 Reciprocal-root three-step multistep family

Take $\sigma (\zeta )=\zeta [(5+\alpha ){\zeta }^2-(4+8\alpha )\zeta +11-5\alpha ]/6$. The root condition for a multistep method holds exactly for $-1<\alpha <1$. Since ${\rho }^{\prime }(1)=\sigma (1)=2(1-\alpha )$, these and only these parameters give a convergent method. Its exponential-symbol order criterion for a multistep method has defect $-(\alpha +5)z^4/12+O(z^5)$, giving formal order three except at $\alpha =-5$, where the first defect is $z^5/10$. None of the convergent family is A-stable: for a parasitic root $r=\alpha +i\sqrt{1-{\alpha }^2}$, perturbation of $\rho (r(z))-z\sigma (r(z))=0$ gives $\mathrm{Re}[r^{\prime }(0)/r]=(\alpha -1)/12<0$. A small negative real $z$ therefore moves it outside the unit disk. At $\alpha =1$, cancellation would remove a double unit root, but the original recurrence retains that unstable parasitic behavior; it is not legitimately replaced by the reduced Backward Euler recurrence for arbitrary starting values.