Distributing labels on infinite trees

Using the well-known formula ∑ d−1i=0 ⌊x + i d ⌋ = ⌊dx⌋, we can compute h v(n + 1):h v (n + 1) ==∑d−1⌊S(n)α + φ i ⌋i=0∑d−1⌊ dn − 1d − 1 α + α + φ + i ⌋di=0= ⌊d( dn − 1d − 1 α + α + φ )⌋d= ⌊S(n + 1)α + φ⌋.Therefore, (11) holds for all n which means that the tree is mechanical.Uniqueness. Now, let A be a mechanical tree of density α. Let v be a node and φ 0 , . . . , φ d−1be the phases of its children. Let i and j be two children and let h i (n) be the number of onesin the ith child subtree (of phase φ i ). We want to prove that either for all n: h i (n) ≤ h j (n) orfor all n: h i (n) ≥ h j (n). If the two nodes are both inferior (resp. superior), this is clearly true:h i (n) ≤ h j (n) if and only if φ i ≤ φ j (resp. φ i ≥ φ j ). If i is inferior and j is superior, it is notdifficult to show that φ i < 1 − φ j implies h i (n) ≤ h j (n) and φ i ≥ 1 − φ j implies h i (n) ≥ h j (n).Therefore we can assume (otherwise we exchange the order of the children) that for all n:h 0 (n) ≤ h 1 (n) ≤ · · · ≤ h d−1 (n).Moreover as h d−1 (n) − h 0 (n) ≤ 1, there exists k such that h 0 (n) = h 1 (n) = · · · = h k (n) φ, ψ is nota phase of the node. Conversely, if the exists δ > 0 such that frac(S(n)α + φ) > δ (resp. < 1 − δ)for all n, then let φ ′ = φ − ε (resp. φ ′ = φ + ε), with ε < δ. Then ⌊S(n)α + φ⌋ = ⌊S(n)α + φ ′ ⌋for all n.Finally, a phase φ is unique if and only if 0 and 1 are accumulation points of the sequence{frac(S(n)α + φ)} n∈N.Let us call x def= 1defd−1α and y = φ − x and let us consider the sequencefrac(S(n)α + φ) = frac(xd n − y).14