## Why an Arithmetic Argument Fails; Why Twins (etc.) Arise

On the way down from the root node of the abstract predecessor tree,  each generation produces disjoint sets in the succeeding generation.  What we'll do here is show that an argument from the magnitude of the integers cannot show that the subsets are disjoint.  As a byproduct, we'll see why there is such a predilection for closely spaced integers to arise in close proximity in the predecessor tree.

Each step in generating children forms new coefficients and new constants in the resulting descriptive formula as determined by the properties of the parental set.  From the expression for the set in the parental node,  dn+c,  the three subsets whose values mod 3 will determine the directions to be taken to the next generation are 3dn+c,  3dn+c+d and 3dn+c+2d.  Since 3dn is congruent to 0 mod 3,  the values of cc+d,  and c+2d determine the residue mod 3 of each subset.  Because d (a power of 2)  is relatively prime to 3 on the way down in the abstract predecessor tree,  cc+d,  and c+2d each have a different residue mod 3,  so one of the three subsets will be leaves,  another will take an s step,  and another a b step.  The predecessor tree is grown by indefinite repetition of this process.

The s steps produce children with one of the constants (2c-1)/3,  (2(c+d)-1)/3,  or (2(c+2d)-1)/3 and the b steps with one of (4c-1)/3,  (4(c+d)-1)/3,  or (4(c+2d)-1)/3.  If the -1 is omitted,  and the floor function employed instead,  these become floor(2c/3),  floor(2(c+d)/3),  floor(2(c+2d)/3),  floor(4c/3),  floor(4(c+d)/3),  and floor(4(c+2d)/3),  where the argument to the floor function always has a fractional part equal to 1/3.  Clearly whatever combination of children arise,  the children will be disjoint from the parent.

Since the tree starts with 8n+5,  in which c<d,  the coefficients of the successive generation are 2d/3 for the s step and 4d/3 for the b step,  and the expressions for the constants always result from applying a floor function to a mixed number,  the new c will always be less than the new d.  Applied iteratively,  this shows that c<d in all formulas for sets.

There are two possible regions of coincidence or overlap in the children sets in the event that c closely approaches d,  at floor(4c/3)  and floor(2(c+d)/3),  and between floor(4(c+d)/3)  and floor(2(c+2d)/3).  The latter is of more concern because there is a range of possible overlap there.  While it is often the case that children reached from 3dn+c+d are smaller than children reached from 3dn+c+2d,  the values can occur in the inverse order when 3dn+c+d takes the b step and 3dn+c+2d takes the s step.  There are two early occurrences of this in the predecessor tree: first,  15 as parent gives 105 from 3dn+c+d using the b step and 95 from 3dn+c+2d using the s step and,  second,  3 as parent gives 25 from 3dn+c+d using the b step and 23 from 3dn+c+2d using the s step.

And what if,  in two successive generations,  one path goes bs and another path goes sb?  Both would reach a constant roughly 8/9 of the original.  No such simple cases appear to exist,  but the multiple instances of closely spaced integers closely positioned in the predecessor tree result from just such an interchange of steps except that they are interspersed with extension steps.  (E.g.  847 reaches 12049 by an eseb path but 12045 by a bese path,  and 385 reaches 9733 by an esbbbe path,  but 9731 by a bebbes path.)  Many more complex paths with interchanged steps would lead to the same point if it were not for the effect of the floor function at each generational level.  (E.g.  583 reaches 7371 by an esbebs path,  but 7369 by a beessb path.  [I'm using e here to denote a single step to a right child in the binary predecessor tree,  so ee means two steps through the right descent.])  This phenomenon is not limited to paths with equal numbers of b and s steps.  (E.g.  1429 reaches 10161 by beb,  but reaches 10163 by eses.)

The argument that all subsets are unique will have to depend on deeper issues than simple numerical ranges.