Question 15
Julian and Caleb alternate removing either one coin or two adjacent coins from a row of $n\ge1$ coins numbered $1$ to $n$. Gaps remain; coins originally separated by a gap are not adjacent. The player removing the last coin wins. Julian starts. Find the sum of all $n\le10$ for which Julian can always win. Example: from $1,2,3$ one may remove any single coin or the pairs $(1,2),(2,3)$; from $1,3$ the pair is forbidden.
