Bijections Φm and Ψm

The map Φm sends {1,…,2m} bijectively onto {0,1}m via n ↦ (c1,…,cm). The map Ψm sends {1,…,6·2m} bijectively onto the paths of length m in the graph G mod 6. Click any row to see the trajectory and the formula for Tm.

ci = Ti−1(n) mod 2 — parity of the i-th iterate (0 = even, 1 = odd)
hi = Ti−1(n) mod 6 — residue mod 6 of the i-th iterate
s = ∑ ci — Hamming weight (number of odd iterates)
T(n) = n/2 if even  ·  (3n+1)/2 if odd — accelerated Collatz map
Binary antisymmetry: n and n+2m share c1…cm; bit cm+1 is always the complement (0↔1)
Mod 6 antisymmetry: n and n+3·2m share h1…hm; residue hm+1 is always the opposite (r↔r+3 mod 6)

Corrective residue R(Cm) and Tm(n)

Explore the formula Tm(n) = (3s·n + R(Cm)) / 2m: three methods to compute the corrective residue (recursive, explicit, matrix), the chain concatenation law and the power formula. Supports n ∈ ℤ.

Fundamental identity
Tm(n)= 3s·n + R(Cm(n))2m s = Σ ci
R(Cm)= 2m·Tm(n) − 3s·n
ci = Ti−1(n) mod 2  ·  r(i) = c1+···+ci  ·  s = r(m)  ·  n ∈ ℤ
Input
n = m =
Computing R
Application: Tm(n)
Concatenation law
Click a · point between ci positions to add or remove a cut.
Chain powers
R(BN) = R(B) · 2pN − 3r0N 2p − 3r0
B = base chain of length p BN = concatenation of B with itself N times p = length(B), r0 = weight(B) = Σ ci
Repeat current chain N = times

Graph G mod 6

Preamble — What does this graph show?
T: accelerated Collatz map — T(n) = n/2 (even), (3n+1)/2 (odd), T(0)=0. The trajectory is T⁰(n), T¹(n), T²(n), …
Graph G mod 6: 6 nodes = residues 0–5, 12 edges. Write n = 6k + r. Each node r has two outgoing edges: ●green (k even) and ●amber (k odd).
Periodicity: all n with the same residue mod 6 and the same parity sequence follow exactly the same path in G for m iterations — i.e., members of a congruence class modulo 6·2m share their trajectory for m steps.
SCCs: V₁={0} fixed point · V₂={3} transient · V₃={1,2,4,5} closed attractor.
G' = G[V₃]: induced subgraph on the attractor. Stationary distribution: π(1)=π(4)=1/6, π(2)=π(5)=1/3.
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