Let k ≥ 2 be an even integer. The iterates x, t(x), t(t(x)),... of the mapping
t(x) | = | (k + 1)x + 1 | if x is odd |
t(x) | = | ⌈x/k⌉ | if x is even, |
t(x) | = | (k + 1)x+ 1 | if x is odd |
t(x) | = | x/k | if x ≡ 0 (mod k) |
t(x) | = | ⌊x/k⌋ + 1 | if x ≡ 0 (mod 2) and x ≢ 0 (mod k), |
It seems certain that if x ≠ 0, then tn(x) will eventually reach one of the cycles 0 → 0, or 1 → k + 2 → 2 → 1, or -(2t + 1) → -2t(k + 1) - k → -(2t + 1), where 0 ≤ t ≤ k/2 - 1.
Last modified 12th December 2020
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