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.

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A003849 The infinite Fibonacci word (start with 0, apply 0->01, 1->0, take limit).

Original entry on oeis.org

0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1
Offset: 0

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Author

Keywords

Comments

A Sturmian word.
Define strings S(0)=0, S(1)=01, S(n)=S(n-1)S(n-2); iterate; sequence is S(infinity). If the initial 0 is omitted from S(n) for n>0, we obtain A288582(n+1).
The 0's occur at positions in A022342 (i.e., A000201 - 1), the 1's at positions in A003622.
Replace each run (1;1) with (1;0) in the infinite Fibonacci word A005614 (and add 0 as prefix) A005614 begins: 1,0,1,1,0,1,0,1,1,0,1,1,... changing runs (1,1) with (1,0) produces 1,0,0,1,0,1,0,0,1,0,0,1,... - Benoit Cloitre, Nov 10 2003
Characteristic function of A003622. - Philippe Deléham, May 03 2004
The fraction of 0's in the first n terms approaches 1/phi (see for example Allouche and Shallit). - N. J. A. Sloane, Sep 24 2007
The limiting mean and variance of the first n terms are 2-phi and 2*phi-3, respectively. - Clark Kimberling, Mar 12 2014, Aug 16 2018
Let S(n) be defined as above. Then this sequence is S(1) + Sum_{n=0..} S(n), where the addition of strings represents concatenation. - Isaac Saffold, May 03 2019
The word is a concatenation of three runs: 0, 1, and 00. The limiting proportions of these are respectively 1 - phi/2, 1/2, and (phi - 1)/2. The mean runlength is (phi + 1)/2. - Clark Kimberling, Dec 26 2010
From Amiram Eldar, Mar 10 2021: (Start)
a(n) is the number of the trailing 0's in the dual Zeckendorf representation of (n+1) (A104326).
The asymptotic density of the occurrences of k (0 or 1) is 1/phi^(k+1), where phi is the golden ratio (A001622).
The asymptotic mean of this sequence is 1/phi^2 (A132338). (End)

Examples

			The word is 010010100100101001010010010100...
Over the alphabet {a,b} this is a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, ...
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
  • Jean Berstel, Fibonacci words—a survey, In The book of L, pp. 13-27. Springer Berlin Heidelberg, 1986.
  • J. C. Lagarias, Number Theory and Dynamical Systems, pp. 35-72 of S. A. Burr, ed., The Unreasonable Effectiveness of Number Theory, Proc. Sympos. Appl. Math., 46 (1992). Amer. Math. Soc. - see p. 64.
  • Wolfdieter Lang, The Wythoff and the Zeckendorf representations of numbers are equivalent, in G. E. Bergum et al. (edts.) Application of Fibonacci numbers vol. 6, Kluwer, Dordrecht, 1996, pp. 319-337. [See A317208 for a link.]
  • G. Melançon, Factorizing infinite words using Maple, MapleTech journal, vol. 4, no. 1, 1997, pp. 34-42, esp. p. 36.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

Crossrefs

There are several versions of this sequence in the OEIS. This one and A003842 are probably the most important. See also A008352, A076662, A288581, A288582.
Positions of 1's gives A003622.
Sequences mentioned in the Allouche et al. "Taxonomy" paper, listed by example number: 1: A003849, 2: A010060, 3: A010056, 4: A020985 and A020987, 5: A191818, 6: A316340 and A273129, 18: A316341, 19: A030302, 20: A063438, 21: A316342, 22: A316343, 23: A003849 minus its first term, 24: A316344, 25: A316345 and A316824, 26: A020985 and A020987, 27: A316825, 28: A159689, 29: A049320, 30: A003849, 31: A316826, 32: A316827, 33: A316828, 34: A316344, 35: A043529, 36: A316829, 37: A010060.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

Programs

  • Haskell
    a003849 n = a003849_list !! n
    a003849_list = tail $ concat fws where
       fws = [1] : [0] : (zipWith (++) fws $ tail fws)
    -- Reinhard Zumkeller, Nov 01 2013, Apr 07 2012
    
  • Magma
    t1:=[ n le 2 select ["0","0,1"][n] else Self(n-1) cat "," cat Self(n-2) : n in [1..12]]; t1[12];
    
  • Maple
    z := proc(m) option remember; if m=0 then [0] elif m=1 then [0,1] else [op(z(m-1)),op(z(m-2))]; fi; end; z(12);
    M:=19; S[0]:=`0`; S[1]:=`01`; for n from 2 to M do S[n]:=cat(S[n-1], S[n-2]); od:
    t0:=S[M]: l:=length(t0); for i from 1 to l do lprint(i-1,substring(t0,i..i)); od: # N. J. A. Sloane, Nov 01 2006
  • Mathematica
    Nest[ Flatten[ # /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 10] (* Robert G. Wilson v, Mar 05 2005 *)
    Flatten[Nest[{#, #[[1]]} &, {0, 1}, 9]] (* IWABUCHI Yu(u)ki, Oct 23 2013 *)
    Table[Floor[(n + 2) #] - Floor[(n + 1) #], {n, 0, 120}] &[2 - GoldenRatio] (* Michael De Vlieger, Aug 15 2016 *)
    SubstitutionSystem[{0->{0,1},1->{0}},{0},{10}][[1]] (* Harvey P. Dale, Dec 20 2021 *)
  • PARI
    a(n)=my(k=2);while(fibonacci(k)<=n,k++);while(n>1,while(fibonacci(k--)>n,); n-=fibonacci(k)); n==1 \\ Charles R Greathouse IV, Feb 03 2014
    
  • PARI
    M3849=[2,2,1,0]/*L(k),S(k),L(k-1),S(k-1)*/; A003849(n)={while(n>M3849[1],M3849=vecextract(M3849,[1,2,1,2])+[M3849[3],M3849[4]<M. F. Hasler, Apr 07 2021
    
  • Python
    def fib(n):
        """Return the concatenation of A003849(0..F-1) where F is the smallest
           Fibonacci number > n, so that the result contains a(n) at index n."""
        a, b = '10'
        while len(b)<=n:
            a, b = b, b + a
        return b # Robert FERREOL, Apr 15 2016, edited by M. F. Hasler, Apr 07 2021
    
  • Python
    from math import isqrt
    def A003849(n): return 2-(n+2+isqrt(m:=5*(n+2)**2)>>1)+(n+1+isqrt(m-10*n-15)>>1) # Chai Wah Wu, Aug 25 2022

Formula

a(n) = floor((n+2)*r) - floor((n+1)*r) where r=phi/(1+2*phi) and phi is the Golden Ratio. - Benoit Cloitre, Nov 10 2003
a(n) = A003714(n) mod 2 = A014417(n) mod 2. - Philippe Deléham, Jan 04 2004
The first formula by Cloitre is just one of an infinite family of formulas. Using phi^2=1+phi, it follows that r=phi/(1+2*phi)=2-phi. Then from floor(-x)=-floor(x)-1 for non-integer x, it follows that a(n)=2-A014675(n)=2-(floor((n+2)* phi)-floor((n+1)*phi)). - Michel Dekking, Aug 27 2016
a(n) = 1 - A096270(n+1), i.e., A096270 is the complement of this sequence. - A.H.M. Smeets, Mar 31 2024

Extensions

Revised by N. J. A. Sloane, Jul 03 2012

A005614 The binary complement of the infinite Fibonacci word A003849. Start with 1, apply 0->1, 1->10, iterate, take limit.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0
Offset: 0

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Author

Keywords

Comments

Previous name was: The infinite Fibonacci word (start with 1, apply 0->1, 1->10, iterate, take limit).
Characteristic function of A022342. - Philippe Deléham, May 03 2004
a(n) = number of 0's between successive 1's (see also A003589 and A007538). - Eric Angelini, Jul 06 2005
With offset 1 this is the characteristic sequence for Wythoff A-numbers A000201=[1,3,4,6,...].
Eric Angelini's comment made me think that if 1 is defined to be the number of 0's between successive 1's in a string of 0's and 1's, then this string is 101. Applying the same operation to the digits of 101 leads to 101101, the iteration leads to successive palindromes of lengths given by A001911, up to a(n). - Rémi Schulz, Jul 06 2010
For generalized Fibonacci words see A221150, A221151, A221152, ... - Peter Bala, Nov 11 2013
The limiting mean of the first n terms is phi - 1; the limiting variance is phi (A001622). - Clark Kimberling, Mar 12 2014
Apply the difference operator to every column of the Wythoff difference array, A080164, to get an array of Fibonacci numbers, F(h). Replace each F(h) with h, and apply the difference operator to every column. In the resulting array, every column is A005614. - Clark Kimberling, Mar 02 2015
Binary expansion of the rabbit constant A014565. - M. F. Hasler, Nov 10 2018

Examples

			The infinite word is 101101011011010110101101101011...
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
  • G. Melançon, Factorizing infinite words using Maple, MapleTech journal, vol. 4, no. 1, 1997, pp. 34-42, esp. p. 36.

Crossrefs

Binary complement of A003849, which is the standard form of this sequence.
Two other essentially identical sequences are A096270, A114986.
Subwords: A178992, A171676.
Cf. A000045 (Fibonacci numbers), A001468, A001911, A005206 (partial sums), A014565, A014675, A022342, A036299, A044432, A221150, A221151, A221152.
Cf. A339051 (odd bisection), A339052 (even bisection).
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

Programs

  • Haskell
    a005614 n = a005614_list !! n
    a005614_list = map (1 -) a003849_list
    -- Reinhard Zumkeller, Apr 07 2012
    
  • Magma
    [Floor((n+1)*(-1+Sqrt(5))/2)-Floor(n*(-1+Sqrt(5))/2): n in [1..100]]; // Vincenzo Librandi, Jan 17 2019
    
  • Maple
    Digits := 50; u := evalf((1-sqrt(5))/2); A005614 := n->floor((n+1)*u)-floor(n*u);
  • Mathematica
    Nest[ Flatten[ # /. {0 -> {1}, 1 -> {1, 0}}] &, {1}, 10] (* Robert G. Wilson v, Jan 30 2005 *)
    Flatten[Nest[{#, #[[1]]} &, {1, 0}, 9]] (* IWABUCHI Yu(u)ki, Oct 23 2013 *)
    SubstitutionSystem[{0 -> {1}, 1 -> {1, 0}}, {1, 0}, 9] // Last (* Jean-François Alcover, Feb 06 2020 *)
  • PARI
    a(n,w1,s0,s1)=local(w2); for(i=2,n,w2=[ ]; for(k=1,length(w1),w2=concat(w2, if(w1[ k ],s1,s0))); w1=w2); w2
    for(n=2,10,print(n" "a(n,[ 0 ],[ 1 ],[ 1,0 ]))) \\ Gives successive convergents to sequence
    
  • PARI
    /* for m>=1 compute exactly A183136(m+1)+1 terms of the sequence */
    r=(1+sqrt(5))/2;v=[1,0];for(n=2,m,v=concat(v,vector(floor((n+1)/r),i,v[i]));a(n)=v[n];) /* Benoit Cloitre, Jan 16 2013 */
    
  • Python
    from math import isqrt
    def A005614(n): return (n+isqrt(m:=5*(n+2)**2)>>1)-(n+1+isqrt(m-10*n-15)>>1) # Chai Wah Wu, Aug 17 2022

Formula

Define strings S(0)=1, S(1)=10, thereafter S(n)=S(n-1)S(n-2); iterate. Sequence is S(oo). The individual S(n)'s are given in A036299.
a(n) = floor((n+2)*u) - floor((n+1)*u), where u = (-1 + sqrt(5))/2.
Sum_{n>=0} a(n)/2^(n+1) = A014565. - R. J. Mathar, Jul 19 2013
From Peter Bala, Nov 11 2013: (Start)
If we read the present sequence as the digits of a decimal constant c = 0.101101011011010 ... then we have the series representation c = Sum_{n >= 1} 1/10^floor(n*phi). An alternative representation is c = Sum_{n >= 1} 1/10^floor(n/phi) - 10/9.
The constant 9*c has the simple continued fraction representation [0; 1, 10, 10, 100, 1000, ..., 10^Fibonacci(n), ...]. See A010100.
Using this result we can find the alternating series representation c = 1/9 - 9*Sum_{n >= 1} (-1)^(n+1)*(1 + 10^Fibonacci(3*n+1))/( (10^(Fibonacci(3*n - 1)) - 1)*(10^(Fibonacci(3*n + 2)) - 1) ). The series converges very rapidly: for example, the first 10 terms of the series give a value for c accurate to more than 5.7 million decimal places. Cf. A014565. (End)
a(n) = A005206(n+1) - A005206(n). a(2*n) = A339052(n); a(2*n+1) = A339051(n+1). - Peter Bala, Aug 09 2022

Extensions

Corrected by Clark Kimberling, Oct 04 2000
Name corrected by Michel Dekking, Apr 02 2019

A014701 Number of multiplications to compute n-th power by the Chandah-sutra method.

Original entry on oeis.org

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

Views

Author

James Kilfiger (jamesk(AT)maths.warwick.ac.uk)

Keywords

Comments

In other words, number of steps to reach 1 starting from n and using the process: x -> x-1 if n is odd and x -> x/2 otherwise.
a(n) = number of 0's + twice number of 1's (disregarding the leading digit 1) in the binary expansion of n, i.e., A007088(n). - Lekraj Beedassy, May 28 2010
From Daniel Forgues, Jul 31 2012: (Start)
For the binary Fibonacci rabbits sequence (A036299) (cf. OEIS Wiki link below) we have the substitution/concatenation rule: a(n), n >= 3, may be obtained by the concatenation of a(n-1) and a(n-2), with a(1) = 0, a(2) = 1. Thus, using . (dot) as the concatenation operator, we have the recursive substitution/concatenation
a(n) = a(n-0)
a(n) = a(n-1).a(n-2)
a(n) = a(n-2).a(n-3).a(n-3).a(n-4)
a(n) = a(n-3).a(n-4).a(n-4).a(n-5).a(n-4).a(n-5).a(n-5).a(n-6)
which suggests the sequence
{0}
{1, 2}
{2, 3, 3, 4}
{3, 4, 4, 5, 4, 5, 5, 6}
whose concatenation gives A014701 (this sequence).
Number of multiplications to compute n-th power by the Chandah-sutra method, also called left-to-right binary exponentiation:
x^1 = x^( 1_2) = (x) (0 prod)
x^2 = x^( 10_2) = (x^2) (1 prod)
x^3 = x^( 11_2) = (x^2) * (x) (2 prod)
x^4 = x^( 100_2) = (x^2)^2 (2 prod)
x^5 = x^( 101_2) = (x^2)^2 * (x) (3 prod)
x^6 = x^( 110_2) = (x^2)^2 * (x^2) (3 prod)
x^7 = x^( 111_2) = (x^2)^2 * (x^2) * (x) (4 prod)
x^8 = x^(1000_2) = ((x^2)^2)^2 (3 prod) (End)
From Ya-Ping Lu, Mar 03 2021: (Start)
Index at which record m occurs is A052955(m).
First appearance of m in the sequence (or the record value m) is at n = 2^(m/2 + 1) - 1 for even m, and at n = 3*2^((m - 1)/2) - 1 for odd m.
The last appearance of m in the sequence is at n = 2^m. (End)
a(n) is the digit sum of n-1 in bijective base-2. Since the Fibonacci number F(m) can be defined as the number of ways to compose m as the sum of 1s and 2s, we get that m appears F(m) times in the sequence. - Oscar Cunningham, Apr 14 2024
Conjecture: a(n+1) is the minimal number of steps to go from 0 to n, by choosing before each step, after the first step, whether to keep the same step length or double it. The initial step length is 1. - Jean-Marc Rebert, May 15 2025

Examples

			5 -> 4 -> 2 -> 1 so 3 steps are needed to reach 1 hence a(5)=3; 9 -> 8 -> 4 -> 2 -> 1 hence a(9)=4.
		

Crossrefs

Programs

  • Haskell
    a014701 1 = 0
    a014701 n = a007953 $ a007931 (n - 1)
    -- Reinhard Zumkeller, Oct 26 2012
    
  • Maple
    A014701 := proc(n) local j,k; j := n; k := 0; while(j>1) do if j mod 2=1 then j := j-1 else j := j/2 fi; k := k+1 od end;
    # second Maple program:
    a:= n-> add(i+1, i=Bits[Split](n))-2:
    seq(a(n), n=1..128);  # Alois P. Heinz, Aug 30 2021
  • Mathematica
    a[n_] := DigitCount[n, 2] /. {x_, y_} -> 2x + y - 2; Array[a, 100] (* Robert G. Wilson v, Jul 31 2012 *)
  • PARI
    a(n)=hammingweight(n)+logint(n,2)-1 \\ Charles R Greathouse IV, Dec 29 2016
    
  • Python
    def a(n):
        if n==1:
            return 0
        return a(n//2)+1+n%2
    for i in range(1,60):
        print(a(i), end=", ")
    # Pablo Hueso Merino, Oct 28 2020

Formula

a(n) = A056792(n) - 1 = A056791(n) - 2.
a(n) = floor(log_2(n)) + (number of 1's in binary representation of n) - 1. - Corrected (- 1 at end) by Daniel Forgues, Aug 01 2012
a(2^n) = n, a(2^n-1) = 2*(n-1), and for n >= 2, log_2(n) <= a(n) <= 2*log_2(n) - 1. - Robert FERREOL, Oct 01 2014
Let u(1) = 1, u(2*n) = u(n)+1, u(2*n+1) = u(2*n)+1; then a(1) = 0 and a(n) = u(n-1). - Benoit Cloitre, Dec 19 2002
G.f.: -2/(1-x) + (1/(1-x)) * Sum_{k>=0} (2*x^2^k + x^2^(k+1))/(1+x^2^k). - Ralf Stephan, Aug 15 2003
From {0}, apply the substitution rule (n -> n+1, n+2) repeatedly, giving {{0}, {1, 2}, {2, 3, 3, 4}, {3, 4, 4, 5, 4, 5, 5, 6}, ...} and concatenate. - Daniel Forgues, Jul 31 2012
For n > 1: a(n) = A007953(A007931(n-1)). - Reinhard Zumkeller, Oct 26 2012
a(n) >= A003313(n). - Charles R Greathouse IV, Jan 03 2018
a(n) = a(floor(n/2)) + 1 + (n mod 2) for n > 1. - Pablo Hueso Merino, Oct 28 2020
a(n+1) = max_{1<=i<=n} (H(i) + H(n-i)) where H(n) denotes the Hamming weight of n (A000120(n)). See Lemma 8 in Gruber/Holzer 2021 article. - Hermann Gruber, Jun 26 2024

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

Views

Author

Alois P. Heinz, Sep 17 2008

Keywords

Examples

			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,  ...
		

Crossrefs

Rows n=1-3 give: A000012, A001477, A002522.
Main diagonal gives A144288.

Programs

  • Maple
    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);
  • Mathematica
    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 *)

Formula

See program.

A061107 a(0) = 0, a(1) = 1, a(n) is the concatenation of a(n-2) and a(n-1) for n > 1.

Original entry on oeis.org

0, 1, 10, 101, 10110, 10110101, 1011010110110, 101101011011010110101, 1011010110110101101011011010110110, 1011010110110101101011011010110110101101011011010110101
Offset: 0

Views

Author

Amarnath Murthy, Apr 20 2001

Keywords

Comments

Original name was: In the Fibonacci rabbit problem we start with an immature pair 'I' which matures after one season to 'M'. This mature pair after one season stays alive and breeds a new immature pair and we get the following sequence I, MI, MIM, MIMMI, MIMMIMIM, MIMMIMIMMIMMI... if we replace 'I' by a '0' and 'M' by a '1' we get the required binary sequence.

Examples

			a(0) = 0, a(1) = 1, a(2) = a(1)a(0)= 10, etc.
		

References

  • Amarnath Murthy, Smarandache Reverse auto correlated sequences and some Fibonacci derived sequences, Smarandache Notions Journal Vol. 12, No. 1-2-3, Spring 2001.
  • Ian Stewart, The Magical Maze.

Crossrefs

Cf. A063896, A131242. See A005203 for the sequence version converted to decimal.
Column k=10 of A144287.

Programs

  • Maple
    A[0]:= 0: A[1]:= 1: A[2]:= 10:
    for n from 3 to 20 do
    A[n]:= 10^(ilog10(A[n-2])+1)*A[n-1]+A[n-2]
    od:
    seq(A[n],n=0..10); # Robert Israel, Apr 30 2015
  • Mathematica
    nxt[{a_,b_}]:={b,FromDigits[Join[IntegerDigits[b],IntegerDigits[a]]]}; Transpose[NestList[nxt,{0,1},10]][[1]] (* Harvey P. Dale, Jul 05 2015 *)
  • PARI
    { default(realprecision, 100); L=log(10); for (n=0, 15, if (n>2, a=a1*10^(log(a2)\L + 1) + a2; a2=a1; a1=a, if (n==0, a=0, if (n==1, a=a2=1, a=a1=10))); write("b061107.txt", n, " ", a) ) } \\ Harry J. Smith, Jul 18 2009

Formula

a(0) = 0, a(1) =1, a(n) = concatenation of a(n-1) and a(n-2).
a(n) = a(n-1)*2^floor(log_2(a(n-2))+1)+a(n-2), for n>2, a(2)=10 (base 2). - Hieronymus Fischer, Jun 26 2007
a(n) = A036299(n-1), n>0. - R. J. Mathar, Oct 02 2008
a(n) can be transformed by a(n-1) when you change every single "1"(from a(n-1)) into "10" and every single "0"(from a(n-1)) into "1". [YuJiping and Sirius Caffrey, Apr 30 2015]

Extensions

More terms from Hieronymus Fischer, Jun 26 2007

A005205 Coding Fibonacci numbers.

Original entry on oeis.org

1, 3, 10, 93, 2521, 612696, 4019900977, 6409020585966267, 67040619014505181883304178, 1118048584563024433220786501983631190591549, 195042693446883195450571898296824337898272003171567594807867055549521
Offset: 1

Views

Author

Keywords

Comments

Binary Fibonacci (or rabbit) sequence A036299, read in base 3, then converted to decimal. - Jonathan Vos Post, Oct 19 2007

Examples

			a(0) = 1 because A036299(0) = "1" and 1 base 3 = 1 base 10.
a(1) = 3 because A036299(1) = "10" and 10 base 3 = 3 base 10.
a(2) = 10 because A036299(2) = "101" and 101 base 3 = 10 base 10.
a(3) = 93 because A036299(3) = "10110" and 10110 base 3 = 93 base 10.
a(4) = 2521 because A036299(4) = "10110101" and 10110101 base 3 = 2521 base 10.
a(5) = 612696 because A036299(5) = "1011010110110" and 1011010110110 base 3 = 612696 base 10.
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Column k=3 of A144287.

Programs

  • Maple
    b:= proc(n) option remember; `if` (n<2, [n, n], [b(n-1)[1] *3^b(n-1)[2] +b(n-2)[1], b(n-1)[2] +b(n-2)[2]]) end: a:= n-> b(n)[1]: seq(a(n), n=1..11);  # Alois P. Heinz, Sep 17 2008
  • Mathematica
    b[0] = {1}; b[1] = {1, 0}; b[n_] := b[n] = Join[b[n-1], b[n-2]]; a[n_] := FromDigits[b[n], 3]; Table[a[n], {n, 0, 10}] (* Jean-François Alcover, Apr 24 2014 *)

Extensions

More terms from Jonathan Vos Post, Oct 19 2007
Corrected (a(4) was missing) and extended by Alois P. Heinz, Sep 17 2008

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

Views

Author

Alois P. Heinz, Sep 17 2008

Keywords

Crossrefs

Programs

  • Maple
    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);
  • Mathematica
    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 *)

Formula

See program.

A065353 Decimal representation of palindromes extracted from the Golden String using ever increasing Fibonacci-style subdivisions.

Original entry on oeis.org

1, 0, 3, 2, 27, 90, 7003, 744282, 14687099739, 12786682083105626, 529158535306496354546309979, 7914572860144723898900437268660641289952090
Offset: 0

Views

Author

Patrick De Geest, Oct 31 2001

Keywords

Comments

A zero must be prefixed to the 2n (n>0) terms when converting back to binary.

Examples

			Bin (Dec) -> 1 (1); 0 (0); 11 (3); 010 (2); 11011 (27); 01011010 (90); 1101101011011 (7003); 010110101101101011010 (744282); etc.
		

Crossrefs

A065354 Decimal representation of binary palindromes extracted from the Golden String using ever-increasing Lucas-style subdivisions.

Original entry on oeis.org

1, 5, 6, 107, 730, 224091, 190536538, 120316721060699, 26815615903949132618586, 9090874414162652716478489106641017691, 285152539069955354427985396951391834474389843433339258362714
Offset: 1

Views

Author

Patrick De Geest, Oct 31 2001

Keywords

Comments

A zero must be prefixed to the 2n+1 terms (n>0) when converting back to binary.

Examples

			Bin (Dec) -> {10} skipped -> start: 1 (1); 101 (5); 0110 (6); 1101011 (107); 01011011010 (730); 110110101101011011 (224091); etc.
		

Crossrefs

A135500 Generating function for Viswanath's constant, using the golden string.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0
Offset: 0

Views

Author

Shane Findley, Feb 19 2008

Keywords

Comments

For each bit of the golden string that is a 1, write seven consecutive bits and for each 0 of the golden string write eight consecutive bits.
Alternate between 0 and 1. Exclude the first seven bits since we are based at zero, just like the golden string. Heads = 1, Tails = 0.

Examples

			Bit 1 of the golden string is a 1, so bits 8-14(seven consecutive) are the same. Bit 2 of the golden string is 0, so bits 15-22(eight consecutive) are the same. Odd bits of the golden string mean to write 1 and even bits of the golden string mean to write 0.
		

Crossrefs

Formula

Each power of Viswanath's constant (1.131988248...) And V^3 has a corresponding coin flip. Simply take the absolute values of the two possible(+Heads or -Tails) outcomes and choose the flip closest to the value of the power.
Showing 1-10 of 10 results.