\hline

    \framebox{$1 0 1$} &\to \framebox{$0 0 0$} & p^3

\end{align*}

With this we can write a recursive formula for the expected number of resamples from $101$:

\begin{align*}

    R_{101} &= (1-3p+3p^2 - p^3)(1) + (3p -6p^2 +3p^3) (1+R_{101}) \\

            &\quad + (p^2 - p^3) (1+R_{10101}) + (2p^2-2p^3) (1+R_{1001}) \\

			&= 1 + 3 p + 7 p^2 + 14.6667 p^3 + 29 p^4 + 55.2222 p^5 + 102.444 p^6 + 186.36 p^7 \\

            &\quad + 333.906 p^8 + 590.997 p^9 + 1035.58 p^{10} + 1799.39 p^{11} + 3104.2 p^{12} \\

            &\quad+ 5322.18 p^{13} + 9075.83 p^{14} + 15403.6 p^{15} + 26033.4 p^{16} + 43833.5 p^{17} \\

            &\quad+ 73555.2 p^{18} + 123053 p^{19} + 205290 p^{20} + 341620 p^{21} + 567161 p^{22} \\

            &\quad+ 939693 p^{23} + 1.5537\cdot10^{6} p^{24} + 2.56158\cdot10^{6} p^{25} + \mathcal{O}(p^{26})

\end{align*}

where the recursion steps were done with a computer. This assumes $n$ to be much larger than the largest power of $p$ considered.

Note: in the first line at the second term it uses that with probability $(3p-6p^2)$ the state goes to $\framebox{$101$}$ and then the expected number of resamplings is $1+R_{101}$. I (Tom) believe this requires the assumption $p_\mathrm{tot} := \sum_{\xi\in\paths{b}} \mathbb{P}[\xi] = 1$. To see why this is required, note that the actual term in the recursive formula should be $$(3p-6p^2)\cdot\left( \sum_{\xi\in\paths{101}} \mathbb{P}[\xi] \cdot \left( 1 + |\xi|\right) \right) = (3p-6p^2)\left( p_\mathrm{tot} + R_{101} \right)$$

When there would be a non-zero probability of never stopping the resample process then $p_\mathrm{tot}$ (the probability of ever reaching $\mathbf{1}$) could be less than one. Therefore I assume that $R^{(n)}(p)$ is finite which implies that the probability of ever reaching $\mathbf{1}$ is 1.

\newpage

\subsection{Cancellation of gapped configurations}

Here we prove claim \ref{claim:weakcancel}, the weaker version of the claim. We require the following definition

\begin{definition}[Path independence] \label{def:independence}

	We say two paths $\xi_i\in\paths{b_i}$ ($i=1,2$) of the Markov Chain are \emph{independent} if $\xi_1$ never resamples a site that was ever zero in $\xi_2$ and the other way around. It is allowed that $\xi_1$ resamples a $1$ to a $1$ that was also resampled from $1$ to $1$ by $\xi_2$ and vice versa. If the paths are not independent then we call the paths \emph{dependent}.

\end{definition}

\begin{definition}[Path independence - alternative] \label{def:independence2}

    Equivalently, on the infinite line $\xi_1$ and $\xi_2$ are independent if there is a site `inbetween' them that was never zero in $\xi_1$ and never zero in $\xi_2$. On the circle $\xi_1$ and $\xi_2$ are independent if there are \emph{two} sites inbetween them that are never zero.

\end{definition}

\begin{claim}[Sum of expectation values] \label{claim:expectationsum}

When $b=b_1\land b_2\in\{0,1\}^n$ is a state with two groups ($b_1\lor b_2 = 1^n$) of zeroes with $k$ $1$s inbetween the groups, then we have $R_b(p) = R_{b_1}(p) + R_{b_2}(p) + \mathcal{O}(p^{k})$ where $b_1$ and $b_2$ are the configurations where only one of the groups is present and the other group has been replaced by $1$s. To be precise, the sums agree up to and including order $p^{k-1}$.

\end{claim}

\textbf{Example}: For $b_1 = 0111111$ and $b_2 = 1111010$ we have $b=0111010$ and $k=3$. The claim says that the expected time to reach $\mathbf{1}$ from $b$ is the time to make the first group $1$ plus the time to make the second group $1$, as if they are independent. Simulation shows that

\begin{align*}

    R_{b_1} &= 1 + 3p + 7p^2 + 14.67p^3 + 29p^4 + \mathcal{O}(p^5)\\

    R_{b_2} &= 2 + 5p + 10.67p^2 + 21.11p^3+40.26p^4 + \mathcal{O}(p^5)\\

    R_{b} &= 3 + 8p + 17.67p^2 + 34.78p^3+65.27p^4 + \mathcal{O}(p^5)\\

    R_{b_1} + R_{b_2} &= 3 + 8p + 17.67p^2+35.78p^3 + 69.26p^4 +\mathcal{O}(p^5)

\end{align*}

and indeed the sums agree up to order $p^{k-1}=p^2$. When going up to order $p^{k}$ or higher, there will be terms where the groups interfere so they are no longer independent.

~

\begin{proof}

Consider a path $\xi_1\in\paths{b_1}$ and a path $\xi_2\in\paths{b_2}$ such that $\xi_1$ and $\xi_2$ are independent (Definition \ref{def:independence}). The paths $\xi_1,\xi_2$ induce $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ different paths of total length $|\xi_1|+|\xi_2|$ in $\paths{b_1\land b_2}$. In the sums $R_{b_1}$ and $R_{b_2}$, the contribution of these paths are $\mathbb{P}[\xi_1]\cdot |\xi_1|$ and $\mathbb{P}[\xi_2]\cdot |\xi_2|$. The next diagram shows how these $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ paths contribute to $R_{b_1\land b_2}$. At every step one has to choose between doing a step of $\xi_1$ or a step of $\xi_2$. The number of zeroes in the current state determine the probabilities with which this happens (beside the probabilities associated to the two original paths already). The grid below shows that at every point one can choose to do a step of $\xi_1$ with probability $p_i$ or a step of $\xi_2$ with probability $1-p_i$. These $p_i$ could in principle be different at every point in this grid. 

\begin{center}

\includegraphics{diagram_paths.pdf}

\end{center}

The weight of such a new path is the weight of the path in the diagram, multiplied by $\mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2]$. By induction one can show that the sum over all $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ paths in the grid is $1$. Hence the contribution of all $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ paths together to $R_{b_1\land b_2}$ is given by

\[

\mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2]\cdot(|\xi_1|+|\xi_2|) = \mathbb{P}[\xi_2]\cdot\mathbb{P}[\xi_1]\cdot|\xi_1| \;\; + \;\; \mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2]\cdot|\xi_2|.

\]

Ideally we would now like to sum this expression over all possible paths $\xi_1,\xi_2$ and use $p_\mathrm{tot}:=\sum_{\xi\in\paths{b_i}} \mathbb{P}[\xi] = 1$ (which also holds up to arbitrary order in $p$). The above expression would then become $R_{b_1} + R_{b_2}$. However, not all paths in the sum would satisfy the independence condition so it seems we can't do this. We now argue that it works up to order $p^{k-1}$.

For all $\xi\in\paths{b_1\land b_2}$ we have that \emph{either} $\xi$ splits into two independent paths $\xi_1,\xi_2$ as above, \emph{or} it does not. In the latter case, when $\xi$ can not be split like that, we know $\mathbb{P}[\xi]$ contains a power $p^k$ or higher because there is a gap of size $k$  and the paths must have moved at least $k$ times `towards each other' (for example one path moves $m$ times to the right and the other path moves $k-m$ times to the left). So the total weight of such a combined path is at least order $p^k$. Therefore we have

\[

	R_{b_1\land b_2} = \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1| + \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_1]\mathbb{P}[\xi_2]|\xi_2| + \sum_{\mathclap{\xi\;\mathrm{dependent}}} \mathbb{P}[\xi]|\xi|.

\]

where last sum only contains only terms of order $p^{k}$ or higher. Now for the first sum, note that

\[

	\sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|

    = \sum_{\xi_1\in\paths{b_1}} \sum_{\substack{\xi_2\in\paths{b_2}\\ \text{independent of }\xi_1}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|

\]

where the sum over independent paths could be empty for certain $\xi_1$. Now we replace this last sum by a sum over \emph{all} paths $\xi_2\in\paths{b_2}$. This will change the sum but only for terms where $\xi_1,\xi_2$ are dependent. For those terms we already know that $\mathbb{P}[\xi_1]\mathbb{P}[\xi_2]$ contains a factor $p^k$ and hence we have 

\begin{align*}

    \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|

    &= \sum_{\xi_1\in\paths{b_1}} \sum_{\xi_2\in\paths{b_2}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1| + \mathcal{O}(p^k) \\

    &= \sum_{\xi_1\in\paths{b_1}} \mathbb{P}[\xi_1]|\xi_1| + \mathcal{O}(p^k) \\

    &= R_{b_1} + \mathcal{O}(p^k)

\end{align*}

we can do the same with the second term and this proves the claim.

\end{proof}

~\\

\textbf{Proof of claim \ref{claim:weakcancel}}: We can assume $C$ consists of a group on the left with $l$ slots and a group on the right with $r$ slots (so $r+l=|C|$), with a gap of size $k=\mathrm{gap}(C)$ between these groups. Then on the left we have strings in $\{0,1'\}^l$ as possibilities and on the right we have strings in $\{0,1'\}^r$. The combined configuration can be described by strings $f=(a,b)\in\{0,1'\}^{l+r}$. The initial probability of such a state $C(a,b)$ is $\rho_{C(a,b)} = (-1)^{|a|+|b|} p^{r+l}$ and by claim \ref{claim:expectationsum} we know $R_{C(a,b)} = R_{C(a)} + R_{C(b)} + \mathcal{O}(p^k)$ where $C(a)$ indicates that only the left slots have been filled by $a$ and the other slots are filled with $1$s. The total contribution of these configurations is therefore

\begin{align*}

    \sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R_{C(f)}

    &= \sum_{a\in\{0,1'\}^l} \sum_{b\in\{0,1'\}^r} (-1)^{|a|+|b|}p^{r+l} \left( R_{C(a)} + R_{C(b)} + \mathcal{O}(p^k) \right) \\

    &=\;\;\; p^{r+l}\sum_{a\in\{0,1'\}^l} (-1)^{|a|} R_{C(a)} \sum_{b\in\{0,1'\}^r} (-1)^{|b|} \\

    &\quad + p^{r+l}\sum_{b\in\{0,1'\}^r} (-1)^{|b|} R_{C(b)} \sum_{a\in\{0,1'\}^l} (-1)^{|a|}

        + \mathcal{O}(p^{r+l+k})\\

    &= 0 + \mathcal{O}(p^{|C|+k})

\end{align*}

where we used the identity $\sum_{a\in\{0,1\}^l} (-1)^{|a|} = 0$.

~

It is useful to introduce some new notation: for any event $A$ (where an event is a set of paths), define

\begin{align*}

    \mathbb{P}_b(A) &= \mathbb{P}(A \;|\; \text{start in }b) \\

    R_{b,A} &= \mathbb{E}( \#resamples \;|\; A\;,\; \text{start in }b)

\end{align*}

Denote by $\mathrm{Z}_j$ the event that site $j$ becomes zero at any point in time before the Markov Chain terminates. Denote the complement by $\mathrm{NZ}_j$, i.e. the event that site $j$ does \emph{not} become zero before it terminates.

The proof of claim \ref{claim:expectationsum} also proves the following claim

\begin{claim}[Conditional independence] \label{claim:eventindependence}

    As in \ref{claim:expectationsum}, let $b=b_1\land b_2\in\{0,1\}^n$ be a state with two groups ($b_1\lor b_2 = 1^n$) of zeroes. Let $j_1$, $j_2$ be indices `inbetween' the groups (or only one index in case of the infinite line). Then we have

    \begin{align*}

        &=

        \; \cdot \;

        &=

        \; + \;

    \end{align*}

\end{claim}

\begin{proof}

    Note that any path $\xi\in\paths{b} \cap \mathrm{NZ}_j$ can be split into paths $\xi_1\in\paths{b_1}\cap \mathrm{NZ}_j$ and $\xi_2\in\paths{b_2}\cap\mathrm{NZ}_j$ and by the same reasoning as in the proof of claim \ref{claim:expectationsum}, we obtain

    \begin{align*}

        \mathbb{P}_b(\mathrm{NZ}_j)

        = \sum_{\substack{\xi\in\paths{b}\\\xi \in \mathrm{NZ}_j}} \mathbb{P}[\xi]

        &= \sum_{\substack{\xi_1\in\paths{b_1}\\\xi_1 \in \mathrm{NZ}_j}}

          \sum_{\substack{\xi_2\in\paths{b_1}\\\xi_2 \in \mathrm{NZ}_j}}

        \mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2] \\

        &=

        \mathbb{P}_{b_1}(\mathrm{NZ}_j)

        \; \cdot \;

        \mathbb{P}_{b_2}(\mathrm{NZ}_j).

    \end{align*}

    For the second equality, note that again by the same reasoning as in the proof of claim \ref{claim:expectationsum} we have

    \begin{align*}

        \mathbb{P}_b(\mathrm{NZ}_j) R_{b,\mathrm{NZ}_j}

        := \sum_{\substack{\xi\in\paths{b}\\\xi \in \mathrm{NZ}_j}} \mathbb{P}[\xi] |\xi| 

        &= \sum_{\substack{\xi_1\in\paths{b_1}\\\xi_1 \in \mathrm{NZ}_j}}

          \sum_{\substack{\xi_2\in\paths{b_2}\\\xi_2 \in \mathrm{NZ}_j}}

        \mathbb{P}[\xi_1]\mathbb{P}[\xi_2] (|\xi_1| + |\xi_2|) \\

        &=

        \mathbb{P}_{b_2}(\mathrm{NZ}_j) \mathbb{P}_{b_1}(\mathrm{NZ}_j) R_{b_1,\mathrm{NZ}_j}

        \; + \;

        \mathbb{P}_{b_1}(\mathrm{NZ}_j) \mathbb{P}_{b_2}(\mathrm{NZ}_j) R_{b_2,\mathrm{NZ}_j} .

    \end{align*}

    Dividing by $\mathbb{P}_b(\mathrm{NZ}_j)$ and using the first equality gives the desired result.

\end{proof}

~

Because of that, we have

\begin{align*}

    \mathbb{P}_{b_1}(\mathrm{PZ}_j) &= \mathbb{P}_{b_1}(\mathrm{Z}_{j-1} \cap \mathrm{NZ}_j) = \mathcal{O}(p^{j-1}) \\

\end{align*}

Now observe that

\begin{align*}

    R_b &= \sum_{j=1}^k \mathbb{P}_b(\mathrm{PZ}_j) R_{b,\mathrm{PZ}_j} + \mathbb{P}_b(\mathrm{AZ}) R_{b,\mathrm{AZ}} \\

        + \mathcal{O}(p^k) \\

        &= \sum_{j=1}^k \mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}

        - \sum_{j=1}^k \mathbb{P}_{b_2}(\mathrm{Z}_j)\mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}

        + \mathcal{O}(p^k) \\

        &= \sum_{j=1}^k \mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}

        + \mathcal{O}(p^k) \\

        &= R_{b_1}

        + \mathcal{O}(p^k) \\

\end{align*}

\newpage

    \subsection{Sketch of the (false) proof of the linear bound \ref{it:const}}

    Let us interpret $[n]$ as the vertices of a length-$n$ cycle, and interpret operations on vertices mod $n$ s.t. $n+1\equiv 1$ and $1-1\equiv n$.

    %\begin{definition}[Resample sequences]

    %	A sequence of indices $(r_\ell)=(r_1,r_2,\ldots,r_k)\in[n]^k$ is called resample sequence if our procedure performs $k$ consequtive resampling, where the first resampling of the procedure resamples around the mid point $r_1$ the second around $r_2$ and so on. Let $RS(k)$ the denote the set of length $k$ resample sequences, and let $RS=\cup_{k\in\mathbb{N}}RS(k)$.

    %\end{definition}

    %\begin{definition}[Constrained resample sequence]\label{def:constrainedRes}

    %	Let $C\subseteq[n]$ denote a slot configuration, and let $a\in\{\text{res},\neg\text{res}\}^{n-|C|}$, where the elements correspond to labels ``resampled" vs. ``not resampled" respectively. 

    %	For $j\in[n-|C|]$ let $i_j$ denote the $j$-th index in $[n]\setminus C$.

    %	We define the set $A^{(C,a)}\subseteq RS$ as the set of resample sequences $(r_\ell)$ such that for all $j$ which has $a_j=\text{res}$ we have that $i_j$ appears in $(r_\ell)$ but for $j'$-s which have $a_{j'}=\neg\text{res}$ we have that $i_{j'}$ never appears in $(r_\ell)$. 

    %\end{definition}    

    \begin{definition}[Conditional expected number of resamples]

    	For a slot configuration $C\subseteq[n]$ and $a\in\{\!\text{ever},\text{ never}\}^{n-|C|}$ we define the event $A^{(C,a)}:=\bigwedge_{j\in[n-|C|]}\{i_j\text{ has }a_j\text{ become }0\text{ before reaching }\mathbf{1}\}$,

    	where $i_j$ is the $j$-th vertex of $[n]\setminus C$.

    	Then we also define

    	$$R^{(C,a)}_b:=\mathbb{E}[\#\{\text{resamplings when started from inital state }b\}|A^{(C,a)}].$$

    \end{definition}     

    As in Mario's proof I use the observation that 

    \begin{align*}

    R^{(n)}(p) &= \frac{1}{n}\sum_{b\in\{0,1,1'\}^{n}} \rho_b \; R_{\bar{b}}(p)\\

    &= \frac{1}{n}\sum_{C\subseteq [n]}\sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R_{C(f)}(p)\\

    &= \frac{1}{n}\sum_{C\subseteq [n]}\sum_{f\in\{0,1'\}^{|C|}}\sum_{a\in\{\!\text{ever},\text{ never}\}^{n-|C|}} \rho_{C(f)} R^{{(C,a)}}_{C(f)}(p)P_{C(f)}(A^{(C,a)})\\

    &= \frac{1}{n}\sum_{C\subseteq [n]}\sum_{a\in\{\!\text{ever},\text{ never}\}^{n-|C|}} \sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R^{{(C,a)}}_{C(f)}(p)P_{C(f)}(A^{(C,a)}), 

    \end{align*}

    where we denote by $C\subseteq[n]$ a slot configuration, whereas $C(f)$ denotes the slots of $C$ filled with the particles described by $f$, while all other location in $[n]\setminus C$ are set to $1$. 

    When we write $R_{C(f)}$ we mean $R_{C(\bar{f})}$, i.e., replace $1'$-s with $1$-s. Since the notation is already heavy we dropped the bar from $f$, as it is clear from the context. Finally by $P_{C(f)}(A^{(C,a)})$ we denote the probability that the event $A^{(C,a)}$ holds.

    As in Definition for $j\in[n-|C|]$ let $i_j$ denote the $j$-th index in $[n]\setminus C$.

    Suppose that $a$ is such that there are two indices $j_1\neq j_2$ such that 

    $a_{j_1}=\text{never}=a_{j_2}$, moreover the sets $\{i_{j_1}+1,\ldots, i_{j_2}-1\}$ and $\{i_{j_2}+1,\ldots, i_{j_1}-1\}$ partition $C$ non-trivially, and we denote by $C_l$,$C_r$ the corresponding partitions. 

    I wanted to prove that

    \begin{equation}\label{eq:conditionalCancellation}

		\sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R^{{(C,a)}}_{C(f)}(p)=0,

    \end{equation}    

    based on the observation that for all $f\in\{0,1'\}^{|C|}$ we have 

    that 

    \begin{equation}\label{eq:keyIndependce}

    R^{{(C,a)}}_{C(f)}(p)=R^{{(C_l,a_l)}}_{C_l(f_l)}(p)+R^{{(C_r,a_r)}}_{C_r(f_r)}(p),

    \end{equation}

    where $f_l\in\{0,1'\}^{|C_l|}$ is defined as taking only the indices (and values) of $f$ corresponding to vertices of $C_l$, also $a_l\in[n-|C_l|]$ is defined such that $a$ and $a_l$ agree on vertices where $a$ is defined, and on the vertices where $a$ is not defined, i.e., the vertices of $C_r$ we define $a_l$ to contain ``never". We define things analogously for $f_r$ and $a_r$. 

    The reason why \eqref{eq:keyIndependce} holds is that as before the two halves of the cycle are conditionally independent because neither $i_{j_1}$ nor $i_{j_2}$ can become $0$. To be more precise each resample sequence $\left(C(f)\rightarrow \mathbf{1} \right)\in A^{(C,a)}$ can be uniquely decomposed to resample sequences $\left(C_l(f_l)\rightarrow \mathbf{1}\right)\in A^{(C_l,a_l)}$ and $\left(C_r(f_r)\rightarrow \mathbf{1}\right)\in A^{(C_r,a_r)}$. The sum of probabilities of the set of resample sequences $\{r\}$ which have decomposition $(r_l,r_r)$ have probability which is the product of the probabilities of $r_l$ and $r_r$ as shown in the proof of Claim~\ref{claim:expectationsum}. This proves that the set of all resample sequences $\left(C(f)\rightarrow \mathbf{1}\right)\in A^{(C,a)}$ for our purposes can be viewed as a product set with product probability distribution. Therefore the halves can be treated independently and so the expectation values just add up. 

    From here I wanted to mimic Mario's proof:

    \begin{align*}

    \sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R^{{(C,a)}}_{C(f)}(p)&=

    \sum_{f_l\in\{0,1'\}^{|C_l|}} \sum_{f_r\in\{0,1'\}^{|C_r|}}  (-1)^{|f_l|+|f_r|}p^{|C_l|+|C_r|} \left( R^{{(C_l,a_l)}}_{C_l(f_l)}(p) + R^{{(C_r,a_r)}}_{C_r(f_l)}(p) \right)\\

    &= p^{|C|}\sum_{f_l\in\{0,1'\}^{|C_l|}} (-1)^{|f_l|} R^{{(C_l,a_l)}}_{C_l(f_l)}(p) \sum_{f_r\in\{0,1'\}^{|C_r|}} (-1)^{|f_r|} \\

    &\quad + p^{|C|}\sum_{f_r\in\{0,1'\}^{|C_r|}} (-1)^{|f_r|} R^{{(C_r,a_r)}}_{C_r(f_r)}(p) \sum_{f_l\in\{0,1'\}^{|C_l|}} (-1)^{|f_l|} \\

    &= 0.

    \end{align*}

    The nasty issue which I did not realise that the missing term $P_{C(f)}(A^{(C,a)})$ is non-constant: even though the event $A^{(C,a)}$ is independent of $f$ the probability $P_{C(f)}(A^{(C,a)})=P_{C(f_l)}(A^{(C_l,a_l)})\cdot P_{C(f_r)}(A^{(C_r,a_r)})$ is not and so the above breaks down.

    Observe that if \eqref{eq:conditionalCancellation} would hold for configurations that cut the slot configuration to two halves it would imply that the only non-zero contribution comes from pairs $(C,a)$ such that $C\cup\{i_j:a_j=\text{ever}\}$ is connected. This is because if this set is not connected, then either we can cut $C$ to two halves non-trivially along ``never" vertices, or there is an island of $\text{ever}$ vertices separated from any slots, and therefore from any $0$-s. This latter case has zero contribution since we cannot set these indices to $0$, without reaching them by some resamplings, and thereby building a path of $0$-s leading there.

    If $|C\cup\{i_j:a_j=\text{ever}\}|\geq k+1$ then all contribution has a power at least $k+1$ in $p$ since $(C,a)$ requires the prior appearance of at least $k+1$ particles. If $n\geq k+1$ than all $(C,a)$ such that $|C\cup\{i_j:a_j=\text{ever}\}|\leq k$ appears exactly $n$ times, since $(C,a)$ cannot be translationally invariant. Moreover the quantity $R^{{(C,a)}}_{C(f)}(p)$ is independent of $n$ due to the conditioning that every resampling happens on a connected component of length at most $k<n$. This would prove that $a_k^{(n)}$ is constant for $n\geq k+1$. The same arguments would directly translate to the torus and other translationally invariant objects, so we could go higher dimensional as Mario suggested.

    Questions:

    \begin{itemize}

    	\item Is it possible to somehow fix this proof?

    	\item In view of this (false) proof, can we better characterise $a_k^{(k+1)}$?

    	\item Why did Mario's and Tom's simulation show that for fixed $C$ the contribution coefficients have constant sign? Is it relevant for proving \ref{it:pos}-\ref{it:geq}?

    	\item Can we prove the conjectured formula for $a_k^{(3)}$?		

    \end{itemize} 

\begin{comment}

    \subsection{Sketch of the proof of the linear bound \ref{it:const}}

    Let us interpret $[n]$ as the vertices of a length-$n$ cycle, and interpret operations on vertices mod $n$ s.t. $n+1\equiv 1$ and $1-1\equiv n$.

    \begin{definition}[Resample sequences]

		A sequence of indices $(r_\ell)=(r_1,r_2,\ldots,r_k)\in[n]^k$ is called resample sequence if our procedure performs $k$ consequtive resampling, where the first resampling of the procedure resamples around the mid point $r_1$ the second around $r_2$ and so on. Let $RS(k)$ the denote the set of length $k$ resample sequences, and let $RS=\cup_{k\in\mathbb{N}}RS(k)$.

    \end{definition}

    \begin{definition}[Constrained resample sequence]\label{def:constrainedRes}

    	Let $C\subseteq[n]$ denote a slot configuration, and let $a\in\{\text{res},\neg\text{res}\}^{n-|C|}$, where the elements correspond to labels ``resampled" vs. ``not resampled" respectively. 

    	For $j\in[n-|C|]$ let $i_j$ denote the $j$-th index in $[n]\setminus C$.

		We define the set $A^{(C,a)}\subseteq RS$ as the set of resample sequences $(r_\ell)$ such that for all $j$ which has $a_j=\text{res}$ we have that $i_j$ appears in $(r_\ell)$ but for $j'$-s which have $a_{j'}=\neg\text{res}$ we have that $i_{j'}$ never appears in $(r_\ell)$. 

    \end{definition}    

    \begin{definition}[Expected number of resamples]

		For $b\in\{0,1\}^n$ we define 

		$$R_b=\mathbb{E}[\#\{\text{resamplings when started from inital state }b\}],$$

		and for $(C,a)$ as in the previous definition we also define

		$$R^{(C,a)}_b=\mathbb{E}[\#\{\text{resamplings }\in A^{(C,a)} \text{ when started from inital state }b\}].$$

		Here we mean by the latter that after each resampling we check whether the sequence of resamplings so far is in $A^{(C,a)}$, if yes we count it, otherwise we do not count.

    \end{definition}     

    As in Mario's proof I use the observation that 

    \begin{align*}

    R^{(n)}(p) &= \frac{1}{n}\sum_{b\in\{0,1,1'\}^{n}} \rho_b \; R_{\bar{b}}(p)\\

    &= \frac{1}{n}\sum_{C\subseteq [n]}\sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R_{C(f)}(p)\\

    &= \frac{1}{n}\sum_{C\subseteq [n]}\sum_{f\in\{0,1'\}^{|C|}}\sum_{a\in\{\text{res},\neg\text{res}\}^{n-|C|}} \rho_{C(f)} R^{{(C,a)}}_{C(f)}(p)\\

    &= \frac{1}{n}\sum_{C\subseteq [n]}\sum_{a\in\{\text{res},\neg\text{res}\}^{n-|C|}} \sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R^{{(C,a)}}_{C(f)}(p), 

    \end{align*}

    where we denote by $C\subseteq[n]$ a slot configuration, whereas $C(f)$ denotes the slots of $C$ filled with the particles described by $f$, while all other location in $[n]\setminus C$ are set to $1$. 

	When we write $R_{C(f)}$ we mean $R_{C(\bar{f})}$, i.e., replace $1'$-s with $1$-s. Since the notation is already heavy we dropped the bar from $f$, as it is clear from the context.