A 6 branch generalized 3x+1 conjecture

The iterates y, t(y), t(t(y)),... of the function \[ t(x)=\left\{\begin{array}{cl} x/6 & \mbox{ if \(x\equiv0\pmod{6}\)}\\ (7x+1)/2 & \mbox{ if \(x\equiv1\pmod{6}\)}\\ x/2 & \mbox{ if \(x\equiv2\pmod{6}\)}\\ x/3 & \mbox{ if \(x\equiv3\pmod{6}\)}\\ x/2 & \mbox{ if \(x\equiv4\pmod{6}\)}\\ (7x+1)/6 & \mbox{ if \(x\equiv5\pmod{6}\)} \end{array} \right. \] are printed and the number of steps taken to reach one of the integers 19, 1, 0, -1, -5, -11 is recorded.

This mapping is of type (b) in Generalized 3x+1 mappings with Markov matrix $$ Q(6)=\left[\begin{array}{cccccc} 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6\\ 0 & 1/2 & 1/2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1/2 & 1/2 & 0 \\ 0 & 1/2 & 1/2 & 0 & 0 & 0 \\ 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6\\ 0 & 1/3 & 0 & 0 & 1/3 & 1/3 \end{array}\right] $$ with stationary vector (2/53, 15/53, 14/53, 9/53, 10/53, 3/53).
Then, as we have the inequality

(1/6)2/53(21/6)15/53(3/6)14/53(2/6)9/53(3/6)10/53(7/6)3/53 < 1,

it is conjectured by Keith Matthews that every trajectory will end in one of the numbers in this list and subsequently cycle. (The cycle lengths are printed in bold type.):

This mapping was communicated on July 16, 2017 by Kevin Lamoreau, who was inspired by Tomás Oliveira e Silva's 5x + 1 and 7x + 1 variants.

Enter M:

Last modified 17th July 2017
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