cp's OEIS Frontend

From Michel Dekking, Oct 29 2018: (Start)

Here is a proof that (a(n)) is fixed point of the morphism 0->102, 1->102, 2->02.

Let alpha:=phi-1. Then alpha*phi = 1. So

g(k) = floor(phi*floor(k*alpha)).

Write k*alpha = floor(k*alpha) + {k*alpha}, i.e., {k*alpha} is the fractional part of k*alpha. Then

g(k) = floor(phi*(k*alpha-{k*alpha})) = k + floor(-phi*{k*alpha}).

Thus

a(n) = n+1 +floor(-phi*{(n+1)*alpha})-n -floor(-phi*{n*alpha}).

It follows that

a(n) = 1 - floor(phi*{(n+1)*alpha}) + floor(phi*{n*alpha}).

The difference -floor(phi*{(n+1)*alpha}) + floor(phi*{n*alpha}) is equal to -1, 0 or 1, since floor(phi*{n*alpha}) is equal to 0 or 1.

In fact, phi*{n*alpha} can only take values between 0 and 1.619, and floor(phi*{n*alpha}) = 0 if and only if

{n*alpha} < 1/phi = alpha.

This is the same (putting rho:=1-alpha) as requiring

{n*alpha+rho} < 1-alpha.

Via the rotation description of Sturmian sequences (see, e.g., Lothaire), one sees that this sequence is the inhomogeneous Sturmian sequence s(alpha, rho), but with offset 1, and with 0 and 1 exchanged. Since rho+alpha=1, it follows that s(alpha, rho) with offset 2 equals s(1-alpha, 1-alpha), the classical Fibonacci sequence xF:=A003849, fixed point of 0->01, 1->0. We have found that

a(n+1)=0 iff xF(n)=0, xF(n+1)=1,

a(n+1)=1 iff xF(n)=0, xF(n+1)=0,

a(n+1)=2 iff xF(n)=1, xF(n+1)=0.

This means that (a(n+1)) equals the 3-symbol Fibonacci sequence A270788 on the alphabet {0,2,1}. Then Proposition 5 in "Morphisms, Symbolic Sequences, and Their Standard Forms" yields that (a(n)) is fixed point of the morphism 0->102, 1->102, 2->02. (End)