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Attaching Blur–Consistent Step Functions to 3xand 3x+1 Making the Collatz Drift Mechanism Obvious Aleksandar Perišić September 2025 Abstract We give a blur–consistent decomposition of the accelerated Collatz step that makes the descent mechanism explicit. Using the Mellin/Fourier lens distinction and a soft lens–switch with a blur floor θ∈ (0 , δ + ρ ), we attach two nonnegative functions: a residue function mθ on odd classes modulo 2 k (multiplicative side) and an integer function aθ on odd n (additive nudge), where aθ(n) = max nlog1 + 1 3n−θ, 0o,⇒aθ(n)=0for all n>Nθ=l1 3(eθ−1)m. Assuming a finite max–plus residue certificate a ( r )+ ρ + ϕ ( Fk ( r )) ≤ϕ ( r ) −δ with potential ϕ (and the tight bar ρ + δ < log (4 / 3)), the scale functional S ( n ) = log n + ϕ ( nmod 2 k )satisfies the split S(n)−S(T(n)) ≥mθ(nmod 2k)−aθ(n), hence for all n>Nθ one has the uniform drift S ( T ( n )) ≤S ( n ) − ( δ−θ ). Past this threshold the additive contribution vanishes and the decrease is carried purely by the multiplicative component, making the proof strategy transparent; the construction is finite, checkable, and reflects the unavoidable “channel–switch toll” when mixing addition with multiplication. Setup (accelerated odd map). For odd nlet T(n) := odd(3n+ 1) = 3n+ 1 2v2(3n+1) . Write r≡n(mod 2k)for an odd residue and Fk(r)≡odd(3r+1) (mod 2k). Define a(r) = (log 3 −v2(3r+1) log 2, r =r∗, log 3 −klog 2, r =r∗(the unique exceptional class),(1) and ε ( n ) := log 1 + 1 3n∈ (0 ,log (4 / 3)]. Assume we already have a residue potential ϕ and constants δ > 0,ρ≥0satisfying the certificate a(r)+ρ+ϕ(Fk(r)) ≤ϕ(r)−δ(∀rodd mod 2k).(2) Then for the scale functional S(n) := log n+ϕ(nmod 2k)we have the one–step drift S(T(n)) ≤S(n)−(δ+ρ)−ε(n)(nodd).(3) Goal Define two nonnegative functions mθ:{odd residues mod 2k}→R≥0and aθ:Nodd →R≥0 1
“attached” respectively to the multiplicative part (3 x with the 2–adic division) and to the additive nudge (+1), such that S(n)−S(T(n)) ≥mθ(nmod 2k)−aθ(n),(4) and moreover aθ(n)=0for all n > Nθfor an explicit threshold Nθdetermined by θ. Construction (blur threshold split) Fix any blur floor θwith 0< θ < δ +ρ. Define: aθ(n) := max{ε(n)−θ, 0}= max nlog1 + 1 3n−θ, 0o,(5) mθ(r) := ϕ(r)−ϕ(Fk(r))−a(r)−ρ−θ. (6) Lemma 1 (Basic bounds).For every odd residue rand odd nwith r≡n(mod 2k), mθ(r)≥δ−θand 0≤aθ(n)≤ε(n). Proof. From (2) , ϕ ( r ) −ϕ ( Fk ( r )) ≥a ( r ) + ρ + δ , so mθ ( r ) ≥δ−θ . The bound for aθ is by definition. Proposition 2 (Per–step split).For every odd n with residue r≡n ( mod 2 k ), the decomposition (4)holds: S(n)−S(T(n)) ≥mθ(r)−aθ(n). In particular, using Lemma 1, S(n)−S(T(n)) ≥(δ−θ)−aθ(n). Proof. Let r≡n(mod 2k). From (6) we can rewrite ϕ(r)−ϕ(Fk(r)) = a(r)+ρ+θ+mθ(r). Since log n−log T(n)=−ε(n)−a(r), we obtain S(n)−S(T(n)) = log n−log T(n)+ϕ(r)−ϕ(Fk(r))=mθ(r)+ρ+θ−ε(n). Because aθ(n) = max{ε(n)−θ, 0}, we have θ−ε(n)≥ − aθ(n), hence S(n)−S(T(n)) ≥mθ(r)+ρ−aθ(n)≥mθ(r)−aθ(n), using ρ≥0. The final inequality in the statement follows from Lemma 1. Vanishing of the additive part for large n.From (5), aθ(n)=0whenever ε(n)≤θ, i.e. log1 + 1 3n≤θ⇐⇒ n≥1 3 (eθ−1). Thus we may take the explicit threshold Nθ:= l1 3 (eθ−1) m,(7) and then aθ(n)=0for all n > Nθ. What this buys us. Beyond Nθthe decomposition (4) simplifies to S(n)−S(T(n)) ≥mθ(nmod 2k)≥δ−θ > 0, so the entire per–step decrease comes transparently from the multiplicative side. The additive side has been blurred away by the floor θ. 2
How to tune the knobs (practical recipe) 1. Choose k(e.g., k= 13) and solve the residue certificate (2) once to get ϕ, δ, ρ. 2. Pick any θ∈ (0 , δ + ρ ). Larger θ means a smaller Nθ in (7) (the additive nudge vanishes sooner). 3. Use (6) and (5): attach mθto 3xand aθto (3x+1). Remark 3 (Concrete numbers).If (illustratively) δ≈ 0 . 10 and ρ≈ 0 . 106, you may take θ = 0 . 06. Then Nθ = 1 / 3(e 0.06 − 1) = 6. Thus aθ ( n )=0for every odd n≥ 7, and S drops by at least δ−θ≈0.04 at each odd step from that point on. Why this matches the “fun” intuition The pair (mθ, aθ)are exactly the two knobs one could ask for: •mθ depends only on the residue (the 3 x side, with the 2–adic division). Its uniform lower bound δ−θis guaranteed by the finite residue certificate. •aθ is the detectable part of the +1 nudge after blur. By choosing the floor θ , you set a detection threshold: beyond Nθ the additive nudge is too small to see and is treated as zero. This makes the per–step decrease both explicit and obvious: above Nθ , only the multiplicative component matters, so the scale potential Sfalls at a uniform rate. References for context. The soft lens–switch and the finite residue certificate are developed in: References [1] A. Perišić. Soft-Addition and Soft-Multiplication and the Channel–Switch Error. Zenodo (2025). [2] A. Perišić. A Lyapunov Certificate for the Accelerated Collatz Map. Zenodo (2025). [3] A. Perišić. Collatz Without the Mystique of Addition and Multiplication. Zenodo (2025). 3
