### Gonçalves - Greenfield - Madrid 12-branched generalized 3x+1 conjecture

We consider the d-branched mapping $$T: \mathbb{Z}\to\mathbb{Z}$$ given by $$\displaystyle T(x) = \left\lfloor\frac{m_ix}{d}\right\rfloor + x_i, \mbox{ if x \equiv i \pmod d},$$
with $$m_0=1, m_i=15, 1\le i\le11; x_0=0, x_1=6, x_2=4, x_3=3, x_4=-2, x_5=6, x_6=5, x_7=0, x_8=3, x_9=-1, x_{10}=0, x_{11}=-1$$:
$T(x)= \small \left\{ \begin{array}{cl} \frac{x}{12} & \mbox{ if x\equiv 0\pmod{12}}\\ \frac{5x+23}{4} & \mbox{ if x\equiv 1\pmod{12}}\\ \frac{5x+14}{4} & \mbox{ if x\equiv 2\pmod{12}}\\ \frac{5x+9}{4} & \mbox{ if x\equiv 3\pmod{12}}\\ \frac{5x-8}{4} & \mbox{ if x\equiv 4\pmod{12}}\\ \frac{5x+23}{4} & \mbox{ if x\equiv 5\pmod{12}}\\ \frac{5x+18}{4} & \mbox{ if x\equiv 6\pmod{12}}\\ \frac{5x-3}{4} & \mbox{ if x\equiv 7\pmod{12}}\\ \frac{5x+12}{4} & \mbox{ if x\equiv 8\pmod{12}}\\ \frac{5x-5}{4} & \mbox{ if x\equiv 9\pmod{12}}\\ \frac{5x-2}{4} & \mbox{ if x\equiv 10\pmod{12}}\\ \frac{5x-7}{4} & \mbox{ if x\equiv 11\pmod{12}} \end{array} \right.$ The iterates $$x, T(X), T(T(x)),\ldots$$ of the mapping are conjectured to eventually reach one of the cycles
• 0;
• 1, 7, 8, 13, 22, 27, 36, 3, 6 12; (length 10)
• 226, 282, 357, 445, 562, 702, 882, 1107, 1386, 1737, 2170, 2712; (length 12)
• -3, -5, -7;
• -9;
• -18;
• -23;
• -54, -63, -80, -102, -123, -155, -188, -237, -294, -363, -455, -563, -698, -873, -1089, -1359, -1700, -2127, -2660, -3327, -4160, -5202, -6498, -8118, -10143, -12680, -15852, -1321, -1653, -2064, -172, -212, -267, -335, -413, -517, -648; (length 37)
(This mapping is equivalent to one on page 32 of the paper https://arxiv.org/pdf/2111.06170.pdf.)

The mapping can be regarded as a 12-branched example of type (b).

The associated Markov matrix $Q(12)= \small \left[ \begin{array}{cccccccccccc} 1/12& 1/12& 1/12& 1/12 & 1/12& 1/12& 1/12& 1/12& 1/12& 1/12& 1/12& 1/12\\ 0 & 1/4 & 1/4 & 0 & 0 & 0 & 0 & 1/4 & 0 & 0 & 1/4 & 0\\ 1/4 & 0 & 0 & 0 & 1/4 & 0 & 1/4 & 0 & 0 & 1/4 & 0 & 0\\ 1/4 & 0 & 0 & 0 & 1/4 & 0 & 1/4 & 0 & 0 & 1/4 & 0 & 0\\ 1/4 & 0 & 0 & 0 & 1/4 & 0 & 1/4 & 0 & 0 & 1/4 & 0 & 0\\ 1/4 & 0 & 0 & 0 & 1/4 & 0 & 1/4 & 0 & 0 & 1/4 & 0 & 0\\ 1/4 & 0 & 0 & 0 & 1/4 & 0 & 1/4 & 0 & 0 & 1/4 & 0 & 0\\ 0 & 0 & 0 & 1/4 & 0 & 1/4 & 0 & 0 & 1/4 & 0 & 0 & 1/4\\ 0 & 1/4 & 1/4 & 0 & 0 & 0 & 0 & 1/4 & 0 & 0 & 1/4 & 0\\ 0 & 1/4 & 1/4 & 0 & 0 & 0 & 0 & 1/4 & 0 & 0 & 1/4 & 0\\ 1/4 & 0 & 0 & 0 & 1/4 & 0 & 1/4 & 0 & 0 & 1/4 & 0 & 0\\ 1/4 & 0 & 0 & 0 & 1/4 & 0 & 1/4 & 0 & 0 & 1/4 & 0 & 0 \end{array} \right]$ has stationary vector $$p = \small\frac{1}{228}(33, 17, 17, 7, 33, 7, 33, 17, 7, 33, 17, 7)$$ and we have the inequality

$\prod_{i=0}^{11}\left(\frac{m_i}{12}\right)^{p_i}=\left(\frac{1}{12}\right)^{33/228}\left(\frac{15}{12}\right)^{1-33/228}=\frac{15^{195/228}}{12}<1,$ thereby predicting everywhere eventual cycling.

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