diff --git a/main.tex b/main.tex
index de43fef3939747b2ed9a2620bf3de3a12122e15f..6cf2e851f04da8964d15316dda774f132d7c905f 100644
--- a/main.tex
+++ b/main.tex
@@ -57,6 +57,7 @@
 
 \newcommand{\diam}[1]{\mathcal{D}\left(#1\right)}
 \newcommand{\paths}[1]{\mathcal{P}\left(#1\to\mathbf{1}\right)}
+\newcommand{\gapsum}[1]{\mathrm{gapsum}\left(#1\right)}
 
 \long\def\ignore#1{}
 
@@ -147,7 +148,9 @@
 			8 &	0 & 1 & 2 & 3+2/3 & 6.44 & 11.0 & 18.7 & \cellcolor{blue!25}31.44 & 52.08 & 84.95 & 136.0 & 213.6 & 328.9 & 496.5 & 735.6 & 1070.7 & 1532.5 & 2159.5 & 2998.8 & 4108.1 & 5556.7 \\
 			9 &	0 & 1 & 2 & 3+2/3 & 6.44 & 11.0 & 18.7 & 31.44 & \cellcolor{blue!25}52.30 & 86.27 & 140.7 & 226.3 & 358.4 & 558.4 & 855.4 & 1289.0 & 1911.5 & 2791.4 & 4017.2 & 5701.4 & 7985.9 \\
 			10&	0 & 1 & 2 & 3+2/3 & 6.44 & 11.0 & 18.7 & 31.44 & 52.30 & \cellcolor{blue!25}86.49 & 142.1 & 231.6 & 373.4 & 594.8 & 934.4 & 1447.1 & 2209.0 & 3324.6 & 4934.8 & 7226.9 & 10447. \\
-		\end{tabular}
+            \vdots \\
+            15& 0 & 1 & 2 & 3+2/3 & 6.44 & 11.08 & 18.76 & 31.45 & 52.31 & 86.49 & 142.33 & 233.31 & 381.17 & 621.02 & 1009.38 & \cellcolor{blue!25}1637.13 & % 2650.74 & 4285.68 & 6913.55 & 11171.2 & 18052.2
+        \end{tabular}
 	}
 	\end{table}
 
@@ -161,7 +164,7 @@
 		\item $\forall k\in\mathbb{N}, \forall n,m\geq \max(k,3) : a^{(n)}_k=a^{(m)}_k$ \label{it:const}		
   		\item $\exists p_c=\lim\limits_{k\rightarrow\infty}1\left/\sqrt[k]{a_{k}^{(k+1)}}\right.$ \label{it:lim}			
 	\end{enumerate}
-	We also conjecture that $p_c\approx0.62$, see Figure~\ref{fig:coeffs_conv_radius}.
+	We also conjecture that $p_c\approx0.61$, see Figure~\ref{fig:coeffs_conv_radius}.
 
 	\begin{figure}[!htb]\centering
 	\includegraphics[width=0.5\textwidth]{coeffs_conv_radius.pdf}
@@ -269,13 +272,13 @@ and indeed the lowest order is $\diam{C}=9$.
 
 ~
 
-A weaker version of the claim is that if $C$ contains a gap of size $k$, then the sum is zero up to and including order $p^{k-1}$.
+A weaker version of the claim is that if $C$ contains a gap of size $k$, then the sum is zero up to and including order $p^{|C|+k-1}$.
 \begin{claim}[Weak cancellation claim] \label{claim:weakcancel}
 	For $C\subseteq[n]$ a configuration of slot positions, the lowest order term in
     \begin{align*}
         \sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R_{C(f)} ,
     \end{align*}
-	is at least $p^{\mathrm{gap}(C)}$ when $n$ is large enough. Here $\mathrm{gap}(C)$ is defined as in Figure \ref{fig:diametergap}, its the size of the largest gap of $C$ within the diameter of $C$. All lower order terms cancel out.
+	is at least $p^{|C|+\mathrm{gap}(C)}$ when $n$ is large enough. Here $\mathrm{gap}(C)$ is defined as in Figure \ref{fig:diametergap}, its the size of the largest gap of $C$ within the diameter of $C$. All lower order terms cancel out.
 \end{claim}
 This weaker version would imply \ref{it:const} but for $\mathcal{O}(k^2)$ as opposed to $k+1$.
 
@@ -338,13 +341,23 @@ The key ingredient of the proof is the following claim:
 \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}
-
-For example for $b_1 = 10111111$ and $b_2 = 11111000$ we have $b=10111000$ 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. When going up to order $p^{k}$ or higher, there will be terms where the groups interfere so they are no longer independent.
+\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 path 1 or a step of path 2. The number of zeroes in the current state determine probabilities with which this happens (aside from 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 path 1 with probability $p_i$ or a step of path 2 with probability $1-p_i$. These $p_i$ could in principle be different at every point in this grid. The weight of such a new path is the weight of the path in the diagram below, 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
+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
 \[
 \mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2]\cdot(|\xi_1|+|\xi_2|) = \mathbb{P}[\xi_2]\cdot\mathbb{P}[\xi_1]\cdot|\xi_1| \;\; + \;\; \mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2]\cdot|\xi_2|.
 \]
@@ -367,20 +380,20 @@ where the sum over independent paths could be empty for certain $\xi_1$. Now we
 \end{align*}
 we can do the same with the second term and this proves the claim.
 \end{proof}
-\begin{center}
-\includegraphics{diagram_paths.pdf}
-\end{center}
 
-\textbf{Proof of claim \ref{claim:weakcancel}}: Say we have a group on the left with $l$ slots and a group on the right with $r$ slots, with enough space between the 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 $(a,b)\in\{0,1'\}^{l+r}$. Such a configuration has probability $(-1)^{|a|+|b|} p^{r+l}$ in $\rho$ and by claim \ref{claim:expectationsum} we know $R_{(a,b)} = R_a + R_b + \mathcal{O}(p^\mathrm{spacing})$. The total contribution of these configurations is therefore
+~\\
+\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_{a\in\{0,1'\}^l} \sum_{b\in\{0,1'\}^r} (-1)^{|a|+|b|}p^{r+l} \left( R_a + R_b \right) + \mathcal{O}(p^\mathrm{spacing})
-    &= p^{r+l}\sum_{a\in\{0,1'\}^l} (-1)^{|a|} R_a \sum_{b\in\{0,1'\}^r} (-1)^{|b|} \\
-    &\quad + p^{r+l}\sum_{b\in\{0,1'\}^r} (-1)^{|b|} R_b \sum_{a\in\{0,1'\}^l} (-1)^{|a|} \\
-    &\quad + \mathcal{O}(p^\mathrm{spacing})\\
-    &= 0 + \mathcal{O}(p^\mathrm{spacing})
+    \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 $\sum_{a\in\{0,1\}^l} (-1)^{|a|} = 0$.
 
+\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]
