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A014675
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The infinite Fibonacci word (start with 1, apply 1->2, 2->21, take limit).
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44
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2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2
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OFFSET
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0,1
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COMMENTS
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The limiting mean and variance of the first n terms are both equal to the golden ratio (A001622). - Clark Kimberling, Mar 12 2014
Let F = A000045 (Fibonacci numbers). For n >= 3, the first F(n)-2 terms of A014675 form a palindrome; see A001911. If k is not one of the numbers F(n)-2, then the first k terms of A014675 do not form a palindrome. - Clark Kimberling, Jul 14 2014
The word is a concatenation of three runs: 1, 2, and 22. The limiting proportions of these are respectively 1/2, 1 - phi/2, and (phi - 1)/2, where phi = golden ratio. The mean runlength is (phi + 1)/2. - Clark Kimberling, Dec 26 2010
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REFERENCES
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D. Gault and M. Clint, "Curiouser and curiouser" said Alice. Further reflections on an interesting recursive function, Internat. J. Computer Math., 26 (1988), 35-43. See Table 2.
D. E. Knuth, The Art of Computer Programming, Vol. 4A, Section 7, p. 36.
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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FORMULA
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Define strings S(0)=1, S(1)=2, S(n)=S(n-1).S(n-2) for n>=2. Sequence is S(infinity).
a(n) = floor((n+2)*phi) - floor((n+1)*phi) = A000201(n+2) - A000201(n+1), phi = (1 + sqrt(5))/2.
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MAPLE
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Digits := 50: t := evalf( (1+sqrt(5))/2); A014675 := n->floor((n+2)*t)-floor((n+1)*t);
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MATHEMATICA
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SubstitutionSystem[{1->{2}, 2->{2, 1}}, {1}, {11}][[1]] (* Harvey P. Dale, Jan 01 2023 *)
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PROG
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(PARI) first(n)=my(v=[1], u); while(#v<n, u=List(); for(i=1, #v, listput(u, 2); if(v[i]==2, listput(u, 1))); v=Vec(u)); v[1..n] \\ Charles R Greathouse IV, Jun 21 2017
(PARI) apply( {A014675(n, r=quadgen(5)-1)=(n+2)\r-(n+1)\r}, [0..99]) \\ - M. F. Hasler, Apr 07 2021, improved on suggestion from Kevin Ryde, Apr 23 2021
(Python)
from math import isqrt
def A014675(n): return (n+2+isqrt(m:=5*(n+2)**2)>>1)-(n+1+isqrt(m-10*n-15)>>1) # Chai Wah Wu, Aug 10 2022
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CROSSREFS
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This is the {2,1} version. The standard form is A003849 (alphabet {0,1}). See also A005614 (alphabet {1,0}), A003842 (alphabet {1,2} instead of {2,1}).
Equals A001468 except for initial term.
Differs from A025143 in many entries starting at entry 8.
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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