A061107 -id:A061107 - cp's OEIS frontend

A063896 a(n) = 2^Fibonacci(n) - 1.

0, 1, 1, 3, 7, 31, 255, 8191, 2097151, 17179869183, 36028797018963967, 618970019642690137449562111, 22300745198530623141535718272648361505980415

Offset: 0

Robert G. Wilson v, Aug 29 2001

The recurrence can also be written a(n)+1 = (a(n-1)+1)*(a(n-2)+1) or log_p(a(n)+1) = log_p(a(n-1)+1) + log_p(a(n-2)+1), respectively. Setting a(1)=p-1 for any natural p>1, it follows that log_p(a(n)+1)=Fibonacci(n). Hence any other sequence p^Fibonacci(n)-1 could also serve as a valid solution to that recurrence, depending only on the value of the term a(1). - Hieronymus Fischer, Jun 27 2007
Written in binary, a(n) contains Fibonacci(n) 1's. Thus the sequence converted to base-2 is A007088(a(n)) = 0, 1, 1, 11, 111, 11111, 11111111, ... . - Hieronymus Fischer, Jun 27 2007
In general, if b(n) is defined recursively by b(0) = p, b(1) = q, b(n) = b(n-1)*b(n-2) + b(n-1) + b(n-2) for n >= 2 then b(n) = p^Fibonacci(n-1) * q^Fibonacci(n) - 1. - Rahul Goswami, Apr 15 2020
a(n) is also the numerator of the continued fraction [2^F(0), 2^F(1), 2^F(2), 2^F(3), ..., 2^F(n-2)] for n>0. For the denominator, see A005203. - Chinmay Dandekar and Greg Dresden, Sep 19 2020

Alois P. Heinz, Table of n, a(n) for n = 0..18
M. Tamba and Y. S. Valaulikar, A nonlinear extension of Fibonacci sequence, Turkish Journal of Analysis and Number Theory, Vol. 4, No. 4, 2016.

Cf. A000045 (Fibonacci), A000225, A000301, A005203, A061107.

See A131293 for a base-10 analog with Fib(n) 1's.

a:= n-> 2^(<<0|1>, <1|1>>^n)[1,2]-1:
seq(a(n), n=0..15);  # Alois P. Heinz, Aug 12 2017

2^Fibonacci[Range[0,15]]-1 (* Harvey P. Dale, May 20 2014 *)
RecurrenceTable[{a[0] == 0, a[1] == 1, a[n] == (a[n - 1] + 1)*(a[n - 2] + 1) - 1}, a[n], {n, 0, 12}] (* Ray Chandler, Jul 30 2015 *)

a(n) = 2^fibonacci(n) - 1 \\ Charles R Greathouse IV, Oct 03 2016

The solution to the recurrence a(0) = 0; a(1) = 1; a(n) = a(n-1)*a(n-2) + a(n-1) + a(n-2).
a(n) = A000301(n) - 1. - R. J. Mathar, Apr 26 2007
a(n) = a(n-2)*2^ceiling(log_2(a(n-1))) + a(n-1) for n>1. - Hieronymus Fischer, Jun 27 2007
a(n) = A000225(A000045(n)). - Alois P. Heinz, Mar 19 2020

A144287 Square array A(n,k), n>=1, k>=1, read by antidiagonals: A(n,k) = Fibonacci rabbit sequence number n coded in base k.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 5, 3, 1, 4, 10, 22, 5, 1, 5, 17, 93, 181, 8, 1, 6, 26, 276, 2521, 5814, 13, 1, 7, 37, 655, 17681, 612696, 1488565, 21, 1, 8, 50, 1338, 81901, 18105620, 4019900977, 12194330294, 34, 1, 9, 65, 2457, 289045, 255941280, 1186569930001, 6409020585966267, 25573364166211253, 55

Offset: 1

Alois P. Heinz, Sep 17 2008

			Square array begins:
  1,   1,    1,     1,     1,  ...
  1,   2,    3,     4,     5,  ...
  2,   5,   10,    17,    26,  ...
  3,  22,   93,   276,   655,  ...
  5, 181, 2521, 17681, 81901,  ...

Alois P. Heinz, Antidiagonals n = 1..16, flattened
H. W. Gould, J. B. Kim and V. E. Hoggatt, Jr., Sequences associated with t-ary coding of Fibonacci's rabbits, Fib. Quart., 15 (1977), 311-318.

Columns k=1-10 give: A000045, A005203, A005205, A320986, A320987, A320988, A320989, A320990, A320991, A061107 and A036299.
Rows n=1-3 give: A000012, A001477, A002522.
Main diagonal gives A144288.

f:= proc(n,b) option remember; `if`(n<2, [n,n], [f(n-1, b)[1]*
       b^f(n-1, b)[2] +f(n-2, b)[1], f(n-1, b)[2] +f(n-2, b)[2]])
    end:
A:= (n,k)-> f(n,k)[1]:
seq(seq(A(n, 1+d-n), n=1..d), d=1..11);

f[n_, b_] := f[n, b] = If[n < 2, {n, n}, {f[n-1, b][[1]]*b^f[n-1, b][[2]] + f[n-2, b][[1]], f[n-1, b][[2]] + f[n-2, b][[2]]}]; t[n_, k_] := f[n, k][[1]]; Flatten[ Table[t[n, 1+d-n], {d, 1, 11}, {n, 1, d}]] (* Jean-François Alcover, translated from Maple, Dec 09 2011 *)

See program.

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.

A144288 Fibonacci rabbit sequence number n coded in base n, also diagonal of A144287.

Original entry on oeis.org

1, 2, 10, 276, 81901, 2247615258, 81658169024988865, 644986443956439734064225751112, 3427833941153173630835645403655873661712817810325122

Offset: 1

Alois P. Heinz, Sep 17 2008

Alois P. Heinz, Table of n, a(n) for n = 1..14

Cf. A000045, A005203, A005205, A061107, A036299, A144287.

f:= proc(n, b) option remember; `if`(n<2, [n, n], [f(n-1, b)[1] *b^f(n-1, b)[2] +f(n-2, b)[1], f(n-1, b)[2] +f(n-2, b)[2]]) end: a:= n-> f(n, n)[1]: seq(a(n), n=1..11);

f[n_, b_] := f[n, b] = If[n < 2, {n, n}, {f[n-1, b][[1]]*b^f[n-1, b][[2]] + f[n-2, b][[1]], f[n-1, b][[2]] + f[n-2, b][[2]]}]; a[n_] := f[n, n][[1]]; Table[a[n], {n, 1, 9}] (* Jean-François Alcover, Jan 03 2013, translated from Maple *)

See program.

A248425 Number of "squares" (repeated identical blocks) in the n-th Fibonacci word.

Original entry on oeis.org

0, 0, 0, 0, 1, 4, 11, 26, 57, 118, 235, 454, 857, 1588, 2899, 5228, 9333, 16520, 29031, 50702, 88077, 152290, 262239, 449930, 769461, 1312104, 2231591, 3786456, 6410857, 10832908, 18272195, 30769154, 51733857, 86859598, 145642579, 243907918, 408005393, 681773980, 1138094971, 1898045252, 3162632157, 5265345680

Offset: 1

Jeffrey Shallit, Oct 06 2014

Here the Fibonacci words are given by X_0 = 0, X_1 = 1, and X_n = X_{n-1} X_{n-2} where juxtaposition means concatenation.

			The 5th Fibonacci word is 10110101, which has the following four squares: 11 starting at position 3, 1010 at position 4, 0101 at position 5, and 101101 at position 1.

G. C. Greubel, Table of n, a(n) for n = 1..1000
A. S. Fraenkel and J. Simpson, The exact number of squares in Fibonacci words, Theoret. Comput. Sci. 218 (1999), 95-106. Note: the formula given there has a small error, which has been corrected below.
A. S. Fraenkel and J. Simpson, Corrigendum to “The exact number of squares in Fibonacci words”, Theoret. Comput. Sci. 218 (1999), 95-106.
Index entries for linear recurrences with constant coefficients, signature (4,-4,-2,4,0,-1).

Cf. A000045, A061107.

A248425:= func< n | n le 3 select 0 else (2/5)*(2*(n-6)*Fibonacci(n) - (n-5)*Fibonacci(n-1)) + n >;
[A248425(n): n in [1..50]]; // G. C. Greubel, Oct 02 2024

A248425[n_]:= 2*(2*(n-6)*Fibonacci[n] -(n-5)*Fibonacci[n-1])/5 +n +3*Boole[n ==1] + Boole[n==3];
Table[A248425[n], {n,50}] (* G. C. Greubel, Oct 02 2024 *)

def A248425(n): return (2/5)*(2*(n-6)*fibonacci(n) - (n-5)*fibonacci(n-1)) + n + 3*int(n==1) + int(n==3)
[A248425(n) for n in range(1,51)] # G. C. Greubel, Oct 02 2024

a(n) = (4/5)*(n-1)*F(n) - (2/5)*(n+5)*F(n-1) - 4*F(n-2) + n, for n >= 4, where F(n) = Fibonacci(n).
G.f.: x^5*(1-x^2+x^4)/((1-x)*(1-x-x^2))^2. - Colin Barker, Oct 07 2014
E.g.f.: 2*exp(x/2)*(5*(5 + 2*x)*cosh(sqrt(5)*x/2) - 29*sqrt(5)*sinh(sqrt(5)*x/2))/25 + x^3/6 + (3 + exp(x))*x - 2. - Stefano Spezia, May 23 2025

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 *)

A339713 a(n) = (a(n-2) concatenate a(n-1)) for n > 2, with a(1)=1, a(2)=10.

Original entry on oeis.org

1, 10, 110, 10110, 11010110, 1011011010110, 110101101011011010110, 1011011010110110101101011011010110, 1101011010110110101101011011010110110101101011011010110, 10110110101101101011010110110101101101011010110110101101011011010110110101101011011010110

Offset: 1

Wesley Ivan Hurt, Apr 24 2021

Number of digits in a(n) = A000045(n+1). - Michael S. Branicky, Apr 24 2021

a(n) and a(n+1) contain Fibonacci(n) 1's and Fibonacci(n) 0's respectively.

Cf. A000045, A111061 (in decimal), A061107, A131293.

def aupton(terms):
  alst = [1, 10]
  for n in range(3, terms+1): alst.append(int(str(alst[-2])+str(alst[-1])))
  return alst[:terms]
print(aupton(10)) # Michael S. Branicky, Apr 24 2021

cp's OEIS Frontend

A063896 a(n) = 2^Fibonacci(n) - 1.

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A144287 Square array A(n,k), n>=1, k>=1, read by antidiagonals: A(n,k) = Fibonacci rabbit sequence number n coded in base k.

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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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A144288 Fibonacci rabbit sequence number n coded in base n, also diagonal of A144287.

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A248425 Number of "squares" (repeated identical blocks) in the n-th Fibonacci word.

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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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Mathematica

A339713 a(n) = (a(n-2) concatenate a(n-1)) for n > 2, with a(1)=1, a(2)=10.

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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.