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.

    As in Definition~\ref{def:constrainedRes} 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 

    $a_{j_1}=\neg\text{res}=a_{j_2}$, 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. 

    We claim that 

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

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

    that 

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

    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 $\neg\text{res}$. We define things analogously for $f_r$ and $a_r$.

    The reason why \eqref{eq:keyIndependceWrong} holds is as before that the two halves of the cycle are conditionally independent because neither $i_{j_1}$ nor $i_{j_2}$ are resampled. One could probably also argue similarly as Tom's grid figure shows. 

    From here the proof goes just as in 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*}

    Observe that it implies that the only non-zero contribution comes from pairs $(C,a)$ such that $C\cup\{i_j:a_j=\text{res}\}$ is connected. This is because if this set is not connected, then either we can cut $C$ to two halves non-trivially along $\neg\text{res}$ vertices, or there is an island of $\text{res}$ vertices separated from any slots, and therefore from any $0$-s. This latter case has zero contribution since we cannot resample these indices without first setting them to $0$, but under the conditions they can be never reached by any resampling, therefore they remain $1$ always.

    If $|C\cup\{i_j:a_j=\text{res}\}|\geq k+1$ 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 $|C\cup\{i_j:a_j=\text{res}\}|\leq k$ 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 proves that $a_k^{(n)}$ is constant for $n\geq k+1$.

    Note that the heart of the proof is \eqref{eq:keyIndependceWrong}, so this is what we should double check.    

	The same arguments directly translate to the torus and other translationally invariant objects, so we can go higher dimensional :-) as Mario suggested.

	Questions:

	\begin{itemize}

		\item In view of this 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} 

\end{comment}

\begin{comment}    

    \begin{definition}[Neighborhood]

	   	For the length-$n$ cycle we identify sites with $[n]$. 

	   	For a subset $S\subseteq [n]$ we define the $k$ neighborhood of $S$ as

	   	$N_k(S):=\cup_{s\in S} \{s-k,s-k+1,\ldots,s+k\}$ where numbers are interpreted mod $n$ and we represent the $\equiv 0$ class by $n$).

	\end{definition}

	\begin{definition}[Blocks and Gaps]

	   	For a configuration $C\subseteq [n]$ we call the connected components of $[n]\setminus N_1(C)$ the gaps. We denote by $m_C$ the number of gaps.

	   	We call a non-empty subset $B\subset C$ a block if $N_3(B)\cap C=B$ and $B$ is minimal, i.e., there is no proper subset $\emptyset\neq B'\subsetneq B$ satisfying $N_3(B')\cap C=B'$. 

	   	Observe that whenever $m_C\geq 2$ the number of blocks is the same as the number of gaps.  

    \end{definition}

    \begin{definition}[Crossings]

    	We say that a run (path) of the resampling procedure crosses $i\in[n]$ if there is ever a $0$ in $N_1({i})$ during the run.

    \end{definition}

    \begin{definition}[Enumerating gaps and mid points]

		Let $G_1,G_2,\ldots, G_{m_C}$ be an enumeration of the gaps respecting the cyclic ordering, and let $g_i$ be the middle element of $G_i$, if there are two middle elements we choose the smaller according to the cyclic ordering. (If $m_C=1$ and $G_1=[n]$ let $g_1=1$.)

		If $m_C\geq 2$ then for all $i\in[m_C]$ let $B_i$ be the block between $G_i$ and $G_{i+1}$.

    \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_b(p)\\

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

    For $a\in\{\text{crossed},\text{not crossed}\}^m$ we also introduce the notation $R^a_{C(f)}(p):=\mathbb{E}(\#\{\text{resamples before reaching }\mathbbm{1} \text{ from } C(f)\}|\bigwedge_{j\in[m_C]}g_j \text{ is } a_j)\cdot\mathbb{P}(\bigwedge_{j\in[m_C]}g_j \text{ is } a_j)$, which we define as $0$ if the conditioning event has $0$ probability. 

    Since $$R_{C(f)}(p)=\sum_{a\in\{\text{crossed},\text{not crossed}\}^{m_C}}R^a_{C(f)}(p),$$ we can further rewrite the expectation as

    \begin{align*}

	    R^{(n)}(p) &= \frac{1}{n}\sum_{C\subseteq [n]}\sum_{a\in\{\text{crossed},\text{not crossed}\}^{m_C}}\sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R^a_{C(f)}(p).

    \end{align*}

    Suppose that $a$ contains at least two ``not crossed'', the we claim that $\sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R^a_{C(f)}(p)=0$. Let $j_1\neq j_2$ be two distinct indexes such that $a_{j_1}$ and $a_{j_2}$ are both saying ``not crossed''. Let $B_l:=B_{j_1}\cup B_{j_1+1}\cup\cdots\cup B_{j_2-1}$ and $B_r:=B_{j_2}\cup B_{j_2+1}\cup\cdots\cup B_{j_1-1}$ (again we interpret indexes mod $m_C$).

    Then we claim that for all $f\in\{0,1'\}^{|C|}$ we have 

    that 

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

		R^a_{C(f)}(p)=R^a_{B_l(f)}(p)+R^a_{B_r(f)}(p).

    \end{equation} 

    The reason is as before that the halves are independent because neither $g_{j_1}$ nor $g_{j_2}$ is crossed. One could probably similarly prove it as the grid figure shows. 

    From here the proof goes just as in Mario's proof:

    \begin{align*}

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

    \sum_{f_l\in\{0,1'\}^{|B_l|}} \sum_{f_r\in\{0,1'\}^{|B_r|}}  (-1)^{|f_l|+|f_r|}p^{|B_l|+|B_r|} \left( R^a_{B_l(f)} + R^a_{B_r(f)} \right)\\

    &= p^{|C|}\sum_{f_l\in\{0,1'\}^{|B_l|}} (-1)^{|f_l|} R^a_{B_l(f)} \sum_{f_r\in\{0,1'\}^{|B_r|}} (-1)^{|f_r|} \\

    &\quad + p^{|C|}\sum_{f_r\in\{0,1'\}^{|B_r|}} (-1)^{|f_r|} R^a_{B_r(f)} \sum_{f_l\in\{0,1'\}^{|B_l|}} (-1)^{|f_l|} \\

    &= 0 

    \end{align*}

    From this it follows that the only contribution comes from paths that cross all but one (or all) of the mid gaps. This then implies that it is enough to consider $\mathcal{O}(k)$ length configurations. (We define the length of a configuration $C$ as $n-\max_{j\in[m_C]}|G_j|$.)

    Note that the heart of the proof is \eqref{eq:keyIndependceOld}, so this is what we should double check.

    In fact I think the independence that we use in \eqref{eq:keyIndependceOld} can be also proven when we define a crossing as crossing the actual point, and not its $1$-neighborhood. It then would make it possible to define blocks as consecutive slacks. Also then we could actually use all points of the gaps not only the mid points. The requirement for the cancellation would be that there are ``not crossed'' labels from at least two distinct gaps. This would probably lead to the optimal $k+1$ bound giving the actual statement \ref{it:const}. 

    Speculation: The $n=k$ case would then probably not work because the all $0$ starting configuration is invariant under rotations.

    To actually go below $2k$ one needs to be careful, because there are periodic configurations that are invariant under some rotations causing double counting issues. This can be probably resolved by showing that when a pattern becomes periodic for some $n$ it actually produces periodicity times more expectation due to symmetry. But this is all just speculation.

\end{comment}

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