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Tom Bannink - 8 years ago 2017-09-10 16:10:16
tombannink@gmail.com
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@@ -167,27 +167,29 @@
 
            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 & \cellcolor{blue!25}1009.38 & 1637.13 & % 2650.74 & 4285.68 & 6913.55 & 11171.2 & 18052.2
 
        \end{tabular}
 
	}
 
	\end{table}
 

	
 
	We observe that this is a power series in $p$. We discovered a very regular structure in this power series. It seems that for all $k\in\mathbb{N}$ and for all $n>k$ we have that $a^{(n)}_k$ is constant, this conjecture we verified using a computer up to $n=14$. 
 
	\newpage
 
	\noindent Based on our calculations presented in Table~\ref{tab:coeffs} and Figure~\ref{fig:coeffs_conv_radius} we make the following conjectures:
 
	\begin{enumerate}[label=(\roman*)]
 
		\item $\forall k\in\mathbb{N}, \forall n\geq 3 : a^{(n)}_k\geq 0$	\label{it:pos}	
 
        (A simpler version: $\forall k>0: a_k^{(3)}=(k+1)(k+2)/6$)
 
		\item $\forall k\in\mathbb{N}, \forall n>m\geq 3 : a^{(n)}_k\geq a^{(m)}_k$ \label{it:geq}		
 
		\item $\forall k\in\mathbb{N}, \forall n,m\geq \max(k,3) : a^{(n)}_k=a^{(m)}_k$ \label{it:const}		
 
		\item $\forall k\in\mathbb{N}, \forall n,m > \max(k,3) : a^{(n)}_k=a^{(m)}_k$ \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}
 
	\colorbox{red}{\ref{it:pos}-\ref{it:geq} is false since $a_{1114}^{(10)}<0$ -- needs to be double checked!}
 
	
 
	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}
 
	%\includegraphics[width=0.5\textwidth]{log_coeffs.pdf}	
 
	\caption{$1\left/\sqrt[k]{a_{k}^{(k+1)}}\right.$} %$\frac{1}{\sqrt[k]{a_k^{(k+1)}}}$
 
	\label{fig:coeffs_conv_radius}
 
	\end{figure}
 
    
 
    For reference, we also explicitly give formulas for $R^{(n)}(p)$ for small $n$. We also give them in terms of $q=1-p$ because they sometimes look nicer that way.
 
    \begin{align*}
 
    	R^{(3)}(p) &= \frac{1-(1-p)^3}{3(1-p)^3}
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