From Michel Dekking, Sep 27 2019: (Start)
This sequence is a morphic sequence, i.e, the lettertoletter image of a fixed point of a morphism.
Proof: We note first that the mapping SR: 00>0010, 10>011, is an algorithmic procedure given by StringReplace in Mathematica. This makes it hard to describe iterates of it. However, in this particular case one can deal with this.
There are three types of 0's occurring in (a(n)). The first 0 in a pair 00, which we code by the letter F, the second 0 in a pair 00 which we code by the letter G, and the single 0 occurring between two 1's, coded by G. The 1 at the end of the iterates SR^n, we code by D ('dead'). The other 1's we code by A ('active').
So, if we denote this code by psi, then for example,
psi(SR(00)) = psi(0010) = FGAD,
psi(SR^2(00)) = psi(00100110) = FGAFGAAD.
Now consider the morphism sigma on the alphabet {F,G,D,A} given by
sigma(F) = FGA, sigma(G) = F, sigma(D) = D, sigma(A) = GAA.
Let pi be the inverse of psi: pi(F) = pi(G) = pi(D) = 0, pi(A) = 1.
Any word with prefix 00 and without occurrences of 000, is a concatenation of the three words u=00, v=1 and w=101. The SRprocedure acts contextfree on such concatenations. One has
SR(u) = 0010, SR(v) = 011, SR(w) = 0110011.
The psiimages are
psi(u) = FG, psi(v) = A, psi(w) = AGA.
The sigmaimages of these are
sigma(FG) = FGAF, sigma(A) = GAA, sigma(AGA) = GAAFGAA.
The piimages of these are
pi sigma psi(u) = 0010, pi sigma psi(v) = 011, pi sigma psi(w) = 0110011.
We see that pi sigma psi acts exactly in the same way as SR.
It follows that (a(n)) = pi(x), where x = FGAFGAA... is the unique fixed point of sigma. QED
(End)
From Michel Dekking, Sep 27 2019: (Start)
Now that we know that this sequence is a morphic sequence, we may prove Kimberling's conjecture that the lengths of the words SR^n(00) are given by A182780.
Here b:=A182780 has b(0) = 2, b(1) = 4, and satisfies the linear recurrence b(n) = 3*b(n1)  b(n2)  b(n3) for n>2.
The number of letters (F's, G's, D's and A's) in the word psi(SR^n(00)) is equal to the vector/matrix/vector product
(1,1,1,1) M^n (1,0,1,0)^T,
where (1,0,1,0)^T is the transpose of (1,0,1,0), and M is the incidence matrix of the morphism sigma, i.e., M equals
1 1 0 0
1 0 0 1
0 0 1 0
1 0 0 2.
The characteristic polynomial of M is equal to
chi(u) = u^4  4u^3 + 4u^2  1 = (u^3  3u^2 + u + 1)(u  1).
Thus the conjecture is proved by an application of the CayleyHamilton theorem: M^3 = 3M^2  M  Id.
(End)
From Michel Dekking, Sep 28 2019: (Start)
The application of CayleyHamilton is more subtle than suggested above. In fact chi(u) has a second factor u1: chi(u) = (u^22u1)(u1)^2, but b=A182780 does not satisfy the Pell recurrence a(n) = 2*a(n1) + a(n2).
Removing the letter D from {F,G,D,A} gives a morphism tau defined by
tau(F) = FGA, tau(G) = F, tau(A) = GAA.
The iterates tau^n(F) are equal to the iterates sigma^n(FD), with the suffix D removed. So
SR^n(00)= sigma^n(FD)=tau^n(F)+1,
where . denotes length.
The characteristic polynomial of the incidence matrix of tau is equal to u^33u^2+u+1.
So, by the CayleyHamilton theorem, the sequence
(tau^n(F)) = 1,3,7,17,41,99,... = A078057=:c
satisfies the recurrence c(n) = 3c(n1)c(n2)c(n3). This implies that the sequence (SR^n(00)) = (c(n)+1) satisfies the recurrence
b(n) = 3b(n1)  b(n2)  b(n3). QED
(End)
