Search: a003849 -id:a003849
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A005614
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The binary complement of the infinite Fibonacci word A003849. Start with 1, apply 0->1, 1->10, iterate, take limit.
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+20
83
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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
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OFFSET
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0,1
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COMMENTS
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Previous name was: The infinite Fibonacci word (start with 1, apply 0->1, 1->10, iterate, take limit).
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
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
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REFERENCES
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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.
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LINKS
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Glen Joyce C. Dulatre, Jamilah V. Alarcon, Vhenedict M. Florida, and Daisy Ann A. Disu, On Fractal Sequences, DMMMSU-CAS Science Monitor (2016-2017) Vol. 15 No. 2, 109-113.
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FORMULA
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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.
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)
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EXAMPLE
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The infinite word is 101101011011010110101101101011...
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MAPLE
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Digits := 50; u := evalf((1-sqrt(5))/2); A005614 := n->floor((n+1)*u)-floor(n*u);
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MATHEMATICA
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Nest[ Flatten[ # /. {0 -> {1}, 1 -> {1, 0}}] &, {1}, 10] (* Robert G. Wilson v, Jan 30 2005 *)
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PROG
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(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 */
(Haskell)
a005614 n = a005614_list !! n
a005614_list = map (1 -) a003849_list
(Magma) [Floor((n+1)*(-1+Sqrt(5))/2)-Floor(n*(-1+Sqrt(5))/2): n in [1..100]]; // Vincenzo Librandi, Jan 17 2019
(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
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CROSSREFS
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Binary complement of A003849, which is the standard form of this sequence.
Cf. A000045 (Fibonacci numbers), A001468, A001911, A005206 (partial sums), A014565, A014675, A022342, A036299, A044432, A221150, A221151, A221152.
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
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A316342
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Fibonacci word A003849 with first two terms replaced by 2's.
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+20
23
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2, 2, 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
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OFFSET
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0,1
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COMMENTS
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A word that is morphic; but neither pure morphic, uniform morphic, primitive morphic, nor recurrent.
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LINKS
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CROSSREFS
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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.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A316825
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Fibonacci word A003849 with its initial term changed to 2.
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+20
23
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2, 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
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OFFSET
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0,1
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COMMENTS
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A word that is pure morphic, but neither uniform morphic, primitive morphic, nor recurrent.
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LINKS
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CROSSREFS
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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.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A085357
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Common residues of binomial(3n,n)/(2n+1) modulo 2: relates ternary trees (A001764) to the infinite Fibonacci word (A003849).
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+20
16
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1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
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OFFSET
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0,1
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COMMENTS
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The n-th runs of ones is given by: 3 - A003849(n) (infinite Fibonacci word) = A076662(n+1). Runs of zeros are given by: A085358 and are also directly related to the Fibonacci sequence. Coefficients of A(x)^3 are found in A085359.
a(n) = 0 iff some binary digit of n is 1 while the corresponding binary digit of 3*n is 0. - Robert Israel, Jul 12 2016
The Run Length Transform of [0,1,0,0,0,...], A063524, the characteristic function of 1. (See A227349 for the definition). - Antti Karttunen, Oct 15 2016
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LINKS
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FORMULA
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G.f.: 1 + x*A(x)^3 = A(x) (Mod 2); a(n) = A001764(n) (Mod 2).
a(n) = binomial(3n, n) (mod 2). Characteristic function of Fibbinary numbers (i.e. a(n)=1 iff n is in A003714). - Benoit Cloitre, Nov 15 2003
Recurrence: a(0) = 1, a(2n) = a(4n+1) = a(n), a(4n+3) = 0.
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MAPLE
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f:= proc(n) local L, Lp;
L:= convert(n, base, 2);
Lp:= convert(3*n, base, 2);
if has(L-Lp[1..nops(L)], 1) then 0 else 1 fi
end proc:
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MATHEMATICA
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PROG
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A182028
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Take first n bits of the infinite Fibonacci word A003849, regard them as a binary number, then convert it to base 10.
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+20
8
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0, 1, 2, 4, 9, 18, 37, 74, 148, 297, 594, 1188, 2377, 4754, 9509, 19018, 38036, 76073, 152146, 304293, 608586, 1217172, 2434345, 4868690, 9737380, 19474761, 38949522, 77899045, 155798090, 311596180, 623192361, 1246384722, 2492769444, 4985538889, 9971077778
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OFFSET
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0,3
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COMMENTS
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LINKS
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FORMULA
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a(n) = 2*a(n-1) + A003849(n) for n > 0, a(0) = 0.
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EXAMPLE
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0 -> 0 -> a(0) = 0,
0,1 -> 01 -> a(1) = 1,
0,1,0 -> 010 -> a(2) = 2,
0,1,0,0 -> 0100 -> a(3) = 4,
0,1,0,0,1 -> 01001 -> a(4) = 9,
0,1,0,0,1,0 -> 010010 -> a(5) = 18,
0,1,0,0,1,0,1 -> 0100101 -> a(6) = 37
0,1,0,0,1,0,1,0 -> 01001010 -> a(7) = 74
0,1,0,0,1,0,1,0,0 -> 010010100 -> a(8) = 148,
0,1,0,0,1,0,1,0,0,1 -> 0100101001 -> a(9) = 297.
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MATHEMATICA
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nesting = 7; A003849 = Flatten[Nest[{#, #[[1]]}&, {0, 1}, nesting]]; a[n_] := FromDigits[Take[A003849, n+1], 2]; Table[a[n], {n, 0, Length[A003849] - 1}] (* Jean-François Alcover, Feb 13 2016 *)
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PROG
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(Haskell)
a182028 n = a182028_list !! n
a182028_list = scanl1 (\v b -> 2 * v + b) a003849_list
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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A187200
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Parse the Fibonacci word A003849 into distinct phrases 0, 1, 00, 10, 100, 1001, 01, 001, 010, ...; a(n) = length of n-th phrase.
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+20
7
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1, 1, 2, 2, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 5, 4, 5, 5, 6, 6, 5, 5, 6, 6, 7, 7, 7, 8, 7, 6, 8, 9, 6, 6, 10, 7, 8, 8, 7, 8, 7, 11, 7, 9, 8, 9, 8, 10, 9, 8, 9, 8, 9, 9, 9, 12, 10, 10, 9, 10, 9, 11, 10, 13, 12, 10, 10, 11, 11, 11, 10, 11, 11, 14, 11, 12, 12, 11, 13, 10, 12, 12, 11, 12, 13, 12, 13, 14, 13, 11, 13, 14, 15, 14, 13, 16, 15, 12, 12, 17
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OFFSET
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1,3
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LINKS
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A284749
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{001->2}-transform of the infinite Fibonacci word A003849.
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+20
7
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0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 2, 0, 1, 2
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OFFSET
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1,3
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COMMENTS
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This word is the fixed point of the morphism 0->2, 1->01, 2->2201 with the finite string 0, 1, 2, 0, 1 appended to the beginning. This morphism comes from taking the 'proper' version of the Fibonacci morphism 0->01, 1->1, given by 0->001, 1->01 (A189661 but with the rightmost 0 moved to the left of each image word), then replacing 001 with 2 and noting that the new symbol 2 should map to 00100101 = 2201 in order to be consistent.
The finite string appended to the beginning comes from the process of finding a proper version of the Fibonacci morphism using a return word encoding and taking conjugates which causes a shift of the respective fixed points.
(End)
This sequence is the unique fixed point of the morphism 0->01, 1->2, 2->0122. See the paragraph following Lemma 23 in the paper by Allouche and me. - Michel Dekking, Oct 05 2018
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LINKS
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EXAMPLE
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As a word, A003849 = 01001010010010100..., and replacing each 001 by 2 gives 01201220120...
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MATHEMATICA
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s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 13] (* A003849 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"001" -> "2"}]
st = ToCharacterCode[w1] - 48 (* A284749 *)
Flatten[Position[st, 0]] (* A214971 *)
Flatten[Position[st, 1]] (* A284624 *)
Flatten[Position[st, 2]] (* A284625 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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A284620
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{00->2}-transform of the infinite Fibonacci word A003849.
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+20
5
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0, 1, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1, 2, 1, 0, 1, 2, 1
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OFFSET
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1,3
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COMMENTS
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This sequence is a morphic sequence, i.e., the letter-to-letter image of the fixed point of a morphism mu. In fact, one can take the alphabet {A,B,C,D} with the morphism
mu: A->AB, B->CD, C->ABCD, D->CD,
and the letter-to-letter map lambda defined by
lambda: A->0, B->1, C->2, D->1.
Then (a(n)) = lambda(x), where x = ABCDABCDCD... is the unique fixed point of the morphism mu.
How does one see this? The infinite Fibonacci word
can be written as a concatenation of the two words 01 and 001.
In fact, if beta is the Fibonacci morphism 0->01, 1->0, then beta(01)=010, beta(001)=01010, from which this can be read off.
This can also be found in Lemma 23 in Allouche and Dekking, which gives, moreover, that if we define the morphism g on the alphabet {a,b} by
g(a) = ab, g(b) =abb
then a(n) = delta(x_G(n)), where
x_G = ababbababb...
is the unique fixed point of g, and delta is the map
delta(a) = 01, delta(b) = 21.
In my paper "Morphic words,..." such a map delta is called a decoration.
It is well-known that decorated fixed points are morphic sequences, and the 'natural' algorithm to achieve this splits a in AB, and b in CD. This gives the morphism mu on the alphabet {A,B,C,D}, and the letter-to-letter map lambda.
We next prove the first conjecture by Kimberling. We easily see from the form of mu that the letters B and D occur, and only occur, at even positions in the fixed point x of mu. Since lambda(B)=lambda(D)=1, and A and C are not mapped to 1, it follows immediately that the positions of 1 in (a(n)) are given by A005843 = (2n).
For a proof of Kimberling's second conjecture, note that a(n)=2 iff x(n)=C, where x is the fixed point of mu. The return words of C in x are CD and CDAB. Coding these two return words by their lengths, mu induces a descendant morphism tau given by
tau(2) = 24, tau(4) = 244.
We see that tau is just an alphabet change of the morphism g. Let t = 2424424244... be the unique fixed point of tau. It is well-known (see, e.g., Lemma 12 in "Morphic words,..."), that t = 2 x_F, where x_F = x_F(4,2) now is the Fibonacci word on the alphabet {4,2}. The partial sums of x_F(4,2) are equal to the generalized Beatty sequence V given by
V(n) = 2*floor(n*phi) + 1,
according to Lemma 8 in the Allouche and Dekking paper.
This proves Kimberling's conjecture, provided we give the sequence A130568 offset 1, as we should.
(End)
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LINKS
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EXAMPLE
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As a word, A003849 = 01001010010010100..., and replacing each 00 by 2 gives 012101212101210...
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MATHEMATICA
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s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 13] (* A003849 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"00" -> "2"}]
st = ToCharacterCode[w1] - 48 (* A284620 *)
Flatten[Position[st, 0]] (* A284621 *)
Flatten[Position[st, 1]] (* A005843 - conjectured *)
Flatten[Position[st, 2]] (* A130568 - conjectured *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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A287773
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{0->010, 1->101}-transform of the infinite Fibonacci word A003849.
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+20
5
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0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1
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OFFSET
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1
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LINKS
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FORMULA
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EXAMPLE
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As a word, A003849 = 0100101001001010010100100..., and replacing each 0 by 010 and each 1 by 101 gives 01010101001010101010101001010101001...
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MATHEMATICA
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s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 10] (* A003849 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"0" -> "010", "1" -> "101"}]
st = ToCharacterCode[w1] - 48 (* A287773 *)
Flatten[Position[st, 0]] (* A287774 *)
Flatten[Position[st, 1]] (* A287777 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0
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LINKS
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FORMULA
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a(n) = 2 - [(2n+2)r] + [(2n+1)r], where [ ] = floor and r = golden ratio (A001622).
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EXAMPLE
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A003849 = (0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, ...), so that
A339824 = (0, 0, 1, 1, 0, 0, 1, ...), the even bisection, and
A339825 = (1, 0, 0, 0, 1, 0, 0, ...), the odd bisection.
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MATHEMATICA
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r = (1 + Sqrt[5])/2; z = 300;
f[n_] := 2 - Floor[(n + 2) r] + Floor[(n + 1) r]; (* A003849 *)
Table[2 - Floor[(2 n + 2) r] + Floor[(2 n + 1) r], {n, 0, Floor[z/2]}](* A339824 *)
Table[2 - Floor[(2 n + 3) r] + Floor[(2 n + 2) r], {n, 0, Floor[z/2]}](* A339825 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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