Perfect out-shuffle
= Perfect out-shuffle
{title2=$i\mapsto2i\bmod(n-1)$}
Interleave the two equal halves of an even finite <list>, placing the first half's next entry before the second half's next entry. With zero-based positions and length $n>2$, position $i<n-1$ moves to $2i\bmod(n-1)$, and the last position stays fixed. For an index in the first half the new position is $2i$; in the second it is $2i-(n-1)$, proving the formula. The least positive period of the whole <permutation> is the <multiplicative order> of $2$ modulo $n-1$: equality of the multiplier to one fixes every position, and fixing position one requires that equality. In particular, a 52-entry <list> has period eight. The empty and two-entry cases are identity <permutations> with least positive period one.