Since the forest is so hard to see from the trees in the extensive discussion
of various attributes of the (3n+3^{j})/2^{i}
systems, it seems worthwhile to summarize some key behaviors in a tabular
form.
It was conjectured that there might be some revealing features in the extended
family which would suggest an approach to a proof of the Collatz
conjecture, but
the features exhibited in the following table are so disparate that it is hard
to find any such helpful one.
for the case: | (3n+1)/2^i (j=0) | (3n+3)/2^i (j=1) | (3n+3^j)/2^i (j>=2) |
steps to predecessors: | both "s"(i=1) and "b"(i=2) | "s" steps (i=1) only (Note 1)
(Note 2) | i even or odd (Note 3) |
1st predecessor extensions: | at i=4,i=3 | at i=2 | i may be large (Note 3) |
further extensions from n at: | 4n+1 | 2n+1 | 2n+3^(j-1) |
leaf nodes are: | 0[3] | 1[3],2[3] | 1[3],2[3] |
internal nodes are: | 1[3],2[3] | 0[3] | 0[3] |
left descent headers are: | 5[8] | 0[3] extended from 1[3] | 0[3] extended from 0[3] |
l.d.a. element size changes: | both decrement ("s") and increment ("b") | decrement only | decrement only |
extensions are: | from all: 0[3],1[3],2[3] | from 0[3]:1[3], from 1[3]:0[3]; from 2[3]:2[3] | from 0[3]:0[3]; from 1[3]:1[3],2[3]; from 2[3]:1[3].2[3] |
Note 1: It is a misleading usage to refer to steps in the predecessor graph when j>0 as "s" steps since the whole structure built to describe the (3n+1)/2^{i} predecessors fails in one or more aspects when j≥1. In this instance, "s" means only that i is one and that predecessors are smaller than their successors.
Note 2: The only exception to the value of i being 1 is at the root of the tree, where (3*3+3)/2^i requires i=2 to return to the value 3 in the tree-rooting trivial loop.
Note 3: These odd behaviors are illustrated in the predecessor tree for (3n+3^12)/2^i. We need only look at the left-most branch of the developing tree. The predecessor formula (using p and s for predecessor and successor, respectively) is p=(2^i*s-3^12)/3. This will result in negative numbers wherever 2^i*s>3^12.
531441 (1) / \ (3) 177147 1240029 (1) / \ (2) \ (4) -59049 531441 2657205 \ (2) \ (3) \ (5) /----------------- 59049 1240029 5491557 (1) -> (2) -> (3) \ (3) \ (4) -137781 -98415 -19683 \ 295245 2657205 (4) \ (4) \ (5) /---------- 137781 767637 5491557 (1) -> (2) \ (5) \ (5) -85293 6561 452790 1712421 \ (3) \ (6) 190269 1082727 \ (4) \ (7) 557685 2342601 \ (5) 1292517
Although the predecessor of 3^12 itself is a normal "s" step product with its infinite set of extensions, the leftmost element in each of the next four generations of predecessors are small enough that one or more negative numbers would be reached, with the result that a value of i>1 is necessary to reach the smallest legitimate predecessor. The parenthesized integer above each predecessor indicates the value of i required to reach it. All the extensions (right descent elements) are carried out to the same vertical column. Even the root itself has its infinite set of extensions. Note that the first extension of 177147 has the value of the root itself; that produces the trivial loop at the head of the predecessor tree. The left descendants of all the nodes after the first of a set of extensions are omitted.
Those negative numbers need not concern us. Their presence is in some sense an artifact of our definition of the binary tree representation of the predecessor trees. Remember that when the Collatz itinerary is pursued from a number deep in a right descent, the itinerary will skip over all the elder entries in the right descent and over the right descent header itself to the next odd integer. So the appearance of negative numbers in the table in no way implies that negative numbers are visited in the itinerary.