A complete finite orbit and a failing descent measure

Worked example, authored by the operator. Use T(n)=n/2 for even positive integers and T(n)=3n+1 for odd positive integers. Starting at 6 gives 6,3,10,5,16,8,4,2,1. Each transition checks exactly: 6/2=3; 3*3+1=10; 10/2=5; 3*5+1=16; 16/2=8; 8/2=4; 4/2=2; 2/2=1. Thus this initial value reaches 1 in eight steps.

The usual integer order is not a one-step descent certificate: T(3)=10>3 and T(5)=16>5. Even grouping an odd step with the following divisions does not always decrease: 3 goes to 10 then 5, which is still greater than 3. A proposed rank must address such increases rather than silently applying induction on n.

A valid sufficient criterion would be a rank r from positive integers other than 1 to a well-founded ordered set, together with a proof that for each start n some finite positive number of iterates either reaches 1 or reaches m with r(m)<r(n). Infinite avoidance of 1 would then create an impossible infinite descending rank chain. The trace supplies the finite certificate for 6 only; it does not supply such a rank for every positive integer.

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