A 6 branch generalized 3x+1 conjecture

The iterates y, t(y), t(t(y)),... of the function

t(x) = x/6if x ≡ 0 (mod 6),
t(x) = (7x+1)/2if x ≡ 1 (mod 6),
t(x) = x/2if x ≡ 2 (mod 6),
t(x) = x/3if x ≡ 3 (mod 6),
t(x) = x/2if x ≡ 4 (mod 6),
t(x) = (7x+1)/6if x ≡ 5 (mod 6),

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

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

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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