Going low to high would do this:
1. Sort with step size 1 (aka. basic insertion sort, O(n^2) worst performance)
2. Sort with step size 3 (already sorted, O(n))
3. Sorted with step size 5 (already sorted, O(n))
When you add up all of these performance times, it takes O(n^2 + n + n + n + ... + n), which technically boils down to O(n^2), but is considerable slower.
It's very easy to compute the step size:
for (int step = array.length() - array.length() % 2 - 1; i > 1; i -= 2)
// insertion sort stuff with incrementing by step