint(n, b));
    (isok(n) = leadingdigit(fibonacci(n))==2);
    (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 17 2023

Formula

a(n) = #{k: A008963(k) = 2 and 0<=k<=n};
a(A105502(n)) = a(A105502(n) - 1) + 1;
n = A105511(n) + a(n) + A105513(n) + A105514(n) + A105515(n) + A105516(n) + A105517(n) + A105518(n) + A105519(n).
a(n) ~ log_10(3/2) * n. - Amiram Eldar, Jan 12 2023

A105513 Number of times 3 is the leading digit of the first n+1 Fibonacci numbers in decimal representation.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11
Offset: 0

Views

Author

Reinhard Zumkeller, Apr 11 2005

Keywords

Crossrefs

Programs

  • Mathematica
    Accumulate[Table[If[IntegerDigits[Fibonacci[n]][[1]] == 3, 1, 0], {n, 0, 100}]] (* Amiram Eldar, Jan 12 2023 *)
  • PARI
    (leadingdigit(n, b=10) = n \ 10^logint(n, b));
    (isok(n) = leadingdigit(fibonacci(n))==3);
    (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 17 2023

Formula

a(n) = #{k: A008963(k) = 3 and 0<=k<=n};
a(A105503(n)) = a(A105503(n) - 1) + 1;
n = A105511(n) + A105512(n) + a(n) + A105514(n) + A105515(n) + A105516(n) + A105517(n) + A105518(n) + A105519(n).
a(n) ~ log_10(4/3) * n. - Amiram Eldar, Jan 12 2023

A105514 Number of times 4 is the leading digit of the first n+1 Fibonacci numbers in decimal representation.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9
Offset: 0

Views

Author

Reinhard Zumkeller, Apr 11 2005

Keywords

Crossrefs

Programs

  • Mathematica
    Accumulate[Table[If[IntegerDigits[Fibonacci[n]][[1]] == 4, 1, 0], {n, 0, 100}]] (* Amiram Eldar, Jan 12 2023 *)
  • PARI
    (leadingdigit(n, b=10) = n \ 10^logint(n, b));
    (isok(n) = leadingdigit(fibonacci(n))==4);
    (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 17 2023

Formula

a(n) = #{k: A008963(k) = 4 and 0<=k<=n};
a(A105504(n)) = a(A105504(n) - 1) + 1;
n = A105511(n) + A105512(n) + A105513(n) + a(n) + A105515(n) + A105516(n) + A105517(n) + A105518(n) + A105519(n).
a(n) ~ log_10(5/4) * n. - Amiram Eldar, Jan 12 2023

A105515 Number of times 5 is the leading digit of the first n+1 Fibonacci numbers in decimal representation.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 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, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9
Offset: 0

Views

Author

Reinhard Zumkeller, Apr 11 2005

Keywords

Crossrefs

Programs

  • Mathematica
    Accumulate[If[First[IntegerDigits[#]]==5,1,0]&/@Fibonacci[Range[0,110]]] (* Harvey P. Dale, Nov 02 2014 *)
  • PARI
    (leadingdigit(n, b=10) = n \ 10^logint(n, b));
    (isok(n) = leadingdigit(fibonacci(n))==5);
    (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 17 2023

Formula

a(n) = #{k: A008963(k) = 5 and 0<=k<=n};
a(A105505(n)) = a(A105505(n) - 1) + 1;
n = A105511(n) + A105512(n) + A105513(n) + A105514(n) + a(n) + A105516(n) + A105517(n) + A105518(n) + A105519(n).
a(n) ~ log_10(6/5) * n. - Amiram Eldar, Jan 12 2023

A105516 Number of times 6 is the leading digit of the first n+1 Fibonacci numbers in decimal representation.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 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, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6
Offset: 0

Views

Author

Reinhard Zumkeller, Apr 11 2005

Keywords

Crossrefs

Programs

  • Mathematica
    Prepend[Accumulate[If[First[IntegerDigits[#]]==6,1,0]&/@Fibonacci[ Range[ 110]]],0] (* Harvey P. Dale, Feb 18 2011 *)
  • PARI
    (leadingdigit(n, b=10) = n \ 10^logint(n, b));
    (isok(n) = leadingdigit(fibonacci(n))==6);
    (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 17 2023

Formula

a(n) = #{k: A008963(k) = 6 and 0<=k<=n};
a(A105506(n)) = a(A105506(n) - 1) + 1;
n = A105511(n) + A105512(n) + A105513(n) + A105514(n) + A105515(n) + a(n) + A105517(n) + A105518(n) + A105519(n).
a(n) ~ log_10(7/6) * n. - Amiram Eldar, Jan 12 2023

A105517 Number of times 7 is the leading digit of the first n+1 Fibonacci numbers in decimal representation.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 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, 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, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
Offset: 0

Views

Author

Reinhard Zumkeller, Apr 11 2005

Keywords

Crossrefs

Programs

  • Mathematica
    Accumulate[Table[If[IntegerDigits[Fibonacci[n]][[1]]==7,1,0],{n,0,120}]] (* Harvey P. Dale, Apr 29 2018 *)
  • PARI
    (leadingdigit(n, b=10) = n \ 10^logint(n, b));
    (isok(n) = leadingdigit(fibonacci(n))==7);
    (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 17 2023

Formula

a(n) = #{k: A008963(k) = 7 and 0<=k<=n};
a(A105507(n)) = a(A105507(n) - 1) + 1;
n = A105511(n) + A105512(n) + A105513(n) + A105514(n) + A105515(n) + A105516(n) + a(n) + A105518(n) + A105519(n).
a(n) ~ log_10(8/7) * n. - Amiram Eldar, Jan 12 2023

A105518 Number of times 8 is the leading digit of the first n+1 Fibonacci numbers in decimal representation.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 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, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7
Offset: 0

Views

Author

Reinhard Zumkeller, Apr 11 2005

Keywords

Crossrefs

Programs

  • Mathematica
    Accumulate[Table[If[IntegerDigits[Fibonacci[n]][[1]] == 8, 1, 0], {n, 0, 100}]] (* Amiram Eldar, Jan 12 2023 *)
  • PARI
    (leadingdigit(n, b=10) = n \ 10^logint(n, b));
    (isok(n) = leadingdigit(fibonacci(n))==8);
    (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 17 2023

Formula

a(n) = #{k: A008963(k) = 8 and 0<=k<=n};
a(A105508(n)) = a(A105508(n) - 1) + 1;
n = A105511(n) + A105512(n) + A105513(n) + A105514(n) + A105515(n) + A105516(n) + A105517(n) + a(n) + A105519(n).
a(n) ~ log_10(9/8) * n. - Amiram Eldar, Jan 12 2023

A141053 Most-significant decimal digit of Fibonacci(5n+3).

Original entry on oeis.org

2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6, 7, 8, 8, 9, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6, 7, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2
Offset: 0

Views

Author

Paul Curtz, Aug 01 2008

Keywords

Comments

Leading digit of A134490(n).
From Johannes W. Meijer, Jul 06 2011: (Start)
The leading digit d, 1 <= d <= 9, of A141053 follows Benford’s Law. This law states that the probability for the leading digit is p(d) = log_10(1+1/d), see the examples.
We observe that the last digit of A134490(n), i.e. F(5*n+3) mod 10, leads to the Lucas sequence A000032(n) (mod 10), i.e. a repetitive sequence of 12 digits [2, 1, 3, 4, 7, 1, 8, 9, 7, 6, 3, 9] with p(0) = p(5) = 0, p(1) = p(3) = p(7) = p(9) = 1/6 and p(2) = p(4) = p(6) = p(8) = 1/12. This does not obey Benford’s Law, which would predict that the last digit would satisfy p(d) = 1/10, see the links. (End)

Examples

			From _Johannes W. Meijer_, Jul 06 2011: (Start)
d     p(N=2000) p(N=4000) p(N=6000) p(Benford)
1      0.29900   0.29950   0.30033   0.30103
2      0.17700   0.17675   0.17650   0.17609
3      0.12550   0.12525   0.12517   0.12494
4      0.09650   0.09675   0.09700   0.09691
5      0.07950   0.07950   0.07933   0.07918
6      0.06700   0.06675   0.06700   0.06695
7      0.05800   0.05825   0.05800   0.05799
8      0.05150   0.05125   0.05100   0.05115
9      0.04600   0.04600   0.04567   0.04576
Total  1.00000   1.00000   1.00000   1.00000 (End)
		

Crossrefs

Cf. A000045 (F(n)), A008963 (Initial digit F(n)), A105511-A105519, A003893 (F(n) mod 10), A130893, A186190 (First digit tribonacci), A008952 (Leading digit 2^n), A008905 (Leading digit n!), A045510, A112420 (Leading digit Collatz 3*n+1 starting with 1117065), A007524 (log_10(2)), A104140 (1-log_10(9)). - Johannes W. Meijer, Jul 06 2011

Programs

  • Maple
    A134490 := proc(n) combinat[fibonacci](5*n+3) ; end proc:
    A141053 := proc(n) convert(A134490(n),base,10) ; op(-1,%) ; end proc:
    seq(A141053(n),n=0..70) ; # R. J. Mathar, Jul 04 2011
  • Mathematica
    Table[IntegerDigits[Fibonacci[5n+3]][[1]],{n,0,70}] (* Harvey P. Dale, Jun 22 2025 *)

Formula

a(n) = floor(F(5*n+3)/10^(floor(log(F(5*n+3))/log(10)))). - Johannes W. Meijer, Jul 06 2011
For n>0, a(n) = floor(10^{alpha*n+beta}), where alpha=5*log_10(phi)-1, beta=log_10(1+2/sqrt(5)), {x}=x-floor(x) denotes the fractional part of x, log_10(phi) = A097348, and phi = (1+sqrt(5))/2 = A001622. - Hans J. H. Tuenter, Aug 27 2025

Extensions

Edited by Johannes W. Meijer, Jul 06 2011
Showing 1-10 of 10 results.