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:
- OEIS A006697
- J.-P. Allouche and J. Shallit, "On the subword complexity of the fixed point of a → aab, b → b, and generalizations," arXiv:1605.02361 [math.CO], 2016.
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995.
The n-th iterate of the morphism applied to [a].
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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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- OEIS.A006697.infiniteWord i = (OEIS.A006697.finiteWord (i + 1))[i]
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A subword (factor) of length n starting at position i.
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- OEIS.A006697.subwordAt i n = List.map (fun (j : ℕ) => OEIS.A006697.infiniteWord (i + j)) (List.range n)
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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.