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FormalConjectures.OEIS.«006697»

Subword complexity of the morphism a → aab, b → b #

Let $a_n$ be the number of distinct subwords (contiguous factors) of length $n$ in the infinite word generated by the morphism $\sigma: a \mapsto aab, b \mapsto b$, starting from $a$.

The conjectured generating function is: $$\sum_{n \geq 0} a_n x^n = 1 + \frac{1}{1-x} + \frac{1}{(1-x)^2}\left(\frac{1}{1-x} - \sum_{k \geq 1} x^{2^k + k - 1}\right).$$

References:

The morphism σ on {a, b} defined by a ↦ aab, b ↦ b, represented on Bool where false = a and true = b.

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    @[simp]
    theorem OEIS.A006697.length_finiteWord (n : ) :
    (finiteWord n).length = 2 ^ (n + 1) - 1
    noncomputable def OEIS.A006697.infiniteWord (i : ) :

    The infinite word w is the fixed point of the morphism starting from 'a'. We define it as the limit: w(i) is the i-th symbol, which stabilizes after sufficiently many iterations.

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      noncomputable def OEIS.A006697.subwordAt (i n : ) :

      A subword (factor) of length n starting at position i.

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        The set of all distinct subwords of length n in the infinite word.

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          noncomputable def OEIS.A006697.a (n : ) :

          The subword complexity function: a(n) is the number of distinct subwords of length n.

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            Conjecture (A006697): The generating function for the number of subwords of length $n$ in the infinite word generated by $a \mapsto aab, b \mapsto b$ is $$\sum_{n \geq 0} a_n x^n = 1 + \frac{1}{1-x} + \frac{1}{(1-x)^2}\left(\frac{1}{1-x} - \sum_{k \geq 0} x^{2^{k+1} + k}\right).$$

            Equivalently, a(n) equals the n-th coefficient of this generating function.