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Entropy and Computation: The Landauer-Bennett Thesis Reexamined

Entropy 2013, 15(9), 3312-3324; https://doi.org/10.3390/e15093312

Article
On the Entropy of a Two Step Random Fibonacci Substitution
Department of Mathematics, Universität Bielefeld, Postfach 100131, D–33501 Bielefeld, Germany
Received: 8 July 2013; in revised form: 30 July 2013 / Accepted: 19 August 2013 / Published: 23 August 2013

Abstract

:
We consider a random generalization of the classical Fibonacci substitution. The substitution we consider is defined as the rule mapping, a baa and b ab , with probability p, and b ba , with probability 1 p for 0 < p < 1 , and where the random rule is applied each time it acts on a b . We show that the topological entropy of this object is given by the growth rate of the set of inflated random Fibonacci words, and we exactly calculate its value.
Keywords:
combinatorics on words; asymptotic enumeration; symbolic dynamics

1. Introduction

In [1], Godrèche and Luck define the random Fibonacci chain by the generalized substitution:
θ : a b b ab w i t h   p r o b a b i l i t y   p ba w i t h   p r o b a b i l i t y   1 p
for 0 < p < 1 and where the random rule is applied each time θ acts on a b . They introduce the random Fibonacci chain when studying quasi-crystalline structures and tilings in the plane. In their paper, it is claimed (without proof) that the topological entropy of the random Fibonacci chain is given by the growth rate of the set of inflated random Fibonacci words. This was later, with a combinatorial argument, proven in a more general context in [2].
The renewed interest in this system, and in possible generalizations, stems from the observation that the natural geometric generalization of the symbolic sequences by tilings of the line had to be Meyer sets with entropy and interesting spectra [3]. There is now a fair understanding of systems that emerge from the local mixture of inflation rules that each define the same hull. However, little is known so far about more general mixtures. Here, we place our attention to one such generalization. It is still derived from the Fibonacci rule, but mixes inflations that define distinct hulls.
In this paper, we consider the randomized substitution, ϕ, defined by:
ϕ = a baa b ab w i t h   p r o b a b i l i t y   p ba w i t h   p r o b a b i l i t y   1 p
for 0 < p < 1 and where the random rule is applied each time ϕ acts on a b . The substitution, ϕ, is a mixture of two substitutions, whose hulls are different. This is true, since the hull of the substitution, ( a , b ) ( baa , ab ) , contains words with the sub-words, aaa and bb , but neither of these sub-words are to be found in any word of the hull of ( a , b ) ( baa , ba ) . For a more detailed survey of the differences and similarities of the generated hulls of these two substitutions, see [4].
Before we can state our main theorem in detail, we need to introduce some notation. A word, w, over an alphabet, Σ, is a finite sequence, w 1 w 2 w n , of symbols from Σ. We let, here, Σ = { a , b } . We denote a sub-word of w by w [ a , b ] = w a w a + 1 w a + 2 w b 1 w b , and similarly, we let W [ a , b ] = { w [ a , b ] : w W } . By | · | , we mean the length of a word and the cardinality of a set. Note that | w [ a , b ] | = b a + 1 . When indexing the brackets with a letter, α, from the alphabet, | · | α , we shall mean the numbers of occurrences of α in the enclosed word.
For two words, u = u 1 u 2 u 3 u n and v = v 1 v 2 v 3 v m , we denote by u v the concatenation of the two words, that is, u v = u 1 u 2 u 3 u n v 1 v 2 v m . Similarly, we let, for two sets of words, U and V, their product be the set, U V = { u v : u U , v V } , containing all possible concatenations.
Letting ϕ act on the word, a , repeatedly yields an infinite sequence of words, r n = ϕ n 1 ( a ) . We know that r 1 = a and r 2 = baa . However, r 3 is one of the words, abbaabaa or babaabaa , with probability p or 1 p . The sequence, { r n } n = 1 , converges in distribution to an infinite random word, r. We say that r n is an inflated word (under ϕ) in generation n, and we introduce, here, sets that correspond to all inflated words in generation n;
Definition 1.
Let A 1 = { a } , B 1 = { b } , and for n 2 , we define recursively:
A n = B n 1 A n 1 A n 1 B n = A n 1 B n 1 B n 1 A n 1
and we let A : = lim n A n and B : = lim n B n .
The sets, A and B, are indeed well defined. This is a direct consequence of Corollary 6. It is clear from the definition of A n and B n that all their elements have the same length, that is, for all x , y A n (or x , y B n ), we have | x | = | y | . By induction, it easily follows that for a A n , we have | a | = f 2 n and for b B n , we have | b | = f 2 n 1 , where f m is the mth Fibonacci number, defined by f n + 1 = f n + f n 1 with f 0 = 0 and f 1 = 1 .
For a word, w, we say that x is a sub-word of w if there are two words, u , v , such that w = u x v . The sub-word set, F ( S , n ) , is the set of all sub-words of length n of words in S. The combinatorial entropy of the random Fibonacci chain is defined as the limit, lim n 1 n log | F ( A , n ) | . The combinatorial entropy is known to equal the topological entropy for our type of systems; see [5]. The existence of this limit is direct by Fekete’s lemma [6], since we have sub-additivity, log | F ( S , n + m ) | log | F ( S , n ) | + log | F ( S , m ) | . We can now state the main result in this paper.
Theorem 2.
The logarithm of the growth rate of the size of the set of inflated random Fibonacci words equals the topological entropy of the random Fibonacci chain, that is:
lim n log | A n | f 2 n = lim n log | B n | f 2 n 1 = lim n log | F ( C , n ) | n = 1 τ 3 log 2
where τ is the golden mean, τ = 1 + 5 2 and C { A , B } .
The outline of the paper is that we start by studying the sets, A n and B n . Next, we give a finite method for finding the sub-word set, F ( A , n ) , (which, we will see, is the same as F ( B , n ) ). Thereafter, we derive some Diophantine properties of the Fibonacci number that will play a central part when we look at the distribution of the letters in words from F ( A , n ) . Finally, we present an estimate of | F ( A , n ) | , leading up to the proof of Theorem 2.

2. Inflated Words

In this section, we present the sets of inflated words and give an insight to their structure. The results presented here will also play an important role for the results in the coming sections.
Proposition 3.
Let u , v A n (or both in B n ). Then, u v , if and only if { ϕ ( u ) } { ϕ ( v ) } = , where, here, { ϕ ( z ) } denotes the set of all possible words that can be obtained by applying ϕ on z.
Proof. 
Let u v , and assume that w { ϕ ( u ) } { ϕ ( v ) } . Denote by ϕ u and ϕ v the special choices of ϕ, such that w = ϕ u ( u ) = ϕ v ( v ) . Let k be the first position, such that u k v k , where u = u 1 u 2 u m and v = v 1 v 2 v m . Then, we may assume u k = a and v k = b ; otherwise, just swap the names of u and v. Since we have ϕ ( a ) = baa , we see that we must have ϕ v ( v k ) = ϕ v ( b ) = ba . However, then, also, ϕ v ( v k v k + 1 ) = ϕ v ( bb ) = baab . This then implies u k + 1 = b , since, if we have u k + 1 = a , then there must be two consecutive a s in w, and we could not find a continuation in v. Hence, we have ϕ u ( u k u k + 1 ) = ϕ u ( ab ) = baaba . As previously, v must continue with a b . We now see that we are in a cycle, where | ϕ u ( u k u k + 1 u k + s ) | = 3 + 2 s and | ϕ v ( v k v k + 1 v k + s ) | = 2 ( s + 1 ) . Since there is no s N , such that 3 + 2 s = 2 ( s + 1 ) , we conclude that there can be no such w. ☐
We can now turn to the question of counting the elements in the sets, A n and B n .
Proposition 4.
For n 2 , we have:
| A n | = 2 f 2 n 3 1 and | B n | = 2 f 2 n 4 + 1
Proof. 
Let us start with the proof of the the size of A n . From the Definition 1 of A n and B n , it follows by induction that | x | b = f 2 n 2 for x A n . Combining this with Proposition 3, we find the recursion:
| A n | = | A n 1 | · 2 | x | b = | A n 1 | · 2 f 2 n 4
The size of A n now follows from Equation (2) by induction. For the size of B n , we have, by the definition of B n and that we already know the size of A n ,
| B n | = | A n + 1 | | A n | | A n | = 2 f 2 n 1 1 2 f 2 n 3 1 · 2 f 2 n 3 1 = 2 f 2 n 4 + 1
which completes the proof. ☐
From Proposition 4, the statements of the logarithmic limits of the sets, A n and B n , in Theorem 2 follows directly. Our next step is to give some result on sets of prefixes of A n and B n . These results will play a central role when we later look at sets of sub-words.
Proposition 5.
For n 2 , we have:
( 3 ) A n [ 1 , f 2 n 1 ] A n + 1 [ 1 , f 2 n 1 ] ( 4 ) A n [ 1 , f 2 n 1 ] B n A n [ 1 , f 2 n 1 ]
Proof. 
Let us first consider Equation (3). We give a proof by induction on n. For the basis case, n = 2 , we have:
A 2 [ 1 , f 2 · 2 1 ] = A 2 [ 1 , 2 ] = { ab } { ab , ba } = A 3 [ 1 , f 4 1 ]
Now, assume for induction that Equation (3) holds for 2 n p . Then, for n = p + 1 , we have by the induction assumption:
A p + 1 [ 1 , f 2 ( p + 1 ) 1 ] = B p A p A p [ 1 , f 2 ( p + 1 ) 1 ] ( A p B p B p A p ) A p [ 1 , f 2 ( p + 1 ) 1 ] = B p + 1 A p [ 1 , f 2 ( p + 1 ) 1 ] = B p + 1 A p [ 1 , f 2 p 1 ] B p + 1 A p + 1 [ 1 , f 2 p 1 ] = B p + 1 A p + 1 [ 1 , f 2 ( p + 1 ) 1 ] = B p + 1 A p + 1 A p + 1 [ 1 , f 2 ( p + 1 ) 1 ] = A p + 2 [ 1 , f 2 ( p + 1 ) 1 ]
which completes the induction and the proof of Equation (3). Let us turn to the proof of Equation (4). By the help of Equation (3), we have:
A n [ 1 , f 2 n 1 ] = B n 1 A n 1 A n 1 [ 1 , f 2 n 1 ] = B n 1 A n 1 A n 1 [ 1 , f 2 ( n 1 ) 1 ] B n 1 A n 1 A n [ 1 , f 2 ( n 1 ) 1 ] = B n 1 A n 1 A n [ 1 , f 2 n 1 ] B n A n [ 1 , f 2 n 1 ]
which concludes the proof. ☐
From Proposition 5, it is straight forward, by recalling the recursive definition of A n and B n , to derive the following equalities on prefix-sets.
Corollary 6.
For n 3 , we have:
A n [ 1 , f 2 ( n 1 ) 1 ] = A n + 1 [ 1 , f 2 ( n 1 ) 1 ] B n [ 1 , f 2 ( n 1 ) 1 ] = A n [ 1 , f 2 ( n 1 ) 1 ] B n = B n + 1 [ 1 , f 2 n 1 ]
We end the section by proving a result on suffixes of the sets, A n and B n , that we shall make use of in the next sections.
Proposition 7.
For n 2 , we have:
( 5 ) A n [ f 2 n 2 + 2 , f 2 n ] B n [ 2 , f 2 n 1 ] ( 6 ) B n [ 2 , f 2 n 1 ] = B n + 1 [ f 2 n + 2 , f 2 n + 1 ]
Proof. 
We give a proof by induction on n. For the basis case, n = 2 , we have:
A 2 [ f 2 + 2 , f 4 ] = A 2 [ 2 , 3 ] = { a } { a , b } = B 2 [ 2 , 2 ]
Now, assume for induction that Equation (5) holds for 2 n p . Then, for the induction step, n = p + 1 , we have by the induction assumption:
A p + 1 [ f 2 ( p + 1 ) 2 + 2 , f 2 ( p + 1 ) ] = B p A p A p [ f 2 ( p + 1 ) 2 + 2 , f 2 ( p + 1 ) ] = A p A p [ f 2 p 2 + 2 , 2 f 2 p ] = A p [ f 2 p 2 + 2 , f 2 p ] A p B p [ 2 , f 2 p 1 ] A p = B p A p [ 2 , f 2 p + 1 ] B p + 1 [ 2 , f 2 ( p + 1 ) 1 ]
which completes the induction and the proof of Equation (5). For the proof of Equation (6), we have:
B n [ 2 , f 2 n 1 ] = A n B n [ f 2 n + 2 , f 2 n + f 2 n 1 ] B n + 1 [ f 2 n + 2 , f 2 n + f 2 n 1 ]
and for the converse inclusion, we have by Equation (5):
B n + 1 [ f 2 n + 2 , f 2 n + f 2 n 1 ] = A n B n B n A n [ f 2 n 2 , f 2 n + f 2 n 1 ] = B n [ 2 , f 2 n 1 ] A n [ f 2 n 2 + 2 , f 2 n ] B n [ 2 , f 2 n 1 ]
which proves the equality (6). ☐

3. Sets of Sub-Words

Here, we investigate properties of the sets of sub-words, F ( A , m ) and F ( B , m ) . We will prove that they coincide, and moreover, we show how to find them by considering finite sets, which will be central when estimating their size, depending on m.
First, we turn our attention to proving that it is indifferent if we consider sub-words of A n or of B n .
Proposition 8.
For n 1 , we have:
F ( A n + 1 , f 2 n 1 ) = F ( B n + 1 , f 2 n 1 )
Proof. 
Let us first turn to the proof of the inclusion:
F ( A n + 1 , f 2 n 1 ) F ( B n + 1 , f 2 n 1 )
Let x ( k ) A n + 1 [ k , k 1 + f 2 n 1 ] for 1 k f 2 n + 1 + 2 . It is clear that x ( k ) F ( A n + 1 , f 2 n 1 ) for any k. We have to prove that also x ( k ) F ( B n + 1 , f 2 n 1 ) .
For 1 k f 2 n 1 + 2 , we have:
x ( k ) F ( B n A n , f 2 n 1 ) F ( B n + 1 , f 2 n 1 )
For f 2 n 1 + 3 k f 2 n + 1 , we have by Corollary 6, which x ( k ) must be a sub-word of:
A n A n [ 3 , f 2 n + f 2 n 2 1 ] = A n B n [ 3 , f 2 n + f 2 n 2 1 ] = B k + 1 [ 3 , f 2 n + f 2 n 2 1 ]
For f 2 n + 2 k f 2 n + 1 + 2 , we have by Proposition 7:
B n A n A n [ f 2 n + 2 , f 2 n + 2 ] = A n [ f 2 n 2 + 2 , f 2 n ] A n B n [ 2 , f 2 n 1 ] A n B n + 1 [ 2 , f 2 n + 1 ]
which concludes the proof of the inclusion Equation (7). For the converse inclusion, it is enough to consider sub-words of A n B n , since any sub-word of B n A n clearly is a sub-word of A n + 1 . Therefore, let y ( k ) ( A n B n ) [ k , k 1 + f 2 n 1 ] for 1 k f 2 n 1 + 1 . We now proceed as in the case above.
For 1 k f 2 n 2 + 1 , we have:
( A n B n ) [ 1 , f 2 n + f 2 n 2 1 ] = A n B n [ 1 , f 2 n 2 1 ] = A n A n [ 1 , f 2 n 2 1 ] = A n + 1 [ f 2 n + 1 + 1 , f 2 n + 1 + f 2 n 2 1 ]
For f 2 n 2 + 2 k f 2 n 1 + 2 , we have:
( A n B n ) [ f 2 n 2 + 2 , f 2 n 1 + 2 ] = A n [ f 2 n 2 + 2 , f 2 n ] A n = B n [ 2 , f 2 n 1 ] A n = A n + 1 [ 2 , f 2 n + 1 ]
which completes the proof. ☐
The above result shows that the set of sub-words from A n and B n coincide if the sub-words are not chosen too long. If we consider the limit sets, A and B, their sets of sub-words turn out to be the same. We have the following:
Proposition 9.
For m 1 , we have F ( A , m ) = F ( B , m ) .
Proof. 
Let x F ( A , m ) . Then, there is an n, such that:
x F ( A n , m ) F ( A n B n B n A n , m ) = F ( B n + 1 , m ) F ( B , m )
Similarly, if x F ( B , m ) . Then, there is an n, such that:
x F ( B n , m ) F ( B n A n A n , m ) = F ( A n + 1 , m ) F ( A , m )
which completes the proof. ☐
The direct consequence of Proposition 9 is that we find the topological entropy in Equation (1) independent if we look at sub-words from A or B.
Now, let us turn to the question of finding F ( A , m ) from a finite set, A n , and not having to consider the infinite set, A.
Proposition 10.
For n 2 , we have:
F ( A n + 1 , f 2 n f 2 n 3 ) = F ( A n + 2 , f 2 n f 2 n 3 )
Proof. 
It is clear that F ( A n + 1 , f 2 n f 2 n 3 ) F ( A n + 2 , f 2 n f 2 n 3 ) holds for all n 2 . For the reverse inclusion, assume that x F ( A n + 2 , f 2 n f 2 n 3 ) . Note that we can write A n + 1 and A n + 2 on the form:
A n + 1 = B n A n A n , ( 8 ) A n + 2 = B n A n B n A n A n B n A n A n A n B n B n A n A n B n A n A n
From Equation (8), we see that any x is a sub-word of any element in some of the seven sets:
A n A n , A n B n A n , A n B n , A n B n A n , B n B n , A n B n B n , B n B n A n
in such a way that the first letter in x is in the first factor (that is, A n or B n ) of the sets. If x is a sub-word of A n A n or B n A n or completely contained in A n , it is clear that we have x F ( A n + 1 , f 2 n f 2 n 3 ) . For the case when x is a sub-word of A n B n , it follows from Proposition 8 that we have x F ( A n + 1 , f 2 n f 2 n 3 ) .
If x is a sub-word of a word in A n B n A n , such that x begins in the first A n factor and ends in the second, then we have that x is a sub-word of a word in the set:
( A n [ f 2 n 3 + f 2 n 1 + 2 , f 2 n ] ) B n 1 A n 1 A n [ 1 , f 2 n 4 1 ] = A n [ f 2 n 3 + f 2 n 1 + 2 , f 2 n ] B n 1 A n 1 A n 1 [ 1 , f 2 n 4 1 ] = A n [ f 2 n 3 + f 2 n 1 + 2 , f 2 n ] A n [ 1 , f 2 n 1 + f 2 n 4 1 ] = ( A n A n ) [ f 2 n 3 + f 2 n 1 + 2 , f 2 n + f 2 n 1 + f 2 n 4 1 ]
and we see that we have x F ( A n + 1 , f 2 n f 2 n 3 ) .
If x is a sub-word of a word in B n B n , let us first consider the case when it is a sub-word of B n B n 1 A n 1 . Then, it follows that:
B n B n 1 A n 1 B n A n [ 1 , f 2 n 1 ] = B n A n [ 1 , 2 f 2 n 1 ]
so x is a sub-word of a word in A n + 1 . For the the second case, B n A n 1 B n 1 , we have:
B n A n 1 B n 1 = A n 1 B n 1 A n 1 B n 1 B n 1 A n 1 A n 1 B n 1 A n B n A n [ f 2 n 1 + 1 , 3 f 2 n 1 ] ( A n B n ) [ 1 , 2 f 2 n 1 ]
and again, x is a sub-word of a word in A n + 1 , by what we just proved above.
If x is a sub-word of a word in A n B n B n , we have by Corollary 6:
A n B n B n [ f 2 n 1 + f 2 n 3 + 1 , 2 f 2 n f 2 n 3 1 ] = A n [ f 2 n 1 + f 2 n 3 + 1 , f 2 n ] B n B n [ 1 , f 2 n 4 1 ] = A n [ f 2 n 1 + f 2 n 3 + 1 , f 2 n ] B n A n [ 1 , f 2 n 4 1 ]
which shows that x is a sub-word of a word in A n + 1 by what we previously have shown.
Finally, if x is a sub-word of a word in B n B n A n , we first consider the case when x is a sub-word of a word in B n B n 1 A n 1 A n . By Corollary 6, we have:
B n B n A n [ 2 f 2 n 3 + 1 , f 2 n + 1 f 2 n 3 1 ] = B n [ 2 f 2 n 3 + 1 , f 2 n 1 ] B n 1 A n 1 A n [ 1 , f 2 n 4 1 ] = B n [ 2 f 2 n 3 + 1 , f 2 n 1 ] B n 1 A n 1 A n 1 [ 1 , f 2 n 4 1 ] = B n [ 2 f 2 n 3 + 1 , f 2 n 1 ] A n [ 1 , f 2 n 1 + f 2 n 4 1 ]
which, by the help of the previous case, shows that x is a sub-word of a word in A n + 1 . For the last case, B n A n 1 B n 1 A n , we have by Corollary 6 and Proposition 7:
( B n B n A n ) [ 2 f 2 n 3 + 1 , f 2 n + 1 f 2 n 3 1 ] = B n [ 2 f 2 n 3 + 1 , f 2 n 1 ] A n 1 B n 1 A n [ 1 , f 2 n 4 1 ] = B n 1 [ 2 f 2 n 3 f 2 n 2 + 1 , f 2 n 3 ] A n 1 B n 1 A n 1 [ 1 , f 2 n 4 1 ] = B n [ 2 f 2 n 3 f 2 n 2 + 1 , f 2 n 1 ] A n [ 1 , f 2 n 2 1 ]
and again, we see that x is a sub-word of a word in A n + 1 by what we have proven above. ☐
The result of Proposition 10 can be extended to hold for sub-words from elements A n and A n + k , where k 1 . A straight forward argument via induction gives:
F ( A n + 1 , f 2 n f 2 n 3 ) = F ( A n + k , f 2 n f 2 n 3 )
for k 1 . By combining Proposition 10 and Equation (10), we can now prove that to find the factors set, it is sufficient to only consider a finite set.
Proposition 11.
For n 2 , we have:
F ( A n + 1 , f 2 n f 2 n 3 ) = F ( A , f 2 n f 2 n 3 )
Proof. 
It is clear that we have F ( A n + 1 , f 2 n f 2 n 3 ) F ( A , f 2 n f 2 n 3 ) . For the reversed inclusion, let x F ( A , f 2 n f 2 n 3 ) . Then, there is a smallest m n + 1 , such that x is a sub-word of an element of A m . Then, Equation (10) gives:
x F ( A m , f 2 n f 2 n 3 ) = F ( A n + 1 , f 2 n f 2 n 3 )
which shows the desired inclusion. ☐

4. Fibonacci Numbers Revisited

In this section, we shall restate, and adopt for our purposes, some of the Diophantine properties of the Fibonacci numbers and use them to derive results on the distribution of the letters in the words in the sets, A n and B n . Let us introduce the notation:
τ = 1 + 5 2 a n d τ ^ = 1 5 2
for the roots of x 2 x 1 = 0 . It is well known that τ and τ ^ appear in Binet’s formula, the Fibonacci numbers; see [7]:
f n = 1 5 1 + 5 2 n 1 5 1 5 2 n = 1 5 τ n τ ^ n
From Equation (12), it is with induction straight forward to derive:
f n = τ f n 1 + τ ^ n 1 = τ 2 f n 2 + τ ^ n 2
Definition 12.
Let · denote the smallest distance to an integer.
By using the special property, τ 2 = τ + 1 , we have for an integer, k, the following line of equalities:
1 τ 2 k = τ 1 τ k = k 1 τ k = 1 τ k = ( τ 1 ) k = τ k
From Equation (13), it follows that:
τ f n = f n + 1 τ ^ n = 1 τ n
since τ ^ = 1 τ . For an integer, k, which is not a Fibonacci number, we have the following estimate of how far away from an integer τ k is.
Proposition 13.
For a positive integer, k, such that f n 1 < k < f n , we have:
τ k > 1 τ n 2
Proof. 
We give a proof by induction on n. For the basis case, n = 5 , the statement of the proposition follows by an easy calculation. Now, assume for induction that Equation (14) holds for 5 n p . For the induction step, n = p + 1 , let f p < k < f p + 1 . Then, if k f p 1 is not a Fibonacci number, we have:
τ k = τ ( k f p ) + τ f p τ ( k f p ) < f p 1 τ f p > 1 τ p 3 1 τ p > 1 τ p 2
If k f p 1 = f m for some m < p 1 , then:
τ k τ f m τ f p = 1 τ m 1 τ p 1 τ p 2 1 τ p = 1 τ p 1
Proposition 14.
Let x A n [ 1 , k ] for 1 k f 2 n (or x B n [ 1 , k ] for 1 k f 2 n 1 ) and n 2 . Then:
| x | b 1 τ 2 k , 1 τ 2 k
Proof. 
We give a proof by induction on n. The basis case, n = 2 , follows by considering each of the words contained in A 2 and B 2 . To be able to use Proposition 13 in the induction step, we have to consider the basis step, n = 3 , as well, but only for the set, B 3 (since the words in A 2 are of length 3 ). This is, however, seen to hold by a straight forward enumeration of the elements of B 3 .
Now, assume for induction that Equation (15) holds for 2 n p , for words both from A n and B n . For the induction step, n = p + 1 , let us first derive an identity of which we shall later make use. Let q and m be positive integers, such that f m 1 < q < f m . Then, by the help of Proposition 13, we have:
1 τ 2 ( q f m 1 ) = 1 τ 2 q f m 3 τ ^ m 1 = 1 τ 2 q + ( 1 ) m τ m 1 f m 3 ( 16 ) = 1 τ 2 q f m 3
With the same argumentation, we can derive a similar result for · . For the induction step, we consider first the number of b s in prefixes of words in A p + 1 = B p A p A p . It is clear from the induction assumption that Equation (15) holds for 1 k f 2 p 1 . For f 2 p 1 < k < f 2 p or f 2 p < k < f 2 p + 1 , let x = u v A p + 1 [ 1 , k ] , where u B p . By the induction assumption, we may assume that | v | b is given by rounding downwards, (the result is obtained in a similar way for the case with · ). By Equation (16), it now follows that:
| u v | b = | u | b + | v | b = f 2 p 3 + 1 τ 2 ( k f 2 p 1 ) = 1 τ 2 k
For k = f 2 p , we have:
| u v | b = f 2 p 3 + 1 τ 2 ( f 2 p f 2 p 1 ) = 1 τ 2 f 2 p + 1 τ 2 p 1 = 1 τ 2 f 2 p
For f 2 p + 1 < k < f 2 p + 2 , let x = u v w A p + 1 [ 1 , k ] , where u B p and v A p . Then, the induction assumption and Equation (16) gives:
| u v w | b = | u | b + | v | b + | w | b = f 2 p 3 + f 2 p 2 + 1 τ 2 ( k f 2 p 1 f 2 p ) = f 2 p 1 + 1 τ 2 ( k f 2 p + 1 ) = 1 τ 2 k
For the last case, k = f 2 p + 2 , we have:
| x | b = 1 τ 2 ( f 2 p + 2 ) = f 2 p + 1 τ 2 p = f 2 p
The case when we consider words from B p + 1 is treated in the same way, but where we do not need to do the induction step for the case n = 3 . This completes the induction and the proof. ☐
Proposition 15.
Let x F ( A n + 2 , f 2 n ) for n 2 . Then:
f 2 n 2 1 | x | b f 2 n 2 + 1
Proof. 
Let us first turn our attention to the upper bound in Equation (17). In the same way as in the proof of Proposition 10, we consider sub-words of the seven sets, given in Equation (9).
If x is a sub-word, beginning at position 2 < k f 2 n , in an element in A n A n or A n B n , then:
| x | b f 2 n 2 + 1 τ 2 ( k f 2 n ) + f 2 n 1 τ 2 k f 2 n 2 + 1
since the number of b s in a word in A n is f 2 n 2 , and a word in A n is of length f 2 n . The proof of the upper bound in Equation (17) and for the other sets in Equation (9) is obtained in the same way.
For the lower bound, we have:
| x | b 1 τ 2 ( k + f 2 n ) 1 τ 2 k = f 2 n 2 + 1 τ 2 k + 1 τ 2 n 2 1 τ 2 k f 2 n 2 1
for any x F ( A n + 2 , f 2 n ) . ☐

5. Estimating the Size of the Sub-Word Set

We shall in this section give an estimate of the sub-word set, F ( A , f 2 n ) , and give the final part of the proof of Theorem 2. Let us introduce the set:
C n = ϕ F ( A , f 2 n 2 + 1 )
By Proposition 15, we can estimate the number of b s in words in F ( A , f 2 n 2 + 1 ) . This estimate then gives that we have bounds on the length of words in C n . That is, for x C n , we have:
| x | = | x | a + | x | b 3 ( f 2 n 3 1 ) + 2 ( f 2 n 4 + 2 ) = f 2 n + 1
and:
| x | = | x | a + | x | b 3 ( f 2 n 3 + 2 ) + 2 ( f 2 n 4 1 ) = f 2 n + 4
Proposition 16.
For n 2 , we have:
F ( A , f 2 n ) = F C n , f 2 n
Proof. 
The set, F C n , f 2 n , is created by inflating words from F ( A , f 2 n 2 + 1 ) , which are then cut into suitable lengths. This implies that F ( A , f 2 n ) F C n , f 2 n .
For the converse inclusion, let x F ( A , f 2 n ) . Then, there is a word, w A n + 1 , and words, u , v , such that u x v A n + 2 and u x v ϕ ( w ) . For any word, z F { w } , f 2 n 2 + 1 , we have from Equation (18) that any s ϕ ( z ) fulfills f 2 n + 1 | s | . This gives that there is a word, z x F { w } , f 2 n 2 + 1 , such that x is a sub-word of a word in ϕ ( z x ) , which implies x F C n , f 2 n . ☐
Proposition 17.
For n 2 , we have:
| F ( A , f 2 n ) | 2 f 2 n 3 + 2 n · 5 n 1
Proof. 
We give a proof by induction on n. For the basis case, n = 2 , we have:
| F ( A , f 4 ) | = 7 160 = 2 f 1 + 4 · 5
Assume for induction that Equation (20) holds for 2 n p . For the induction step, n = p + 1 , note that from Equations (18) and (19), it follows that | F ( { x } , f 2 p + 2 ) | 5 for x C p + 1 . By Proposition 15, we have that the number of b s in u F ( A , f 2 p + 1 ) is at most f 2 p 2 + 2 . This gives, then, with the help of the induction assumption:
| F A , f 2 p + 2 | | C p + 1 | · 5 | F A , f 2 p | · 2 f 2 p 2 + 2 · 5 2 f 2 p 3 + 2 p · 5 p 1 · 2 f 2 p 2 + 2 · 5 = 2 f 2 ( p + 1 ) 3 + 2 ( p + 1 ) · 5 p
which completes the proof. ☐
We can now turn to proving the last equality in Equation (1) and, thereby, completing the proof of Theorem 2. By Proposition 17, we have:
lim n log | F ( A , f 2 n ) | f 2 n lim n log 2 f 2 n 3 + 2 n · 5 n 1 f 2 n = lim n f 2 n 3 + 2 n f 2 n log 2 + n 1 f 2 n log 5 = 1 τ 3 log 2
which implies the equality in Equation (1).
A further generalization of the random Fibonacci substitutions would be to study the structure occurring when mixing two substitutions with different inflation multipliers. This, however, seems to be a far more complex question.

Acknowledgments

The author wishes to thank M. Baake and M. Moll at Bielefeld University, Germany, for our discussions and for reading drafts of the manuscript. This work was supported by the German Research Council (DFG), via CRC 701.

Conflicts of Interest

The authors declare no conflict of interest.

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