cp's OEIS Frontend

This is the second sequence reached in the infinite process described in A066983 comment line.

(a(n)) is a morphic sequence, i.e., a letter to letter projection of a fixed point of a morphism. The morphism is 1->121,2->3,1,3->313. The fixed point is the fixed point 3131213131213... starting with 3. The letter-to-letter map is 1->1, 2->1, 3->3. See also COMMENTS of A108103. - Michel Dekking, Jan 06 2018

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}}.

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(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)

Terms are palindromic. If b_3(n) denotes the number of 3's in a(n) then b(n) satisfies the recursion: b_3(1)=0, b_3(2)=1 and b_3(n) = b_3(n-1) + b_3(n-2) + (-1)^n so that b_3(2n)=A055588(n) and b_3(2n+1)=A027941(n). If b_1(n) denotes the number of 1's: b_1(1)=1, b_1(2)=0 and b_1(n) = b_1(n-1) + b_1(n-2) - 2*(-1)^n so that b_1(2n)=A004146(n) and b_1(2n+1) = A000032(n-2) - 2.