A perfect binary Collatz tree

To compute the OPI values in the tree…Let q = the quotient of the lower OPI and r = its residue mod 3:

• Upward OPI…upward to the left:
  • if q is odd:
    • if 2q+1 is not ≡ 0 mod 3, then upward OPI = 2q+1
    • if 2q+1 ≡ 0 mod 3, then upward OPI = 8q+5
  • if q is even:
    • if 4q+1 is not ≡ 0 mod 3, then upward OPI = 4q+1
    • if 4q+1 ≡ 0 mod 3, then upward OPI = 16q+5
• Intra-prefix class OPI…upward to the right:
  • if r = 1, then the next OPI to the right (in class) = 4(OPI)+1
  • if r = 2, then the next OPI to the right (in class) = 16(OPI)+5

These rules will generate the identical OPI not ≡ 0 mod 3 found in the graph generated by the original Collatz function and will map onto the odd positive integer results of the original Collatz function precisely. The downward behavior of a terminal OPI ≡ 0 mod 3 will follow the same trajectory as any element not ≡ 0 mod 3 in its prefix class. The absence of all elements ≡ 0 mod 3 from consideration does not change the dependency structure of the OPIs that remain.

The structure is identical to that of the irrational elements >1 in the extension field encoded in the Stern-Brocot tree. In that tree between every two integer values to the right of 1 there is a subtree isomorphic to the whole left side (the subtree between 0 and 1). As encoded in the S-B tree every irrational element >1 in the extension field   has a finite prefix and an infinite alternating tail. The tree is rooted at 1. There are no cycles. The structure is hyperbolic.

In the odds-only Collatz graph every downward path from an OPI will eventually reach 1 under iteration of C_p

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