So you have a correct (d+n) and a starting (d+n) same with (x+n). Look at the differences between the start and correct (d+n) and (x+n). Then look at the difference between these differences. The result is always a-1.
So you have a correct (d+n) and a starting (d+n) same with (x+n). Look at the differences between the start and correct (d+n) and (x+n). Then look at the difference between these differences. The result is always a-1.