Editing Collatz conjecture (section)

=== Iterating on rationals with odd denominators ===
The Collatz map can be extended to (positive or negative) rational numbers which have odd denominators when written in lowest terms.
The number is taken to be 'odd' or 'even' according to whether its numerator is odd or even. Then the formula for the map is exactly the same as when the domain is the integers: an 'even' such rational is divided by 2; an 'odd' such rational is multiplied by 3 and then 1 is added. A closely related fact is that the Collatz map extends to the ring of [[2-adic integers]], which contains the ring of rationals with odd denominators as a subring.

When using the "shortcut" definition of the Collatz map, it is known that any periodic [[#As a parity sequence|parity sequence]] is generated by exactly one rational.<ref>{{Cite journal|last=Lagarias|first=Jeffrey|date=1990|title=The set of rational cycles for the 3x+1 problem|url=https://eudml.org/doc/206298|journal=Acta Arithmetica|volume=56|issue=1|pages=33–53|issn=0065-1036|doi=10.4064/aa-56-1-33-53|doi-access=free}}</ref> Conversely, it is conjectured that every rational with an odd denominator has an eventually cyclic parity sequence (Periodicity Conjecture<ref name="Lagarias (1985)"/>).

If a parity cycle has length {{mvar|n}} and includes odd numbers exactly {{mvar|m}} times at indices {{math|''k''<sub>0</sub> < ⋯ < ''k''<sub>''m''−1</sub>}}, then the unique rational which generates immediately and periodically this parity cycle is
{{NumBlk|:|<math>\frac{3^{m-1} 2^{k_0} + \cdots + 3^0 2^{k_{m-1}}}{2^n - 3^m}.</math>|{{EquationRef|1}}}}

For example, the parity cycle {{nowrap|(1 0 1 1 0 0 1)}} has length 7 and four odd terms at indices 0, 2, 3, and 6. It is repeatedly generated by the fraction
<math display="block">\frac{3^3 2^0 + 3^2 2^2 + 3^1 2^3 + 3^0 2^6}{2^7 - 3^4} = \frac{151}{47}</math>
as the latter leads to the rational cycle
<math display="block">\frac{151}{47} \rightarrow \frac{250}{47} \rightarrow \frac{125}{47} \rightarrow \frac{211}{47} \rightarrow \frac{340}{47} \rightarrow \frac{170}{47} \rightarrow \frac{85}{47} \rightarrow \frac{151}{47} .</math>

Any cyclic permutation of {{nowrap|(1 0 1 1 0 0 1)}} is associated to one of the above fractions. For instance, the cycle {{nowrap|(0 1 1 0 0 1 1)}} is produced by the fraction
<math display="block">\frac{3^3 2^1 + 3^2 2^2 + 3^1 2^5 + 3^0 2^6}{2^7 - 3^4} = \frac{250}{47} . </math>

For a one-to-one correspondence, a parity cycle should be ''irreducible'', that is, not partitionable into identical sub-cycles. As an illustration of this, the parity cycle {{nowrap|(1 1 0 0 1 1 0 0)}} and its sub-cycle {{nowrap|(1 1 0 0)}} are associated to the same fraction {{sfrac|5|7}} when reduced to lowest terms.

In this context, assuming the validity of the Collatz conjecture implies that {{nowrap|(1 0)}} and {{nowrap|(0 1)}} are the only parity cycles generated by positive whole numbers (1 and 2, respectively).

If the odd denominator {{mvar|d}} of a rational is not a multiple of 3, then all the iterates have the same denominator and the sequence of numerators can be obtained by applying the "{{math|3''n'' + ''d''}}&nbsp;" generalization<ref name="Belaga (1998a)"/> of the Collatz function
<math display="block"> T_d(x) = \begin{cases}
 \frac{x}{2} &\text{if } x \equiv 0 \pmod{2},\\
 \frac{3x+d}{2} & \text{if } x\equiv 1 \pmod{2}.
\end{cases}</math>