A108047 -id:A108047 - cp's OEIS frontend

A214679 A(n,k) = Fibonacci(n) represented in bijective base-k numeration; square array A(n,k), n>=1, k>=1, read by antidiagonals.

1, 1, 1, 1, 1, 11, 1, 1, 2, 111, 1, 1, 2, 11, 11111, 1, 1, 2, 3, 21, 11111111, 1, 1, 2, 3, 12, 112, 1111111111111, 1, 1, 2, 3, 11, 22, 221, 111111111111111111111, 1, 1, 2, 3, 5, 14, 111, 1221, 1111111111111111111111111111111111

Offset: 1

Alois P. Heinz, Jul 25 2012

The digit set for bijective base-k numeration is {1, 2, ..., k}.

			Square array A(n,k) begins:
:                     1,    1,   1,   1,   1,  1,  1,  1,  1, ...
:                     1,    1,   1,   1,   1,  1,  1,  1,  1, ...
:                    11,    2,   2,   2,   2,  2,  2,  2,  2, ...
:                   111,   11,   3,   3,   3,  3,  3,  3,  3, ...
:                 11111,   21,  12,  11,   5,  5,  5,  5,  5, ...
:              11111111,  112,  22,  14,  13, 12, 11,  8,  8, ...
:         1111111111111,  221, 111,  31,  23, 21, 16, 15, 14, ...
: 111111111111111111111, 1221, 133, 111,  41, 33, 27, 25, 23, ...

Alois P. Heinz, Antidiagonals n = 1..13
R. R. Forslund, A logical alternative to the existing positional number system, Southwest Journal of Pure and Applied Mathematics, Vol. 1, 1995, 27-29.
Eric Weisstein's World of Mathematics, Zerofree
Wikipedia, Bijective numeration

Columns k=1-9 give: A108047, A085652, A282234, A282235, A282236, A282237, A282238, A282239, A282240.

Cf. A000045, A214676.

with(combinat):
A:= proc(n, b) local d, l, m; m:= fibonacci(n); l:= NULL;
      while m>0 do  d:= irem(m, b, 'm');
        if d=0 then d:=b; m:=m-1 fi;
        l:= d, l
      od; parse(cat(l))
    end:
seq(seq(A(n, 1+d-n), n=1..d), d=1..10);

A[n_, b_] := Module[{d, l, m}, m = Fibonacci@n; l = Nothing; While[m > 0, {m, d} = QuotientRemainder[m, b]; If[d == 0, d = b; m--]; l = {d, l}]; FromDigits @ Flatten @ l];
Table[A[n, d-n+1], {d, 1, 10}, {n, 1, d}] // Flatten (* Jean-François Alcover, May 28 2019, from Maple *)

A(n,k) = A214676(A000045(n),k).

A131293 Concatenate a(n-2) and a(n-1) to get a(n); start with a(0)=0, a(1)=1, delete the leading zero. Also: concatenate Fibonacci(n) 1's.

Original entry on oeis.org

0, 1, 1, 11, 111, 11111, 11111111, 1111111111111, 111111111111111111111, 1111111111111111111111111111111111, 1111111111111111111111111111111111111111111111111111111

Offset: 0

Hieronymus Fischer, Jun 26 2007

Interpreted as base-2 numbers the result is A063896.

This sequence differs from A108047 by the leading a(0) = 0. - Jason Kimberley, Dec 15 2012

			a(3)=11, a(4)=111, so a(5) = a(4)*a(3) = 11111.

Vincenzo Librandi, Table of n, a(n) for n = 0..18

Cf. A000045, A061107.

Cf. A037842, A108047.

import Data.Function (on)
a131293 n = a131293_list !! n
a131293_list = 0 : 1 : map read
               (zipWith ((++) `on` show) a131293_list $ tail a131293_list)
-- Reinhard Zumkeller, Oct 05 2015

[(10^Fibonacci(n)-1)/9: n in [0..10]]; // Vincenzo Librandi, Aug 29 2011

a:= n-> parse(cat(0, 1$combinat[fibonacci](n))):
seq(a(n), n=0..11);  # Alois P. Heinz, Apr 17 2020

Join[{0},FromDigits/@(PadLeft[{},#,1]&/@Fibonacci[Range[10]])] (* Harvey P. Dale, Aug 28 2011 *)

a(n) = a(n-2)*10^ceiling(log_10(a(n-1))) + a(n-1) for n > 1.

a(n) = (10^Fibonacci(n) - 1)/9.

A037842 Fibonacci numbers in base 1.

Original entry on oeis.org

1, 11, 111, 11111, 11111111, 1111111111111, 111111111111111111111, 1111111111111111111111111111111111, 1111111111111111111111111111111111111111111111111111111

Offset: 0

N. J. A. Sloane, Feb 27 2000

Cf. A108047, A131293.

FromDigits[Table[1,{#}]]&/@Fibonacci[Range[2,10]] (* Harvey P. Dale, Apr 06 2013 *)

a(n+2) = concat( a(n+1), a(n) ), a(0) = 1, a(1) = 11. - corrected by Jason Kimberley, Dec 17 2012

More terms from James Sellers, Mar 01 2000

A214326 Square array read by antidiagonals in which T(n,b) gives the n-th Fibonacci number written in base b with n,b >= 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 11, 1, 1, 10, 111, 1, 1, 2, 11, 11111, 1, 1, 2, 10, 101, 11111111, 1, 1, 2, 3, 12, 1000, 1111111111111, 1, 1, 2, 3, 11, 22, 1101, 111111111111111111111, 1, 1, 2, 3, 10, 20, 111, 10101, 1111111111111111111111111111111111, 1, 1, 2, 3, 5, 13, 31, 210, 100010

Offset: 1

Alois P. Heinz, Jul 24 2012

For b > 10, some terms cannot be properly notated using only decimal characters.

			Square array A(n,b) begins:
              1,    1,   1,  1,  1,  1,  1,  1,  1,  1,  1,  1, ...
              1,    1,   1,  1,  1,  1,  1,  1,  1,  1,  1,  1, ...
             11,   10,   2,  2,  2,  2,  2,  2,  2,  2,  2,  2, ...
            111,   11,  10,  3,  3,  3,  3,  3,  3,  3,  3,  3, ...
          11111,  101,  12, 11, 10,  5,  5,  5,  5,  5,  5,  5, ...
       11111111, 1000,  22, 20, 13, 12, 11, 10,  8,  8,  8,  8, ...
  1111111111111, 1101, 111, 31, 23, 21, 16, 15, 14, 13, 12, 11, ...

Alois P. Heinz, Antidiagonals n = 1..13

Columns b=1-10, 12, 13 give: A108047, A004685, A004686, A004687, A004688, A004689, A004690, A004691, A004692, A000045, A004693, A004694.

Cf. A037842, A131293.

A:= proc(n, b) local f, l; f:= combinat[fibonacci](n);
      if b=1 then parse(cat(1$f))
    else l:= NULL;
         while f>0 do l:= irem(f, b, 'f'), l od;
         parse(cat(l))
      fi
    end:
seq(seq(A(n, 1+d-n), n=1..d), d=1..10);

A121389 a(n) = 10^Fibonacci(n) - 1.

Original entry on oeis.org

0, 9, 9, 99, 999, 99999, 99999999, 9999999999999, 999999999999999999999, 9999999999999999999999999999999999, 9999999999999999999999999999999999999999999999999999999

Offset: 0

Rick L. Shepherd, Jul 26 2006

nonn

Each a(n) has Fibonacci(n) (trailing) 9's. In general, if the same recurrence below is used with any a(0), a(1) >= 0, then, for all k >= 2, a(k) has the same number of trailing 9's as a(k-2) and a(k-1) have altogether (see for example A121390).

Cf. A000045, A063896, A108047, A121390.

10^Fibonacci[Range[0,10]]-1 (* Harvey P. Dale, Dec 24 2022 *)

a(n) = 10^Fibonacci(n) - 1 = 10^A000045(n) - 1 (= 9*A108047(n) for n>=1). a(0) = 0; a(1) = 9; a(n) = a(n-2)*a(n-1) + a(n-2) + a(n-1).

A334025 a(0)=0, a(1)=1; and a(n) = {2a(n-2), 2a(n-1)}, where {x,y} is the concatenation of x and y.

Original entry on oeis.org

0, 1, 2, 24, 448, 48896, 89697792, 97792179395584, 179395584195584358791168, 195584358791168358791168391168717582336, 358791168391168717582336391168717582336717582336782337435164672

Offset: 0

Jamie Robert Creasey, Apr 14 2020

This sequence, due to the process of concatenating one number with another, bears similarities to A131293 and other familiar sequences. However, unlike A131293, this sequence increases at a faster rate. It happens due to the multiplier applied to the existing terms, which increases the number of digits present in the successive term drastically (see a(7) and a(8)). a(11) is too large to include here and has 102 digits.

			a(2) = {2*a(2-2), 2*a(2-1)} = {2*0, 2*1} = 02 = 2.
a(5) = {2*a(5-2), 2*a(5-1)} = {2*24, 2*448} = 48896.

Cf. A002605, A061107, A108047, A131293, A008352, A113526, A104458.

a[0] = 0; a[1] = 1; a[n_] := a[n] = FromDigits @ Join[IntegerDigits[2*a[n - 2]], IntegerDigits[2*a[n - 1]]]; Array[a, 11, 0] (* Amiram Eldar, Apr 18 2020 *)

cp's OEIS Frontend

A214679 A(n,k) = Fibonacci(n) represented in bijective base-k numeration; square array A(n,k), n>=1, k>=1, read by antidiagonals.

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A131293 Concatenate a(n-2) and a(n-1) to get a(n); start with a(0)=0, a(1)=1, delete the leading zero. Also: concatenate Fibonacci(n) 1's.

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A037842 Fibonacci numbers in base 1.

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A214326 Square array read by antidiagonals in which T(n,b) gives the n-th Fibonacci number written in base b with n,b >= 1.

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A121389 a(n) = 10^Fibonacci(n) - 1.

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A334025 a(0)=0, a(1)=1; and a(n) = {2*a(n-2), 2*a(n-1)}, where {x,y} is the concatenation of x and y.

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A334025 a(0)=0, a(1)=1; and a(n) = {2a(n-2), 2a(n-1)}, where {x,y} is the concatenation of x and y.