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For mathematicians

The open statements, exactly

The period-two problem and the route we hold, in the words a proof would use, most important first; then Mahler's problem and Erdős's question in the same terms (sections 9 and 10). Each statement says whether it is proved (a written proof with a second reader inside the project, no outside review), measured (an exact computation in the trove) or open, and names its section of the record. Nothing here is new; it is the record, compressed and typeset.

1. The rule, and what a proof must show

Rule 30 acts on bi-infinite rows $x(\cdot,t)\in\{0,1\}^{\mathbb Z}$ by

$$x(i,\,t+1)\;=\;x(i-1,\,t)\;\oplus\;\bigl(x(i,\,t)\;\vee\;x(i+1,\,t)\bigr).$$

It is left-permutive: solving for the left neighbour,

$$x(i-1,\,t)\;=\;x(i,\,t+1)\;\oplus\;\bigl(x(i,\,t)\vee x(i+1,\,t)\bigr),$$

so columns $i$ and $i+1$ over time determine every column to their left. A configuration is finite if it has finitely many black cells; the single-seed pattern is one. Problem 1 (Wolfram, 2019) asks whether the centre column $t\mapsto x(0,t)$ of the single-seed pattern is eventually periodic. Period one is closed:

Condrey (2026), proved

No column of a nonzero finite configuration is eventually constant.

Period two is the open case. Shift time so that the centre reads $0101\ldots$ from $t=0$ and call it the wall, $x(0,t)=t \bmod 2$. Column 1 is the trace $c(t)=x(1,t)$. The target of all the work:

The statement to prove

$$\text{No finite configuration has } x(0,t)=t\bmod 2 \text{ for all } t\ge 0.$$

This is more than the prize asks (the prize is about one configuration) and is where the work went, because the constant case was closed for all finite configurations and the tools are the same.

Sources. RULE30-PRIZE.md sections 8.13 and 8.54; Condrey, arXiv:2609.09431; PERIOD-TWO.md section 2.

2. What the left half can see: masking and the visible word

Beside the wall, the first left column is

$$x(-1,t)\;=\;x(0,t+1)\oplus\bigl(x(0,t)\vee c(t)\bigr)\;=\;\begin{cases}1-c(t) & t \text{ even (white wall)}\\ 1 & t \text{ odd (black wall)}\end{cases}$$

so only the samples at white-wall ticks reach the left half-plane. The visible word is $q=c(0)\,c(2)\,c(4)\cdots$, and the whole left half-plane is a function of $q$ alone. In the left row at time 0, the cell at depth $k$ depends on $c(t)$ for even $t\le k-1$ only.

A trace is real if some configuration beside the wall produces it; the real visible words form a factorial language $L$ (one per phase). Its minimal forbidden words are the cuts: every cut of length at most 44 is known and certified (1,130 in phase 0, 1,199 in phase 1).

Sources. CHAT-LEDGER.14.md CL356 (the masking identity, checked); L640, L641 (the language and its certified cuts).

3. The record route, and the missing lemma

Suppose a finite configuration kept the wall for ever. Its left half at time 0 is white beyond its support, so its real trace has a left image white on an infinite interval. Define the record at depth $d$:

$$R(d)\;=\;\sup\Bigl\{\,L\;:\;\exists\ \text{real } c \text{ with } x(-d,0)=x(-d-1,0)=\cdots=x(-d-L+1,0)=0\Bigr\}.$$

Target (open)

$$R(d)<\infty\quad\text{for every } d\ge 1 .$$

Finiteness at each depth, not a uniform bound: the data grow with $d$. This single statement refutes period two for every finite configuration.

Theorem B (proved), the engine of every solved case

If columns 0 and 1 are both $P$-periodic, every white run in row 0 of the forced left half has length at most $2P-2$. Corollary (checked exhaustively for $m\le 5$, tight at $m=2$): a visible word periodic with period $m$ over a window fixes the left cells the window determines, and their white gaps are at most $4m-2$.

The rings (proof sketches for the pure traces)

Column 1 can run in two travelling waves, the S ring (84 cells, column-1 word $110100$, largest white gap 5) and the L ring (155 cells, word $1101000100$, gap 6). For a white interval $[d,\,d+L-1]$ at a white anchor $a$ with $N$ samples of the S trace,

$$L\le 2a+5\quad\text{or}\quad N\le d+a+5,$$

and the same with 6 for the L trace. A mixed prefix $LSLSL$ forces seven white left cells at depths 30 to 36.

Two endpoints, explained exactly (proved for those two records)

At depth 139 the record is 17 and its endpoint is forced by masking: the cell at depth 156 is black from the 78 visible samples through tick 154, and no change at sites 156 and beyond can move it. At depth 132 the record is 17 and the white extension would force the certified cut $100100001001000010000101$ into the visible word at index 51 (tick 102).

The missing lemma (open)

$$\forall d\ \exists L(d):\quad \bigl\{\,q\in L\;:\;\text{the depth-}d\text{ image of } q \text{ is white on } [d,\,d+L(d))\bigr\}=\varnothing .$$

A white band of length $L$ at depth $d$ forces a pattern in the visible samples over a window growing with $L$, and every such pattern long enough contains a cut. Theorem B is this lemma for periodic samples; the two endpoints are what the pattern looks like in two real cases; the general argument is open.

Data. $R(d)$ is exact at every depth $20\le d\le 120$, at $d=125$ (21), at $d=132,139,142,143$ (17 at each) and at $d=147$ (18); at most 17 at $121\le d\le 124$, $d=140$, $144\le d\le 146$ and $148\le d\le 150$; at least 18 at the other depths $126\le d\le 141$ (at least 23 at $d=133$); by sampling at least 20 at $d=200$, 28 at 279, 38 at 1{,}035. Every witness is in the trove, except the one for 18 at depth 147, which is in rule30_exact_cone_cert.py. Drawn depth by depth.

Sources. PERIOD-TWO.md section 2 and row Q6; CHAT-LEDGER.md CL355, GC1148, GC1151, GC1153, GC1154 with CL356, GC1155 to GC1157 with CL356 and CL359; L643 (the census).

4. What is known about column 1 beside the wall

Jen (1990), proved

In a finite configuration, two adjacent columns are never both eventually periodic. So beside the wall, column 1 of a finite configuration is never eventually periodic.

Theorem A, proved (Jen with a clock)

Columns 0 and 1 of a finite configuration can agree with a periodic pair only over a bounded window; the kicks that break column 1's periodicity cannot thin out faster than geometrically.

Theorem E, proved

Column 1 cannot be a Sturmian word (the coding of an irrational rotation by an arc whose length is the rotation angle). With Jen's theorem, no Sturmian coding of a rotation, rational or irrational, is column 1; codings by other arcs are excluded only for almost every angle (Theorem E″). Corollary F, Theorems E″ and A⁗ extend this to near-squares at unbounded periods, block codes of Sturmian words, resets that come further apart than geometrically, and to Thue–Morse and paperfolding for every left edge up to 15,868 cells.

The channel bound, proved (certified in exact arithmetic)

$$\limsup_{n\to\infty}\frac{\log_2 \#\{\text{real visible words of length } n\}}{n}\;\le\;\log_2\lambda'_{28}\;=\;0.1236 \text{ bits per visible bit.}$$

The wheel, measured

Between kicks, column 1 is the two-arc coding of the rotation $t\mapsto 17t \bmod 56$, the word

$$U=\underbrace{0001001101}_{10}\ \underbrace{0001001101}_{10}\ \underbrace{0001001101}_{10}\ \underbrace{0001001101}_{10}\ \underbrace{0001001101}_{10}\ \underbrace{001101}_{6},$$

seventeen turns in 56 steps. After a settled wheel, kicks can occur at only four classes with short size ranges, and at three after 140 steps (Propositions 13 and 14). No other column of the single-seed pattern, $-40\le i\le 40$, shows any such structure (exploratory).

Conjecture LR (open for the real column)

No column 1 beside the wall makes the left half eventually white. Proved for eventually periodic column 1 (Jen) and for Sturmian column 1 (Theorem E); open for the kicked wheel.

Sources. RULE30-PRIZE.md sections 8.4, 8.13, 8.20, 8.33, 8.54, 8.57; PROOFS.md entry 26; PERIOD-TWO.md row Q7.

5. The counting form and the one sentence

Let $N(w,T)$ be the number of configurations on $w$ cells whose centre reads $0101\ldots$ for $T$ steps.

Open

$$N(w,T)\;\le\;2^{\,w-\alpha T+c}\quad\text{for some }\alpha>0 .$$

Measured: $\alpha$ close to 1 at $w=24$. Proved: the part paid by the seed's left part. Open: the right part. The same statement for Collatz, with the free bits in place of the left part, is COLLATZ-PRIZE.md section 1.

Every form of the problem ends at one sentence: keeping the wall's conditions costs real information. The delivery side is the channel bound. The cost side is proved only beside a constant wall (Condrey); beside $0101\ldots$ it is a coin model, measured and not proved. A proof must use what is special about the alternating wall, as Jen's and Condrey's do; the one such structure found is the wheel.

Sources. RULE30-PRIZE.md section 8.51; PERIOD-TWO.md sections 2 and 5.

6. Exact machinery at fixed depth

The depth-$k$ left images of cut-avoiding traces form a regular language. A seeded decoder presents them exactly in both directions with at most

$$2\,S\,2^{\lceil (k-1)/2\rceil}$$

live states, $S$ the cut automaton's live states; checked exact by exhaustive enumeration for $k\le 5$. For $k\ge 4$ in the white-start phase, under the gate that $11$ never occurs in a visible word, the first visible bit $q_0$ leaves the image unchanged (at time 0 the depth-3 and depth-4 values are $1-q_1$ and $q_0q_1$), so it can be quotiented away provided the cut obligation of both origins is carried. The state space grows with $k$: none of this is a closure across depths.

Sources. CHAT-LEDGER.md GC1158 with CL361, GC1159 with CL364; RULE30-GPT.md GC971, GC1158, GC1159.

7. The carry

Rule 30 is the Gray step plus a correction, and the correction is a two-time AND:

$$x(i,t+1)=x(i-1,t)\oplus x(i,t)\oplus c(i,t),\qquad c(i,t)=\bigl(1-x(i,t)\bigr)\,x(i+1,t)=x(i+1,t)\,x(i+1,t+1).$$

Under the fair-coin measure, which Rule 30 preserves, every column is a sequence of independent fair coins, and the carry feeding a column is the sliding AND of fair coins $a(t)a(t+1)$: correlation $\tfrac13$ at lag 1 and $0$ beyond, not Markov of any finite order, with

$$\Pr\bigl(c(t)=1\,\big|\,c(t-1)=\cdots=c(t-m)=0\bigr)=\frac{F_{m+1}}{2F_{m+3}},\qquad h=1-\frac1{16}\sum_{g\ge1}\frac{F_g\log_2 F_g}{2^{g}}=0.69924320338768\ \text{bits}.$$

Multiplying by three is the same Gray step plus carries: bit $k$ of $3n$ is $n_k\oplus n_{k-1}\oplus\text{carry}$. Drawn and checked live.

Sources. RULE30-PRIZE.md section 8.90; KIMI-QUESTIONS-5.md; COLLATZ-PRIZE.md section 8.

8. Where the problem sits

Three problems share one skeleton: a free side that is a full shift, a constrained side that is a thin language, and the requirement that they agree for ever. A fourth, Erdős's ternary question, has it too (section 10), and in all four the missing step is the same: one specific object must be excluded from a thin set (the one missing method).

The common mechanism: a finite constrained side plus an injective free side gives periodicity and a contradiction. The common gap: every sparsity theorem is about sets, and each prize is about one object. The three side by side, with the theorems.

Where problem 2 sits (added 15 October 2026). The prize's second problem asks that the centre column have equal frequencies of 0 and 1: simple normality of order 1, the weakest finite-state observer. A bounded search of the literature found no column of any elementary (two-state, radius-one) rule from a finite seed that is proved normal, or proved simply normal without being eventually periodic or automatic; the linear rules' finite-seed columns are automatic (Litow and Dumas; Rowland and Yassawi) and so never normal, and for the nonlinear rules nothing is proved beyond a Bernoulli initial row. Not found in that search, not a theorem that none exists. So problem 2 sits in a named open family: a finite-seed column of a local rule with a proved statistical property that no shortcut supplies.

Sources. RULE30-PRIZE.md sections 8.45, 8.47 and 8.116 (door D2); CHAT-LEDGER.md L740; COLLATZ-PRIZE.md; PRIOR-ART.md ("8.116 door D2").

9. Mahler's 3/2 problem: the exact form and the October 2026 results

The exact form (proved; G50)

Write $\xi(\tfrac32)^j=n_j+u_j$ with $n_j\in\mathbb Z_{\ge0}$ and $u_j\in[0,\tfrac12)$, and $b_j=n_j \bmod 2$. Then $$n_{j+1}=\Big\lceil \tfrac{3n_j}{2}\Big\rceil=\tfrac{3n_j+b_j}{2},\qquad u_j=\sum_{k\ge0} b_{j+k}\,\frac{2^k}{3^{k+1}}.$$ A Z-number is exactly a positive integer $n_0$ whose $\lceil 3x/2\rceil$ parity code keeps every tail $u_j<\tfrac12$; then $\xi=n_0+u_0$. On $[0,\tfrac12)$ the parity is forced by $u$, so the fractional side alone is free: the admissible codes are the greedy $\beta$-shift at $\beta=\tfrac32$.

Two norms, one series (proved; G293)

The same series converges in $\mathbb Z_2$, to $-n_0$. Let $S\subset\mathbb Z_2$ be the starts with every tail strictly below $\tfrac12$. $S$ is not closed; its closure $K$ adds the starts reaching the critical code $d(1)$. Both have Hausdorff dimension $\log_2\tfrac32$, and $$\text{Mahler's conjecture}\iff S\cap\mathbb Z_{>0}=\varnothing .$$

The gap-$g$ ladder (proved; G294)

For $g\ge3$ put $\alpha_g=\dfrac{3^{g-1}}{3^g-2^g}$ and $\beta_g=\dfrac{1+(2/3)^{g-1}}{3}$. For every $a\in[\alpha_g,\beta_g]$, $$\exists\,\xi>0:\ \{\xi(\tfrac32)^n\}\in[0,a)\ \text{ for all } n$$ if and only if $$\exists\,n\ge1:\ \text{every odd iterate } T^j(n)\equiv-\tfrac13 \pmod{2^g},$$ with $\xi=n+u_0(n)$. At $g=3$: windows $[\tfrac{9}{19},\tfrac{13}{27}]$ and odd iterates $\equiv5\pmod 8$. The mirror (proved): $\{\eta(\tfrac32)^n\}\in(1-a,1)$ for all $n$ iff some $n\ge1$ has every even iterate $\equiv-\tfrac23\pmod{2^g}$. An upward ladder (proved, GC1230) covers the windows $[0,a_M)$, $a_M\uparrow\tfrac12$, by finite residue conditions around the 2-adic start $D$ of $d(1)$. Neither ladder reaches Mahler: a Z-number with $\sup_j u_j=\tfrac12$ lies on no rung.

Computed exclusions and one existence window

Sources. MAHLER.md sections 1 and 2; RULE30-GPT.md G50, GC1221 (G293), GC1228 (G294), GC1230, GC1233, GC1235 and GC1237; the census.

10. A fourth figure: Erdős's ternary question

The statement and its free part

Erdős (1979): for $n>8$ the base-3 expansion of $2^n$ contains the digit 2. Let $u_k=2\cdot3^{k-1}$, the order of 2 modulo $3^k$. Because 2 is a primitive root, exactly $2^{k-1}$ residues $n \bmod u_k$ give a last-$k$-digit block of $2^n$ free of 2 (proved for every $k$; Saye 2022, Lemma 1), and each surviving residue has exactly two surviving lifts. If $m_k$ is the least $n\ge1$, $n\notin\{2,8\}$, whose last $k$ ternary digits avoid 2, then $$m_k>\frac{k}{\log_3 2}\ \text{ for every } k \quad\Longrightarrow\quad \text{Erdős's conjecture}.$$ Computed: $m_k$ reaches 1134 at $k=16$ and 25458 at $k=22$, against a needed bound near $1.6k$. Chance predicts growth of order $(3/2)^k$. Verified for every $n\le2\cdot3^{45}$ (Saye 2022).

Sources. CONSTELLATION.md section G.1; Saye, J. Integer Seq. 25 (2022); Lagarias, J. London Math. Soc. 79 (2009); RULE30-GPT.md GC1240 and GC1241.