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An explicit formula is given in this page which is given in the OEIS page you have referred to.

The explicit formula for the Graham-Pollak sequence is $a_n=\lfloor{\tau(m)(2^{\frac{n}{2}}+2^{\frac{(n-1)}{2}})}\rfloor$. Hence, $b_n(m)=a_{2n+1}(m)-2a_{2n-1}(m)=\lfloor{\tau(m)(\sqrt{2}+1)2^n}\rfloor-\lfloor{\tau(m)(\sqrt{2}+1)2^{n-1}}\rfloor$.

Now, the sequence $\tau(m)$ is the set $\mathbb N \cup \sqrt{2} \mathbb N$ arranged in ascending order.

Claim 1: We can write $\tau(m)=\chi(m)\sqrt{\eta(m)}$, where $\chi(m)$ is a sequence of natural numbers with each natural number appearing exactly twice.

The set $\chi(m)$ is such that,if for some $m$, $\chi(m)=k$ is the first appearence $\eta(m)=1$ and if it's second appearence $\eta(m)=2$.

For the sequence $\{b_n\}(m)=a_{2n+1}(m)-2a_{2n-1}(m)$ we would show that $b_n(m)=b_n(m'),\forall n\in \mathbb N$, where $(m,m')$ is such that $\chi(m)=k=\chi(m'), m'>m,k \in \mathbb N$.

We have, $b_n(m)=\lfloor{\Lambda_n(m)}\rfloor-2\lfloor{\Lambda_n(m)/2}\rfloor$ where $\lfloor{\Lambda_n(m)}\rfloor=\lfloor{\tau(m)(\sqrt{2}+1)2^n}\rfloor$. If $\lfloor{\Lambda_n(m)}\rfloor$ is even $b_n(m)=0$, if it is odd, $b_n(m)=1$.

Now, $\Lambda(m')=\sqrt(2)\chi(m)\sqrt{\eta(m)}(\sqrt{2}+1)2^n$. For, $m,m'$ pair $\eta(m)=1, \chi(m) \in \mathbb N$. Hence, $\Lambda(m')=\Lambda(m)+\chi(m)2^n$. Hence, $\lfloor{\Lambda(m)}\rfloor$ and $\lfloor{\Lambda(m')}\rfloor$ has same parity, implying that $\{b_n\}(m)=\{b_n\}(m') \forall n \in \mathbb N$....$(1)$

(There is no $q$ other than $m'$ such that $\{b_n\}(m)=\{b_n\}(q)$ for all $n$. Because the parity can't be same for all n when $\tau(q) = \sqrt{2}$and $\tau(m)$ and $\eta(m)=1$ isn't true simultaneously).

As the sequence $a(m)$ (which is asked) contains all the natural numbers in ascending (meaning that $p$ appears before $q$ if $p<q$), $(1)$ implies that $\{a(m)\}=\{\chi(m)\}$. As,the asked sequence $\{b(n)\}=\{\tau(n)^2\}$, it just requiers to prove Claim 1.

To do so, we have to prove $l_{m+1}-l_{m}=\eta(m)+1$, where $l_i$ is such that $\eta(l_i)$ is the $i$-th $2$ in the $\eta$ sequence.

Edit: The proof of Claim-1: Let's assume $l_i$ be a sequence such that $l_{m+1}-l_{m}=\eta(m)+1\Rightarrow l_m=m-1+\sum_{i}^{m-1}\eta(i)+2$. We will show that $l_m$ is the $m$-th $2$ in the $\eta(n)$ sequence.

First of all we have, $\sum_{i=1}^{m-1}\eta(i)=\lfloor{m\sqrt{2}}\rfloor-1$. Also assume,$m\sqrt{2}=a+r, a\in \mathbb N, 0<r<1$.

Hence, $l_m=m+a$.So, $\eta(l_m)=\lfloor{(m+a+1)\sqrt{2}}\rfloor-\lfloor(m+a)\sqrt{2}\rfloor$.

or, $\eta(l_m)=\lfloor{\sqrt2+m\sqrt{2}+(m\sqrt2-r)\sqrt2}\rfloor-\lfloor m\sqrt{2}+(m\sqrt2-r)\sqrt2\rfloor$.

Using $m\sqrt{2}=a+r, a\in \mathbb N, 0<r<1$, we get $\eta(l_m)=\lfloor{\sqrt2-(\sqrt2-1)r}\rfloor-\lfloor-r(\sqrt2-1)\rfloor=2$ as $0<r<1$

Now, we just need to show these are the only numbers having $\eta=2$.

We have, $l_{m+1}-l_m=\eta(m)+1$. We would show that $\eta(l_m+c)=1$ for all $c\leq \eta(m)$. If $\eta(m)=1, c=1$. In that case like previous steps we get $\eta(l_m+1)=\lfloor{2\sqrt2-(\sqrt2-1)}\rfloor-\lfloor\sqrt2-r(\sqrt2-1)r\rfloor$, as $r<1$ we have $\eta(l_m+c)=1$.

If $\eta(m)=2, c=1 \text{or} 2$. We have similarly $\eta(l_m+2)=\lfloor{3\sqrt2-(\sqrt2-1)r}\rfloor-\lfloor{2\sqrt2-r(\sqrt2-1)}\rfloor=1$ because $\eta(m)=2 \Rightarrow 1>r>2(\sqrt2-1)$ and so, $3<3\sqrt2-r(\sqrt2-1)<4$ while $2<2\sqrt2-r(\sqrt2-1)<3$.

This implies that $l_m$ is the $m$-th $2$ in $\eta$ sequence, proving the Claim-1, and so the asked conjecture.

An explicit formula is given in this page which is given in the OEIS page you have referred to.

The explicit formula for the Graham-Pollak sequence is $a_n=\lfloor{\tau(m)(2^{\frac{n}{2}}+2^{\frac{(n-1)}{2}})}\rfloor$. Hence, $b_n(m)=a_{2n+1}(m)-2a_{2n-1}(m)=\lfloor{\tau(m)(\sqrt{2}+1)2^n}\rfloor-\lfloor{\tau(m)(\sqrt{2}+1)2^{n-1}}\rfloor$.

Now, the sequence $\tau(m)$ is the set $\mathbb N \cup \sqrt{2} \mathbb N$ arranged in ascending order.

Claim 1: We can write $\tau(m)=\chi(m)\sqrt{\eta(m)}$, where $\chi(m)$ is a sequence of natural numbers with each natural number appearing exactly twice.

The set $\chi(m)$ is such that,if for some $m$, $\chi(m)=k$ is the first appearence $\eta(m)=1$ and if it's second appearence $\eta(m)=2$.

For the sequence $\{b_n\}(m)=a_{2n+1}(m)-2a_{2n-1}(m)$ we would show that $b_n(m)=b_n(m'),\forall n\in \mathbb N$, where $(m,m')$ is such that $\chi(m)=k=\chi(m'), m'>m,k \in \mathbb N$.

We have, $b_n(m)=\lfloor{\Lambda_n(m)}\rfloor-2\lfloor{\Lambda_n(m)/2}\rfloor$ where $\lfloor{\Lambda_n(m)}\rfloor=\lfloor{\tau(m)(\sqrt{2}+1)2^n}\rfloor$. If $\lfloor{\Lambda_n(m)}\rfloor$ is even $b_n(m)=0$, if it is odd, $b_n(m)=1$.

Now, $\Lambda(m')=\sqrt(2)\chi(m)\sqrt{\eta(m)}(\sqrt{2}+1)2^n$. For, $m,m'$ pair $\eta(m)=1, \chi(m) \in \mathbb N$. Hence, $\Lambda(m')=\Lambda(m)+\chi(m)2^n$. Hence, $\lfloor{\Lambda(m)}\rfloor$ and $\lfloor{\Lambda(m')}\rfloor$ has same parity, implying that $\{b_n\}(m)=\{b_n\}(m') \forall n \in \mathbb N$....$(1)$

(There is no $q$ other than $m'$ such that $\{b_n\}(m)=\{b_n\}(q)$ for all $n$. Because the parity can't be same for all n when $\tau(q) = \sqrt{2}$and $\tau(m)$ and $\eta(m)=1$ isn't true simultaneously).

As the sequence $a(m)$ (which is asked) contains all the natural numbers in ascending (meaning that $p$ appears before $q$ if $p<q$), $(1)$ implies that $\{a(m)\}=\{\chi(m)\}$. As,the asked sequence $\{b(n)\}=\{\tau(n)^2\}$, it just requiers to prove Claim 1.

To do so, we have to prove $l_{m+1}-l_{m}=\eta(m)+1$, where $l_i$ is such that $\eta(l_i)$ is the $i$-th $2$ in the $\eta$ sequence.

Edit: The proof of Claim-1: Let's assume $l_i$ be a sequence such that $l_{m+1}-l_{m}=\eta(m)+1\Rightarrow l_m=m-1+\sum_{i}^{m-1}\eta(i)+2$. We will show that $l_m$ is the $m$-th $2$ in the $\eta(n)$ sequence.

First of all we have, $\sum_{i=1}^{m-1}\eta(i)=\lfloor{m\sqrt{2}}\rfloor-1$. Also assume,$m\sqrt{2}=a+r, a\in \mathbb N, 0<r<1$.

Hence, $l_m=m+a$.So, $\eta(l_m)=\lfloor{(m+a+1)\sqrt{2}}\rfloor-\lfloor(m+a)\sqrt{2}\rfloor$.

or, $\eta(l_m)=\lfloor{\sqrt2+m\sqrt{2}+(m\sqrt2-r)\sqrt2}\rfloor-\lfloor m\sqrt{2}+(m\sqrt2-r)\sqrt2\rfloor$.

Using $m\sqrt{2}=a+r, a\in \mathbb N, 0<r<1$, we get $\eta(l_m)=\lfloor{\sqrt2-(\sqrt2-1)r}\rfloor-\lfloor-r(\sqrt2-1)\rfloor=2$ as $0<r<1$

Now, we just need to show these are the only numbers having $\eta=2$.

We have, $l_{m+1}-l_m=\eta(m)+1$. We would show that $\eta(l_m+c)=1$ for all $c\leq \eta(m)$. If $\eta(m)=1, c=1$. In that case like previous steps we get $\eta(l_m+1)=\lfloor{2\sqrt2-(\sqrt2-1)}\rfloor-\lfloor\sqrt2-r(\sqrt2-1)r\rfloor$, as $r<1$ we have $\eta(l_m+c)=1$.

If $\eta(m)=2, c=1 \text{or} 2$. We have similarly $\eta(l_m+2)=\lfloor{3\sqrt2-(\sqrt2-1)r}\rfloor-\lfloor{2\sqrt2-r(\sqrt2-1)}\rfloor=1$ (because $\eta(m)=2 \Rightarrow 1>r>2(\sqrt2-1)$ and so, $3<3\sqrt2-r(\sqrt2-1)<4$ while $2<2\sqrt2-r(\sqrt2-1)<3$).

This implies that $l_m$ is the $m$-th $2$ in $\eta$ sequence, proving the Claim-1, and so the asked conjecture.

This is not an answer,I have tried to point out some correspondences.

An explicit formula is given in this page which is given in the OEIS page you have referred to.

The explicit formula for the Graham-Pollak sequence is $a_n=\lfloor{\tau(m)(2^{\frac{n}{2}}+2^{\frac{(n-1)}{2}})}\rfloor$. Hence, $b_n(m)=a_{2n+1}(m)-2a_{2n-1}(m)=\lfloor{\tau(m)(\sqrt{2}+1)2^n}\rfloor-\lfloor{\tau(m)(\sqrt{2}+1)2^{n-1}}\rfloor$.

Now, the sequence $\tau(m)$ is the set $\mathbb N \cup \sqrt{2} \mathbb N$ arranged in ascending order.

Claim 1: We can write $\tau(m)=\chi(m)\sqrt{\eta(m)}$, where $\chi(m)$ is a sequence of natural numbers with each natural number appearing exactly twice.

The set $\chi(m)$ is such that,if for some $m$, $\chi(m)=k$ is the first appearence $\eta(m)=1$ and if it's second appearence $\eta(m)=2$.

For the sequence $\{b_n\}(m)=a_{2n+1}(m)-2a_{2n-1}(m)$ we would show that $b_n(m)=b_n(m'),\forall n\in \mathbb N$, where $(m,m')$ is such that $\chi(m)=k=\chi(m'), m'>m,k \in \mathbb N$.

We have, $b_n(m)=\lfloor{\Lambda_n(m)}\rfloor-2\lfloor{\Lambda_n(m)/2}\rfloor$ where $\lfloor{\Lambda_n(m)}\rfloor=\lfloor{\tau(m)(\sqrt{2}+1)2^n}\rfloor$. If $\lfloor{\Lambda_n(m)}\rfloor$ is even $b_n(m)=0$, if it is odd, $b_n(m)=1$.

Now, $\Lambda(m')=\sqrt(2)\chi(m)\sqrt{\eta(m)}(\sqrt{2}+1)2^n$. For, $m,m'$ pair $\eta(m)=1, \chi(m) \in \mathbb N$. Hence, $\Lambda(m')=\Lambda(m)+\chi(m)2^n$. Hence, $\lfloor{\Lambda(m)}\rfloor$ and $\lfloor{\Lambda(m')}\rfloor$ has same parity, implying that $\{b_n\}(m)=\{b_n\}(m') \forall n \in \mathbb N$....$(1)$

(There is no $q$ other than $m'$ such that $\{b_n\}(m)=\{b_n\}(q)$ for all $n$. Because the parity can't be same for all n when $\tau(q) = \sqrt{2}$and $\tau(m)$ and $\eta(m)=1$ isn't true simultaneously).

As the sequence $a(m)$ (which is asked) contains all the natural numbers in ascending (meaning that $p$ appears before $q$ if $p<q$), $(1)$ implies that $\{a(m)\}=\{\chi(m)\}$. As,the asked sequence $\{b(n)\}=\{\tau(n)^2\}$, it just requiers to prove Claim 1.

To do so, we have to prove $l_{m+1}-l_{m}=\eta(m)+1$, where $l_i$ is such that $\eta(l_i)$ is the $i$-th $2$ in the $\eta$ sequence. I will try to prove this part.

An explicit formula is given in this page which is given in the OEIS page you have referred to.

The explicit formula for the Graham-Pollak sequence is $a_n=\lfloor{\tau(m)(2^{\frac{n}{2}}+2^{\frac{(n-1)}{2}})}\rfloor$. Hence, $b_n(m)=a_{2n+1}(m)-2a_{2n-1}(m)=\lfloor{\tau(m)(\sqrt{2}+1)2^n}\rfloor-\lfloor{\tau(m)(\sqrt{2}+1)2^{n-1}}\rfloor$.

Now, the sequence $\tau(m)$ is the set $\mathbb N \cup \sqrt{2} \mathbb N$ arranged in ascending order.

Claim 1: We can write $\tau(m)=\chi(m)\sqrt{\eta(m)}$, where $\chi(m)$ is a sequence of natural numbers with each natural number appearing exactly twice.

The set $\chi(m)$ is such that,if for some $m$, $\chi(m)=k$ is the first appearence $\eta(m)=1$ and if it's second appearence $\eta(m)=2$.

For the sequence $\{b_n\}(m)=a_{2n+1}(m)-2a_{2n-1}(m)$ we would show that $b_n(m)=b_n(m'),\forall n\in \mathbb N$, where $(m,m')$ is such that $\chi(m)=k=\chi(m'), m'>m,k \in \mathbb N$.

We have, $b_n(m)=\lfloor{\Lambda_n(m)}\rfloor-2\lfloor{\Lambda_n(m)/2}\rfloor$ where $\lfloor{\Lambda_n(m)}\rfloor=\lfloor{\tau(m)(\sqrt{2}+1)2^n}\rfloor$. If $\lfloor{\Lambda_n(m)}\rfloor$ is even $b_n(m)=0$, if it is odd, $b_n(m)=1$.

Now, $\Lambda(m')=\sqrt(2)\chi(m)\sqrt{\eta(m)}(\sqrt{2}+1)2^n$. For, $m,m'$ pair $\eta(m)=1, \chi(m) \in \mathbb N$. Hence, $\Lambda(m')=\Lambda(m)+\chi(m)2^n$. Hence, $\lfloor{\Lambda(m)}\rfloor$ and $\lfloor{\Lambda(m')}\rfloor$ has same parity, implying that $\{b_n\}(m)=\{b_n\}(m') \forall n \in \mathbb N$....$(1)$

(There is no $q$ other than $m'$ such that $\{b_n\}(m)=\{b_n\}(q)$ for all $n$. Because the parity can't be same for all n when $\tau(q) = \sqrt{2}$and $\tau(m)$ and $\eta(m)=1$ isn't true simultaneously).

As the sequence $a(m)$ (which is asked) contains all the natural numbers in ascending (meaning that $p$ appears before $q$ if $p<q$), $(1)$ implies that $\{a(m)\}=\{\chi(m)\}$. As,the asked sequence $\{b(n)\}=\{\tau(n)^2\}$, it just requiers to prove Claim 1.

To do so, we have to prove $l_{m+1}-l_{m}=\eta(m)+1$, where $l_i$ is such that $\eta(l_i)$ is the $i$-th $2$ in the $\eta$ sequence.

Edit: The proof of Claim-1: Let's assume $l_i$ be a sequence such that $l_{m+1}-l_{m}=\eta(m)+1\Rightarrow l_m=m-1+\sum_{i}^{m-1}\eta(i)+2$. We will show that $l_m$ is the $m$-th $2$ in the $\eta(n)$ sequence.

First of all we have, $\sum_{i=1}^{m-1}\eta(i)=\lfloor{m\sqrt{2}}\rfloor-1$. Also assume,$m\sqrt{2}=a+r, a\in \mathbb N, 0<r<1$.

Hence, $l_m=m+a$.So, $\eta(l_m)=\lfloor{(m+a+1)\sqrt{2}}\rfloor-\lfloor(m+a)\sqrt{2}\rfloor$.

or, $\eta(l_m)=\lfloor{\sqrt2+m\sqrt{2}+(m\sqrt2-r)\sqrt2}\rfloor-\lfloor m\sqrt{2}+(m\sqrt2-r)\sqrt2\rfloor$.

Using $m\sqrt{2}=a+r, a\in \mathbb N, 0<r<1$, we get $\eta(l_m)=\lfloor{\sqrt2-(\sqrt2-1)r}\rfloor-\lfloor-r(\sqrt2-1)\rfloor=2$ as $0<r<1$

Now, we just need to show these are the only numbers having $\eta=2$.

We have, $l_{m+1}-l_m=\eta(m)+1$. We would show that $\eta(l_m+c)=1$ for all $c\leq \eta(m)$. If $\eta(m)=1, c=1$. In that case like previous steps we get $\eta(l_m+1)=\lfloor{2\sqrt2-(\sqrt2-1)}\rfloor-\lfloor\sqrt2-r(\sqrt2-1)r\rfloor$, as $r<1$ we have $\eta(l_m+c)=1$.

If $\eta(m)=2, c=1 \text{or} 2$. We have similarly $\eta(l_m+2)=\lfloor{3\sqrt2-(\sqrt2-1)r}\rfloor-\lfloor{2\sqrt2-r(\sqrt2-1)}\rfloor=1$ because $\eta(m)=2 \Rightarrow 1>r>2(\sqrt2-1)$ and so, $3<3\sqrt2-r(\sqrt2-1)<4$ while $2<2\sqrt2-r(\sqrt2-1)<3$.

This implies that $l_m$ is the $m$-th $2$ in $\eta$ sequence, proving the Claim-1, and so the asked conjecture.

This is not an answer,I have tried to point out some correspondences.

An explicit formula is given in this page which is given in the OEIS page you have referred to.

The explicit formula for the Graham-Pollak sequence is $a_n=\lfloor{\tau(m)(2^{\frac{n}{2}}+2^{\frac{(n-1)}{2}})}\rfloor$. Hence, $b_n(m)=a_{2n+1}(m)-2a_{2n-1}(m)=\lfloor{\tau(m)(\sqrt{2}+1)2^n}\rfloor-\lfloor{\tau(m)(\sqrt{2}+1)2^{n-1}}\rfloor$.

Now, the sequence $\tau(m)$ is the set $\mathbb N \cup \sqrt{2} \mathbb N$ arranged in ascending order.

Claim 1: We can write $\tau(m)=\chi(m)\sqrt{\eta(m)}$, where $\chi(m)$ is a sequence of natural numbers with each natural number appearing exactly twice.

The set $\chi(m)$ is such that,if for some $m$, $\chi(m)=k$ is the first appearence $\eta(m)=1$ and if it's second appearence $\eta(m)=2$.

For the sequence $\{b_n\}(m)=a_{2n+1}(m)-2a_{2n-1}(m)$ we would show that $b_n(m)=b_n(m'),\forall n\in \mathbb N$, where $(m,m')$ is such that $\chi(m)=k=\chi(m'), m'>m,k \in \mathbb N$.

We have, $b_n(m)=\lfloor{\Lambda_n(m)}\rfloor-2\lfloor{\Lambda_n(m)/2}\rfloor$ where $\lfloor{\Lambda_n(m)}\rfloor=\lfloor{\tau(m)(\sqrt{2}+1)2^n}\rfloor$. If $\lfloor{\Lambda_n(m)}\rfloor$ is even $b_n(m)=0$, if it is odd, $b_n(m)=1$.

Now, $\Lambda(m')=\sqrt(2)\chi(m)\sqrt{\eta(m)}(\sqrt{2}+1)2^n$. For, $m,m'$ pair $\eta(m)=1, \chi(m) \in \mathbb N$. Hence, $\Lambda(m')=\Lambda(m)+\chi(m)2^n$. Hence, $\lfloor{\Lambda(m)}\rfloor$ and $\lfloor{\Lambda(m')}\rfloor$ has same parity, implying that $\{b_n\}(m)=\{b_n\}(m') \forall n \in \mathbb N$....$(1)$

(There is no $q$ other than $m'$ such that $\{b_n\}(m)=\{b_n\}(q)$ for all $n$. Because the parity can't be same for all n when $\tau(q) = \sqrt{2}$and $\tau(m)$ and $\eta(m)=1$ isn't true simultaneously).

As the sequence $a(m)$ (which is asked) contains all the natural numbers in ascending (meaning that $p$ appears before $q$ if $p<q$), $(1)$ implies that $\{a(m)\}=\{\chi(m)\}$. As,the asked sequence $\{b(n)\}=\{\tau(n)^2\}$, it just requiers to prove Claim 1.

To do so, we have to prove $l_{m+1}-l_{m}=\eta(m)$, where $l_i$ is such that $\eta(l_i)$ is the $i$-th $2$ in the $\eta$ sequence. I will try to prove this part.

This is not an answer,I have tried to point out some correspondences.

An explicit formula is given in this page which is given in the OEIS page you have referred to.

The explicit formula for the Graham-Pollak sequence is $a_n=\lfloor{\tau(m)(2^{\frac{n}{2}}+2^{\frac{(n-1)}{2}})}\rfloor$. Hence, $b_n(m)=a_{2n+1}(m)-2a_{2n-1}(m)=\lfloor{\tau(m)(\sqrt{2}+1)2^n}\rfloor-\lfloor{\tau(m)(\sqrt{2}+1)2^{n-1}}\rfloor$.

Now, the sequence $\tau(m)$ is the set $\mathbb N \cup \sqrt{2} \mathbb N$ arranged in ascending order.

Claim 1: We can write $\tau(m)=\chi(m)\sqrt{\eta(m)}$, where $\chi(m)$ is a sequence of natural numbers with each natural number appearing exactly twice.

The set $\chi(m)$ is such that,if for some $m$, $\chi(m)=k$ is the first appearence $\eta(m)=1$ and if it's second appearence $\eta(m)=2$.

For the sequence $\{b_n\}(m)=a_{2n+1}(m)-2a_{2n-1}(m)$ we would show that $b_n(m)=b_n(m'),\forall n\in \mathbb N$, where $(m,m')$ is such that $\chi(m)=k=\chi(m'), m'>m,k \in \mathbb N$.

We have, $b_n(m)=\lfloor{\Lambda_n(m)}\rfloor-2\lfloor{\Lambda_n(m)/2}\rfloor$ where $\lfloor{\Lambda_n(m)}\rfloor=\lfloor{\tau(m)(\sqrt{2}+1)2^n}\rfloor$. If $\lfloor{\Lambda_n(m)}\rfloor$ is even $b_n(m)=0$, if it is odd, $b_n(m)=1$.

Now, $\Lambda(m')=\sqrt(2)\chi(m)\sqrt{\eta(m)}(\sqrt{2}+1)2^n$. For, $m,m'$ pair $\eta(m)=1, \chi(m) \in \mathbb N$. Hence, $\Lambda(m')=\Lambda(m)+\chi(m)2^n$. Hence, $\lfloor{\Lambda(m)}\rfloor$ and $\lfloor{\Lambda(m')}\rfloor$ has same parity, implying that $\{b_n\}(m)=\{b_n\}(m') \forall n \in \mathbb N$....$(1)$

(There is no $q$ other than $m'$ such that $\{b_n\}(m)=\{b_n\}(q)$ for all $n$. Because the parity can't be same for all n when $\tau(q) = \sqrt{2}$and $\tau(m)$ and $\eta(m)=1$ isn't true simultaneously).

As the sequence $a(m)$ (which is asked) contains all the natural numbers in ascending (meaning that $p$ appears before $q$ if $p<q$), $(1)$ implies that $\{a(m)\}=\{\chi(m)\}$. As,the asked sequence $\{b(n)\}=\{\tau(n)^2\}$, it just requiers to prove Claim 1.

To do so, we have to prove $l_{m+1}-l_{m}=\eta(m)+1$, where $l_i$ is such that $\eta(l_i)$ is the $i$-th $2$ in the $\eta$ sequence. I will try to prove this part.