That’s what I figured. But let’s say that you gave each of your numbers (the six you split the original into) to a different person, and each person sMerits another non-intersecting person in a linear manner. As you say, there will be six people with one Merit at the leaf of each linear chain. What we cannot do is then scoop those six units together, generate a batch of 6 sMerits, and generate four more after two steps. What I mean by this is that the last part of the algorithm does not seem to apply in practice to sMerits, because we cannot scoop the ones together...