Another "fun" thing regarding this.
Compute a cell (cell_0) with (x+n) odd and n even and d even.
Then create a new cell with a = n, b = 2*d + n, with n and d from cell_0 and call this cell_1.
x+n from cell_1 is equal to d from cell_0. And the difference between d/2 - x is the same as n - d/2.
Sorry for derailing a bit, I know this stuff belongs back in thread #8-9, but I haven't noticed this pattern before.