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Given the four functions $P_1$, $P_2$, $N_1$ and $N_2$ (which together is a piecewise function) each with domain and range as shown above:

  1. Is there an explanation as to why starting at any integer (excluding 0) and continuously composing among these functions tend to always end in the cycle -1, 1, -1?
  2. Can it be shown that starting at any integer will always end in the cycle -1, 1, -1?
  3. The digraph is provided to show the possible paths an integer in the domain of any of the functions may take. I am also hoping that by posting the digraph a link can be made to graph theory; specifically, does the digraph suggest that 1 can be reached by any element in the domain of any of the functions?

For example, the trajectory of 4 is: 4, 6, 9, -13, 7, -10, -5, 3, -4, -2, -1, 1, -1…; while the trajectory of 14 is: 21,-31,16,24,36,54,81,-121,61,-91,46,69,-103,52,78,117,-175,88,132,198,297,-445, 223,-334,-167,84,126,189,-283,142,213,-319,160,240,360,540,810,1215,-1822,-911,456,684,1026,1539, -2308, -1154, -577, 289, -433, 217, -325, 163, -244, -122, -61, 31, -46, -23, 12, 18, 27, -40, -20,-10, -5, 3, -4, -2, -1, 1, -1.

In addition to composing among $P_1$ , $P_2$,$N_1$ and $N_2$, the trajectory of any integer (excluding 0) can be obtained using the sequence: 1,-1, 2, -2, 3, -3, 4, -4, 5, -5, 6, -6, 7, -7, 8, -8, 9, -9, 10, -10, 11, -11, 12, -12, 13, -13, 14, -14, 15, -15 ... Starting at the number whose trajectory one wishes to find, if the number is positive, move forward that number of terms; if the number is negative, move backward that number of terms, and continue this forward-backward movement indefinitely.

Part of the 'tree' showing the trajectories of a few integers is shown below. Trajectories are followed from the bottom up.

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Feel free to point me to any related resource.

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  • $\begingroup$ The immediate association is the famous $\ \phi:\mathbb N\to\mathbb N\ $ given by $\ \phi(n):=3\cdot n+1\ $ for odd $n,\ $ and $\phi(n)=\frac n2\ $ for even $\ n.$ $\endgroup$
    – Wlod AA
    Commented Dec 8, 2020 at 21:32

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Let $f$ denote the function described in the question. The assertion that every trajectory of $f$ except for the one starting at 0 ends in the cycle -1, 1, -1 is equivalent to the Collatz conjecture since the mapping $$ g: \ \mathbb{Z} \setminus \{0\} \rightarrow \mathbb{N}, \ \ n \ \mapsto \ \begin{cases} -2n & \text{if} \ n < 0, \\ 2n-1 & \text{if} \ n > 0 \end{cases} $$ induces a natural bijection between trajectories of $f$ and trajectories of the Collatz mapping.

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