Per Lauge Buresø Holst · with Claude (Anthropic AI)
4 June 2026
A structural exploration of why 3n+1 trajectories tend downward — and why “tend” cannot yet be upgraded to “must.”
Collaboration & process
This document is a collaboration between Per Lauge Buresø Holst (who conceived and directed the investigation) and Claude, an AI assistant from Anthropic. The work split along its natural seam: the human supplied the ideas, intuitions, and direction — the binary lens, the “firewall” / “fuel” / “glider” framings, the cycle-climb questions, and the persistent skepticism that caught more than one overstatement; Claude supplied verification and formalization — running every claim in code, producing closed forms and counterexamples, correcting errors, cross-checking attributions against the literature (Lagarias, Terras, Tao, Conway, Eliahou), and writing the prose, code, figures, and this document. Every quantitative statement here was checked computationally and every “known result” traced to a source. The honest conclusion that this is expository, not novel (§14–15) is itself an output of that process — as is the recurring theme that each elementary line of attack halts at the same wall. Remaining errors are ours, jointly.
1. The Problem
For a positive integer n, apply: if even, \(n \to n/2\); if odd, \(n \to 3n+1\). The Collatz conjecture asserts every starting value eventually reaches 1. Verified to ~2??; unproven in general.
This writeup follows one line of attack: read the dynamics in binary, where \(\div 2\) is a right shift and \(3n+1 = n + (n\ll 1) + 1\) is an add-with-carry. Dividing by 2 does no real work — it merely pans the bit pattern through a fixed positional frame, a change of reference (Figure 1); all the dynamics live in the odd step \(3n+1\). The goal was to find structure that forces convergence. We found a great deal of structure — enough to explain why trajectories descend on average — but each structural fact stops exactly at the same wall, which turns out to be the conjecture itself.
Notation & Symbols
| Symbol | Meaning |
|---|---|
| \(T(n)\) | shortcut map on odd \(n\): \(T(n) = (3n+1) / 2^\nu \) — does \(3n+1\) then strips all trailing zeros, landing on the next odd number. |
| \(\nu \) (= \(\nu _{2}(3n+1)\)) | 2-adic valuation of \(3n+1\): the number of trailing zero bits = how many times you halve in one shortcut step. \(\nu =1 \iff n \equiv 3 \bmod 4\). (Greek “nu”.) |
| \(v_{3}(m)\) | 3-adic valuation of \(m\): the exponent of the highest power of 3 dividing \(m\). E.g. \(v_{3}(54) = 3\) because \(54 = 2\cdot 3^{3}\) (so \(3^{3} \mid 54\) but \(3^{4} \nmid 54\)); \(v_{3}(102)=1\) since \(102 = 2\cdot 3\cdot 17\); \(v_{3}(80)=0\). (Latin “v” — not the same as \(\nu \) above.) |
| \(\equiv a (\bmod b)\) | congruence: \(n \equiv 3 \bmod 4\) means \(n\) leaves remainder 3 on division by 4, i.e. its low two bits are \(\dots 11\). |
| \(M\) | the maximum odd element of a hypothetical non-trivial cycle. |
| \(P, Q, R\) | the predecessors (all odd numbers) climbing down from \(M\): \(P\) maps to \(M\), \(Q\) to \(P\), \(R\) to \(Q\). Generally \(x_j\) is the \(j\)-th rung, \(x_{0}=M, x_{1}=P, x_{2}=Q, x_{3}=R\), with \(x_j + 1 = (2/3)^j (M+1)\). |
| \(a\), \(S\) | for a whole cycle: \(a\) = number of odd (\(3n+1\)) steps; \(S = \sum \nu \) = total halvings. The balance is \(3^a \approx 2^S\). |
| \(H, L\) | high / low parts of a number split across a zero-run “firewall”: \(n = H\cdot 2^{m+b} + L\). |
| \(A_j\), \(r_m\) | alternating numbers: \(A_j = (4^{j+1}-1)/3 = 101\dots 01\); \(r_m = (2^m-1)/3\) (same family). |
| \(\mathrm{octave}\) | a factor-of-2 band of magnitude (one bit of length); “log-uniform” = roughly equal count per octave. |
| \(\ll \), \(\gg \) | left / right bit shift (\(n\ll 1 = 2n\)). |
| \(2^k-1\) | “all-ones” number (\(k\) consecutive 1-bits); the \(-1\) of the 2-adic integers \(Z_{2}\) is the infinite all-ones \(\dots 1111\). |
2. The Shortcut Map and the Two Residue Classes
Work odd-to-odd via the shortcut map \(T(n) = (3n+1) / 2^\nu \), where ? = number of trailing zeros of \(3n+1\). Even numbers are irrelevant — they just halve down. So only odd residues mod 4 matter:
| n mod 4 | 3n+1 | ? | net per step |
|---|---|---|---|
| ? 1 (\(\dots 01\)) | 4(3k+1) | ? ? 2 | shrinks (descends below n immediately) |
| ? 3 (\(\dots 11\)) | 2(6k+5) | ? = 1 | grows (×3/2) |
Immediate consequences:
- Any cycle’s minimum element is odd, and cannot be ? 1 mod 4 (it would drop below itself). So a non-trivial cycle’s minimum must be ? 3 mod 4 — binary \(\dots 11\).
- \(n \equiv 1 \bmod 4\) provably descends in one step. All difficulty lives in the \(\equiv 3 \bmod 4\) (“sticky”) class.
Figure 2 shows a full trajectory — that of 27 — read in binary: the heatmap (rows = steps), the value curve, the pattern metrics, and the carry depth \(\nu \) at each odd step.
3. The Carry Mechanism
In \(3n = n + (n\ll 1)\), classify each bit position by its pair:
- Generator (\(11\)): always emits a carry.
- Propagator (\(10\)/\(01\)): passes a carry if one arrives.
- Absorber (\(00\)): stops the carry.
A carry travels exactly as far as the trailing run of 1s in \(3n\), i.e. \(\nu _{2}(3n+1)\) = length of that run. For a generic number the carry fizzles in 1–2 bits. The exception is the alternating pattern \(1010\dots 1\), which is all propagators: \(3 \times (1010\dots 1) = (111\dots 1)\), so a single triggering carry conducts across its entire length.
\(\times 3\) as a 3-state transducer. Processing \(n\) in 4-bit chunks from LSB to MSB, each chunk computes \(3x + c\) where \(c\) is the carry-in from below; its low 4 bits stay, and \(\mathrm{cout} = (3x+c)\gg 4\) passes up. Since \(3\cdot 15 + 2 = 47\) and \(47\gg 4 = 2\), the carry-out never exceeds 2 when the carry-in is \(\le 2\) — so the carry state lives in a closed set \({0,1,2}\) (carry-in 3 never arises). Multiplication by 3 is thus a finite-state transducer with three carry states; the chunks split into three bands by \(\lfloor 3x/16\rfloor \) (\(0000-0101 \to 0\), \(0110-1010 \to 1\), \(1011-1111 \to 2\)), and the carry-in only matters at the band edges:
| chunk | behaviour | role |
|---|---|---|
| \(0101\) | \(c=0\to \mathrm{cout} 0\), \(c\ge 1\to \mathrm{cout} 1\) | conductor — the unique band-0 chunk that lets a carry escape (\(3\cdot 5 = 1111\)) |
| \(1010\) | \(c\le 1\to \mathrm{cout} 1\), \(c=2\to \mathrm{cout} 2\) | promoter — carries a 2 onward |
| \(1111\) | \(\mathrm{cout} = 2\) for every \(c\) | generator — a saturated carry source |
| all others | \(\mathrm{cout}\) fixed by \(3x\) alone | absorber — incoming carry shifts bits but cannot escape |
This is the finite, local object behind the whole carry story (and the cellular-automaton analogy of Section 11): a long alternating run \(\dots 0101 0101\dots \) is a chain of conductors that walks a bottom-injected carry (the \(+1\)) all the way up, while a long all-ones run \(\dots 1111\dots \) is a wall of generators forcing \(\mathrm{cout} = 2\) at every chunk — exactly the two motifs (alternating/firewall, all-ones/\(-1\)) that dominate both obstacles. Figure 3 shows this toggle: the pure alternating number 1365 has carry-reach 0, but ×3 turns it all-ones, flipping carry-reach to full width.
The ?-distribution is geometric, \(P(\nu =k) = 1/2^k\), giving \(E[\nu ] = 2\). Hence the famous heuristic: each odd step multiplies by ~\(3/2^{2} = 3/4\) on average. Figure 4 shows the per-step carry roles along 27’s shortcut trajectory.
4. The Firewall: Isolation by a Zero Run
Write \(n = H\cdot 2^{m+b} + L\), with a low part L (b bits) separated from a high part H by m zeros. Because \(3L+1 < 2^{b+2}\), the carry from L cannot cross a sufficiently wide zero run. While the firewall holds, H and L evolve independently:
\[T^j(n) = 3^j \cdot H \cdot 2^{m+b-\sum \nu } + T^j(L)\]
H is purely passive (just ×3 each step); L alone drives the dynamics. If L converges, n converges, reducing by a factor ? 1/L.
Two conditions (verified in code) make this exact:
- \(\mathrm{max}(L-\mathrm{trajectory}) < 2^m\) — carry never reaches H.
- \(m + b \ge \sum \nu (L-\mathrm{trajectory})\) — H stays at a positive bit position.
Condition (2) binds: the required firewall scales with L’s stopping time, not its peak. This is a rigorous convergence lemma for an infinite class of numbers — but it only covers numbers already containing such a structure. Whether every trajectory acquires one is the open part.
(This is a concrete restatement of the standard 2-adic view: a wide zero run means n is 2-adically close to L, and the map is Lipschitz there.)
5. The Average Argument, Made Rigorous — and Its Ceiling
The \(\times 3/4\) drift is provable for explicit subclasses:
- Trivial: \(n \equiv 1 \bmod 4\) descends in one step (half of all odds).
- Terras / Everett (1976–77): The first k steps depend only on \(n \bmod 2^k\), and the parity-vector map is a bijection. A class with j odd-steps contracts iff \(3^j < 2^k\), i.e. \(j < k\cdot \mathrm{log}2/\mathrm{log}3 \approx 0.631k\). Since odd-steps are Binomial(k, ½) with mean k/2 < 0.631k, the density of contracting classes ? 1. Theorem: the set of integers with finite stopping time has natural density 1.
- Tao (2019): Almost all orbits (logarithmic density) attain almost bounded values.
Why it is believed (the even-distribution heuristic). The drift is governed by one statistic: how often a step has \(\nu =1\). And \(\nu =1 \iff n \equiv 3 \bmod 4\), i.e. the number simply ends in \(\dots 11\) — half of all odd numbers. If the trailing bits along an orbit are unbiased (each two-bit ending at its natural rate), then
P(?=1)=½, P(?=2)=¼, … ? E[?]=2 > log?3 ? 1.585 ? bit-length drifts down ? n ? 1.This is borne out empirically: over ~570,000 odd steps sampled from ~80-bit starts, \(\nu =1\) occurs 49.97% of the time and average \(\nu = 1.998\) — exactly the unbiased prediction. *The evenness of the trailing bits is the contraction. Divergence would require a sustained skew: with \(\nu =1\) at fraction \(p\) and the rest following the natural tail (\(E[\nu |\nu \ge 2]=3\)), \(E[\nu ] = 3 – 2p\), so neutrality needs \(p = (3 – \mathrm{log}_{2}3)/2 \approx 0.71\) and growth needs more — a \(\equiv 3 \bmod 4\) rate of about 71%, a ~5:2 ratio, versus the natural 50%. (Not to be confused with the Terras threshold \(\mathrm{log}2/\mathrm{log}3 \approx 0.631\), which counts odd steps among all* steps, a different ratio.) There is no apparent mechanism for such a skew — the bits look balanced and Hamming density hovers at ½. The \(\nu \) distribution that drives all this matches the geometric \(1/2^k\) prediction to several orders of magnitude (Figure 5).
Where the fuel comes from (and why the split stays even). A step lands on the growth class (\(T(n) \equiv 3 \bmod 4\), “fuel”) according to the bottom 3 bits of \(n\):
| \(n\) ends in | \(T(n) \equiv 3 \bmod 4\)? | role |
|---|---|---|
| \(\dots 111\) (\(7 \bmod 8\)) | always (\(\nu =1\), \(T=(3n+1)/2\)) | definite feeder |
| \(\dots 011\) (\(3 \bmod 8\)) | never (? \(\dots 01\)) | definite exit |
| \(\dots x01\) (\(1 \bmod 4\)) | ~50%, \(T = \mathrm{oddpart}(3\cdot (n\gg 2)+1)\) | recursive in the high bits |
(Careful: this table is about which \(n\) produce a fuel successor. A \(\dots 011\) number is itself \(\equiv 3 \bmod 4\), so it is a growth step and counts toward the rate — \(\dots 011\) and \(\dots 111\) both grow \(\times 3/2\). The distinction is only in their successor: \(\dots 111\) chains to another growth step, \(\dots 011\) exits to a descending one. So \(\dots 011\) is not “unusable” — it just doesn’t perpetuate the chain.)
The only deterministic fuel source is a \(\dots 111\) ending — but a trailing-1 run erodes one bit per step (\(\dots 111 \to \dots 11 \to \dots 1\)) and cannot be regenerated as structure (Sections 7–8). Once spent, the orbit falls into the \(\dots 01\) case, which is a fresh ~50% coin flip on the high bits. So the guaranteed source is self-limiting and everything else averages ½ — which is exactly why \(P(\equiv 3 \bmod 4) \approx \tfrac{1}{2}\) with no bias. A sustained skew would have to come from the recursive \(\dots 01\) branch landing on \(\equiv 3 \bmod 4\) more than half the time forever — a conspiracy in the high bits that nothing local forbids. That branch is precisely where an un-rule-out-able bias could hide.
The ceiling: “evenly distributed” is provable in the wrong senses and unprovable in the needed one.
| sense of “trailing bits are even” | status |
|---|---|
| across all odd numbers (static): exactly ½ are \(\equiv 3 \bmod 4\) | proven, trivial — but an orbit visits one specific sequence, not “all numbers” |
| along almost all orbits ? those contract | proven (Terras/Everett density-1; Tao almost-all) |
| along every orbit (no biased exception) | = the conjecture |
The trap is that an orbit is deterministic: the trailing bits of \(n_{i+1}\) are a fixed function of \(n_i\), so consecutive endings are correlated, not independent. “Looks random” is not “is random” — a deterministic sequence may carry any bias. Equivalently, a density-1 set is not forward-invariant: when \(n\) descends to \(n’\), the residue class of \(n’\) is uncorrelated with the contraction just used, so “almost all descend once” cannot be iterated into “all reach 1.” Proving no orbit’s mod-4 classes lock into a self-reinforcing \(\equiv 3 \bmod 4\) skew is exactly the open problem.
6. Cycle Structure
The conjecture has two independent obstacles: ruling out non-trivial cycles would not rule out escape to infinity, and vice versa. Neither is known. This section maps what the binary lens forces about any hypothetical non-trivial cycle; Sections 7–12 then turn to divergence.
Everything follows from one exact per-element rule (verified): under T,
\[n \equiv 3 \bmod 4 \implies T(n) > n\qquad \text{(ascends, $\nu$=1, $\times$3/2)}\]
\[n \equiv 1 \bmod 4 \implies T(n) < n\qquad \text{(descends, $\nu$$\ge$2; n > 1)}\]
Minimum and maximum (a dual pair).
- The minimum of a non-trivial cycle ascends out of itself, so it is ? 3 mod 4 (it cannot be \(\equiv 1 \bmod 4\) — that would drop below itself; the trivial cycle’s \(1 = 4\cdot 0+1\) is the \(k=0\) boundary case).
- The maximum descends out of itself, so it is ? 1 mod 4.
- Hence both residue classes are mandatory: a cycle cannot be all-\(\equiv 3\) (strictly increasing) nor all-\(\equiv 1\) (strictly decreasing). \(4k+1\) numbers are not excludable from cycles — they are required, with the maximum among them.
Two-arc decomposition. The elements are distinct, so the minimum \(c_\mathrm{min}\) and maximum \(c_\mathrm{max}\) are unique; cutting the loop at these two points splits it into exactly two arcs — a rise \(c_\mathrm{min} \to \dots \to c_\mathrm{max}\) and a fall \(c_\mathrm{max} \to \dots \to c_\mathrm{min}\). Every element lies on exactly one arc, both arcs stay within \([c_\mathrm{min}, c_\mathrm{max}]\), and the net multipliers are \(c_\mathrm{max}/c_\mathrm{min} > 1\) (rise) and \(c_\mathrm{min}/c_\mathrm{max} < 1\) (fall), product 1 around the loop. The endpoints are pinned as above: \(c_\mathrm{min} \equiv 3 \bmod 4\) opens the rise, \(c_\mathrm{max} \equiv 1 \bmod 4\) opens the fall. Within an arc the path may wander up and down (local peaks and valleys); it need not be monotone. In fact, for every cycle but one it cannot be: a cycle whose rise is a single clean run of increases and whose fall is a single clean run of decreases is a 1-circuit, and Steiner (1977) [14] proved the only 1-circuit is the trivial \(1\to 4\to 2\to 1\). So \({1,2,4}\) is the unique non-choppy cycle; any other cycle must zig-zag — multiple local maxima on the rise, multiple local minima on the fall (the circuit structure below).
The rise tends to be the longer arc. The climb is rate-limited — the only way up is \(\times 3/2\) per step (\(+0.585\) bits) — while the fall can plunge (one large-\(\nu \) step drops many bits at once), so covering the same height needs more up-steps than down-steps. The one informative \(\times 3\) sibling cycle bears this out sharply: \(3n-1\)’s \({17\to 25\to 37\to 55\to 41\to 61\to 91}\) has rise 6, fall 1 (the fall is the single drop \(91\to 17\), \(\nu =4\)). It is only a tendency, not a law: a fall built from gentle \(\nu =2\) steps (\(-0.415\) bits each, slower than the climb) would be longer, and in \(5n+1\) — where the up-step is the faster \(\times 5/2\) — the cycle \({17\to 43\to 27}\) has the fall longer (rise 1, fall 2). And it carries no leverage for the bounds: growth (\(\equiv 3 \bmod 4\)) steps occur on both arcs, so the split neither isolates them nor sharpens the \(5:2\) count — it only makes the cycle’s shape vivid (a slow rate-limited crawl up, punctuated by deep plunges).
The cousins do cycle. The maps \(3n-1\), \(5n+1\), and \(3n+1\) on the negative integers all share the \(\mathrm{an}+b\) shape, yet each has genuine non-trivial cycles (\(3n-1\): \({5,7}\), \({17,\dots ,91}\); \(5n+1\): \({13,33,83}\), \({17,43,27}\)) — which is exactly what let us read real \(\times 3\) cycle data off them above. So the absence of a non-trivial cycle is special to \(3n+1\), not a property of the form: nearby members of the family cycle freely. That sensitivity is the elementary face of Conway’s theorem (1972 [15]) that the general \(\mathrm{an}+b\)-type iteration is Turing-complete and its reachability undecidable — the family is as wild as computation itself, and \(3n+1\) is just one (conjecturally tame) point in it.
The maximum is ? 5 (mod 12). Its odd predecessor \(P\) satisfies \(T(P)=M\); since \(M\) is the max, \(P<M\) forces the ascending case, giving exactly
\[P = (2M-1)/3, P \equiv 3 \bmod 4, P \approx \tfrac{2}{3}M.\]
For \(P\) to be a positive integer, \(2M \equiv 1 \bmod 3\), i.e. \(M \equiv 2 \bmod 3\). (If \(M \not\equiv 2 \bmod 3\), its only odd predecessor has \(\nu \ge 2\) and exceeds \(M\) — impossible for the max.) Combined with \(M \equiv 1 \bmod 4\): \(M \equiv 5 (\bmod 12)\), verified (e.g. \(13 \equiv 1 \bmod 3\) cannot be a peak; \(5, 17, 29, 41, \dots \) can).
And the maximum is never ? 9 (mod 16). A forward version of the same idea: \(M \equiv 9 \bmod 16\) forces \(\nu = 2\) exactly, so the successor \(T(M) = (3M+1)/4 \equiv 3 \bmod 4\) ascends, giving \(T^{2}(M) = (9M+7)/8 > M\) — impossible for the maximum (the cycle is closed, so \(T^{2}(M) \le M\)). The overshoot is special to \(\nu =2\): for \(\nu =3\) (\(M \equiv 13 \bmod 16\)) one gets \(T^{2}(M)=(9M+11)/16 < M\), and smaller for \(\nu \ge 4\), so only \(9 \bmod 16\) self-destructs. Combining with \(M \equiv 5 \bmod 12\): the maximum is \(\equiv 5, 17, \mathrm{or} 29 (\bmod 48)\) — never \(41 \bmod 48\) — sharpening the sieve from 1-in-12 to 3-in-48.
The predecessor’s low bits are \(\dots 011\). Since \(T(P) = M \equiv 1 \bmod 4\), and the mod-8 rule gives \(P \equiv 3 \bmod 8 \implies T(P) \equiv 1 \bmod 4\) while \(P \equiv 7 \bmod 8 \implies T(P) \equiv 3 \bmod 4\), the peak forces \(P \equiv 3 \bmod 8\) — three pinned bits. The 4th bit is free (\(P \equiv 3 \mathrm{or} 11 \bmod 16\), flipping with \(M \bmod 16\)). Structurally, \(3 \bmod 8\) is the exit residue and \(7 \bmod 8\) the stay-sticky residue (Section 7, below), so the maximum caps an ascending sticky-run, and \(P\) is the run’s terminal exit element — the same mechanism as the divergence-side fuel burns, which also peak at \(2\cdot 3^{k-1}-1 \equiv 1 \bmod 4\). The free 4th bit encodes the (invisible-to-the-peak) shape of the run that climbed in.
The predecessor must not be a multiple of 3 — excluding \(M \equiv 5 \bmod 9\). \(P\) is itself a cycle element, so it needs its own predecessor; but an odd multiple of 3 has none (\(3Q+1 \equiv 1 \bmod 3\), never \(\equiv 0\)), so odd multiples of 3 lie in no cycle. When \(M \equiv 5 \bmod 9\), \(P = (2M-1)/3\) comes out an odd multiple of 3 (e.g. \(M=5\to P=3\), \(M=23\to P=15\)) — contradiction. Hence the maximum is \(\equiv 2\) or \(8 (\bmod 9)\), never \(5 \bmod 9\) (an independent sieve, stacking with the mod-4 and mod-16 ones).
\(M \bmod 9\) fixes how the peak is approached. \(P \bmod 3\) (inherited from \(M \bmod 9\)) decides whether \(P\)’s own predecessor \(Q\) ascends or descends into it:
| \(M \bmod 9\) | \(P \bmod 3\) | \(Q\) (predecessor of \(P\)) | shape before the peak |
|---|---|---|---|
| \(2\) | \(1\) | \((4P-1)/3 > P\) (descends in, \(\nu =2\)) | V-dip: \(Q \to P \to M\) — trajectory drops to \(P\), then jumps to the max |
| \(8\) | \(2\) | \((2P-1)/3 < P\) (ascends in, \(\nu =1\)) | steady climb: \(\dots \to Q \to P \to M\), unbroken ascent into the max |
So the maximum either crowns a smooth ascending run (\(M \equiv 8 \bmod 9\)) or sits atop a V-shaped spike where the trajectory first dipped to the local minimum \(P\) (\(M \equiv 2 \bmod 9\)).
The surviving residues, and their binary shapes. Intersecting just the mod-16 and mod-9 sieves leaves 6 residues mod 144, which group into the three allowed \(\bmod 16\) classes (their low 4 bits), each paired across \(\bmod 9 \in {2,8}\):
| \(\bmod 16\) (low 4 bits) | residues | small-rep shape | regime |
|---|---|---|---|
| \(0001\) (\(1\)) | \(17, 65\) | \(2^k+1\) — sparse, lone high bit | \(\nu =2\), gentlest drop |
| \(1101\) (\(13\)) | \(29, 125\) | \(2^k-3 = (k-2 \mathrm{ones})01\) — near all-ones | \(\nu =3\), steeper |
| \(0101\) (\(5\)) | \(53, 101\) | alternating tail \(\dots 0101\) + cap | \(\nu \ge 4\), steepest (firewall) |
The clean \(2^k\pm c\) forms are an artifact of taking the smallest representatives (a genuine — astronomically large — maximum only shares the low 4 bits and the mod-9 class). But the three shapes are not artifacts: they are the three binary motifs that recur throughout this whole analysis — sparse, all-ones, and alternating. The \(2^k-3\) class is the cycle-side echo of the \(-1\) glider (Section 12, below); the \(0101\) class is the cycle-side echo of the explosive/firewall fuel (Section 9, below). The deeper maximality sieve (mod 32, 64, …) then carves into the higher bits of each.
The peak is a sharp spike: ascend in from \(P \approx \tfrac{2}{3}M\), drop out to the successor \(T(M) \le (3M+1)/4 \approx \tfrac{3}{4}M\).
An ascending-free zone below the peak. No cycle element in the interval \((P, M) = (\tfrac{2}{3}M, M)\) can be \(\equiv 3 \bmod 4\): an ascending step there gives \(T(x) = (3x+1)/2 > M\) (overshoot), since \(x > P \iff T(x) > M\). So that band is reachable only by descending (\(\equiv 1 \bmod 4\)) steps — the lone ascending entry is the final jump \(P \to M\) onto the peak itself. (When \(\nu (M)=2\), i.e. \(M \equiv 1 \bmod 16\), the successor \(T(M) = (3M+1)/4 \in (P,M)\) is one such descending occupant.) It is tempting to read this as “a third of the numbers just below \(M\) are barred from being growth steps, so the 5:2 ratio is harder.” It is not — the exclusion only relocates growth, it does not thin it. The band \((P, M]\) is just the top \(\mathrm{log}_{2}(M/P) = \mathrm{log}_{2}(3/2) \approx 0.585\) octave (\(M\) and \(P \approx \tfrac{2}{3}M\) share a leading magnitude), and all it asserts is that every \(\equiv 3 \bmod 4\) element sits at or below \(P\). The \(5:2\) count is therefore taken over \([c_\mathrm{min}, P]\) with \(P\) as its ceiling — exactly the band the next bound packs — so the exclusion is already paid for there, not lost. (Note this is a real width, not a negligible sliver: the range \(c_\mathrm{max}/c_\mathrm{min}\) itself is forced \(\ge ~3.23\) below, so \((P, M]\) is a sizeable top fraction, not 1% of some imagined 70-octave spread.) The constraint shapes where growth sits (at or below \(\tfrac{2}{3}M\), never in the top sliver), not how much of it there is.
The climb into the peak has length at most \(v_{3}(M+1)\). Trace ascending predecessors back from \(M\) via \(x_{j+1} = (2x_j – 1)/3\). Since that map has fixed point \(-1\), it has a closed form:
\[x_j + 1 = (2/3)^j \cdot (M+1).\]
This is an integer for exactly \(j \le v_{3}(M+1)\) steps (the 3-adic valuation of \(M+1\)) and then stops — \(3\) runs out of \(M+1\). So the maximal ascending chain that could end at \(M\) has length \(v_{3}(M+1)\), with rungs \(P, Q, R, \dots = (2/3)^j(M+1) – 1\). The actual ascending run feeding \(M\) in an orbit is a suffix of this chain — the preceding descent may land partway up — so its length is \(\le v_{3}(M+1)\), with equality only when the descent lands on the bottom rung. (Verified over 390,193 ascending runs from small and random ~80-bit starts: zero violations of the bound; equality \(\mathrm{runlen} = v_{3}(\mathrm{peak}+1)\) in 65.8% of runs, with the shortfall almost always just 1 — slack \(0/1/2 = 65.8\% / 31.2\% / 2.4\%\). So the bound is tight about two-thirds of the time. E.g. peak \(377\) runs \(2\) though \(v_{3}(378)=3\).) The bottom rung \(2^{v}(M+1)/3^{v} – 1\) is all-ones \(2^k-1\) (the §7 fuel reserve, below, viewed from the top) exactly when \(M+1 = 2^a\cdot 3^b\); otherwise generic. Either way it has no ascending predecessor, so in a cycle it is entered by a descent — a local minimum — giving the Steiner “circuit” shape: descent ? local minimum ? ascending run (length \(\le v_{3}(M+1)\)) ? peak \(M\) ? descent.
The mod-3 floor — and the source of the slack. The rungs have forced residues mod 3: every intermediate rung \(x_{1} \dots x_{v-1}\) is \(\equiv 2 \bmod 3\) (because \(x_j+1 = 2^j(M+1)/3^j\) still carries a factor of 3, so \(x_j \equiv -1\)), while only the bottom \(x_v\) can be \(\equiv 0 \bmod 3\). A \(\equiv 2 \bmod 3\) rung always has predecessors; a \(\equiv 0 \bmod 3\) number has none at all (\(3w+1 \equiv 1 \bmod 3 \ne 0\)), so it lies in no cycle and appears in no orbit except as a start. All-ones numbers \(2^k-1\) are \(\equiv 0 \bmod 3\) exactly when \(k\) is even (\(3, 15, 63, \dots \)) — so the “all-ones reserve” is unreachable precisely in the even-\(k\) case. The consequence: when the maximal bottom \(x_v\) is \(\equiv 0 \bmod 3\), the climb cannot reach it — the run must stop at a higher \(\equiv 2 \bmod 3\) rung fed by a descent. This is the slack measured above: the ~31% of runs that fall one short of \(v_{3}(\mathrm{peak}+1)\) are largely the cases where \(x_v \equiv 0 \bmod 3\) blocks the bottom rung. So the all-ones-÷3 obstruction (and \(\equiv 0 \bmod 3\) generally) doesn’t forbid \(M\); it sets the climb’s floor one or more rungs up, at the lowest \(\not\equiv 0 \bmod 3\) rung.
Climb grammar near the peak. The rungs obey a residue rule. A generator (ascending, \(\equiv 3 \bmod 4\)) strictly inside a band \((Q, P)\) maps into \((P, M)\), the ascending-free zone — so its image must be \(\equiv 1 \bmod 4\), forcing the generator to be \(\equiv 3 \bmod 8\) (the “exit” residue). A \(\equiv 7 \bmod 8\) (“stay-sticky”) generator strictly inside is forbidden: its image would be \(\equiv 3 \bmod 4\) in \((P, M)\) and overshoot \(M\) one step later. The \(\equiv 7 \bmod 8\) rungs are therefore exactly the main climb spine (\(P \equiv 3 \bmod 8\) is its terminal exit rung onto \(M\); the rungs below are \(\equiv 7 \bmod 8\)), and any extra generators must be \(\equiv 3 \bmod 8\) “exit” rungs that peel off into descending side-branches. Verified, \(M=53\): spine \(15\to 23\to 35\to 53\); the off-spine candidate \(27 (\equiv 3 \bmod 8)\) is allowed (\(\to 41\), descends) while \(31 (\equiv 7 \bmod 8)\) is forbidden (\(\to 47\to 71 > 53\)).
The descent off the peak is graded by \(M \bmod 16\) (the three surviving residues \(1, 5, 13\); \(9\) is excluded above). The successor’s depth is fixed by \(\nu \), even though its low bits are not:
| \(M \bmod 16\) | \(\nu \) | \(T(M)/M \approx 3/2^\nu \) | character of the drop |
|---|---|---|---|
| \(1\) | 2 | \(\approx \tfrac{3}{4}\) | gentlest; \(T(M) \equiv 1 \bmod 4\) (descends again), fully fixed by 4 bits |
| \(13\) | 3 | \(\approx \tfrac{3}{8}\) | steeper; \(\nu \) fixed, one more bit (bit 4) sets the successor’s class |
| \(5\) | ?4 | \(\le 3/16\), mean \(\approx \tfrac{1}{8}\) | steepest and recursive — \(T(M) = \mathrm{oddpart}(3H+2)\) (a 3n+1 step on the high part), unbounded \(\nu \) |
Each extra halving roughly halves the successor, so \(13\) and \(5\) are the steep-fall maxima. The depth is determined by the residue; the successor’s low bits are not — they scatter across \({01,11}\) as the higher bits vary (only special sub-cases pin them).
\(M \equiv 5 \bmod 16\) is the alternating/firewall regime, governed by one identity. Writing \(M = H\cdot 2^{2j+1} + A_j\) with \(A_j = (4^{j+1}-1)/3\) the pure alternating tail (\(101\dots 01\)), and using \(3A_j+1 = 2^{2j+2}\):
\[3M+1 = 2^{2j+1} \cdot (3H+2).\]
- Uncapped (\(H=0\), never-ending alternating, the \(r_m=(2^m-1)/3\) residues): \(3M+1 = 2^{2j+2}\), a strict power of two ? \(T(M)=1\), total collapse in one step.
- Capped (\(H\ge 1\)): \(3M+1 = [3H+2]\| [2j+1 \mathrm{zeros}]\) — the alternating run is spent to make a \(2j+1\)-zero firewall, and the cap survives as \(3H+2\) (so \(T(M)=\mathrm{oddpart}(3H+2)\), recursing on the cap). E.g. \(110101\): \(H=1\), \(3H+2=5\), \(3M+1 = 101\| 00000 = 160\); successor \(5 = \dots 01\). This is the §9 conservation law (below) made exact for cycle maxima: the alternating tail converts entirely to zeros, only the cap propagates.
What it gives — cycle lower bounds. Around a cycle with \(a\) odd steps and \(S = \sum \nu \) halvings, dropping the \(+1\)s would demand \(3^a = 2^S\) — impossible (coprime). Cycles can exist only via the \(+1\) corrections:
\[3^a \cdot \prod (1 + 1/(3x_{i})) = 2^S.\]
For large elements the correction product is tiny, forcing \(3^a \approx 2^S\) to extreme precision — so \(S/a\) must approximate \(\mathrm{log}_{2}3 \approx 1.585\) extraordinarily well. Good rational approximations to an irrational require large denominators (continued fractions / Baker’s theorem on linear forms in logarithms), forcing \(a\) large, hence the cycle long. Combined with the verified-to-\(2^68\) frontier, this yields the known bounds: a non-trivial cycle’s minimum element must exceed 2?? (? 3×10²?), and its length is bounded below — historically \(~1.7\times 10^{7}\) terms (Eliahou [12], at the verification of its era), far larger now (Simons–de Weger [13]).
Capping the range doesn’t work — but \(c_\mathrm{min}\) in the balance does (the explicit bound). A direct cap on the range is hopeless: the rise multiplies by at most \(3/2\) per step, so only \(c_\mathrm{max}/c_\mathrm{min} \le (3/2)^a\), astronomically loose. The leverage is instead that every element is \(\ge c_\mathrm{min}\), which pins the correction product:
\[1 < 2^S/3^a = \prod (1 + 1/(3x_{i})) \le (1 + 1/(3c_\mathrm{min}))^a \approx 1 + a/(3c_\mathrm{min}).\]
With \(c_\mathrm{min} > 2^{68}\), the right side is \(1 + ~10^{-13}\), so \(2^S/3^a\) lies within \(~10^{-13}\) of 1 — i.e. \(S/a\) must hit \(\mathrm{log}_{2}3\) to ~13–14 digits. Writing \(\delta = S – a\cdot \mathrm{log}_{2}3 > 0\), the requirement is \(\delta < a/(3c_\mathrm{min}\cdot \mathrm{ln}2)\), so a length \(a\) is viable only if its “upward gap” \(\delta \) is that small — which happens only at large continued-fraction denominators of \(\mathrm{log}_{2}3\). Marching through the convergents against the \(2^{68}\) frontier, the smallest viable length is \(a \approx 7.96 \times 10^{19}\) odd steps — the convergent denominator 79,641,170,620,168,673,833 (its upward gap is \(\delta \approx 0.017\)). So a non-trivial cycle with minimum above \(2^{68}\) must have on the order of \(10^{20}\) odd elements (total terms ~2.6× that) — the concrete version of “long.” Note what is and isn’t bounded: the range \(c_\mathrm{max}/c_\mathrm{min}\) stays unbounded; it is the length that the irrationality of \(\mathrm{log}_{2}3\) forces to be enormous. (This estimate captures the mechanism and order of magnitude via the convergents; the rigorous published bounds sharpen the constants with the same engine.)
The growth numbers occupy a sub-band — which can’t be flat. The \(\equiv 3 \bmod 4\) (ascending) elements — the ones the \(5:2\) count tracks — are confined to \([c_\mathrm{min}, P]\), where \(P = (2M-1)/3 \approx \tfrac{2}{3}M\) is the largest of them (the ascending-free zone \((P, M]\) holds none) and \(c_\mathrm{min}\) the smallest. So the relevant span is \(P/c_\mathrm{min} = \tfrac{2}{3}\cdot c_\mathrm{max}/c_\mathrm{min}\), not the full range. Now count the usable slots: a cycle element is coprime to 3 (a multiple of 3 has no predecessor), so among \(\equiv 3 \bmod 4\) integers — residues \(3, 7, 11 \bmod 12\) — only \(7, 11 \bmod 12\) are eligible, density \(\tfrac{1}{6}\), not \(\tfrac{1}{4}\). Packing the \((5/7)a\) ascending elements as distinct such integers needs
\[P – c_\mathrm{min} \ge 6\cdot (5/7)a = (30/7)a.\]
With \(a \ge 8\times 10^{19}\) and \(c_\mathrm{min} > 2^{68}\), that forces \(P \ge 6.4\times 10^{20}\), hence \(c_\mathrm{max} = (3/2)P\) and the clean \(c_\mathrm{min}\)-independent bound
\[c_\mathrm{max} \ge (3/2)c_\mathrm{min} + (45/7)a.\]
Equivalently \(c_\mathrm{max}/c_\mathrm{min} \ge 3/2 + (45/7)(a/c_\mathrm{min})\), which is \(\approx 3.23\) at the frontier (\(c_\mathrm{min} \approx 2^{68}\)). So a non-trivial cycle cannot be flat: its span exceeds \(~3.23\times \) near \(c_\mathrm{min} \approx 2^{68}\), relaxing toward the irreducible \(3/2\) (since \(c_\mathrm{max} = (3/2)P \ge (3/2)c_\mathrm{min}\)) for taller cycles. Density \(\tfrac{1}{6}\) is where this stops — beyond “not a multiple of 3” there is no further universal exclusion on a generic ascending element (the finer residue bounds apply only to the max/predecessor). And it remains a lower bound: like every counting fact here it sets a floor on the shape, not a ceiling that closes the question.
The floor \(c_\mathrm{min}\), by contrast, is verification-bound — packing cannot lift it. Note the asymmetry: the band argument constrains the width (\(P – c_\mathrm{min} \ge (30/7)a\)), which pushes \(c_\mathrm{max}\) up given \(c_\mathrm{min}\) — but it says nothing about how high the band must start. Nothing in the dynamics forces elements low (they may sit arbitrarily high), so density yields no rigorous floor on \(c_\mathrm{min}\); the only lower bound is direct verification (\(c_\mathrm{min} > 2^{68}\): every smaller \(n\) is computed to reach 1), a per-number fact stronger than any count. The log-uniform bottom-octave heuristic merely recovers \(~2^{68}\) — and circularly, since the length \(a\) was itself derived from \(c_\mathrm{min} > 2^{68}\). So the counting/coprimality machinery is a width / \(c_\mathrm{max}\) / ratio tool; the floor is a computation. Raising \(c_\mathrm{min}\) needs more verification, not more counting.
Why this makes cycles heuristically impossible. \(S/a \approx \mathrm{log}_{2}3\) means the cycle’s average \(\nu \) equals \(1.585\) — i.e. \(\nu =1\) at the ~71% (5:2) rate of Section 5, versus the natural 50%. For the probabilistic model (\(\nu \) iid geometric, mean 2, variance 2), holding the sample mean down at \(1.585\) over \(a\) steps is a deviation of \(\approx 0.415\cdot \surd (a/2)\) standard deviations. At the forced length \(a \sim 8\times 10^{19}\) (previous paragraph) that is ~2.6×10? standard deviations, a Gaussian-tail probability on the order of \(10^{-1.5\times 10^{18}}\). Worse, the exponent grows linearly with the length the bounds force — so the longer a cycle must be, the more impossible it becomes, and summed over all candidate lengths the heuristic expected number of non-trivial cycles is essentially zero. (This is the cycle-side analogue of the even-distribution heuristic in Section 5; it rests on the same unprovable pseudorandomness assumption — the \(\nu \) sequence of a deterministic orbit need not behave like iid draws.)
Ceiling. The congruence sieves (\(M \equiv 1 \bmod 4\), \(M \equiv 5 \bmod 12\), \(P \equiv 3 \bmod 8\)) and the spike bounds each remove a constant fraction of candidates and feed the size bound — but \(\mathrm{log}_{2}3\) can be approximated arbitrarily well given enough steps, so they push the minimum cycle size upward without ever reaching zero. Same character as the divergence side: every constraint is sharp and real; none closes the door.
7. Fuel I — Consecutive 1s (Growth, but Weak and Finite)
A consecutive run of growth requires staying in the \(\equiv 3 \bmod 4\) class (?=1) step after step. (Note: net divergence does not need this — only an above-average frequency of ?=1, scattered, which needs no structure; see Section 13. This section is about the structured, consecutive route.) Which numbers stay ?=1 in a row?
Uniqueness theorem (verified): the only residue mod \(2^{j+1}\) that stays ?=1 for j consecutive steps is the all-ones residue \(2^{j+1} – 1\). Concretely: stay-sticky-1-step = \(3 \bmod 4\), 2-steps = \(7 \bmod 8\), 3-steps = \(15 \bmod 16\), … There is no alternative structure — a length-j growth run requires j+1 trailing 1s.
For the all-ones number \(n = 2^k – 1\), the run has an exact closed form:
\[a_i = 3^i \cdot 2^{k-i} – 1\qquad \text{(eats one trailing 1 per step)}\]
It grows ×3/2 for exactly k?1 steps, then exits at \(2\cdot 3^{k-1} – 1 \equiv 1 \bmod 4\) and descends. Maximal fuel is a finite reserve: k ones buy k?1 growth steps.
8. Fuel I, continued — “No Free Fuel”
To reach \(2^k – 1\) (not start there), the nearest odd predecessor has exactly k+1 bits (it is ? 4/3 larger, pushed over the \(2^k\) boundary). And the all-ones numbers split:
- k even (\(3, 15, 63, 255, \dots \)): divisible by 3. Since \(3n+1\) is never divisible by 3, these have no odd predecessor at all — isolated islands, reachable only on their own doubling chain.
- k odd (\(7, 31, 127, \dots \)): reachable, but only by descending into them from a strictly longer (k+1-bit) ancestor.
Conclusion: you can never climb into maximal fuel. Pure fuel is always downhill; assembling a length-k all-ones block costs a number of length k+1 — you compact and lose ?1 bit. Fuel cannot be bootstrapped upward.
9. Fuel II — Alternating Structure (Explosive, but Self-Emptying)
The other way to get “power” is a long carry cascade, which needs the alternating pattern. But propagation converts 1s into 0s:
| n | ones(n) | ones(3n+1) | firewall created |
|---|---|---|---|
| 1365 = \(10101010101\) | 6 | 1 | 12 zeros |
| 1367 = \(10101010111\) | 7 | 3 | 9 zeros |
| 341 = \(101010101\) | 5 | 1 | 10 zeros |
Conservation law: carry distance = 1s destroyed = zeros created. The cascade spends the alternating fuel to manufacture the very firewall (Section 4) that then isolates and collapses the number. Power converts directly into void. The alternating “residues” \(r_m = (2^m-1)/3\) are the pure form of this fuel (Figure 6): each is a \(101\dots 01\) pattern, and \(3\cdot r_m + 1 = 2^m\) exactly — a clean power of two.
10. The Complete Structural Picture
There are exactly two sources of power in the binary string, and the carry mechanism dooms both:
| Fuel | Behaviour | Why it cannot sustain divergence |
|---|---|---|
| Consecutive 1s | slow ×3/2 climb | weak, finite (burns out in k?1 steps), cannot be bootstrapped (downhill to reach) |
| Alternating | one large carry cascade | the explosion is a firewall — it destroys its own 1s to create the void that collapses it |
No third option: a carry either fizzles (no effect) or propagates (destroying the 1s that enabled it). You cannot have a large cascade that preserves its fuel — that is contradictory by construction. This is the structural reason behind the average \(\times 3/4\) drift, with both tails of the ?-distribution explained: small ? = weak bounded growth, large ? = self-emptying collapse.
This dooms every structured route to divergence. What it does not address is the unstructured one: a scattered, above-average frequency of ?=1 steps (numbers merely ending in \(\dots 11\), no all-ones block, no run) sustained over infinitely many steps. That carries no fuel to preserve and so dodges this argument entirely — it is the lone survivor of Section 13.
11. The Game of Life Analogy
The binary trajectory updated by a fixed local-ish rule invites comparison to cellular automata (CA) — and in particular to Conway’s Game of Life, where a glider is a finite pattern that, after a fixed number of steps, reappears unchanged but translated: a self-sustaining structure that neither dies nor blows up. The Collatz analogue of a glider is a binary pattern that perpetually reconstructs the conditions for its own growth — i.e. a divergent trajectory or a non-trivial cycle. The conjecture, restated, is: 3n+1 has no glider.
The comparison is sharper than a slogan:
- In 1-D cellular automata (Wolfram’s elementary CAs), whether gliders exist is decidable for any fixed period and speed, because the rule is local (radius 1) — a glider’s influence spreads at bounded speed, so a finite enumeration (de Bruijn diagrams / transfer matrices) settles it. Wolfram’s Class I/II rules provably have none; Class III/IV (e.g. Rule 110) do.
- Collatz looks Class III/IV — complex, sensitive, plausibly Turing-complete — yet behaves Class I (everything dies to one fixed point). The obstruction to importing the CA decision procedure is exactly that Collatz has no locality: a single carry can, in principle, couple bit 0 to bit 60. The “firewall” of Section 4 is an attempt to manufacture locality — and it works conditionally, but proving every trajectory creates one is the open problem.
So the right question is not “does a glider exist?” but “can a finite configuration sustain glider-like behaviour without the carry mechanism eventually destroying it?” — which Sections 7–10 answer “not by either available fuel,” and Section 12 answers precisely in the 2-adics.
12. The Glider That Exists — in the 2-adics
A perfect glider does exist, but not as a positive integer. In the 2-adic integers, \(-1 = \dots 11111\) is a fixed point of the shortcut map:
\[3\cdot (-1)+1 = -2 = \dots 11110, \div 2 \to -1.\]
The finite all-ones numbers \(2^k – 1\) are its truncations — which is exactly why they are the maximal growth runs: they track ?1 for k?1 steps before their finiteness leaks at the top. So divergence isn’t logically impossible; the obstruction is arithmetic approximation — whether a finite integer can chase ?1 to ever-greater lengths. (In CA terms: the glider lives on the infinite tape \(\dots 1111\), and every finite truncation of it decays.)
Gun vs glider. The distinction matters. A Life glider conserves mass and merely translates; a Life glider gun creates, growing without bound — possible only because Life has no conservation law. Collatz does (Section 9: carry converts 1s?0s, fuel is never created on net), so the gun is forbidden — no configuration perpetually manufactures fresh trailing-1 fuel. The glider is not killed by conservation alone, but a finite one is squeezed by two opposing constraints:
- Anchored below. Growth needs trailing 1s (2-adic closeness to ?1). Erode that anchor to a single 1 (\(\equiv 1 \bmod 4\)), or collapse the low part to a power of 2, and the number detaches from ?1 and falls — to 1.
- In vacuum above. A carry cascade into high structure destroys 1s and creates a zero firewall (Section 9). So a glider cannot grow through structure, only into emptiness — its top must climb into leading-zero vacuum.
The all-ones run shows the geometry exactly: \(111 \to 1011 \to 10001\), the top climbing into vacuum while the anchor erodes (3?2?1 trailing 1s), then detaching. The anchor is finite fuel; refilling it would need the forbidden gun. The one crack: the anchor need not be steady — it could flicker, short anchors re-established by chance often enough to net-grow. Conservation bars a deterministic gun, not a statistically lucky infinite sequence of re-anchorings. That such luck cannot run forever is exactly the conjecture.
13. Where the Wall Stands
Every structural fact above is true per step, per run, or per residue class. Both obstacles fragment into mechanisms that are individually dead, with one survivor each:
Divergence:
| Mechanism | Status |
|---|---|
| One infinite growth run | Dead — capped by bit-length |
| Climbing into ever-longer all-ones runs | Dead — entered only by descent / isolated (Section 8) |
| A finite integer being the glider | Dead — the glider is ?1, not an integer (Section 12) |
| Many short, impure runs regenerating in aggregate | Open = half the conjecture |
Cycles:
| Constraint | Effect |
|---|---|
| min ? 3 mod 4; max ? 5 mod 12, ? 9 mod 16, ? 5 mod 9; predecessor ? 3 mod 8, ? 0 mod 3 | congruence sieves — thin candidates by constant factors (Section 6) |
| \(3^a \ne 2^S\) exactly; only \(\approx \) via \(+1\) | forces \(S/a \approx \mathrm{log}_{2}3\), hence huge \(a\) |
| A small non-trivial cycle | Dead — bounded out to enormous size |
| A large non-trivial cycle (good \(\mathrm{log}_{2}3\) approximation) | Open = half the conjecture |
Both survivors are infinite-aggregate properties: for divergence, enough ?=1 steps to grow without ever exploding or exhausting the impure fuel, forever; for cycles, a step-count ratio approximating \(\mathrm{log}_{2}3\) well enough to close a loop, which only ever requires more steps. Sharp local facts — per step, per run, per residue — do not compose into infinite-aggregate conclusions. That gap is precisely where Collatz has stood since 1937.
14. Honest Assessment
This is expository, not novel. The binary/carry framing is a concrete shadow of standard 2-adic Collatz analysis [4, 5, 6]; the density results are Terras/Everett [2, 3]; the state of the art is Tao 2019 [7]. What the binary lens adds is intuition — making visible, via fuel and firewalls and carry conservation, why the average drift is downward and why that cannot be promoted to a proof. The value is pedagogical: a tangible account of a problem usually stated abstractly.
15. Relation to Known Results
None of the mathematics here is new; the binary lens is a re-derivation of established theory. The correspondence, made explicit:
| In this document | Established source |
|---|---|
| Shortcut map \(T(n)=(3n+1)/2^\nu \); parity of \(\nu \) | standard; Terras [2], Lagarias survey [1] |
| Binary “firewall” = independence across a zero run | 2-adic continuity of the Collatz map [4, 5, 6] |
| \(-1 = \dots 111\) as the all-ones fixed point (the “glider”) | textbook 2-adic fact [5, 6] |
| “Conservation law” (carry spends 1s ? zeros) | a re-description of the \(\nu _{2}(3n+1)\) arithmetic |
| Density-1 descent; the \(\mathrm{log}2/\mathrm{log}3 \approx 0.631\) break-even | Terras [2], Everett [3] |
| “Almost all orbits attain almost bounded values” | Tao [7] |
| Cycle balance \(3^a \approx 2^S\) ? \(\mathrm{log}_{2}3\) approximation ? size bounds | Crandall, Steiner, Eliahou [12], Simons–de Weger [13] |
| Cycle min \(\equiv 3 \bmod 4\); no element \(\equiv 0 \bmod 3\); congruence sieves on the extrema | elementary folklore in the cycle literature |
The specific small artifacts — \(\mathrm{max} \not\equiv 9 \bmod 16\), \(\mathrm{max} \in {17,29,53,65,101,125} \bmod 144\), the binary-shape grouping of those residues — may not appear verbatim elsewhere, but they are immediate consequences of the standard residue/maximality method, not new facts. They are new expressions, not new results.
What the exercise does offer, honestly stated: (i) a from-scratch reconstruction of the 2-adic structure using only elementary binary reasoning; (ii) a unified narrative in which the same three binary motifs — sparse, all-ones, alternating — govern both the divergence and cycle obstacles; and (iii) a clear demonstration of why every elementary structural attack hits the same wall: each constraint is a constant-factor sieve, and constant-factor sieves push the floor up without ever reaching zero, because sharp local facts do not compose into infinite-aggregate conclusions. The lesson is about the shape of the obstruction, not about new ground gained.
16. References
Collatz / 3n+1
- J. C. Lagarias, The 3x+1 problem and its generalizations, American Mathematical Monthly 92 (1985), 3–23. (The standard survey; see also Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, AMS, 2010.)
- R. Terras, A stopping time problem on the positive integers, Acta Arithmetica 30 (1976), 241–252. (Parity vectors; density-1 finite stopping time.)
- C. J. Everett, Iteration of the number-theoretic function f(2n)=n, f(2n+1)=3n+2, Advances in Mathematics 25 (1977), 42–45.
- H. Möller, Über Hasses Verallgemeinerung des Syracuse-Algorithmus (Kakutanis Problem), Acta Arithmetica 34 (1978), 219–226. (2-adic formulation.)
- G. J. Wirsching, The Dynamical System Generated by the 3n+1 Function, Lecture Notes in Mathematics 1681, Springer, 1998.
- K. R. Matthews, Generalized 3x+1 mappings: Markov chains and ergodic theory, in [1, The Ultimate Challenge]. (2-adic/ergodic background.)
- T. Tao, Almost all orbits of the Collatz map attain almost bounded values, Forum of Mathematics, Pi 10 (2022), e12; arXiv:1909.03562 (2019).
Cellular automata / Game of Life
- M. Gardner, Mathematical Games: The fantastic combinations of John Conway’s new solitaire game “life”, Scientific American 223 (Oct. 1970), 120–123.
- E. R. Berlekamp, J. H. Conway, R. K. Guy, Winning Ways for Your Mathematical Plays, 2nd ed., A K Peters, 2001–2004. (Life: gliders, guns, universality.)
- S. Wolfram, Computation theory of cellular automata, Communications in Mathematical Physics 96 (1984), 15–57. (de Bruijn diagrams, symbolic dynamics.)
- S. Wolfram, A New Kind of Science, Wolfram Media, 2002. (Elementary CA classification; full text at wolframscience.com.)
Cycle bounds
- S. Eliahou, The 3x+1 problem: new lower bounds on nontrivial cycle lengths, Discrete Mathematics 118 (1993), 45–56.
- J. Simons, B. de Weger, Theoretical and computational bounds for m-cycles of the 3n+1 problem, Acta Arithmetica 117 (2005), 51–70. (See also R. Crandall, Math. Comp. 32 (1978), 1281–1292.)
- R. P. Steiner, A theorem on the Syracuse problem, in Proc. 7th Manitoba Conf. on Numerical Mathematics (1977), 553–559. (No non-trivial 1-circuit cycle.)
Undecidability of the general family
- J. H. Conway, Unpredictable iterations, in Proc. 1972 Number Theory Conference (Univ. of Colorado, Boulder, 1972), 49–52. (Reachability of generalized \(\mathrm{an}+b\) maps is undecidable; basis of FRACTRAN.)
17. Artifacts
collatz_explorer.py— trajectory analysis, ? / Hamming / zero-run / alternating-run metrics, carry roles & reach, firewall-isolation verifier, r_m residues.collatz_viz.py— overview heatmaps, 3-panel shortcut heatmap (bits / carry roles / ? & reach), ?-distribution, r_m patterns, multi-trajectory.collatz_carry.py— carry-cycle visualizations (generic vs alternating).collatz_discussion.md— the full chronological Q&A log this writeup summarizes. (not shared)
https://github.com/perlauge/collatz
Key reproducible facts: python3 collatz_explorer.py 27, python3 collatz_viz.py 1365.
18. One-Paragraph Summary
In binary, Collatz descent comes from two facts about carries: consecutive-1 “fuel” produces only weak (×3/2), finite, un-bootstrappable growth, and the alternating structure needed for a far-reaching carry cascade is destroyed by that very cascade (1s ? 0s, creating a collapsing firewall). Together these explain the average ×3/4 drift completely. The perfect self-sustaining structure exists only as the 2-adic integer ?1, which no finite integer can be. On the cycle side, the same per-element rule pins any non-trivial cycle’s extremes — minimum ? 3 mod 4, maximum ? 5 mod 12 with immediate predecessor ? 3 mod 8 (the terminus of an ascending sticky-run, identical to the divergence-side fuel burns) — and the impossibility of \(3^a = 2^S\) forces the step-count ratio to approximate \(\mathrm{log}_{2}3\), making any cycle astronomically large. Both obstacles fragment the same way: every mechanism is provably dead except one infinite-aggregate survivor each — impure short runs regenerating forever (divergence), or an arbitrarily-good \(\mathrm{log}_{2}3\) approximation closing a loop (cycles). Each is exactly half the conjecture, unreachable because sharp per-step, per-run, and per-residue facts never compose into an infinite-aggregate guarantee.
