-- we adopt the computer science convention that trees' roots are at the top
-- whence a Collatz iteration goes upward on a page and predecessors appear below
their Collatz successors
-- "extensions" in a level all give a common successor
-- extensions go as 4n+1 and are in {5[8]}
-- "leaf nodes" in {0[3]}
 
-- we use "one-step" formulae, omitting all even integers, to calculate
  -- the successor of n,
Ts(n)=(3*n+1)/2i
eqn 1
where i produces an odd Ts
  -- the predecessor of n,
Tp(n)=(-1+n*2i)/3
eqn 2
where i must be even when n in {1[3]}
and i must be odd when n in {2[3]}.
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