## Formula Development (Descent Portion)

### -- starting from e.g. {5[8]} divide into 3 subsets == {(0[3]}, {1[3]}, and {2[3]}

-- {5[8]} is completely composed of {5[24]},{13[24]}, and {21[24]}

-- in general, {c[d]} becomes {c[3d]}, {(c+d)[3d]}, and {(c+2d)[3d]}

-- only *c* of *c[d]* determines the next step (because 3d[3] is 0)

-- whence {5[24]} gives *s*-step predecessors; {13[24]} gives *b*-step
predecessors; {21[24]} is a leaf set

-- apply this same process to each subset to generate
successive predecessor sets

-- examine a table which goes 3 levels deep

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