{"id":360,"date":"2026-06-04T19:23:46","date_gmt":"2026-06-04T17:23:46","guid":{"rendered":"https:\/\/serverdude.dk\/?p=360"},"modified":"2026-06-04T19:23:46","modified_gmt":"2026-06-04T17:23:46","slug":"the-collatz-conjecture-through-a-binary-lens","status":"publish","type":"post","link":"https:\/\/serverdude.dk\/?p=360","title":{"rendered":"The Collatz Conjecture Through a Binary Lens"},"content":{"rendered":"\n<script>MathJax={tex:{inlineMath:[[\"\\\\(\",\"\\\\)\"]],displayMath:[[\"\\\\[\",\"\\\\]\"]]}};<\/script>\n<script async src=\"https:\/\/cdn.jsdelivr.net\/npm\/mathjax@3\/es5\/tex-chtml.js\"><\/script>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Per Lauge Bures\u00f8 Holst<\/strong> \u00a0\u00b7\u00a0 <em>with Claude (Anthropic AI)<\/em><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">4 June 2026<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><em>A structural exploration of why 3n+1 trajectories tend downward \u2014 and why &#8220;tend&#8221; cannot yet be upgraded to &#8220;must.&#8221;<\/em><\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Collaboration &amp; process<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">This document is a collaboration between <strong>Per Lauge Bures\u00f8 Holst<\/strong> (who conceived and directed the investigation) and <strong>Claude, an AI assistant from Anthropic<\/strong>. The work split along its natural seam: the <strong>human supplied the ideas, intuitions, and direction<\/strong> \u2014 the binary lens, the &#8220;firewall&#8221; \/ &#8220;fuel&#8221; \/ &#8220;glider&#8221; framings, the cycle-climb questions, and the persistent skepticism that caught more than one overstatement; <strong>Claude supplied verification and formalization<\/strong> \u2014 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 &#8220;known result&#8221; traced to a source. The honest conclusion that this is <em>expository, not novel<\/em> (\u00a714\u201315) is itself an output of that process \u2014 as is the recurring theme that each elementary line of attack halts at the same wall. Remaining errors are ours, jointly.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">1. The Problem<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">For a positive integer n, apply: if even, \\(n \\to  n\/2\\); if odd, \\(n \\to  3n+1\\). The <strong>Collatz conjecture<\/strong> asserts every starting value eventually reaches 1. Verified to ~2??; unproven in general.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This writeup follows one line of attack: read the dynamics in <strong>binary<\/strong>, 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 \u2014 it merely <em>pans the bit pattern through a fixed positional frame<\/em>, 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 \u2014 enough to explain <em>why<\/em> trajectories descend on average \u2014 but each structural fact stops exactly at the same wall, which turns out to be the conjecture itself.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><a href=\"https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_binary_viewport-2.png\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"420\" src=\"https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_binary_viewport-2-1024x420.png\" alt=\"\" class=\"wp-image-381\" srcset=\"https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_binary_viewport-2-1024x420.png 1024w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_binary_viewport-2-300x123.png 300w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_binary_viewport-2-768x315.png 768w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_binary_viewport-2-1536x629.png 1536w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_binary_viewport-2.png 1635w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/a><figcaption class=\"wp-element-caption\"><strong>Figure 1.<\/strong> Dividing by 2 is a change of reference frame: a right shift just pans the bit pattern through a fixed positional &#8220;viewport,&#8221; dropping the bit that falls off the units end. No information is created or destroyed \u2014 the real work happens only at 3n+1.<\/figcaption><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Notation &amp; Symbols<\/h2>\n\n\n\n<figure class=\"wp-block-table\"><table><thead><tr><th>Symbol<\/th><th>Meaning<\/th><\/tr><\/thead><tbody><tr><td>\\(T(n)\\)<\/td><td><strong>shortcut map<\/strong> on odd \\(n\\): \\(T(n) = (3n+1) \/ 2^\\nu \\) \u2014 does \\(3n+1\\) then strips all trailing zeros, landing on the next odd number.<\/td><\/tr><tr><td>\\(\\nu \\)  (= \\(\\nu _{2}(3n+1)\\))<\/td><td><strong>2-adic valuation<\/strong> 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 &#8220;nu&#8221;.)<\/td><\/tr><tr><td>\\(v_{3}(m)\\)<\/td><td><strong>3-adic valuation<\/strong> 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 &#8220;v&#8221; \u2014 <em>not<\/em> the same as \\(\\nu \\) above.)<\/td><\/tr><tr><td>\\(\\equiv  a (\\bmod  b)\\)<\/td><td>congruence: \\(n \\equiv  3 \\bmod  4\\) means \\(n\\) leaves remainder 3 on division by 4, i.e. its low two bits are \\(\\dots 11\\).<\/td><\/tr><tr><td>\\(M\\)<\/td><td>the <strong>maximum<\/strong> odd element of a hypothetical non-trivial cycle.<\/td><\/tr><tr><td>\\(P, Q, R\\)<\/td><td>the <strong>predecessors<\/strong> (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)\\).<\/td><\/tr><tr><td>\\(a\\), \\(S\\)<\/td><td>for a whole cycle: \\(a\\) = number of odd (\\(3n+1\\)) steps; \\(S = \\sum \\nu \\) = total halvings. The balance is \\(3^a \\approx  2^S\\).<\/td><\/tr><tr><td>\\(H, L\\)<\/td><td>high \/ low parts of a number split across a zero-run &#8220;firewall&#8221;: \\(n = H\\cdot 2^{m+b} + L\\).<\/td><\/tr><tr><td>\\(A_j\\), \\(r_m\\)<\/td><td>alternating numbers: \\(A_j = (4^{j+1}-1)\/3 = 101\\dots 01\\); \\(r_m = (2^m-1)\/3\\) (same family).<\/td><\/tr><tr><td>\\(\\mathrm{octave}\\)<\/td><td>a factor-of-2 band of magnitude (one bit of length); &#8220;log-uniform&#8221; = roughly equal count per octave.<\/td><\/tr><tr><td>\\(\\ll \\), \\(\\gg \\)<\/td><td>left \/ right bit shift (\\(n\\ll 1 = 2n\\)).<\/td><\/tr><tr><td>\\(2^k-1\\)<\/td><td>&#8220;all-ones&#8221; number (\\(k\\) consecutive 1-bits); the \\(-1\\) of the 2-adic integers \\(Z_{2}\\) is the infinite all-ones \\(\\dots 1111\\).<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">2. The Shortcut Map and the Two Residue Classes<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Work odd-to-odd via the <strong>shortcut map<\/strong> \\(T(n) = (3n+1) \/ 2^\\nu \\), where ? = number of trailing zeros of \\(3n+1\\). Even numbers are irrelevant \u2014 they just halve down. So only odd residues mod 4 matter:<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table><thead><tr><th>n mod 4<\/th><th>3n+1<\/th><th>?<\/th><th>net per step<\/th><\/tr><\/thead><tbody><tr><td>? 1 (\\(\\dots 01\\))<\/td><td>4(3k+1)<\/td><td>? ? 2<\/td><td>shrinks (descends below n immediately)<\/td><\/tr><tr><td>? 3 (\\(\\dots 11\\))<\/td><td>2(6k+5)<\/td><td>? = 1<\/td><td>grows (\u00d73\/2)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Immediate consequences:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Any cycle&#8217;s minimum element is odd, and cannot be ? 1 mod 4 (it would drop below itself). So <strong>a non-trivial cycle&#8217;s minimum must be ? 3 mod 4<\/strong> \u2014 binary \\(\\dots 11\\).<\/li>\n\n\n\n<li>\\(n \\equiv  1 \\bmod  4\\) provably descends in one step. All difficulty lives in the \\(\\equiv  3 \\bmod  4\\) (&#8220;sticky&#8221;) class.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Figure 2 shows a full trajectory \u2014 that of 27 \u2014 read in binary: the heatmap (rows = steps), the value curve, the pattern metrics, and the carry depth \\(\\nu \\) at each odd step.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><a href=\"https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_27_overview-2.png\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"747\" src=\"https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_27_overview-2-1024x747.png\" alt=\"\" class=\"wp-image-383\" srcset=\"https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_27_overview-2-1024x747.png 1024w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_27_overview-2-300x219.png 300w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_27_overview-2-768x560.png 768w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_27_overview-2-1536x1120.png 1536w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_27_overview-2.png 1735w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/a><figcaption class=\"wp-element-caption\"><strong>Figure 2.<\/strong> The trajectory of 27. Top-left: binary heatmap (rows = steps, blue = 1). Top-right: value (log scale). Bottom-left: Hamming density \/ longest zero-run \/ longest alternating-run. Bottom-right: carry depth ??(3n+1) at each odd step. Red lines mark &#8220;detonation&#8221; steps.<\/figcaption><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">3. The Carry Mechanism<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">In \\(3n = n + (n\\ll 1)\\), classify each bit position by its pair:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Generator<\/strong> (\\(11\\)): always emits a carry.<\/li>\n\n\n\n<li><strong>Propagator<\/strong> (\\(10\\)\/\\(01\\)): passes a carry if one arrives.<\/li>\n\n\n\n<li><strong>Absorber<\/strong> (\\(00\\)): stops the carry.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">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 <strong>fizzles<\/strong> in 1\u20132 bits. The exception is the <strong>alternating pattern<\/strong> \\(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.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>\\(\\times 3\\) as a 3-state transducer.<\/strong> 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\\) \u2014 so the carry state lives in a <strong>closed set \\({0,1,2}\\)<\/strong> (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:<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table><thead><tr><th>chunk<\/th><th>behaviour<\/th><th>role<\/th><\/tr><\/thead><tbody><tr><td>\\(0101\\)<\/td><td>\\(c=0\\to \\mathrm{cout} 0\\), \\(c\\ge 1\\to \\mathrm{cout} 1\\)<\/td><td><strong>conductor<\/strong> \u2014 the unique band-0 chunk that lets a carry escape (\\(3\\cdot 5 = 1111\\))<\/td><\/tr><tr><td>\\(1010\\)<\/td><td>\\(c\\le 1\\to \\mathrm{cout} 1\\), \\(c=2\\to \\mathrm{cout} 2\\)<\/td><td><strong>promoter<\/strong> \u2014 carries a 2 onward<\/td><\/tr><tr><td>\\(1111\\)<\/td><td>\\(\\mathrm{cout} = 2\\) for every \\(c\\)<\/td><td><strong>generator<\/strong> \u2014 a saturated carry source<\/td><\/tr><tr><td>all others<\/td><td>\\(\\mathrm{cout}\\) fixed by \\(3x\\) alone<\/td><td><strong>absorber<\/strong> \u2014 incoming carry shifts bits but cannot escape<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">This is the finite, local object behind the whole carry story (and the cellular-automaton analogy of Section 11): a long <strong>alternating<\/strong> run \\(\\dots 0101 0101\\dots \\) is a chain of <em>conductors<\/em> that walks a bottom-injected carry (the \\(+1\\)) all the way up, while a long <strong>all-ones<\/strong> run \\(\\dots 1111\\dots \\) is a wall of <em>generators<\/em> forcing \\(\\mathrm{cout} = 2\\) at every chunk \u2014 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 \u00d73 turns it all-ones, flipping carry-reach to full width.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><a href=\"https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_carry_cycle_1365-2.png\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"518\" src=\"https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_carry_cycle_1365-2-1024x518.png\" alt=\"\" class=\"wp-image-384\" srcset=\"https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_carry_cycle_1365-2-1024x518.png 1024w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_carry_cycle_1365-2-300x152.png 300w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_carry_cycle_1365-2-768x388.png 768w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_carry_cycle_1365-2-1536x777.png 1536w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_carry_cycle_1365-2-2048x1035.png 2048w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/a><figcaption class=\"wp-element-caption\"><strong>Figure 3.<\/strong> \u00d73 applied repeatedly to the alternating number 1365 (= a conductor chain). Red = generator chunk, orange = propagator, gray = absorber; the bar shows how far the carry reaches. The alternating pattern (reach 0) becomes all-ones under \u00d73 (reach = full width), then alternating again \u2014 a two-state toggle.<\/figcaption><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">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\\) <strong>on average<\/strong>. Figure 4 shows the per-step carry roles along 27&#8217;s shortcut trajectory.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><a href=\"https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_27_shortcut-2.png\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"646\" src=\"https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_27_shortcut-2-1024x646.png\" alt=\"\" class=\"wp-image-385\" srcset=\"https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_27_shortcut-2-1024x646.png 1024w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_27_shortcut-2-300x189.png 300w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_27_shortcut-2-768x485.png 768w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_27_shortcut-2-1536x969.png 1536w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_27_shortcut-2-2048x1293.png 2048w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/a><figcaption class=\"wp-element-caption\"><strong>Figure 4.<\/strong> Shortcut (odd-only) steps of 27. Left: bit values. Centre: carry roles per bit position (gray = absorb, orange = propagate, red = generate) with carry-reach markers. Right: ? (the firewall the +1 creates) vs carry-reach. Step 1367 is wall-to-wall propagators \u2014 &#8220;loaded&#8221; \u2014 just before the firewall at 3077.<\/figcaption><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">4. The Firewall: Isolation by a Zero Run<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">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 &lt; 2^{b+2}\\), the carry from L cannot cross a sufficiently wide zero run. While the firewall holds, H and L evolve <strong>independently<\/strong>:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\\[T^j(n) = 3^j \\cdot  H \\cdot  2^{m+b-\\sum \\nu } + T^j(L)\\]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">H is purely passive (just \u00d73 each step); <strong>L alone drives the dynamics<\/strong>. If L converges, n converges, reducing by a factor ? 1\/L.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Two conditions<\/strong> (verified in code) make this exact:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>\\(\\mathrm{max}(L-\\mathrm{trajectory}) &lt; 2^m\\) \u2014 carry never reaches H.<\/li>\n\n\n\n<li>\\(m + b \\ge  \\sum \\nu (L-\\mathrm{trajectory})\\) \u2014 H stays at a positive bit position.<\/li>\n<\/ol>\n\n\n\n<p class=\"wp-block-paragraph\">Condition (2) binds: the required firewall scales with L&#8217;s <strong>stopping time<\/strong>, not its peak. This is a rigorous convergence lemma for an infinite class of numbers \u2014 but it only covers numbers <em>already<\/em> containing such a structure. Whether every trajectory acquires one is the open part.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><em>(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.)<\/em><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">5. The Average Argument, Made Rigorous \u2014 and Its Ceiling<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">The \\(\\times 3\/4\\) drift is provable for explicit subclasses:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Trivial:<\/strong> \\(n \\equiv  1 \\bmod  4\\) descends in one step (half of all odds).<\/li>\n\n\n\n<li><strong>Terras \/ Everett (1976\u201377):<\/strong> 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 &lt; 2^k\\), i.e. \\(j &lt; k\\cdot \\mathrm{log}2\/\\mathrm{log}3 \\approx  0.631k\\). Since odd-steps are Binomial(k, \u00bd) with mean k\/2 &lt; 0.631k, the density of contracting classes ? 1. <strong>Theorem: the set of integers with finite stopping time has natural density 1.<\/strong><\/li>\n\n\n\n<li><strong>Tao (2019):<\/strong> Almost all orbits (logarithmic density) attain almost bounded values.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Why it is believed (the even-distribution heuristic).<\/strong> 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 <strong>ends in \\(\\dots 11\\)<\/strong> \u2014 half of all odd numbers. If the trailing bits along an orbit are <em>unbiased<\/em> (each two-bit ending at its natural rate), then<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>P(?=1)=\u00bd, P(?=2)=\u00bc, \u2026  ?  E&#91;?]=2 &gt; log?3 ? 1.585  ?  bit-length drifts down  ?  n ? 1.<\/code><\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">This is borne out empirically: over ~570,000 odd steps sampled from ~80-bit starts, \\(\\nu =1\\) occurs <strong>49.97%<\/strong> of the time and average \\(\\nu  = 1.998\\) \u2014 exactly the unbiased prediction. *<em>The evenness of the trailing bits <\/em>is<em> the contraction.<\/em><em> Divergence would require a <\/em>sustained skew<em>: with \\(\\nu =1\\) at fraction \\(p\\) and the rest following the natural tail (\\(E[\\nu |\\nu \\ge 2]=3\\)), \\(E[\\nu ] = 3 &#8211; 2p\\), so neutrality needs \\(p = (3 &#8211; \\mathrm{log}_{2}3)\/2 \\approx  0.71\\) and growth needs more \u2014 a <strong>\\(\\equiv 3 \\bmod  4\\) rate of about 71%, a ~5:2 ratio<\/strong>, 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 <\/em>all* steps, a different ratio.) There is no apparent mechanism for such a skew \u2014 the bits look balanced and Hamming density hovers at \u00bd. The \\(\\nu \\) distribution that drives all this matches the geometric \\(1\/2^k\\) prediction to several orders of magnitude (Figure 5).<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><a href=\"https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_nu2_distribution-2.png\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"681\" src=\"https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_nu2_distribution-2-1024x681.png\" alt=\"\" class=\"wp-image-386\" srcset=\"https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_nu2_distribution-2-1024x681.png 1024w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_nu2_distribution-2-300x200.png 300w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_nu2_distribution-2-768x511.png 768w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_nu2_distribution-2.png 1046w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/a><figcaption class=\"wp-element-caption\"><strong>Figure 5.<\/strong> Distribution of \\(\\nu  = \\nu _{2}(3n+1)\\) over odd n, compared with the geometric prediction \\(P(\\nu =k)=1\/2^k\\). The match holds across four orders of magnitude \u2014 the basis of the &#8220;even bits ? \u00d73\/4 drift&#8221; heuristic.<\/figcaption><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Where the fuel comes from (and why the split stays even).<\/strong> A step lands on the growth class (\\(T(n) \\equiv  3 \\bmod  4\\), &#8220;fuel&#8221;) according to the bottom 3 bits of \\(n\\):<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table><thead><tr><th>\\(n\\) ends in<\/th><th>\\(T(n) \\equiv  3 \\bmod  4\\)?<\/th><th>role<\/th><\/tr><\/thead><tbody><tr><td>\\(\\dots 111\\) (\\(7 \\bmod  8\\))<\/td><td><strong>always<\/strong> (\\(\\nu =1\\), \\(T=(3n+1)\/2\\))<\/td><td>definite <strong>feeder<\/strong><\/td><\/tr><tr><td>\\(\\dots 011\\) (\\(3 \\bmod  8\\))<\/td><td><strong>never<\/strong> (? \\(\\dots 01\\))<\/td><td>definite <strong>exit<\/strong><\/td><\/tr><tr><td>\\(\\dots x01\\) (\\(1 \\bmod  4\\))<\/td><td><strong>~50%<\/strong>, \\(T = \\mathrm{oddpart}(3\\cdot (n\\gg 2)+1)\\)<\/td><td><strong>recursive<\/strong> in the high bits<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">(Careful: this table is about which \\(n\\) <em>produce<\/em> a fuel successor. A \\(\\dots 011\\) number is itself \\(\\equiv 3 \\bmod  4\\), so it <em>is<\/em> a growth step and counts toward the rate \u2014 \\(\\dots 011\\) and \\(\\dots 111\\) both grow \\(\\times 3\/2\\). The distinction is only in their <strong>successor<\/strong>: \\(\\dots 111\\) chains to another growth step, \\(\\dots 011\\) exits to a descending one. So \\(\\dots 011\\) is not &#8220;unusable&#8221; \u2014 it just doesn&#8217;t perpetuate the chain.)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The only <em>deterministic<\/em> fuel source is a \\(\\dots 111\\) ending \u2014 but a trailing-1 run <strong>erodes one bit per step<\/strong> (\\(\\dots 111 \\to  \\dots 11 \\to  \\dots 1\\)) and cannot be regenerated as structure (Sections 7\u20138). Once spent, the orbit falls into the \\(\\dots 01\\) case, which is a fresh ~50% coin flip on the <em>high<\/em> bits. So the guaranteed source is self-limiting and everything else averages \u00bd \u2014 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 <strong>recursive \\(\\dots 01\\) branch<\/strong> landing on \\(\\equiv 3 \\bmod  4\\) more than half the time forever \u2014 a conspiracy in the high bits that nothing local forbids. That branch is precisely where an un-rule-out-able bias could hide.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The ceiling:<\/strong> &#8220;evenly distributed&#8221; is provable in the wrong senses and unprovable in the needed one.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table><thead><tr><th>sense of &#8220;trailing bits are even&#8221;<\/th><th>status<\/th><\/tr><\/thead><tbody><tr><td>across <em>all<\/em> odd numbers (static): exactly \u00bd are \\(\\equiv 3 \\bmod  4\\)<\/td><td>proven, trivial \u2014 but an orbit visits one <em>specific<\/em> sequence, not &#8220;all numbers&#8221;<\/td><\/tr><tr><td>along <em>almost all<\/em> orbits ? those contract<\/td><td>proven (Terras\/Everett density-1; Tao almost-all)<\/td><\/tr><tr><td>along <em>every<\/em> orbit (no biased exception)<\/td><td><strong>= the conjecture<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">The trap is that an orbit is <strong>deterministic<\/strong>: the trailing bits of \\(n_{i+1}\\) are a fixed function of \\(n_i\\), so consecutive endings are <em>correlated<\/em>, not independent. &#8220;Looks random&#8221; is not &#8220;is random&#8221; \u2014 a deterministic sequence may carry any bias. Equivalently, a density-1 set is <strong>not forward-invariant<\/strong>: when \\(n\\) descends to \\(n&#8217;\\), the residue class of \\(n&#8217;\\) is uncorrelated with the contraction just used, so &#8220;almost all descend once&#8221; cannot be iterated into &#8220;all reach 1.&#8221; Proving no orbit&#8217;s mod-4 classes lock into a self-reinforcing \\(\\equiv 3 \\bmod  4\\) skew is exactly the open problem.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">6. Cycle Structure<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">The conjecture has <strong>two independent obstacles<\/strong>: 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\u201312 then turn to divergence.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Everything follows from one exact per-element rule (verified): under T,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\\[n \\equiv  3 \\bmod  4  \\implies   T(n) &gt; n\\qquad \\text{(ascends, $\\nu$=1, $\\times$3\/2)}\\]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\\[n \\equiv  1 \\bmod  4  \\implies   T(n) &lt; n\\qquad \\text{(descends, $\\nu$$\\ge$2; n &gt; 1)}\\]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Minimum and maximum (a dual pair).<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>minimum<\/strong> of a non-trivial cycle ascends out of itself, so it is <strong>? 3 mod 4<\/strong> (it cannot be \\(\\equiv 1 \\bmod  4\\) \u2014 that would drop below itself; the trivial cycle&#8217;s \\(1 = 4\\cdot 0+1\\) is the \\(k=0\\) boundary case).<\/li>\n\n\n\n<li>The <strong>maximum<\/strong> descends out of itself, so it is <strong>? 1 mod 4<\/strong>.<\/li>\n\n\n\n<li>Hence both residue classes are <strong>mandatory<\/strong>: a cycle cannot be all-\\(\\equiv 3\\) (strictly increasing) nor all-\\(\\equiv 1\\) (strictly decreasing). \\(4k+1\\) numbers are not excludable from cycles \u2014 they are required, with the maximum among them.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Two-arc decomposition.<\/strong> 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 \u2014 a <strong>rise<\/strong> \\(c_\\mathrm{min} \\to  \\dots  \\to  c_\\mathrm{max}\\) and a <strong>fall<\/strong> \\(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} &gt; 1\\) (rise) and \\(c_\\mathrm{min}\/c_\\mathrm{max} &lt; 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. <strong>In fact, for every cycle but one it cannot be:<\/strong> a cycle whose rise is a single clean run of increases and whose fall is a single clean run of decreases is a <em>1-circuit<\/em>, and <strong>Steiner (1977) [14] proved the only 1-circuit is the trivial \\(1\\to 4\\to 2\\to 1\\)<\/strong>. So \\({1,2,4}\\) is the unique non-choppy cycle; any other cycle must zig-zag \u2014 multiple local maxima on the rise, multiple local minima on the fall (the circuit structure below).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><em>The rise tends to be the longer arc.<\/em> The climb is rate-limited \u2014 the only way up is \\(\\times 3\/2\\) per step (\\(+0.585\\) bits) \u2014 while the fall can <strong>plunge<\/strong> (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\\)&#8217;s \\({17\\to 25\\to 37\\to 55\\to 41\\to 61\\to 91}\\) has <strong>rise 6, fall 1<\/strong> (the fall is the single drop \\(91\\to 17\\), \\(\\nu =4\\)). It is only a <em>tendency<\/em>, 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\\) \u2014 where the up-step is the faster \\(\\times 5\/2\\) \u2014 the cycle \\({17\\to 43\\to 27}\\) has the fall <em>longer<\/em> (rise 1, fall 2). And it carries no leverage for the bounds: growth (\\(\\equiv 3 \\bmod  4\\)) steps occur on <em>both<\/em> arcs, so the split neither isolates them nor sharpens the \\(5:2\\) count \u2014 it only makes the cycle&#8217;s <em>shape<\/em> vivid (a slow rate-limited crawl up, punctuated by deep plunges).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><em>The cousins do cycle.<\/em> The maps \\(3n-1\\), \\(5n+1\\), and \\(3n+1\\) on the <strong>negative<\/strong> integers all share the \\(\\mathrm{an}+b\\) shape, yet each has genuine <strong>non-trivial cycles<\/strong> (\\(3n-1\\): \\({5,7}\\), \\({17,\\dots ,91}\\); \\(5n+1\\): \\({13,33,83}\\), \\({17,43,27}\\)) \u2014 which is exactly what let us read real \\(\\times 3\\) cycle data off them above. So the <em>absence<\/em> 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 <strong>Conway&#8217;s theorem (1972 [15])<\/strong> that the general \\(\\mathrm{an}+b\\)-type iteration is Turing-complete and its reachability <strong>undecidable<\/strong> \u2014 the family is as wild as computation itself, and \\(3n+1\\) is just one (conjecturally tame) point in it.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The maximum is ? 5 (mod 12).<\/strong> Its odd predecessor \\(P\\) satisfies \\(T(P)=M\\); since \\(M\\) is the max, \\(P&lt;M\\) forces the ascending case, giving exactly<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\\[P = (2M-1)\/3,    P \\equiv  3 \\bmod  4,    P \\approx  \\tfrac{2}{3}M.\\]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">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\\) \u2014 impossible for the max.) Combined with \\(M \\equiv  1 \\bmod  4\\): <strong>\\(M \\equiv  5 (\\bmod  12)\\)<\/strong>, verified (e.g. \\(13 \\equiv  1 \\bmod  3\\) cannot be a peak; \\(5, 17, 29, 41, \\dots \\) can).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>And the maximum is never ? 9 (mod 16).<\/strong> 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\\) <em>ascends<\/em>, giving \\(T^{2}(M) = (9M+7)\/8 &gt; M\\) \u2014 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 &lt; M\\), and smaller for \\(\\nu \\ge 4\\), so only \\(9 \\bmod  16\\) self-destructs. Combining with \\(M \\equiv  5 \\bmod  12\\): <strong>the maximum is \\(\\equiv  5, 17, \\mathrm{or} 29 (\\bmod  48)\\) \u2014 never \\(41 \\bmod  48\\)<\/strong> \u2014 sharpening the sieve from 1-in-12 to 3-in-48.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The predecessor&#8217;s low bits are \\(\\dots 011\\).<\/strong> 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 <strong>\\(P \\equiv  3 \\bmod  8\\)<\/strong> \u2014 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 <em>exit<\/em> residue and \\(7 \\bmod  8\\) the <em>stay-sticky<\/em> residue (Section 7, below), so <strong>the maximum caps an ascending sticky-run, and \\(P\\) is the run&#8217;s terminal exit element<\/strong> \u2014 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.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The predecessor must not be a multiple of 3 \u2014 excluding \\(M \\equiv  5 \\bmod  9\\).<\/strong> \\(P\\) is itself a cycle element, so it needs its <em>own<\/em> 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\\)) \u2014 contradiction. Hence <strong>the maximum is \\(\\equiv  2\\) or \\(8 (\\bmod  9)\\), never \\(5 \\bmod  9\\)<\/strong> (an independent sieve, stacking with the mod-4 and mod-16 ones).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>\\(M \\bmod  9\\) fixes how the peak is approached.<\/strong> \\(P \\bmod  3\\) (inherited from \\(M \\bmod  9\\)) decides whether \\(P\\)&#8217;s own predecessor \\(Q\\) ascends or descends into it:<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table><thead><tr><th>\\(M \\bmod  9\\)<\/th><th>\\(P \\bmod  3\\)<\/th><th>\\(Q\\) (predecessor of \\(P\\))<\/th><th>shape before the peak<\/th><\/tr><\/thead><tbody><tr><td>\\(2\\)<\/td><td>\\(1\\)<\/td><td>\\((4P-1)\/3 &gt; P\\) (descends in, \\(\\nu =2\\))<\/td><td><strong>V-dip<\/strong>: \\(Q \\to  P \\to  M\\) \u2014 trajectory drops to \\(P\\), then jumps to the max<\/td><\/tr><tr><td>\\(8\\)<\/td><td>\\(2\\)<\/td><td>\\((2P-1)\/3 &lt; P\\) (ascends in, \\(\\nu =1\\))<\/td><td><strong>steady climb<\/strong>: \\(\\dots  \\to  Q \\to  P \\to  M\\), unbroken ascent into the max<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">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\\)).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The surviving residues, and their binary shapes.<\/strong> Intersecting just the mod-16 and mod-9 sieves leaves <strong>6 residues mod 144<\/strong>, which group into the three allowed \\(\\bmod  16\\) classes (their low 4 bits), each paired across \\(\\bmod  9 \\in  {2,8}\\):<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table><thead><tr><th>\\(\\bmod  16\\) (low 4 bits)<\/th><th>residues<\/th><th>small-rep shape<\/th><th>regime<\/th><\/tr><\/thead><tbody><tr><td>\\(0001\\) (\\(1\\))<\/td><td>\\(17, 65\\)<\/td><td>\\(2^k+1\\) \u2014 sparse, lone high bit<\/td><td>\\(\\nu =2\\), gentlest drop<\/td><\/tr><tr><td>\\(1101\\) (\\(13\\))<\/td><td>\\(29, 125\\)<\/td><td>\\(2^k-3 = (k-2 \\mathrm{ones})01\\) \u2014 near all-ones<\/td><td>\\(\\nu =3\\), steeper<\/td><\/tr><tr><td>\\(0101\\) (\\(5\\))<\/td><td>\\(53, 101\\)<\/td><td>alternating tail \\(\\dots 0101\\) + cap<\/td><td>\\(\\nu \\ge 4\\), steepest (firewall)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">The clean \\(2^k\\pm c\\) forms are an artifact of taking the <em>smallest<\/em> representatives (a genuine \u2014 astronomically large \u2014 maximum only shares the low 4 bits and the mod-9 class). But the three <em>shapes<\/em> are not artifacts: they are the three binary motifs that recur throughout this whole analysis \u2014 <strong>sparse<\/strong>, <strong>all-ones<\/strong>, and <strong>alternating<\/strong>. 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, \u2026) then carves into the higher bits of each.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The peak is a sharp spike<\/strong>: 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\\).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>An ascending-free zone below the peak.<\/strong> 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 &gt; M\\) (overshoot), since \\(x &gt; P \\iff  T(x) &gt; M\\). So that band is reachable only by <em>descending<\/em> (\\(\\equiv 1 \\bmod  4\\)) steps \u2014 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 &#8220;a third of the numbers just below \\(M\\) are barred from being growth steps, so the 5:2 ratio is harder.&#8221; It is not \u2014 the exclusion only <em>relocates<\/em> 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 <strong>every \\(\\equiv 3 \\bmod  4\\) element sits at or below \\(P\\)<\/strong>. The \\(5:2\\) count is therefore taken over \\([c_\\mathrm{min}, P]\\) with \\(P\\) as its ceiling \u2014 exactly the band the next bound packs \u2014 so the exclusion is already paid for there, not lost. (Note this is a real <em>width<\/em>, 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 <em>where<\/em> growth sits (at or below \\(\\tfrac{2}{3}M\\), never in the top sliver), not <em>how much<\/em> of it there is.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The climb into the peak has length at most \\(v_{3}(M+1)\\).<\/strong> Trace ascending predecessors back from \\(M\\) via \\(x_{j+1} = (2x_j &#8211; 1)\/3\\). Since that map has fixed point \\(-1\\), it has a closed form:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\\[x_j + 1 = (2\/3)^j \\cdot  (M+1).\\]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This is an integer for exactly \\(j \\le  v_{3}(M+1)\\) steps (the 3-adic valuation of \\(M+1\\)) and then stops \u2014 \\(3\\) runs out of \\(M+1\\). So the <strong>maximal<\/strong> ascending chain that could end at \\(M\\) has length \\(v_{3}(M+1)\\), with rungs \\(P, Q, R, \\dots  = (2\/3)^j(M+1) &#8211; 1\\). The <em>actual<\/em> ascending run feeding \\(M\\) in an orbit is a <strong>suffix<\/strong> of this chain \u2014 the preceding descent may land partway up \u2014 so its length is <strong>\\(\\le  v_{3}(M+1)\\)<\/strong>, with equality only when the descent lands on the bottom rung. (Verified over <strong>390,193<\/strong> ascending runs from small and random ~80-bit starts: <strong>zero<\/strong> violations of the bound; equality \\(\\mathrm{runlen} = v_{3}(\\mathrm{peak}+1)\\) in <strong>65.8%<\/strong> of runs, with the shortfall almost always just 1 \u2014 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} &#8211; 1\\) is <strong>all-ones \\(2^k-1\\)<\/strong> (the \u00a77 fuel reserve, below, viewed from the top) exactly when \\(M+1 = 2^a\\cdot 3^b\\); otherwise generic. Either way it has <em>no<\/em> ascending predecessor, so in a cycle it is entered by a <strong>descent<\/strong> \u2014 a local minimum \u2014 giving the Steiner &#8220;circuit&#8221; shape: <em>descent ? local minimum ? ascending run (length \\(\\le  v_{3}(M+1)\\)) ? peak \\(M\\) ? descent<\/em>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The mod-3 floor \u2014 and the source of the slack.<\/strong> 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 <strong>\\(\\equiv  0 \\bmod  3\\)<\/strong> number has <em>none at all<\/em> (\\(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 \\)) \u2014 so the &#8220;all-ones reserve&#8221; is unreachable precisely in the even-\\(k\\) case. The consequence: <strong>when the maximal bottom \\(x_v\\) is \\(\\equiv  0 \\bmod  3\\), the climb cannot reach it<\/strong> \u2014 the run must stop at a higher \\(\\equiv  2 \\bmod  3\\) rung fed by a descent. This <em>is<\/em> 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-\u00f73 obstruction (and \\(\\equiv 0 \\bmod  3\\) generally) doesn&#8217;t forbid \\(M\\); it sets the climb&#8217;s <strong>floor<\/strong> one or more rungs up, at the lowest \\(\\not\\equiv 0 \\bmod  3\\) rung.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Climb grammar near the peak.<\/strong> 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 \u2014 so its image must be \\(\\equiv 1 \\bmod  4\\), forcing the generator to be <strong>\\(\\equiv 3 \\bmod  8\\)<\/strong> (the &#8220;exit&#8221; residue). A \\(\\equiv 7 \\bmod  8\\) (&#8220;stay-sticky&#8221;) generator strictly inside is <strong>forbidden<\/strong>: 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 <em>exactly the main climb spine<\/em> (\\(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\\) &#8220;exit&#8221; 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 &gt; 53\\)).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The descent off the peak is graded by \\(M \\bmod  16\\)<\/strong> (the three surviving residues \\(1, 5, 13\\); \\(9\\) is excluded above). The successor&#8217;s <em>depth<\/em> is fixed by \\(\\nu \\), even though its low bits are not:<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table><thead><tr><th>\\(M \\bmod  16\\)<\/th><th>\\(\\nu \\)<\/th><th>\\(T(M)\/M \\approx  3\/2^\\nu \\)<\/th><th>character of the drop<\/th><\/tr><\/thead><tbody><tr><td>\\(1\\)<\/td><td>2<\/td><td>\\(\\approx  \\tfrac{3}{4}\\)<\/td><td>gentlest; \\(T(M) \\equiv  1 \\bmod  4\\) (descends again), fully fixed by 4 bits<\/td><\/tr><tr><td>\\(13\\)<\/td><td>3<\/td><td>\\(\\approx  \\tfrac{3}{8}\\)<\/td><td>steeper; \\(\\nu \\) fixed, one more bit (bit 4) sets the successor&#8217;s class<\/td><\/tr><tr><td>\\(5\\)<\/td><td>?4<\/td><td>\\(\\le  3\/16\\), mean \\(\\approx  \\tfrac{1}{8}\\)<\/td><td>steepest and <strong>recursive<\/strong> \u2014 \\(T(M) = \\mathrm{oddpart}(3H+2)\\) (a 3n+1 step on the high part), unbounded \\(\\nu \\)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">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&#8217;s low bits are <em>not<\/em> \u2014 they scatter across \\({01,11}\\) as the higher bits vary (only special sub-cases pin them).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\\(M \\equiv  5 \\bmod  16\\) is the <strong>alternating\/firewall regime<\/strong>, 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}\\):<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\\[3M+1 = 2^{2j+1} \\cdot  (3H+2).\\]<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Uncapped<\/strong> (\\(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\\), <strong>total collapse<\/strong> in one step.<\/li>\n\n\n\n<li><strong>Capped<\/strong> (\\(H\\ge 1\\)): \\(3M+1 = [3H+2]\\| [2j+1 \\mathrm{zeros}]\\) \u2014 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 \u00a79 conservation law (below) made exact for cycle maxima: <strong>the alternating tail converts entirely to zeros, only the cap propagates.<\/strong><\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>What it gives \u2014 cycle lower bounds.<\/strong> Around a cycle with \\(a\\) odd steps and \\(S = \\sum \\nu \\) halvings, dropping the \\(+1\\)s would demand \\(3^a = 2^S\\) \u2014 impossible (coprime). Cycles can exist <em>only<\/em> via the \\(+1\\) corrections:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\\[3^a \\cdot  \\prod (1 + 1\/(3x_{i})) = 2^S.\\]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">For large elements the correction product is tiny, forcing \\(3^a \\approx  2^S\\) to extreme precision \u2014 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&#8217;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&#8217;s <strong>minimum element must exceed 2?? (? 3\u00d710\u00b2?)<\/strong>, and its length is bounded below \u2014 historically \\(~1.7\\times 10^{7}\\) terms (Eliahou [12], at the verification of its era), far larger now (Simons\u2013de Weger [13]).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Capping the range doesn&#8217;t work \u2014 but \\(c_\\mathrm{min}\\) in the balance does (the explicit bound).<\/strong> A direct cap on the <em>range<\/em> 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 <em>every<\/em> element is \\(\\ge  c_\\mathrm{min}\\), which pins the correction product:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\\[1 &lt; 2^S\/3^a = \\prod (1 + 1\/(3x_{i})) \\le  (1 + 1\/(3c_\\mathrm{min}))^a \\approx  1 + a\/(3c_\\mathrm{min}).\\]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">With \\(c_\\mathrm{min} &gt; 2^{68}\\), the right side is \\(1 + ~10^{-13}\\), so \\(2^S\/3^a\\) lies within \\(~10^{-13}\\) of 1 \u2014 i.e. \\(S\/a\\) must hit \\(\\mathrm{log}_{2}3\\) to ~13\u201314 digits. Writing \\(\\delta  = S &#8211; a\\cdot \\mathrm{log}_{2}3 &gt; 0\\), the requirement is \\(\\delta  &lt; a\/(3c_\\mathrm{min}\\cdot \\mathrm{ln}2)\\), so a length \\(a\\) is viable only if its &#8220;upward gap&#8221; \\(\\delta \\) is that small \u2014 which happens only at <strong>large continued-fraction denominators of \\(\\mathrm{log}_{2}3\\)<\/strong>. Marching through the convergents against the \\(2^{68}\\) frontier, the smallest viable length is \\(a \\approx  7.96 \\times  10^{19}\\) odd steps \u2014 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 <strong>on the order of \\(10^{20}\\) odd elements<\/strong> (total terms ~2.6\u00d7 that) \u2014 the concrete version of &#8220;long.&#8221; Note what is and isn&#8217;t bounded: the <em>range<\/em> \\(c_\\mathrm{max}\/c_\\mathrm{min}\\) stays unbounded; it is the <strong>length<\/strong> 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.)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The growth numbers occupy a sub-band \u2014 which can&#8217;t be flat.<\/strong> The \\(\\equiv 3 \\bmod  4\\) (ascending) elements \u2014 the ones the \\(5:2\\) count tracks \u2014 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 <em>usable<\/em> slots: a cycle element is coprime to 3 (a multiple of 3 has no predecessor), so among \\(\\equiv 3 \\bmod  4\\) integers \u2014 residues \\(3, 7, 11 \\bmod  12\\) \u2014 only \\(7, 11 \\bmod  12\\) are eligible, <strong>density \\(\\tfrac{1}{6}\\), not \\(\\tfrac{1}{4}\\).<\/strong> Packing the \\((5\/7)a\\) ascending elements as distinct such integers needs<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\\[P &#8211; c_\\mathrm{min} \\ge  6\\cdot (5\/7)a = (30\/7)a.\\]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">With \\(a \\ge  8\\times 10^{19}\\) and \\(c_\\mathrm{min} &gt; 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<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\\[c_\\mathrm{max} \\ge  (3\/2)c_\\mathrm{min} + (45\/7)a.\\]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">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 <strong>cannot be flat<\/strong>: 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 \u2014 beyond &#8220;not a multiple of 3&#8221; there is no further <em>universal<\/em> exclusion on a generic ascending element (the finer residue bounds apply only to the max\/predecessor). And it remains a <em>lower<\/em> bound: like every counting fact here it sets a floor on the shape, not a ceiling that closes the question.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The floor \\(c_\\mathrm{min}\\), by contrast, is verification-bound \u2014 packing cannot lift it.<\/strong> Note the asymmetry: the band argument constrains the <em>width<\/em> (\\(P &#8211; c_\\mathrm{min} \\ge  (30\/7)a\\)), which pushes \\(c_\\mathrm{max}\\) up <em>given<\/em> \\(c_\\mathrm{min}\\) \u2014 but it says nothing about how high the band must <em>start<\/em>. Nothing in the dynamics forces elements <em>low<\/em> (they may sit arbitrarily high), so density yields no rigorous floor on \\(c_\\mathrm{min}\\); the only lower bound is <strong>direct verification<\/strong> (\\(c_\\mathrm{min} &gt; 2^{68}\\): every smaller \\(n\\) is <em>computed<\/em> to reach 1), a per-number fact stronger than any count. The log-uniform bottom-octave heuristic merely <em>recovers<\/em> \\(~2^{68}\\) \u2014 and circularly, since the length \\(a\\) was itself derived from \\(c_\\mathrm{min} &gt; 2^{68}\\). So the counting\/coprimality machinery is a <strong>width \/ \\(c_\\mathrm{max}\\) \/ ratio<\/strong> tool; the floor is a <em>computation<\/em>. Raising \\(c_\\mathrm{min}\\) needs more verification, not more counting.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Why this makes cycles heuristically impossible.<\/strong> \\(S\/a \\approx  \\mathrm{log}_{2}3\\) means the cycle&#8217;s <em>average<\/em> \\(\\nu \\) equals \\(1.585\\) \u2014 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 <em>forced<\/em> length \\(a \\sim  8\\times 10^{19}\\) (previous paragraph) that is <strong>~2.6\u00d710? standard deviations<\/strong>, a Gaussian-tail probability on the order of \\(10^{-1.5\\times 10^{18}}\\). Worse, the exponent grows <em>linearly<\/em> with the length the bounds force \u2014 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 \u2014 the \\(\\nu \\) sequence of a <em>deterministic<\/em> orbit need not behave like iid draws.)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Ceiling.<\/strong> 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 \u2014 but \\(\\mathrm{log}_{2}3\\) <em>can<\/em> 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.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">7. Fuel I \u2014 Consecutive 1s (Growth, but Weak and Finite)<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">A <em>consecutive run<\/em> of growth requires staying in the \\(\\equiv 3 \\bmod  4\\) class (?=1) step after step. (Note: net divergence does <strong>not<\/strong> need this \u2014 only an above-average <em>frequency<\/em> of ?=1, scattered, which needs no structure; see Section 13. This section is about the <em>structured, consecutive<\/em> route.) Which numbers stay ?=1 in a row?<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Uniqueness theorem (verified):<\/strong> the <em>only<\/em> residue mod \\(2^{j+1}\\) that stays ?=1 for j consecutive steps is the all-ones residue \\(2^{j+1} &#8211; 1\\). Concretely: stay-sticky-1-step = \\(3 \\bmod  4\\), 2-steps = \\(7 \\bmod  8\\), 3-steps = \\(15 \\bmod  16\\), \u2026 There is <strong>no alternative structure<\/strong> \u2014 a length-j growth run <em>requires<\/em> j+1 trailing 1s.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">For the all-ones number \\(n = 2^k &#8211; 1\\), the run has an exact closed form:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\\[a_i = 3^i \\cdot  2^{k-i} &#8211; 1\\qquad \\text{(eats one trailing 1 per step)}\\]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">It grows \u00d73\/2 for exactly <strong>k?1 steps<\/strong>, then exits at \\(2\\cdot 3^{k-1} &#8211; 1 \\equiv  1 \\bmod  4\\) and descends. <strong>Maximal fuel is a finite reserve: k ones buy k?1 growth steps.<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">8. Fuel I, continued \u2014 &#8220;No Free Fuel&#8221;<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">To <em>reach<\/em> \\(2^k &#8211; 1\\) (not start there), the nearest odd predecessor has <strong>exactly k+1 bits<\/strong> (it is ? 4\/3 larger, pushed over the \\(2^k\\) boundary). And the all-ones numbers split:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>k even<\/strong> (\\(3, 15, 63, 255, \\dots \\)): divisible by 3. Since \\(3n+1\\) is never divisible by 3, these have <strong>no odd predecessor at all<\/strong> \u2014 isolated islands, reachable only on their own doubling chain.<\/li>\n\n\n\n<li><strong>k odd<\/strong> (\\(7, 31, 127, \\dots \\)): reachable, but only by <strong>descending into them<\/strong> from a strictly longer (k+1-bit) ancestor.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Conclusion: you can never climb into maximal fuel.<\/strong> Pure fuel is always downhill; assembling a length-k all-ones block costs a number of length k+1 \u2014 you compact and lose ?1 bit. Fuel cannot be bootstrapped upward.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">9. Fuel II \u2014 Alternating Structure (Explosive, but Self-Emptying)<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">The other way to get &#8220;power&#8221; is a long carry cascade, which needs the alternating pattern. But propagation <strong>converts 1s into 0s<\/strong>:<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table><thead><tr><th>n<\/th><th>ones(n)<\/th><th>ones(3n+1)<\/th><th>firewall created<\/th><\/tr><\/thead><tbody><tr><td>1365 = \\(10101010101\\)<\/td><td>6<\/td><td>1<\/td><td>12 zeros<\/td><\/tr><tr><td>1367 = \\(10101010111\\)<\/td><td>7<\/td><td>3<\/td><td>9 zeros<\/td><\/tr><tr><td>341 = \\(101010101\\)<\/td><td>5<\/td><td>1<\/td><td>10 zeros<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Conservation law: carry distance = 1s destroyed = zeros created.<\/strong> 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 &#8220;residues&#8221; \\(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 \u2014 a clean power of two.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><a href=\"https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_rm_patterns-2.png\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"300\" src=\"https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_rm_patterns-2-1024x300.png\" alt=\"\" class=\"wp-image-387\" srcset=\"https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_rm_patterns-2-1024x300.png 1024w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_rm_patterns-2-300x88.png 300w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_rm_patterns-2-768x225.png 768w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_rm_patterns-2-1536x450.png 1536w, https:\/\/serverdude.dk\/wp-content\/uploads\/collatz_rm_patterns-2.png 1789w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/a><figcaption class=\"wp-element-caption\"><strong>Figure 6.<\/strong> The alternating residues \\(r_m = (2^m-1)\/3\\). Left: their binary patterns (pure \\(101\\dots 01\\)). Right: the firewall width each produces \u2014 \\(3\\cdot r_m+1 = 2^m\\) is an exact power of two, so the whole number collapses to a single 1.<\/figcaption><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">10. The Complete Structural Picture<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">There are exactly two sources of power in the binary string, and the carry mechanism dooms both:<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table><thead><tr><th>Fuel<\/th><th>Behaviour<\/th><th>Why it cannot sustain divergence<\/th><\/tr><\/thead><tbody><tr><td><strong>Consecutive 1s<\/strong><\/td><td>slow \u00d73\/2 climb<\/td><td>weak, finite (burns out in k?1 steps), cannot be bootstrapped (downhill to reach)<\/td><\/tr><tr><td><strong>Alternating<\/strong><\/td><td>one large carry cascade<\/td><td>the explosion <em>is<\/em> a firewall \u2014 it destroys its own 1s to create the void that collapses it<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">No third option: a carry either <strong>fizzles<\/strong> (no effect) or <strong>propagates<\/strong> (destroying the 1s that enabled it). You cannot have a large cascade that preserves its fuel \u2014 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.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This dooms every <em>structured<\/em> route to divergence. What it does <strong>not<\/strong> address is the unstructured one: a scattered, above-average <em>frequency<\/em> 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 \u2014 it is the lone survivor of Section 13.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">11. The Game of Life Analogy<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">The binary trajectory updated by a fixed local-ish rule invites comparison to <strong>cellular automata<\/strong> (CA) \u2014 and in particular to Conway&#8217;s <strong>Game of Life<\/strong>, where a <em>glider<\/em> 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 \u2014 i.e. a divergent trajectory or a non-trivial cycle. <strong>The conjecture, restated, is: 3n+1 has no glider.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The comparison is sharper than a slogan:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>In <strong>1-D cellular automata<\/strong> (Wolfram&#8217;s elementary CAs), whether gliders exist is <em>decidable for any fixed period and speed<\/em>, because the rule is <strong>local<\/strong> (radius 1) \u2014 a glider&#8217;s influence spreads at bounded speed, so a finite enumeration (de Bruijn diagrams \/ transfer matrices) settles it. Wolfram&#8217;s Class I\/II rules provably have none; Class III\/IV (e.g. Rule 110) do.<\/li>\n\n\n\n<li>Collatz <strong>looks<\/strong> Class III\/IV \u2014 complex, sensitive, plausibly Turing-complete \u2014 yet behaves Class I (everything dies to one fixed point). The obstruction to importing the CA decision procedure is exactly that Collatz has <strong>no locality<\/strong>: a single carry can, in principle, couple bit 0 to bit 60. The &#8220;firewall&#8221; of Section 4 is an attempt to <em>manufacture<\/em> locality \u2014 and it works conditionally, but proving every trajectory creates one is the open problem.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">So the right question is not &#8220;does a glider exist?&#8221; but &#8220;can a <em>finite<\/em> configuration sustain glider-like behaviour without the carry mechanism eventually destroying it?&#8221; \u2014 which Sections 7\u201310 answer &#8220;not by either available fuel,&#8221; and Section 12 answers precisely in the 2-adics.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">12. The Glider That Exists \u2014 in the 2-adics<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">A perfect glider <strong>does exist<\/strong>, but not as a positive integer. In the 2-adic integers, \\(-1 = \\dots 11111\\) is a fixed point of the shortcut map:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\\[3\\cdot (-1)+1 = -2 = \\dots 11110,  \\div 2 \\to  -1.\\]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The finite all-ones numbers \\(2^k &#8211; 1\\) are its <strong>truncations<\/strong> \u2014 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&#8217;t <em>logically<\/em> impossible; the obstruction is <strong>arithmetic approximation<\/strong> \u2014 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.)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Gun vs glider.<\/strong> The distinction matters. A Life <em>glider<\/em> conserves mass and merely translates; a Life <em>glider gun<\/em> <strong>creates<\/strong>, growing without bound \u2014 possible only because Life has <em>no<\/em> conservation law. Collatz <em>does<\/em> (Section 9: carry converts 1s?0s, fuel is never created on net), so the <strong>gun is forbidden<\/strong> \u2014 no configuration perpetually manufactures fresh trailing-1 fuel. The <strong>glider is not killed by conservation alone<\/strong>, but a finite one is squeezed by two opposing constraints:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Anchored below.<\/strong> 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 \u2014 to 1.<\/li>\n\n\n\n<li><strong>In vacuum above.<\/strong> A carry cascade into high structure destroys 1s and creates a zero firewall (Section 9). So a glider cannot grow <em>through<\/em> structure, only <em>into<\/em> emptiness \u2014 its top must climb into leading-zero vacuum.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">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. <strong>The one crack:<\/strong> the anchor need not be steady \u2014 it could <em>flicker<\/em>, 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.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">13. Where the Wall Stands<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Every structural fact above is true <strong>per step<\/strong>, <strong>per run<\/strong>, or <strong>per residue class<\/strong>. Both obstacles fragment into mechanisms that are individually dead, with one survivor each:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><em>Divergence:<\/em><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table><thead><tr><th>Mechanism<\/th><th>Status<\/th><\/tr><\/thead><tbody><tr><td>One infinite growth run<\/td><td>Dead \u2014 capped by bit-length<\/td><\/tr><tr><td>Climbing into ever-longer all-ones runs<\/td><td>Dead \u2014 entered only by descent \/ isolated (Section 8)<\/td><\/tr><tr><td>A finite integer <em>being<\/em> the glider<\/td><td>Dead \u2014 the glider is ?1, not an integer (Section 12)<\/td><\/tr><tr><td>Many short, impure runs regenerating in aggregate<\/td><td><strong>Open = half the conjecture<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><em>Cycles:<\/em><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table><thead><tr><th>Constraint<\/th><th>Effect<\/th><\/tr><\/thead><tbody><tr><td>min ? 3 mod 4; max ? 5 mod 12, ? 9 mod 16, ? 5 mod 9; predecessor ? 3 mod 8, ? 0 mod 3<\/td><td>congruence sieves \u2014 thin candidates by constant factors (Section 6)<\/td><\/tr><tr><td>\\(3^a \\ne  2^S\\) exactly; only \\(\\approx \\) via \\(+1\\)<\/td><td>forces \\(S\/a \\approx  \\mathrm{log}_{2}3\\), hence huge \\(a\\)<\/td><\/tr><tr><td>A <em>small<\/em> non-trivial cycle<\/td><td>Dead \u2014 bounded out to enormous size<\/td><\/tr><tr><td>A <em>large<\/em> non-trivial cycle (good \\(\\mathrm{log}_{2}3\\) approximation)<\/td><td><strong>Open = half the conjecture<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Both survivors are <strong>infinite-aggregate<\/strong> properties: for divergence, enough ?=1 steps to grow without ever exploding or exhausting the impure fuel, <em>forever<\/em>; for cycles, a step-count ratio approximating \\(\\mathrm{log}_{2}3\\) well enough to close a loop, which only ever requires <em>more<\/em> steps. Sharp local facts \u2014 per step, per run, per residue \u2014 do not compose into infinite-aggregate conclusions. That gap is precisely where Collatz has stood since 1937.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">14. Honest Assessment<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">This is <strong>expository, not novel<\/strong>. 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 <em>intuition<\/em> \u2014 making visible, via fuel and firewalls and carry conservation, <em>why<\/em> the average drift is downward and <em>why<\/em> that cannot be promoted to a proof. The value is pedagogical: a tangible account of a problem usually stated abstractly.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">15. Relation to Known Results<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">None of the mathematics here is new; the binary lens is a re-derivation of established theory. The correspondence, made explicit:<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table><thead><tr><th>In this document<\/th><th>Established source<\/th><\/tr><\/thead><tbody><tr><td>Shortcut map \\(T(n)=(3n+1)\/2^\\nu \\); parity of \\(\\nu \\)<\/td><td>standard; Terras [2], Lagarias survey [1]<\/td><\/tr><tr><td>Binary &#8220;firewall&#8221; = independence across a zero run<\/td><td>2-adic continuity of the Collatz map [4, 5, 6]<\/td><\/tr><tr><td>\\(-1 = \\dots 111\\) as the all-ones fixed point (the &#8220;glider&#8221;)<\/td><td>textbook 2-adic fact [5, 6]<\/td><\/tr><tr><td>&#8220;Conservation law&#8221; (carry spends 1s ? zeros)<\/td><td>a re-description of the \\(\\nu _{2}(3n+1)\\) arithmetic<\/td><\/tr><tr><td>Density-1 descent; the \\(\\mathrm{log}2\/\\mathrm{log}3 \\approx  0.631\\) break-even<\/td><td>Terras [2], Everett [3]<\/td><\/tr><tr><td>&#8220;Almost all orbits attain almost bounded values&#8221;<\/td><td>Tao [7]<\/td><\/tr><tr><td>Cycle balance \\(3^a \\approx  2^S\\) ? \\(\\mathrm{log}_{2}3\\) approximation ? size bounds<\/td><td>Crandall, Steiner, <strong>Eliahou [12]<\/strong>, Simons\u2013de Weger [13]<\/td><\/tr><tr><td>Cycle min \\(\\equiv  3 \\bmod  4\\); no element \\(\\equiv  0 \\bmod  3\\); congruence sieves on the extrema<\/td><td>elementary folklore in the cycle literature<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">The specific small artifacts \u2014 \\(\\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 \u2014 may not appear <em>verbatim<\/em> elsewhere, but they are immediate consequences of the standard residue\/maximality method, not new facts. They are new <em>expressions<\/em>, not new <em>results<\/em>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">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 \u2014 <strong>sparse<\/strong>, <strong>all-ones<\/strong>, <strong>alternating<\/strong> \u2014 govern <em>both<\/em> the divergence and cycle obstacles; and (iii) a clear demonstration of <em>why<\/em> 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.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">16. References<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Collatz \/ 3n+1<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>J. C. Lagarias, <em>The 3x+1 problem and its generalizations<\/em>, American Mathematical Monthly <strong>92<\/strong> (1985), 3\u201323. (The standard survey; see also Lagarias, ed., <em>The Ultimate Challenge: The 3x+1 Problem<\/em>, AMS, 2010.)<\/li>\n\n\n\n<li>R. Terras, <em>A stopping time problem on the positive integers<\/em>, Acta Arithmetica <strong>30<\/strong> (1976), 241\u2013252. (Parity vectors; density-1 finite stopping time.)<\/li>\n\n\n\n<li>C. J. Everett, <em>Iteration of the number-theoretic function f(2n)=n, f(2n+1)=3n+2<\/em>, Advances in Mathematics <strong>25<\/strong> (1977), 42\u201345.<\/li>\n\n\n\n<li>H. M\u00f6ller, <em>\u00dcber Hasses Verallgemeinerung des Syracuse-Algorithmus (Kakutanis Problem)<\/em>, Acta Arithmetica <strong>34<\/strong> (1978), 219\u2013226. (2-adic formulation.)<\/li>\n\n\n\n<li>G. J. Wirsching, <em>The Dynamical System Generated by the 3n+1 Function<\/em>, Lecture Notes in Mathematics <strong>1681<\/strong>, Springer, 1998.<\/li>\n\n\n\n<li>K. R. Matthews, <em>Generalized 3x+1 mappings: Markov chains and ergodic theory<\/em>, in [1, <em>The Ultimate Challenge<\/em>]. (2-adic\/ergodic background.)<\/li>\n\n\n\n<li>T. Tao, <em>Almost all orbits of the Collatz map attain almost bounded values<\/em>, Forum of Mathematics, Pi <strong>10<\/strong> (2022), e12; arXiv:1909.03562 (2019).<\/li>\n<\/ol>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Cellular automata \/ Game of Life<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>M. Gardner, <em>Mathematical Games: The fantastic combinations of John Conway&#8217;s new solitaire game &#8220;life&#8221;<\/em>, Scientific American <strong>223<\/strong> (Oct. 1970), 120\u2013123.<\/li>\n\n\n\n<li>E. R. Berlekamp, J. H. Conway, R. K. Guy, <em>Winning Ways for Your Mathematical Plays<\/em>, 2nd ed., A K Peters, 2001\u20132004. (Life: gliders, guns, universality.)<\/li>\n\n\n\n<li>S. Wolfram, <em>Computation theory of cellular automata<\/em>, Communications in Mathematical Physics <strong>96<\/strong> (1984), 15\u201357. (de Bruijn diagrams, symbolic dynamics.)<\/li>\n\n\n\n<li>S. Wolfram, <em>A New Kind of Science<\/em>, Wolfram Media, 2002. (Elementary CA classification; full text at wolframscience.com.)<\/li>\n<\/ol>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Cycle bounds<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>S. Eliahou, <em>The 3x+1 problem: new lower bounds on nontrivial cycle lengths<\/em>, Discrete Mathematics <strong>118<\/strong> (1993), 45\u201356.<\/li>\n\n\n\n<li>J. Simons, B. de Weger, <em>Theoretical and computational bounds for m-cycles of the 3n+1 problem<\/em>, Acta Arithmetica <strong>117<\/strong> (2005), 51\u201370. (See also R. Crandall, Math. Comp. <strong>32<\/strong> (1978), 1281\u20131292.)<\/li>\n\n\n\n<li>R. P. Steiner, <em>A theorem on the Syracuse problem<\/em>, in Proc. 7th Manitoba Conf. on Numerical Mathematics (1977), 553\u2013559. (No non-trivial 1-circuit cycle.)<\/li>\n<\/ol>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Undecidability of the general family<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>J. H. Conway, <em>Unpredictable iterations<\/em>, in Proc. 1972 Number Theory Conference (Univ. of Colorado, Boulder, 1972), 49\u201352. (Reachability of generalized \\(\\mathrm{an}+b\\) maps is undecidable; basis of FRACTRAN.)<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">17. Artifacts<\/h2>\n\n\n\n<ul class=\"wp-block-list\">\n<li><code>collatz_explorer.py<\/code> \u2014 trajectory analysis, ? \/ Hamming \/ zero-run \/ alternating-run metrics, carry roles &amp; reach, firewall-isolation verifier, r_m residues.<\/li>\n\n\n\n<li><code>collatz_viz.py<\/code> \u2014 overview heatmaps, 3-panel shortcut heatmap (bits \/ carry roles \/ ? &amp; reach), ?-distribution, r_m patterns, multi-trajectory.<\/li>\n\n\n\n<li><code>collatz_carry.py<\/code> \u2014 carry-cycle visualizations (generic vs alternating).<\/li>\n\n\n\n<li><code>collatz_discussion.md<\/code> \u2014 the full chronological Q&amp;A log this writeup summarizes. (not shared)<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\"><a href=\"https:\/\/github.com\/perlauge\/collatz\">https:\/\/github.com\/perlauge\/collatz<\/a><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Key reproducible facts: <code>python3 collatz_explorer.py 27<\/code>, <code>python3 collatz_viz.py 1365<\/code>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">18. One-Paragraph Summary<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">In binary, Collatz descent comes from two facts about carries: consecutive-1 &#8220;fuel&#8221; produces only weak (\u00d73\/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 \u00d73\/4 drift completely. The perfect self-sustaining structure exists only as the 2-adic integer ?1, which no finite integer can be. On the <strong>cycle<\/strong> side, the same per-element rule pins any non-trivial cycle&#8217;s extremes \u2014 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) \u2014 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 <strong>infinite-aggregate<\/strong> survivor each \u2014 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.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Per Lauge Bures\u00f8 Holst \u00a0\u00b7\u00a0 with Claude (Anthropic AI) 4 June 2026 A structural exploration of why 3n+1 trajectories tend downward \u2014 and why &#8220;tend&#8221; cannot yet be upgraded to &#8220;must.&#8221; Collaboration &amp; process This document is a collaboration between Per Lauge Bures\u00f8 Holst (who conceived and directed the investigation) and Claude, an AI assistant [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-360","post","type-post","status-publish","format-standard","hentry","category-ikke-kategoriseret"],"_links":{"self":[{"href":"https:\/\/serverdude.dk\/index.php?rest_route=\/wp\/v2\/posts\/360","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/serverdude.dk\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/serverdude.dk\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/serverdude.dk\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/serverdude.dk\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=360"}],"version-history":[{"count":9,"href":"https:\/\/serverdude.dk\/index.php?rest_route=\/wp\/v2\/posts\/360\/revisions"}],"predecessor-version":[{"id":390,"href":"https:\/\/serverdude.dk\/index.php?rest_route=\/wp\/v2\/posts\/360\/revisions\/390"}],"wp:attachment":[{"href":"https:\/\/serverdude.dk\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=360"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/serverdude.dk\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=360"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/serverdude.dk\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=360"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}