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A108103
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Fixed point of the square of the morphism: 1->3, 2->1, 3->121, starting with 1.
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7
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1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1
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OFFSET
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1,2
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COMMENTS
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Old name was: A Fibonacci like substitution for three-symbol substitution with characteristic polynomial: x^3-2*x-1.
This sequence gives a three-symbol substitution for A095345.
From Michel Dekking, Jan 06 2018: (Start) What is probably meant by this statement is that A095345 is a morphic sequence, i.e., a letter-to-letter projection of a fixed point of the morphism tau given by tau(1)=121, tau(2)=3, tau(3)=313, followed by the morphism pi given by pi(1)=1, pi(2)=1, pi(3)=3.
This deserves a proof. In fact a proof can only be given if one formulates a joint statement about the sequences v:=A095345=1113111313... and w:=A095346=3131113...., because these two sequences are defined in a loop. Let D be the so-called differentiation operator which maps a word to the lengths of its runs, as studied in [Dekking, 1981]. For example D(1113111) = 313.
The words v and w by definition satisfy D(v)=w, D(w)=v. They are in fact points of period 2 for D (cf. [Dekking,1995]).
Claim: v=A095345 equals pi(x), where x is the fixed point of tau with x(1)=1, and w=A095346 equals pi(y), where y is the fixed point of tau with y(1)=3.
Proof: This is easily shown by induction on n=2,3,..., proving that tau^(n+1)(1) and tau^(n)(3) satisfy D(pi(tau^(n+1)(1)) = pi(tau^n(3)) & D(pi(tau^n(3)) = pi(tau^n(1)).
(End)
Real Salem Roots: {{x -> -1.}, {x -> -0.618034}, {x -> 1.61803}}.
Let tau be the morphism squared: tau(1)=121, tau(2)=3, tau(3)=313.
Then tau(a)=a.
Claims:
(A) a(2n-1) = 1 for n = 1,2,....
(B) a(2n) = A282162(n-1) for n = 1,2,....
Proof of (A): Obviously 2 only occurs in 121, but this implies that also 3 only occurs in 131.
Proof of (B): Let R be the 'remove 1' operator, e.g., R(12131) = 23.
Let psi be the square of the Fibonacci morphism on the alphabet {3,2}: psi(3)=323, psi(2)=32. One proves by induction that R(tau^k(1))3 = 2psi^(k-1)(3) and R(tau^k(2))2 = 3psi^(k-1)(2) for k=1,2,.... This implies (B): see CROSSREFS in A282162.
We give the more complicated induction step of the two:
R(tau^(k+1)(2))2 = R(tau^k(3))2 =
R(tau^(k-1)(3))R(tau^(k-1)(1))R(tau^(k-1)(3))2 =
R(tau^k(2))2psi^(k-2)(3)3^(-1)R(tau^k(2))2 =
3psi^(k-2)(2)psi^(k-2)(3)psi^(k-1)(2) = 3psi^(k-2)(32332)=
3psi^k(2).
(End)
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REFERENCES
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F. M. Dekking: "What is the long range order in the Kolakoski sequence?" in: The Mathematics of Long-Range Aperiodic Order, ed. R. V. Moody, Kluwer, Dordrecht (1997), pp. 115-125.
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LINKS
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FORMULA
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1->121, 2->3, 3->313.
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MATHEMATICA
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s[1] = {3}; s[2] = {1}; s[3] = {1, 2, 1}; t[a_] := Flatten[s /@ a]; p[0] = {1}; p[1] = t[p[0]]; p[n_] := t[p[n - 1]] a = p[12]
Nest[Flatten[# /. {1 -> {3}, 2 -> {1}, 3 -> {1, 2, 1}}] &, {1}, 10] (* Robert G. Wilson v, Nov 05 2015 *)
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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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