Editing Collatz conjecture (section)

===As a parity sequence===
For this section, consider the shortcut form of the Collatz function
<math display="block"> f(n) = \begin{cases} \frac{n}{2} &\text{if } n \equiv 0 \\ \frac{3n + 1}{2} & \text{if } n \equiv 1 \end{cases} \pmod{2}.</math>

If {{math|P(...)}} is the parity of a number, that is {{math|P(2''n'') {{=}} 0}} and {{math|P(2''n'' + 1) {{=}} 1}}, then we can define the Collatz parity sequence (or parity vector) for a number {{mvar|n}} as {{math|''p<sub>i</sub>'' {{=}} P(''a<sub>i</sub>'')}}, where {{math|''a''<sub>0</sub> {{=}} ''n''}}, and {{math|''a''<sub>''i''+1</sub> {{=}} ''f''(''a''<sub>''i''</sub>)}}.

Which operation is performed, {{math|{{sfrac|3''n'' + 1|2}}}} or {{math|{{sfrac|''n''|2}}}}, depends on the parity. The parity sequence is the same as the sequence of operations.

Using this form for {{math|''f''(''n'')}}, it can be shown that the parity sequences for two numbers {{mvar|m}} and {{mvar|n}} will agree in the first {{mvar|k}} terms if and only if {{mvar|m}} and {{mvar|n}} are equivalent modulo {{math|2<sup>''k''</sup>}}. This implies that every number is uniquely identified by its parity sequence, and moreover that if there are multiple Hailstone cycles, then their corresponding parity cycles must be different.<ref name="Lagarias (1985)"/><ref name="Terras (1976)"/>

Applying the {{mvar|f}} function {{mvar|k}} times to the number {{math|''n'' {{=}} 2<sup>''k''</sup>''a'' + ''b''}} will give the result {{math|3<sup>''c''</sup>''a'' + ''d''}}, where {{mvar|d}} is the result of applying the {{mvar|f}} function {{mvar|k}} times to {{mvar|b}}, and {{mvar|c}} is how many increases were encountered during that sequence. For example, for {{math|2<sup>5</sup>''a'' + 1}} there are 3 increases as 1 iterates to 2, 1, 2, 1, and finally to 2 so the result is {{math|3<sup>3</sup>''a'' + 2}}; for {{math|2<sup>2</sup>''a'' + 1}} there is only 1 increase as 1 rises to 2 and falls to 1 so the result is {{math|3''a'' + 1}}. When {{mvar|b}} is {{math|2<sup>''k''</sup> − 1}} then there will be {{mvar|k}} rises and the result will be {{math|3<sup>''k''</sup>''a'' + 3<sup>''k''</sup> − 1}}. The power of 3 multiplying {{mvar|a}} is independent of the value of {{mvar|a}}; it depends only on the behavior of {{mvar|b}}. This allows one to predict that certain forms of numbers will always lead to a smaller number after a certain number of iterations: for example, {{math|4''a'' + 1}} becomes {{math|3''a'' + 1}} after two applications of {{mvar|f}} and {{math|16''a'' + 3}} becomes {{math|9''a'' + 2}} after four applications of {{mvar|f}}. Whether those smaller numbers continue to 1, however, depends on the value of {{mvar|a}}.