= \frac{1-q^3}{3q^3}\\

    	R^{(4)}(p) &= \frac{p(6-12p+10p^2-3p^3)}{6(1-p)^4}

                    = \frac{(1-q)(1+q+q^2+3q^3)}{6q^4}\\

        R^{(5)}(p) &= \frac{p(90-300p+435p^2-325p^3+136p^4-36p^5+6p^6)}{15(1-p)^5(6-2p+p^2)}\\

                   &= \frac{(1-q)(6+5q+6q^2+21q^3+46q^4+6q^6)}{15q^5(5+q^2)}

    \end{align*}

	If statements \ref{it:pos}-\ref{it:lim} are true, then we can define the function 

	$$R^{(\infty)}(p):=\sum_{k=0}^{\infty}a^{(k+1)}_k p^k,$$

	which would then have radius of convergence $p_c$, also it would satisfy for all $p\in[0,p_c)$ that $R^{(n)}(p)\leq R^{(\infty)}(p)$ and $\lim\limits_{n\rightarrow\infty}R^{(n)}(p)=R^{(\infty)}(p)$.

	It would also imply, that for all $p\in(p_c,1]$ we get $R^{(n)}(p)=\Omega\left(\left(\frac{p}{p_c}\right)^{n/2}\right)$.

	This would then imply a very strong critical behaviour. It would mean that for all $p\in[0,p_c)$ the expected number of resamplings is bounded by a constant $R^{(\infty)}(p)$ times $n$, whereas for all $p\in(p_c,1]$ the expected number of resamplings is exponentially growing in $n$.