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Let $W = 01001010010010 \ldots$ be the infinite Fibonacci word, A003849 in the OEIS. Let $B(m)$ be the set of $m+1$ subwords of $W$ that have length $m$, and for each such subword $u$, let $p(u)$ be the sequence of positions in $W$ where $u$ begins. Then $p(u)$ is a composite of the famous Wythoff sequences $A=A000201$ and $B=A001950$. Can someone figure out (or cite a reference) exactly which composites represent the subwords in $B(m)$?

Here's how it looks for $m=4$:

\begin{array}{|c|c|c|c|} \text{subword, } u & \text{positions, } p(u) & \text{composite} & \text{OEIS} \\ 0100 & 1,6,9,14,\ldots & AAA & A134859 \\ 1001 & 2,7,10,15,\ldots & BA & A035336 \\ 0010 & 3,8,11,16,\ldots & AB & A003623 \\ 0101 & 4,12,17,25,\ldots & AAB & A134860 \\ 1010 & 5,13,18,26,\ldots & BB & A101864 \end{array}

(Note that for every $m$, the difference sequence of every $p(u)$ consists of Fibonacci numbers.)

Following Sam Hopkins's note, here's a definition. Start with $0$ and apply the substitutions $0 \rightarrow 01$ and $1 \rightarrow 0$ repeatedly, like this: $$0,01,010,01001,01001010,0100101001001,\ldots.$$ The limiting word is $A003849$, one of several called the infinite Fibonacci word, but this one is regarded as the standard form, according to the Crossrefs section of $A014675$.

Writing those words as $w_0,w_1,w_2,\ldots$, respectively, note that $w_n$ is, for $n \geq 2$, the concatention indicated by $w_n=w_{n-1}w_{n-2}$, so that the length of $w_n$ is a Fibonacci number.

Some more background: suppose that $w$ is a word in $B(m)$. Then at least one of the words $w0$ and $w1$ must be in $B(m+1)$. However, there is only one $w$ in $B(m)$ such that both $w0$ and $w1$ are in $B(m+1)$. Such a "splitter" turns out to be simply a reversal of an initial word of $W$, so that the first few splitters are $,0,10,010,0010,10010,\ldots$. The corresponding Wythoff composites are $$A,B,AA,AB,BA,AAA,BB,AAB,ABA,BAA,AAAA,ABB,\ldots$$ So, if someone can tell specifically how to generate this sequence, the problem will be solved.

Let $W = 01001010010010 \ldots$ be the infinite Fibonacci word, A003849 in the OEIS. Let $B(m)$ be the set of $m+1$ subwords of $W$ that have length $m$, and for each such subword $u$, let $p(u)$ be the sequence of positions in $W$ where $u$ begins. Then $p(u)$ is a composite of the famous Wythoff sequences $A=$A000201 and $B=$A001950. Can someone figure out (or cite a reference) exactly which composites represent the subwords in $B(m)$?

Here's how it looks for $m=4$:

\begin{array}{|c|c|c|c|} \text{subword, } u & \text{positions, } p(u) & \text{composite} & \text{OEIS} \\ 0100 & 1,6,9,14,\ldots & AAA & A134859 \\ 1001 & 2,7,10,15,\ldots & BA & A035336 \\ 0010 & 3,8,11,16,\ldots & AB & A003623 \\ 0101 & 4,12,17,25,\ldots & AAB & A134860 \\ 1010 & 5,13,18,26,\ldots & BB & A101864 \end{array}

(Links: A134859, A035336, A003623, A134860, A101864)

(Note that for every $m$, the difference sequence of every $p(u)$ consists of Fibonacci numbers.)

Following Sam Hopkins's note, here's a definition. Start with $0$ and apply the substitutions $0 \rightarrow 01$ and $1 \rightarrow 0$ repeatedly, like this: $$0,01,010,01001,01001010,0100101001001,\ldots.$$ The limiting word is A003849, one of several called the infinite Fibonacci word, but this one is regarded as the standard form, according to the Crossrefs section of A014675.

Writing those words as $w_0,w_1,w_2,\ldots$, respectively, note that $w_n$ is, for $n \geq 2$, the concatenation indicated by $w_n=w_{n-1}w_{n-2}$, so that the length of $w_n$ is a Fibonacci number.

Some more background: suppose that $w$ is a word in $B(m)$. Then at least one of the words $w0$ and $w1$ must be in $B(m+1)$. However, there is only one $w$ in $B(m)$ such that both $w0$ and $w1$ are in $B(m+1)$. Such a "splitter" turns out to be simply a reversal of an initial word of $W$, so that the first few splitters are $,0,10,010,0010,10010,\ldots$. The corresponding Wythoff composites are $$A,B,AA,AB,BA,AAA,BB,AAB,ABA,BAA,AAAA,ABB,\ldots$$ So, if someone can tell specifically how to generate this sequence, the problem will be solved.

Let $W = 01001010010010 \ldots$ be the infinite Fibonacci word, A003849 in the OEIS. Let $B(m)$ be the set of $m+1$ subwords of $W$ that have length $m$, and for each such subword $u$, let $p(u)$ be the sequence of positions in $W$ where $u$ begins. Then $p(u)$ is a composite of the famous Wythoff sequences $A=A000201$ and $B=A001950$. Can someone figure out (or cite a reference) exactly which composites represent the subwords in $B(m)$?

Here's how it looks for $m=4$:

\begin{array}{|c|c|c|c|} \text{subword, } u & \text{positions, } p(u) & \text{composite} & \text{OEIS} \\ 0100 & 1,6,9,14,\ldots & AAA & A134859 \\ 1001 & 2,7,10,15,\ldots & BA & A035336 \\ 0010 & 3,8,11,16,\ldots & AB & A003623 \\ 0101 & 4,12,17,25,\ldots & AAB & A134860 \\ 1010 & 5,13,18,26,\ldots & BB & A101864 \end{array}

(Note that for every $m$, the difference sequence of every $p(u)$ consists of Fibonacci numbers.)

Following Sam Hopkins's note, here's a definition. Start with $0$ and apply the substitutions $0 \rightarrow 01$ and $1 \rightarrow 0$ repeatedly, like this: $$0,01,010,01001,01001010,0100101001001,\ldots.$$ The limiting word is $A003849$, one of several called the infinite Fibonacci word, but this one is regarded as the standard form, according to the Crossrefs section of $A014675$.

Writing those words as $w_0,w_1,w_2,\ldots$, respectively, note that $w_n$ is, for $n \geq 2$, the concatention indicated by $w_n=w_{n-1}w_{n-2}$, so that the length of $w_n$ is a Fibonacci number.

Let $W = 01001010010010 \ldots$ be the infinite Fibonacci word, A003849 in the OEIS. Let $B(m)$ be the set of $m+1$ subwords of $W$ that have length $m$, and for each such subword $u$, let $p(u)$ be the sequence of positions in $W$ where $u$ begins. Then $p(u)$ is a composite of the famous Wythoff sequences $A=A000201$ and $B=A001950$. Can someone figure out (or cite a reference) exactly which composites represent the subwords in $B(m)$?

Here's how it looks for $m=4$:

\begin{array}{|c|c|c|c|} \text{subword, } u & \text{positions, } p(u) & \text{composite} & \text{OEIS} \\ 0100 & 1,6,9,14,\ldots & AAA & A134859 \\ 1001 & 2,7,10,15,\ldots & BA & A035336 \\ 0010 & 3,8,11,16,\ldots & AB & A003623 \\ 0101 & 4,12,17,25,\ldots & AAB & A134860 \\ 1010 & 5,13,18,26,\ldots & BB & A101864 \end{array}

(Note that for every $m$, the difference sequence of every $p(u)$ consists of Fibonacci numbers.)

Following Sam Hopkins's note, here's a definition. Start with $0$ and apply the substitutions $0 \rightarrow 01$ and $1 \rightarrow 0$ repeatedly, like this: $$0,01,010,01001,01001010,0100101001001,\ldots.$$ The limiting word is $A003849$, one of several called the infinite Fibonacci word, but this one is regarded as the standard form, according to the Crossrefs section of $A014675$.

Writing those words as $w_0,w_1,w_2,\ldots$, respectively, note that $w_n$ is, for $n \geq 2$, the concatention indicated by $w_n=w_{n-1}w_{n-2}$, so that the length of $w_n$ is a Fibonacci number.

Some more background: suppose that $w$ is a word in $B(m)$. Then at least one of the words $w0$ and $w1$ must be in $B(m+1)$. However, there is only one $w$ in $B(m)$ such that both $w0$ and $w1$ are in $B(m+1)$. Such a "splitter" turns out to be simply a reversal of an initial word of $W$, so that the first few splitters are $,0,10,010,0010,10010,\ldots$. The corresponding Wythoff composites are $$A,B,AA,AB,BA,AAA,BB,AAB,ABA,BAA,AAAA,ABB,\ldots$$ So, if someone can tell specifically how to generate this sequence, the problem will be solved.

Let $W = 01001010010010 \ldots$ be the infinite Fibonacci word, A003849 in the OEIS. Let $B(m)$ be the set of $m+1$ subwords of $W$ that have length $m$, and for each such subword $u$, let $p(u)$ be the sequence of positions in $W$ where $u$ begins. Then $p(u)$ is a composite of the famous Wythoff sequences $A=A000201$ and $B=A001950$. Can someone figure out (or cite a reference) exactly which composites represent the subwords in $B(m)$?

Here's how it looks for $m=4$:

\begin{array}{|c|c|c|c|} \text{subword, } u & \text{positions, } p(u) & \text{composite} & \text{OEIS} \\ 0100 & 1,6,9,14,\ldots & AAA & A134859 \\ 1001 & 2,7,10,15,\ldots & BA & A035336 \\ 0010 & 3,8,11,16,\ldots & AB & A003623 \\ 0101 & 4,12,17,25,\ldots & AAB & A134860 \\ 1010 & 5,13,18,26,\ldots & BB & A101864 \end{array}

(Note that for every $m$, the difference sequence of every $p(u)$ consists of Fibonacci numbers.)

Let $W = 01001010010010 \ldots$ be the infinite Fibonacci word, A003849 in the OEIS. Let $B(m)$ be the set of $m+1$ subwords of $W$ that have length $m$, and for each such subword $u$, let $p(u)$ be the sequence of positions in $W$ where $u$ begins. Then $p(u)$ is a composite of the famous Wythoff sequences $A=A000201$ and $B=A001950$. Can someone figure out (or cite a reference) exactly which composites represent the subwords in $B(m)$?

Here's how it looks for $m=4$:

\begin{array}{|c|c|c|c|} \text{subword, } u & \text{positions, } p(u) & \text{composite} & \text{OEIS} \\ 0100 & 1,6,9,14,\ldots & AAA & A134859 \\ 1001 & 2,7,10,15,\ldots & BA & A035336 \\ 0010 & 3,8,11,16,\ldots & AB & A003623 \\ 0101 & 4,12,17,25,\ldots & AAB & A134860 \\ 1010 & 5,13,18,26,\ldots & BB & A101864 \end{array}

(Note that for every $m$, the difference sequence of every $p(u)$ consists of Fibonacci numbers.)

Following Sam Hopkins's note, here's a definition. Start with $0$ and apply the substitutions $0 \rightarrow 01$ and $1 \rightarrow 0$ repeatedly, like this: $$0,01,010,01001,01001010,0100101001001,\ldots.$$ The limiting word is $A003849$, one of several called the infinite Fibonacci word, but this one is regarded as the standard form, according to the Crossrefs section of $A014675$.

Writing those words as $w_0,w_1,w_2,\ldots$, respectively, note that $w_n$ is, for $n \geq 2$, the concatention indicated by $w_n=w_{n-1}w_{n-2}$, so that the length of $w_n$ is a Fibonacci number.