 Two-step family with a third-order member

The exponential-symbol order criterion for a multistep method gives leading defect $-(1+5a)z^3/12$, so the family has order two except at $a=-1/5$, where it has order three. The root condition for a multistep method holds exactly for $-1\le a<1$, and the convergent A-stable members are exactly $0\le a<1$. At $a=0$ it reduces to the trapezoidal rule after removal of a zero root. At $a=1$ a repeated root one destroys convergence of a numerical method of the original recurrence despite a common polynomial factor and a formally vanishing second-order residual.