

A108103


Fixed point of the square of the morphism: 1>3, 2>1, 3>121, starting with 1.


7



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

1,2


COMMENTS

Old name was: A Fibonacci like substitution for threesymbol substitution with characteristic polynomial: x^32*x1.
This sequence gives a threesymbol 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 lettertoletter 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 socalled 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}}.
From Michel Dekking, Dec 27 2017: (Start)
Let tau be the morphism squared: tau(1)=121, tau(2)=3, tau(3)=313.
Then tau(a)=a.
Claims:
(A) a(2n1) = 1 for n = 1,2,....
(B) a(2n) = A282162(n1) 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^(k1)(3) and R(tau^k(2))2 = 3psi^(k1)(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^(k1)(3))R(tau^(k1)(1))R(tau^(k1)(3))2 =
R(tau^k(2))2psi^(k2)(3)3^(1)R(tau^k(2))2 =
3psi^(k2)(2)psi^(k2)(3)psi^(k1)(2) = 3psi^(k2)(32332)=
3psi^k(2).
(End)


REFERENCES

F. M. Dekking: "What is the long range order in the Kolakoski sequence?" in: The Mathematics of LongRange Aperiodic Order, ed. R. V. Moody, Kluwer, Dordrecht (1997), pp. 115125.


LINKS

Michel Dekking, Table of n, a(n) for n = 1..1200
F. M. Dekking, On the structure of selfgenerating sequences, Seminar on Number Theory, 19801981 (Talence, 19801981), Exp. No. 31, 6 pp., Univ. Bordeaux I, Talence, 1981. Math. Rev. 83e:10075.
F. M. Dekking, What Is the Long Range Order in the Kolakoski Sequence?, Report 95100, Technische Universiteit Delft, 1995.


FORMULA

1>121, 2>3, 3>313.


MATHEMATICA

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 *)


CROSSREFS

Cf. A000045, A066983, A282162, A095345, A095346.
Sequence in context: A094959 A162696 A309978 * A111376 A241664 A157226
Adjacent sequences: A108100 A108101 A108102 * A108104 A108105 A108106


KEYWORD

nonn


AUTHOR

Roger L. Bagula, Jun 03 2005


EXTENSIONS

New name from Joerg Arndt, Jan 17 2013
New name from Robert G. Wilson v, Nov 05 2015
Name corrected by Michel Dekking, Dec 27 2017
Offset 1 from Michel Dekking, Jan 01 2020


STATUS

approved



