Greedy Expansions of Rationals as Reciprocal Mersenne Subseries

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The results on this page that are not drawn from a cited source are Plasma AI's, credited to Plasma AI. They are author-recorded and have had no independent review.

Target and standing

Problem 257 asks whether $\sum_{n\in A}\frac{1}{2^n-1}$ is irrational for every infinite set $A\subseteq\mathbb N$. This sketch aims at the negative answer: a rational $q$ and an infinite $A$ with

$$\sum_{n\in A}\frac{1}{2^n-1}=q ,$$

with the base exactly $2$ and every denominator exactly $2^n-1$. Sums with shifted denominators or several bases are a different question (see "What composes" below). The two speculations Kovač and Tao quote (arXiv:2406.17593v4, section 2.1.2) point opposite ways: Erdős wrote on p. 222 of his 1968 paper in Math. Student that a rational value for some base "seems likely", while Erdős and Graham later suggested that the base-$2$ sums are perhaps always irrational, the form recorded as Problem 257.

Lemmas 1–8, 10, 12 and 13 below have elementary author-recorded proofs. Lemmas 9 and 14 import the classical Bang–Zsigmondy primitive-divisor theorem; their deductions are given here, but the external theorem is not re-proved. The construction linked in Lemma 11 uses Bertrand's postulate and applies only to weights with at most four geometric multipliers. Lemma 12 generates candidate digits from already fixed digits by one explicit rule; it follows the greedy expansion until the orbit comes close to $\frac12$. Its never-failure condition is sufficient for Step 4 and is unproved. Lemma 13 replaces a fixed finite multiplier problem by one finite integer seed equation and an explicit extension. Seeds for unbounded initial segments would prove Step 4; that family is also unproved. The route is blocked at Step 4 of "The route": no rational with infinite support is proved to be such a sum, and nothing here proves any sum irrational. The computations are finite and certify only what they state. Throughout, $\mathbb N=\{1,2,\ldots\}$.

Notation

For $n\ge1$ let $w_n=\frac1{2^n-1}$, and for $n\ge0$ let $T_n=\sum_{k>n}w_k$; $E=T_0=1.60669515\ldots$ is the Erdős–Borwein constant. The achievement set is $\Sigma=\{\sum_{n\in A}w_n:A\subseteq\mathbb N\}\subseteq[0,E]$. Put $g_n=w_n-T_n$ and $\sigma_n=2^nT_n$. For $F\subseteq\{1,\ldots,n\}$ let $S_F=\sum_{k\in F}w_k$ and let $C_n(F)=[S_F,S_F+T_n]$ be the level-$n$ cylinder of $F$. Write $A_{<n}=A\cap\{1,\ldots,n-1\}$, $\lambda$ for Lebesgue measure and $\{x\}$ for the fractional part.

Held links

Lemma 1: tails and gaps

For $n\ge0$ and $n\ge1$ respectively,

$$T_n=\sum_{j\ge1}\frac{2^{-jn}}{2^j-1},\qquad g_n=\sum_{j\ge2}\frac{2^j-2}{2^j-1}\,2^{-jn},$$

so that $\frac23\,4^{-n}<g_n<\frac{4^{-n}}{1-2^{-n}}$. Moreover $\sigma_n=1+\frac{2^{-n}}3+\frac{4^{-n}}7+\cdots$ decreases to $1$, and $2\sigma_{n-1}=1+w_n+\sigma_n$.

Proof. Expand $w_k=\sum_{j\ge1}2^{-jk}$, exchange the nonnegative sums and sum the geometric series in $k$. In $g_n=\sum_j2^{-jn}(1-\frac1{2^j-1})$ the term $j=1$ vanishes, the term $j=2$ is $\frac23 4^{-n}$, and the factors $\frac{2^j-2}{2^j-1}$ are below $1$. Finally $2\sigma_{n-1}=2^n(w_n+T_n)$ and $2^nw_n=1+w_n$. $\square$

The inequality $T_n<w_n$ is also Remark 4.1 of Kovač and Tao.

Lemma 2: the Cantor structure

For $F\subseteq\{1,\ldots,n\}$ the cylinder $C_n(F)$ is the disjoint union of $C_{n+1}(F)$, the open gap $(S_F+T_{n+1},S_F+w_{n+1})$ of length $g_{n+1}$, and $C_{n+1}(F\cup\{n+1\})$. Hence the $2^n$ level-$n$ cylinders are pairwise disjoint, $\Sigma=\bigcap_n\bigcup_FC_n(F)$, every point of $\Sigma$ is the sum over exactly one set $A$, and $A\mapsto\sum_{n\in A}w_n$ is an order-preserving homeomorphism from $\{0,1\}^{\mathbb N}$, ordered lexicographically, onto $\Sigma$. Every interval $(S_F+T_n,S_F+w_n)$ with $F\subseteq\{1,\ldots,n-1\}$ is disjoint from $\Sigma$.

Proof. A sum whose set agrees with $F$ below $n+1$ is at most $S_F+T_{n+1}$ when $n+1\notin A$ and at least $S_F+w_{n+1}$ when $n+1\in A$, and $w_{n+1}>T_{n+1}$ by Lemma 1. Induction gives the disjoint cylinders and their nested intersection. Two supports first differing at $n$ have sums separated by at least $g_n>0$, which proves uniqueness and order preservation. The coding map is continuous because agreement through $n$ bounds the difference by $T_n\to0$; its compact domain and Hausdorff image make it a homeomorphism. Flipping a sufficiently late digit gives a distinct arbitrarily nearby point, so the image is perfect. Its cylinder lengths tend to zero and each cylinder has a later gap, so it has empty interior. $\square$

Lemma 3: the measure

For every $n$ and every $F\subseteq\{1,\ldots,n\}$, $\lambda(\Sigma\cap C_n(F))=2^{-n}$. Consequently $\lambda(\Sigma)=1$, $\lambda(\Sigma\cap[0,1))=\frac12$, and the restriction of $\lambda$ to $\Sigma$ is the image of the fair-coin product measure under the coding map. For $N\ge1$,

$$\prod_{n=1}^N\Bigl(1-\frac{g_n}{T_{n-1}}\Bigr)=\frac{2^NT_N}{E} \longrightarrow\frac1E=0.62239560\ldots .$$

Proof. $\Sigma\cap C_n(F)$ is the decreasing intersection over $m\ge n$ of the $2^{m-n}$ disjoint level-$m$ cylinders inside $C_n(F)$, whose total length is $2^{m-n}T_m=2^{-n}\sigma_m\to2^{-n}$. A sum with $1\in A$ is at least $1$ and a sum without is at most $T_1=E-1<1$, so $\Sigma\cap[0,1)=\Sigma\cap C_1(\emptyset)$. For the product, $1-g_n/T_{n-1}=(T_{n-1}-w_n+T_n)/T_{n-1} =2T_n/T_{n-1}$, which telescopes. $\square$

The gap of each level-$(n-1)$ cylinder occupies the fraction $g_n/T_{n-1}=\frac13 2^{-n}(1+O(2^{-n}))$ of it, and the product is the fraction of $[0,E]$ that the level-$N$ cylinders cover. A point placed uniformly at random in $[0,E]$ therefore lies in $\Sigma$ with probability $1/E$; this uniform-position model is a heuristic for rationals, not a statement about them. Conditional on lying in any level-$N$ cylinder, a uniformly placed real point lies in $\Sigma$ with probability $2^{-N}/T_N=1/\sigma_N$. This too is a statement about Lebesgue measure, not a survival probability established for any rational.

Lemma 4: the greedy criterion

For $q\in[0,E]$ put $r_0=q$ and, for $n\ge1$, let $\epsilon_n=1$ if $r_{n-1}\ge w_n$ and $\epsilon_n=0$ otherwise, and $r_n=r_{n-1}-\epsilon_nw_n$. If $0\le r_{n-1}\le T_{n-1}$, then $0\le r_n\le T_n$ unless $T_n<r_{n-1}<w_n$; call such a step a gap at step $n$. At a gap, $q\notin\Sigma$, since $q$ lies in the interval $(S+T_n,S+w_n)$ of Lemma 2 with $S=q-r_{n-1}$. If no step is a gap, then $q=\sum_{\epsilon_n=1}w_n$, and $A=\{n:\epsilon_n=1\}$ is the only set with sum $q$. The point $q$ passes the first $N$ steps exactly when $q\in C_N(F)$ for $F=\{n\le N:\epsilon_n=1\}$.

Proof. When $r_{n-1}\ge w_n$, $r_n\in[0,T_{n-1}-w_n]=[0,T_n]$. Otherwise $r_n=r_{n-1}$, which is admissible exactly when $r_{n-1}\le T_n$. If no step is a gap then $0\le r_n\le T_n\to0$. If $q=\sum_{n\in A}w_n$ for some $A$, then $r_{n-1}=\sum_{k\in A,k\ge n}w_k$ is at least $w_n$ when $n\in A$ and at most $T_n<w_n$ when $n\notin A$, so the greedy choices reproduce $A$. $\square$

A gap is certified by any $F\subseteq\{1,\ldots,n-1\}$ with $T_n<q-S_F<w_n$, whether or not $F$ is the greedy prefix (Lemma 2).

Lemma 5: the scaled and mod-one forms

Put $s_n=2^nr_n$. Then $s_n=2s_{n-1}-\epsilon_n(1+w_n)$, the admissible range of $s_n$ is $[0,\sigma_n]$, step $n$ selects $n$ exactly when $s_{n-1}\ge\frac{1+w_n}2$, and step $n$ is a gap exactly when $s_{n-1}$ lies in

$$W_n=\Bigl(\frac{\sigma_n}2,\ \frac{1+w_n}2\Bigr)\subseteq\Bigl(\frac12,\ \frac12+2^{-n}\Bigr],$$

an interval of length $2^{n-1}g_n$, between $\frac13 2^{-n}$ and $\frac{2^{-n-1}}{1-2^{-n}}$. Moreover

$$s_{n-1}\equiv\{2^{n-1}q\}-\sum_{k\in A_{<n},\,k\ge2} \frac{2^{(n-1)\bmod k}}{2^k-1}\pmod 1 ,$$

and, when no earlier step is a gap, step $n$ is a gap exactly when the fractional part of the right side lies in $W_n$.

Proof. Multiply the recurrence of Lemma 4 by $2^n$ and use $2^nw_n=1+w_n$; the thresholds are those of Lemma 4 multiplied by $2^{n-1}$, and the bounds on $W_n$ come from Lemma 1. For the congruence, $s_{n-1}=2^{n-1}q-\sum_{k\in A_{<n}}2^{n-1}w_k$; the term $k=1$ is an integer, and for $k\ge2$, $2^{n-1}\equiv2^{(n-1)\bmod k}\pmod{2^k-1}$ with $0<2^{(n-1)\bmod k}<2^k-1$. With no earlier gap, $0\le s_{n-1}\le\sigma_{n-1}$. If $s_{n-1}\ge1$, then $\{s_{n-1}\}\le\sigma_{n-1}-1\le\sigma_1-1<\frac12$ for $n\ge2$ (and $\{q\}\le T_1$ when $n=1$), so neither $s_{n-1}$ nor its fractional part lies in $W_n$; otherwise $s_{n-1}=\{s_{n-1}\}$. $\square$

The test at step $n$ therefore depends on $q$ only through $\{2^{n-1}q\}$, which is periodic in $n$ when $q$ has odd denominator, and on the earlier choices only through the residues $(n-1)\bmod k$: each selected $k$ adds to every later test a term of period $k$ in $n$, its echo at the multiples of $k$.

Lemma 6: carries

Let $0\le q<1$ have binary digits $d_m$ (the terminating expansion when $q$ is dyadic). For any $A$ put $\theta(m)=\#\{a\in A:a\mid m\}$, $c_0=0$ and $c_m=2c_{m-1}+d_m-\theta(m)$. Then $\sum_{a\in A}w_a=q$ if and only if $0\le c_m\le m+2$ for every $m$; in that case $c_m=\lfloor H_m\rfloor$ with $H_m=\sum_{j\ge1}\theta(m+j)2^{-j}$ and $H_m-c_m=\{2^mq\}$. The coefficient transform is $\theta_A$, the number of members of $A$ dividing $m$, not $\tau(m)\mathbf 1_A(m)$.

Proof. Telescoping gives $\sum_{m\le N}\theta(m)2^{-m}=\sum_{m\le N}d_m2^{-m}-c_N2^{-N}$, and nonnegative summation gives $\sum_{a\in A}w_a=\sum_m\theta(m)2^{-m}$, which proves sufficiency. Under the identity, $H_m=c_m+\{2^mq\}$, and $0\le H_m\le\sum_j(m+j)2^{-j}=m+2$ because $\theta(m')\le m'$. $\square$

A sum $q\ge1$ forces $1\in A$ (Lemma 3's proof), and the lemma applies to $q-1$ with $A\setminus\{1\}$. The carry and scaled forms describe the same orbit: $H_m=P_m+s_m$, where $P_m=\sum_{j\ge1}\theta_{\le m}(m+j)2^{-j}$ is the pending contribution of the members up to $m$ and $\theta_{\le m}$ counts only those members.

The raw carries of a representation are not bounded, nor eventually periodic, for any infinite $A$: if $m$ is the least common multiple of $K$ members of $A$, then $\theta(m)\ge K$ and, since $c_m\ge0$ under the identity, $2c_{m-1}=\theta(m)+c_m-d_m\ge K-1$.

This excludes periodicity of the literal integer carries. It does not exclude a finite description using unbounded integer parameters, a periodic normalized state, or a nonperiodic arithmetic invariant.

Lemma 7: dyadic targets

Every finite subsum $S_F$ has odd reduced denominator, as a sum of fractions with odd denominators. So for $q=p/2^J$ with $p$ odd and $J\ge1$, membership $q\in\Sigma$ forces $A$ to be infinite. For $n>J$, $\{2^{n-1}q\}=0$ and the test of Lemma 5 involves only the echo sum. For $q=\frac12$: $\epsilon_1=0$, and the first gap, if there is one, is at the least $n\ge2$ with $\{-\Psi_n\}\in W_n$, so that $\frac12\in\Sigma$ exactly when $\{-\Psi_n\}\notin W_n$ for every $n\ge2$, where

$$\Psi_n=\sum_{k\in A_{<n}}\frac{2^{(n-1)\bmod k}}{2^k-1}$$

and $A$ is the set the greedy expansion of $\frac12$ generates.

For a rational with odd denominator, such as $\frac1{21}$, Lemma 9 supplies the infinite-support conclusion instead.

Lemma 8: certificates

For $n,J\ge1$ let $L_{n,J}=\sum_{j=1}^J\frac{2^{-jn}}{2^j-1}$ and $E_{n,J}=\frac{2^{1-(n+1)(J+1)}}{1-2^{-(n+1)}}$. Then $L_{n,J}\le T_n\le L_{n,J}+E_{n,J}$, since each omitted factor $\frac1{2^j-1}$ is at most $2^{1-j}$. A gap at step $n$ is therefore certified in exact rationals by $F\subseteq\{1,\ldots,n-1\}$ and $J$ with $L_{n,J}+E_{n,J}<q-S_F<w_n$, and passing $N$ steps by $0\le q-S_F\le T_N$ for the selected set $F$, which interval enclosures of $S_F$ and $T_N$ certify (Lemma 4).

For example, $q=2/3$, $n=1$, $F=\emptyset$ and $J=2$ give $L_{1,2}+E_{1,2}=7/12+1/24=5/8<2/3<1=w_1$, a strict gap certificate.

Lemma 9: where a finite representation can end

Let $q$ have reduced denominator $Q$, and suppose $q=S_F$ for a finite nonempty $F$ with largest element $n$. Then $n\in\{1,6\}$, or $n$ is the multiplicative order of $2$ modulo some odd prime factor of $Q$. Consequently the greedy expansion of $q$ can terminate (reach residual $0$) only at a step at most $N_Q=\max(\{6\}\cup\{\operatorname{ord}_p2:p\mid Q,\ p\text{ odd prime}\})$, and if $q\in\Sigma$ and its greedy expansion has not terminated by step $N_Q$, its support is infinite.

Proof. Since $w_n=q-S_{F\setminus\{n\}}$, the number $2^n-1$ divides $\operatorname{lcm}(Q,2^k-1:k\in F,\ k<n)$. By the theorem of Bang and Zsigmondy (K. Zsigmondy, Zur Theorie der Potenzreste, Monatsh. Math. Phys. 3 (1892), 265--284), $2^n-1$ has a prime factor $p$ with $\operatorname{ord}_p2=n$ unless $n\in\{1,6\}$. Such a $p$ divides no $2^k-1$ with $k<n$, so it divides $Q$. $\square$

The precise base-two input is also stated in Erdős and Graham, Old and new problems and results in combinatorial number theory, p. 24, footnote 1, in the context of moduli greater than one. That statement was checked against the printed page; the classical theorem's proof is an external dependency, not a proof reconstructed here.

This bounds every possible finite support by $\{1,\ldots,N_Q\}$, so a rational's being a finite subsum is decidable by exact rational arithmetic. The maximum is defined even when $Q$ has no odd prime factor, and $N_Q\le\max(6,Q-1)$, since $\operatorname{ord}_p2\mid p-1$.

For $\frac1{21}$, $\operatorname{ord}_32=2$ and $\operatorname{ord}_72=3$, so $N_{21}=6$. Its greedy expansion skips steps $1$ to $4$, selects $5$ and skips $6$, with residual $1/21-1/31=10/651$ strictly between $0$ and $1/63$. It therefore never reaches residual zero: $\frac1{21}$ is not a finite subsum, and $\frac1{21}\in\Sigma$ exactly when no step is a gap. For every denominator $Q\le200$, $N_Q<200$, so each fraction of the census below that passes step $1000$ would have infinite support if it lies in $\Sigma$.

Lemma 10: rational shadows

Let $m\ge1$, suppose no step up to $m$ is a gap, and write $s_m=y-\delta$, where $y\in(0,1)$ is a rational whose reduced denominator $D$ is not a power of $2$ and $\delta$ is real. Let $e_1e_2\cdots$ be the binary digits of $y$ and $y_i=\{2^iy\}$. If $h\ge1$ satisfies

$$2^{h-1}|\delta|+\frac49\,2^{h-1-m}\le\frac1{2D},$$

then for $1\le i\le h$ step $m+i$ is not a gap and selects $m+i$ exactly when $e_i=1$, and

$$s_{m+i}=y_i-\delta_i,\qquad \delta_i=2^i\delta+\sum_{l=1}^{i}e_l\,2^{i-l}w_{m+l}.$$

Proof. Since $y$ is not dyadic, no $y_i$ is; as $y_{i-1}$ and $\frac12$ both lie in $\frac1{2D}\mathbb Z$, $|y_{i-1}-\frac12|\ge\frac1{2D}$, and $e_i=1$ exactly when $y_{i-1}>\frac12$. Suppose $s_{m+i-1}=y_{i-1}-\delta_{i-1}$, as holds for $i=1$ with $\delta_0=\delta$. From $w_n\le\frac43 2^{-n}$ for $n\ge2$,

$$|\delta_{i-1}|+\frac{w_{m+i}}2\le2^{i-1}|\delta| +\sum_{l=1}^{i}2^{i-1-l}w_{m+l} <2^{i-1}|\delta|+\frac49\,2^{i-1-m}\le\frac1{2D},$$

the last step by the hypothesis, whose left side increases with $h$. If $e_i=1$, then $s_{m+i-1}\ge\frac12+\frac1{2D}-|\delta_{i-1}|$, which is at least $\frac{1+w_{m+i}}2$, so step $m+i$ selects $m+i$, and $s_{m+i-1}$ is not in the open window $W_{m+i}$. If $e_i=0$, then $s_{m+i-1}\le\frac12-\frac1{2D}+|\delta_{i-1}|<\frac12<\frac{\sigma_{m+i}}2$, so $m+i$ is skipped and the step is not a gap. In both cases $s_{m+i}=2s_{m+i-1}-e_i(1+w_{m+i})=y_i-\delta_i$. $\square$

Corollary (dyadic roundings). Let $J\ge1$ and $F\subseteq\{1,\ldots,J\}$ with $S_F\notin\mathbb Z$, let $D$ be the reduced denominator of $S_F$, and put $q=\lceil2^JS_F\rceil/2^J$. Then $q$ is a dyadic rational and not an integer, its greedy expansion selects exactly $F$ among the steps $1,\ldots,J$, and no step $n\le2J+1-\lceil\log_2D\rceil$ is a gap. If $q\in\Sigma$, its support is infinite.

Proof. $D$ is odd and at least $3$ (Lemma 7). Since $0<q-S_F<2^{-J}<T_J$, $q$ lies in $C_J(F)$, so by Lemmas 2 and 4 the first $J$ steps are not gaps and select exactly $F$. Then $s_J=y=\lceil2^JS_F\rceil-2^JS_F\in(0,1)$, whose reduced denominator is $D$ because $2$ is invertible modulo $D$. Put $h=J+1-\lceil\log_2D\rceil$. If $h\le0$, the stated horizon is already covered by the first $J$ steps. If $h\ge1$, Lemma 10 with $m=J$ and $\delta=0$ applies, since $2^h\le2^{J+1}/D$ gives $\frac49 2^{h-1-J}\le\frac4{9D}<\frac1{2D}$. If $1\notin F$, then $0<q<S_F+T_J\le T_1<1$; if $1\in F$, the same holds for $q-1$ and $F\setminus\{1\}$. So $q$ is not an integer, and Lemma 7 gives the infinite support. $\square$

The case $F=\{2\}$, $J=2$ is $q=\frac12$, where the corollary covers only step $3$. For $F=\{2\}$ and every $J\ge2$ it gives the dyadic rationals $\lceil2^J/3\rceil/2^J$, each passing its first $2J-1$ steps by this argument alone. A single member of such a family in $\Sigma$ would answer Problem 257.

How far one shadow reaches. Running the recursion of Lemma 5 backward gives the converse constraint. If step $n$ is a gap and $1\le m<n$, then $s_{n-1}=\frac12+\theta$ with $0<\theta<2^{-n}$, and

$$s_m=2^{m+1-n}s_{n-1}+\sum_{l=m+1}^{n-1}\epsilon_l(1+w_l)2^{m-l} =\frac b{2^{n-m}}+\eta,\qquad 0<\eta<2^{m+1-2n}+\frac49\,2^{-m}<2^{-m},$$

with $b$ an odd integer. An exact shadow ($\delta=0$) of denominator $D$ therefore excludes a gap at step $n$ whenever $2^{n-m}\le2^m/D$, the same horizon as Lemma 10 up to an additive constant. For $n\ge2m+2$ the constraint is empty, because the odd multiples of $2^{m-n}$ are then closer together than $2^{-m}$, so it excludes nothing. Lemma 10 never reaches step $2m$ either, since $D\ge3$ forces $h\le m-1$. A proof by shadows therefore needs fresh shadows at steps $m_1<m_2<\cdots$ with $m_{j+1}<2m_j$, each with denominator and error small relative to $2^{m_j}$.

Shadows from Dirichlet's theorem. Such shadows exist at every step whose state is not close to a dyadic rational, so a chain of them is not an independent mechanism. Let $2\le k<m$. Dirichlet's theorem gives a fraction $a/D$ in lowest terms with $1\le D\le2^k$ and $|s_m-a/D|<1/(D2^k)$. Suppose $s_m$ lies within $2^{-t-k}$ of no fraction $b/2^t$ with $0\le t\le k$. Then $D$ is not a power of $2$; $a/D\in(0,1)$, since $s_m\in[0,\sigma_m]$ and $\frac1D-\frac1{D2^k}>\sigma_m-1$; and Lemma 10 applies with $\delta=a/D-s_m$ and $h=\min(k-1,m-k)$, since $2^{h-1}|\delta|<\frac1{4D}$ and $\frac49 2^{h-1-m}\le\frac29 2^{-k}<\frac1{4D}$. With $k=\lceil m/2\rceil$, every such step $m\ge3$ protects the next $\lceil m/2\rceil-1$ steps. Conversely, by the backward computation, a gap at step $n$ puts $s_m$ within $2^{-m}$ of an odd multiple of $2^{m-n}$, which is such a near-dyadic position, at every step $m$ with $n-m\le\lfloor m/2\rfloor$. The Dirichlet criterion leaves only near-dyadic steps uncovered, and every gap is preceded by an unbroken run of them from about two thirds of its step onward. Proving that a chain of renewals continues means excluding such runs along the orbit, a statement of the same pointwise kind as Step 4 and slightly stronger, not a reduction of it. No such chain is proved here for either rational target; a proof along these lines needs shadows with additional structure that makes their renewal provable.

A different arithmetic protection applies to residuals with numerator one. If $r_{n-1}=1/R$ for a positive integer $R$, step $n$ is not a gap: a gap needs $T_n<1/R<w_n$, that is $2^n-1<R<1/T_n$, and $1/T_n<2^n$ by Lemma 1. The expansion of $\frac12$ selects $2,3,6,7,14$ with residuals $\frac16,\frac1{42},\frac1{126},\frac1{16002},\frac1{688086}$, so its first $20$ steps are not gaps by this remark alone. Step $20$ selects $20$ and leaves the residual $120163/240503259150$, whose numerator is prime, and the remark stops applying. A residual $a/R$ in lowest terms is a gap at step $n$ exactly when $a(2^n-1)<R<a2^n/\sigma_n$, an interval of length below $a$; for $a=2$ it holds exactly when $R=2^{n+1}-1$, the residual $2w_{n+1}$.

Lemma 11: a construction for finitely many multipliers

For every $C=\{1\}\cup B$ with $B\subseteq\{2,3,\ldots\}$ and $|B|\le3$, there is an infinite $A\subseteq\{2,3,\ldots\}$ satisfying

$$\sum_{n\in A}\sum_{a\in C}2^{-an}=\frac12.$$

The construction and proof define each digit using only earlier digits, with carries in $\{1,2\}$. Bertrand's postulate bounds the necessary lookahead by $2n$. When $C\ne\{1\}$, strict domination of each tail identifies the constructed digits with the greedy digits for these weights. These are truncated geometric weights, not the full weights $w_n$.

The same page proves the transfer that would finish this route: half representations for $C=\{1,\ldots,J\}$ along an unbounded sequence of $J$ give a full Mersenne half representation by compactness, since the omitted mass is at most $\frac23 4^{-J}$ uniformly over supports omitting 1. The construction covers only $J\le4$. Its direct two-state extension fails for $C=\{1,2,3,4,6\}$ at step 108, although the actual truncated greedy orbit passes that step by using carry 3. The source-linked finite decision procedure for fixed $J$ also leaves the unbounded sequence of $J$ open. The page also makes the transfer's relation to Step 4 explicit. A half representation for $C=\{1,\ldots,J\}$ already places $\frac12$ outside every gap of $\Sigma$ of level at most $J$, apart from a strip of length $\frac23 4^{-J}$ at its right end. The truncated and full weights also have the same forcing values, for the same digits, at all positions up to $2J+1$. These facts do not rule out a construction using the bounded carries of each finite set, such as the extension in Lemma 13.

Lemma 12: a window rule

One explicit rule generates digits from already fixed digits and follows the greedy expansion of $\frac12$ until its scaled residual comes close to $\frac12$. Put $\epsilon_1=0$ and $c_1=1$. For $n\ge2$, once $\epsilon_1,\ldots,\epsilon_{n-1}$ are fixed, let $f_m$ be the number of $d<m$ with $d\mid m$ and $\epsilon_d=1$; it is determined for every $m\le2n-1$, since a proper divisor of $m$ is at most $m/2$. Define

$$F_n=\sum_{m=n+1}^{2n-1}f_m2^{n-m},\qquad c_n=\max(1,\lceil F_n\rceil),\qquad \epsilon_n=2c_{n-1}-f_n-c_n,$$

and put $\eta_n=f_{2n-2}2^{2-n}+f_{2n-1}2^{1-n}$, the part of $F_n$ from the two positions that enter the window at step $n$.

(a) If every $\epsilon_n$ is $0$ or $1$, then $A=\{n:\epsilon_n=1\}$ satisfies $\sum_{n\in A}w_n=\frac12$, and $A$ is the greedy support of $\frac12$.

(b) $\epsilon_2=\epsilon_3=1$. For $n\ge4$, if $\epsilon_2,\ldots,\epsilon_{n-1}$ are bits, then $\epsilon_n\le1$, and $\epsilon_n<0$ exactly when $\lceil F_{n-1}\rceil-F_{n-1}<\eta_n/2$. Equivalently, let $N_{n-1}=2^{n-2}F_{n-1}$, an integer, and let $k$ be the least nonnegative integer with $N_{n-1}+k\equiv0\pmod{2^{n-2}}$. Then $\epsilon_n=\lfloor(4k-2f_{2n-2}-f_{2n-1})2^{1-n}\rfloor$, and the rule fails at step $n$ exactly when $4k<2f_{2n-2}+f_{2n-1}$.

(c) Let $n\ge4$, and suppose the digits through step $n-1$ are the greedy digits of $\frac12$, no step up to $n-1$ is a gap, and the window rule has agreed with those digits through step $n-1$. Let $\theta_{<n}(m)$ count the selected $d<n$ dividing $m$, and $R_n=\sum_{m\ge2n}\theta_{<n}(m)2^{n-1-m}$. Then

$$s_{n-1}=\lceil F_{n-1}\rceil-F_{n-1}-\frac{\eta_n}2-R_n ,$$

the rule's digit at step $n$ is $\lfloor2(s_{n-1}+R_n)\rfloor$, and it is the greedy digit unless $\frac12-R_n\le s_{n-1}<\frac{1+w_n}2$.

Proof. (a) Since $F_n$ is at most the largest $f_m$ with $n<m<2n$, and $f_m$ is at most the number $\tau(m)\le2\sqrt m$ of divisors of $m$, we have $2^{-n}c_n\to0$. As $\epsilon_n+f_n=\theta_A(n)$ for every $n$, with $\theta_A$ as in Lemma 6, telescoping gives $\sum_{n=2}^N\theta_A(n)2^{-n}=\frac12-2^{-N}c_N$, and $\sum_m\theta_A(m)2^{-m}=\sum_{a\in A}w_a$ as in the proof of Lemma 6. By Lemma 4, $A$ is the greedy support.

(b) $F_2=f_3/2=0$, so $c_2=1$ and $\epsilon_2=1$. Then $f_4=1$ and $f_5=0$, so $F_3=\frac12$, $c_3=1$ and $\epsilon_3=2c_2-f_3-c_3=1$. For $n\ge3$ the window $(n,2n)$ contains an even $m\ge4$, where $f_m\ge\epsilon_2=1$, so $F_n>0$ and $c_n=\lceil F_n\rceil$. For $n\ge4$ the definitions give $F_n=2F_{n-1}-f_n+\eta_n$, so with $x=F_{n-1}$ the digit is $\epsilon_n=2\lceil x\rceil-\lceil2x+\eta_n\rceil$. As $2x+\eta_n>2\lceil x\rceil-2$, $\epsilon_n\le1$, and $\epsilon_n\ge0$ exactly when $2x+\eta_n\le2\lceil x\rceil$. Moreover $\epsilon_n=\lfloor2(\lceil x\rceil-x)-\eta_n\rfloor$. Since $x\in2^{2-n}\mathbb Z$, $\lceil x\rceil-x=k2^{2-n}$, which gives the congruence form.

(c) Write $c^{\mathrm g}_{n-1}$ for the carry computed from the actual greedy digits. By telescoping through step $n-1$, $2^{n-1}\bigl(\frac12-\sum_{m<n}\theta_A(m)2^{-m}\bigr)=c^{\mathrm g}_{n-1}$. The selected $d<n$ contribute $\theta_{<n}(m)$ at each position $m\ge n$, and $\theta_{<n}(m)=f_m$ for $n\le m\le2n-1$. Hence $s_{n-1}=c^{\mathrm g}_{n-1}-F_{n-1}-\eta_n/2-R_n$. Prior agreement with the rule and the common initial carry imply $c^{\mathrm g}_{n-1}=c_{n-1}=\lceil F_{n-1}\rceil$; this is where the agreement hypothesis is used. It gives the displayed formula, and by (b) the rule's digit is $\lfloor2(s_{n-1}+R_n)\rfloor$. The greedy digit is $1$ exactly when $s_{n-1}\ge\frac{1+w_n}2$ (Lemma 5). Since $s_{n-1}\ge0$ and $s_{n-1}+R_n=\lceil F_{n-1}\rceil-F_{n-1}-\eta_n/2<1$, the two digits agree unless $\frac12-R_n\le s_{n-1}<\frac{1+w_n}2$. $\square$

Without prior agreement, the actual-carry identity in the proof still holds, but substituting $\lceil F_{n-1}\rceil$ for $c^{\mathrm g}_{n-1}$ is not justified. In particular, survival of the greedy prefix alone does not supply that substitution. The statement that the rule follows the greedy orbit until the first indicated close approach uses (c) inductively, starting with their common first three digits.

By $\tau(m)\le2\sqrt m$, $R_n\le4\sqrt n\,2^{-n}$ for $n\ge4$. The window $W_n$ of Lemma 5 lies inside the interval of (c). So the rule reproduces the greedy expansion of $\frac12$ while the orbit avoids these intervals, of length at most $(4\sqrt n+1)2^{-n}$, and by (a) it gives a half representation if it never fails. Its success is therefore sufficient for Step 4 and follows from avoidance of the slightly longer intervals at every step $n\ge4$. This gives a sufficient congruence condition on integers computed from the rule's own digits. A proof that this condition holds at every step is missing; equivalence with gap avoidance is not asserted.

For a finite set $C$ of multipliers containing $1$, let $f_m$ count only the $d$ for which $m/d$ lies in $C\setminus\{1\}$. Then (a) holds with $v_n(C)$ in place of $w_n$, with carries at most $\max(1,|C|-1)$, and the computation in (b) applies at every step where $F_{n-1}$ and $F_n$ are positive. For $C=\{1,2,3,4,6\}$ the rule uses the carries $1,2,3,1$ at steps $105$ to $108$. These are the actual truncated greedy carries, at the step where the two-state rule of Lemma 11 fails. Check 6 of the evidence confirms this and the agreements stated there. A rule allowing larger carries therefore exists for every $C$. What it lacks is a proof that it meets no coincidence in (b), and that is a pointwise property of one orbit.

Lemma 13: a finite seed and a doubling extension

Let $C$ be a finite set of positive integers containing $1$, and let $M\ge2$ be a common multiple of its elements. Define

$$Q_C(M,d)=\sum_{\substack{a\in C\\ad\le M}}2^{M-ad} \qquad(2\le d\le M).$$

There exists a support omitting $1$ that represents $\frac12$ for the weights $v_n(C)$ if and only if some $D\subseteq\{2,\ldots,M\}$ satisfies

$$\sum_{d\in D}Q_C(M,d)=2^{M-1}-1.$$

There is at most one seed $D$, and decreasing greedy on these integer weights decides whether it exists. The proof and construction use $c_M=1$: any representation has a positive integer carry there, and $c_M\le1+\sum_{a\in C\setminus\{1\}}(2^a-1)^{-1}<2$. Given a seed ending at $H=2^jM$ with carry $1$, all proper-multiplier forcing $f_m$ for $H<m\le2H$ is known. Set $c_{2H}=1$ and work backward by

$$c_{m-1}=\left\lceil\frac{f_m+c_m}{2}\right\rceil, \qquad \epsilon_m=2c_{m-1}-f_m-c_m\in\{0,1\}.$$

The computed carry at $H$ is $1$, because the positive quantity inside its ceiling is less than $\frac23+2^{-H}<1$. Every block therefore joins the preceding one, retains its divisor contributions and has bounded positive carries. Telescoping proves the half representation. This construction allows $C$ to be arbitrary, but requires its seed.

For $C=\{1,2,3,4,6\}$, $M=12$ has seed $\{2,3,6,7,10\}$; the extension passes the carry-$3$ transition of Lemmas 11 and 12. Adjoining a common multiple of every element of a solved $C$ preserves solvability: at that boundary the added multiplier contributes no seed coefficient. Such extensions do not exhaust all multipliers. The same page proves that an old seed cannot be reused at a common boundary $M\ge2b$ after adding a multiplier $b$: its row sum increases by at least $2^{M-2b}$. The exact $C=\{1,2\}$ and $C'=\{1,2,3\}$ seeds at $M=6$ also show that identical forcing through position $5$ can give different fifth digits, because the later forcing changes the carry. Check 7 checks the finite seeds and four doubling blocks for its stated six sets, not an unbounded family.

For $C_J=\{1,\ldots,J\}$, seeds at $M=\operatorname{lcm}(1,\ldots,J)$ for unbounded $J\ge2$ would finish the full target by Lemma 11's transfer. Their existence remains unproved. No common carry bound as $J\to\infty$ is required.

The multiplier page also writes the doubling blocks for the full weights directly. After step $H$ the block word through $2H$ is one of at most two adjacent integers determined by $s_H$, and the values of $r_H$ admissible for the two words are separated by an interval of length exactly $g_k$, the level-$k$ gap of Lemma 2, where $k$ is the last selected position of the larger word. Choosing between them is Lemma 4 on the whole block, so a selection rule valid at every doubling boundary is equivalent to Step 4. The block cannot restart at carry $1$ either: the carry at $2H$ exceeds the pending mass of the digits through $H$, which for $\frac12$ already exceeds $1$ at $H=16$, so $c_{32}\ge2$. If this pending mass is $E_H$, the multiplier page proves that the only possible terminal carries are $\lfloor E_H\rfloor+1$ and $\lfloor E_H\rfloor+2$. It gives their exact candidate words, proves automatic feasibility at the two word-boundary cases, and identifies the remaining interior failure interval. No uniform avoidance of that interval is proved. Its equivalence to the greedy test does not preclude proving a selection invariant in these variables.

For a discrete induction, the linked multiplier page proves a necessary interface when passing from $C_{b-1}$ to $C_b$: if both half representations exist, their digits agree through $b-1$, and their binary prefix integers through $2b-1$ differ by $0$ or $1$. The corresponding carry difference is $0$ or $1$ in that early window. The unchanged prefix, and the decremented prefix if its borrow starts at or after $b$, satisfy the finite coefficient recurrence. An infinite continuation is not implied. The page states the exact signed repair equation, retaining every later divisor contribution. A repair for every $b$ would supply all the seeds and stabilize every digit, hence prove Step 4. This repair remains unknown. Later carry differences can be negative: the displayed $C_4$ and $C_5$ seeds give carries $2$ and $1$ at position $29$. Thus a nonnegative carry-change invariant cannot prove this induction.

The multiplier page also gives the exact first-failure test for the finite row. If $B_n$ is its remaining capacity and $r_{n-1}$ is the actual greedy remainder, failure first occurs when $B_n<r_{n-1}<Q_C(M,n)$; this is impossible for $n>M/2$, where the remaining row consists of binary powers. Its identity $r_n=2^{M-n}c_n-1-E_n$ keeps the selected prefix's pending contributions $E_n$. Positive bounded carries alone do not control this remainder. Uniform exclusion of the first-failure interval is unproved.

An alternative on that page iterates an order-reversing map on whole binary words. Collapse of its nested integer bounds constructs a seed, but finiteness only gives a fixed point or a two-cycle. An explicit different-target example has both a unique seed and a persistent two-cycle. The page's stable-pair proposition locates every such pair: if the stabilized bounds first differ at position $k$, they agree through position $2k-1$ with the two sides of the dyadic point $Y$ there, and a seed, if one exists, lies within $\frac{16}{9}2^{M-4k}$ of $Y$, so its word has a run of at least $\min(3k-1,M-k)$ digits right after $k$, all different from its digit at $k$. Without a seed the bounds always form such a pair. Collapse therefore implies a seed and is implied by a seed with no such run after any position. Proving collapse for unbounded initial segments would construct their seeds. The necessary long-run condition does not prove collapse or show that this construction cannot work.

At the half target the linked page's mixed carries are positive integers ending at $1$, each at most $\lfloor(|C|-1)/2\rfloor+1$. If their words first differ at $k$, their difference is $1$ through $2k-1$, with a forced binary borrow block. For $C_b$, $b\ge4$, the mixed carries rejoin at $2k$. This is not an ordinary carry splice: the two ordinary prefix carries are $r+1$ and $r-1$ when the mixed carry is $r$. Their old digits also force a nonzero discrepancy at $2k+2$ when that position is in the row. The page records the exact next-block forcing and the coupled carry-sum and carry-difference constraints. These prove collapse only in the already solved case $|C|\le4$; no uniform exclusion or Boolean repair for unbounded $b$ is established.

The same page records that every set $C$ with $1\in C\subseteq\{1,\ldots,12\}$ has its half seed, and asks whether every finite set of multipliers does. That uniform question would contain the initial segments and allows adjoining multipliers in any order. It is open. Two exact examples there show what its proof must use: integer multiplicities, such as the weights $2^{-n}+4^{-n}+3\cdot8^{-n}$, can meet a gap although their carries are integers and each weight dominates its tail, and multiplier sets chosen separately at each index can lose the half target. A proof must use the bit coefficients of one fixed multiplier set, that is, the Dirichlet structure of the coefficients in (1) of that page.

Lemma 14: exact Mersenne renewals

The identities

$$\frac{w_d}{2}=\sum_{n\in G}w_n+\frac{w_L}{2},\qquad 1\le d<L,\qquad G\subseteq\{d+1,\ldots,L\},$$

are exactly

$$(d,L,G)=(1,2,\{2\}),\quad (1,6,\{2,3,6\}),\quad (2,6,\{3,6\}).$$

In particular, no chain of these renewals continues after endpoint $6$.

Proof. Put $e=\mathbf1_{L\in G}$ and multiply the identity by $2$:

$$w_d-2\sum_{n\in G\setminus\{L\}}w_n=(1+2e)w_L.$$

If $L\ge3$ and $L\ne6$, Bang–Zsigmondy, in the precise form used in Lemma 9, supplies a prime $p$ with $\operatorname{ord}_p2=L$. Thus $p\ge L+1>3$. It divides none of the denominators on the left, whereas the right side has negative $p$-adic valuation because $1+2e$ is $1$ or $3$. This is impossible. Only $L=2,6$ remain.

For $L=2$ necessarily $d=1$, and $e=1$ gives the first identity. For $L=6$, a denominator $31$ occurs only at index $5$, and a factor $5$ only at index $4$. Their valuations rule out $d=4,5$ and force $4,5\notin G$ when $d<4$. If $d=3$, multiplication by $63$ would give $9=1+2e$, impossible. If $d=2$, it gives $21=18\mathbf1_{3\in G}+1+2e$, forcing $G=\{3,6\}$. If $d=1$, it gives $63=42\mathbf1_{2\in G}+18\mathbf1_{3\in G}+1+2e$, forcing $G=\{2,3,6\}$. The three displayed identities check directly. $\square$

There is a quantitative restriction on replacing $1/2$ by a varying terminal coefficient. Suppose a half-prefix through $L\ge2$ survives, with selected set $F$, and write

$$\frac{r_L}{w_L}=\frac uv,\qquad 0<u<v,\quad \gcd(u,v)=1, \qquad e=\mathbf1_{L\in F}.$$

The strict range follows from $0<r_L\le T_L<w_L$. For every prime $p$ with $\operatorname{ord}_p2=L$ one has

$$p\nmid v,\qquad p^{v_p(2^L-1)}\mid u+ev.$$

Indeed, $\frac12-S_{F\setminus\{L\}}=(u+ev)/(v(2^L-1))$, whose left side has no $p$ in its reduced denominator. Coprimality gives $\gcd(u+ev,v)=1$, so $p\mid v$ is impossible and the divisibility follows. Writing $P=p^{v_p(2^L-1)}$, this implies $v\ge P+1$ when $e=0$ and $v\ge(P+1)/2$ when $e=1$. Thus, for $L\ne6$, some such $p\ge L+1$ forces $v\ge\lceil(L+2)/2\rceil$.

Consequently, no finite set of fixed rational coefficients $u/v$ can describe terminal residuals $r_L=(u/v)w_L$ at unbounded surviving positions of the half orbit. This excludes only that form of renewal. It does not exclude approximate rational shadows, terminal coefficients with growing denominators, or a full half representation. The variable terminal carries in Lemma 13's full-weight block formulation remain available, with their selection theorem unproved.

The route

1. Take $q=\frac12$, the case $F=\{2\}$, $J=2$ of the corollary to Lemma 10; $\frac1{21}$ is a second test case, with odd denominator. Any member of that corollary's dyadic family would serve in place of $\frac12$: its first $2J+1-\lceil\log_2D\rceil$ steps are then held by the corollary, and its later steps carry the same obligation as Step 4.
2. By Lemmas 4 and 5, $q\in\Sigma$ exactly when no step is a gap. Held.
3. No step $n\le20000$ is a gap, for $\frac12$ and for $\frac1{21}$ (check 2 of the evidence). Held, finite.
4. Gap. No step $n>20000$ is a gap: by Lemma 7, $\{-\Psi_n\}\notin W_n$ for every $n>20000$. Lemma 12 gives a sufficient integer form: the rule there meets no coincidence, which check 6 confirms through step 20000. Alternatively, construct the seeds of Lemma 13 for unbounded initial segments of multipliers and apply the uniform approximation of Lemma 11. Two sufficient approaches are to exclude the exact first-failure interval at every successor or to prove that the half-target whole-word iteration collapses. The mixed-carry rejoin supplies neither conclusion.
5. By Lemma 7, $A$ is then infinite with $\sum_{n\in A}\frac1{2^n-1}=\frac12$, a negative answer to Problem 257. Held, given Step 4. For $\frac1{21}$, Lemma 9 gives the infinite support in the same way.

The steps use one convention throughout: base $2$, the denominators $2^n-1$, and the single greedy sequence of $q$; Lemmas 4–6 are three descriptions of that sequence, and Lemmas 7 and 9 turn membership into an infinite support.

Failure account

• Step. Step 4, for every $n>20000$; any finite range can be checked, with unresolved interval comparisons reported as undecided. The missing obligation is all later steps, not a longer finite prefix.
• Missing input. A proof that the scaled residual of one specific rational avoids, at every step $n$, the window $W_n$ of length about $\frac13 2^{-n}$ just above $\frac12$. For $\frac12$ this is, by Lemma 7, the statement that the sum $\Psi_n$ of the rationals $2^{(n-1)\bmod k}/(2^k-1)$ over the self-generated set $A_{<n}$ never has $\{-\Psi_n\}\in W_n$.
• Why it fails. For $\frac12$ the residual $r_{n-1}$ is a rational whose reduced denominator divides $2\operatorname{lcm}\{2^k-1:k\in A_{<n}\}$. The elementary upper bound obtained by replacing this least common multiple by a product is as large as $2^{1+n(n-1)/2}$, and supplies no separation from an interval of length $g_n<2\cdot4^{-n}$. This does not assert that the reduced denominator attains that bound. The raw-carry obstruction below excludes literal periodic carries; the finite census tests only the stated periodic supports. Neither supplies pointwise avoidance for this orbit. In binary, a gap at step $n$ means that $r_{n-1}$ has a one at position $n$ followed by $n-1$ zeros, so the obligation resembles questions about long zero runs in the binary digits of specific constants. The two arithmetic protections proved here are local. A rational shadow at step $m$ (Lemma 10) excludes gaps only before step $2m$, and the necessary condition that the backward computation after it places on $s_m$ is empty for gaps at steps $n\ge2m+2$; the numerator-one remark covers the first 20 steps of $\frac12$ and then stops. Continuing the remark needs a residual with numerator one after every selection; continuing the shadows needs a renewal at least once per doubling of the step index. By Dirichlet's theorem shadows renew at every step that is not near-dyadic, and each gap is preceded by a run of near-dyadic steps, so proving the renewals is again a pointwise property of the orbit, of the same kind as the missing input. The alternative through geometric multipliers needs representations for unbounded initial segments $\{1,\ldots,J\}$. Lemma 11 supplies only $J\le4$. Lemma 12 supplies a rule with the larger carries for every such segment, but its digits are bits only away from the coincidences of Lemma 12(b), a pointwise condition again. A representation for one such segment already excludes the gaps of $\Sigma$ of level at most $J$, apart from a strip at each right end. Lemma 13 supplies a different completion from a finite integer seed. Its block recurrence is proved, but existence of the seeds for unbounded $J$ is missing. Adjoining common multiples cannot fill arbitrary missing multipliers, and finitely many seeds do not approximate the full weights to arbitrarily small error. Keeping an old seed at a common boundary fails by the positive row increment in Lemma 13; keeping earlier digits merely because their forcing agrees fails in the explicit position-$5$ example. A replacement induction must control the changed digits and their carries. Convex interpolation does not supply that control: the multiplier page gives a gap at position $4$ for weights $2^{-n}+\frac79 4^{-n}$, between two solvable Boolean endpoints. The necessary two-prefix interface for adjoining $b$ only controls the first $2b-1$ positions. Its signed repair equation has no proved uniform solution. Requiring later carry changes to stay nonnegative is false, even from $C_4$ to $C_5$, where the carry decreases at position $29$. This refutes that proposed invariant, not Boolean seed extension. The exact remaining-capacity test retains the pending contributions through $r_n=2^{M-n}c_n-1-E_n$, but supplies no uniform avoidance bound. The alternative whole-word iteration has no proof of collapse at the half target. Its stable-pair proposition gives a necessary long-run condition when a seed coexists with a nontrivial pair, but does not exclude all such pairs. The special half-target mixed carries force a borrow block through $2k$ and rejoin there; using that rejoin as a splice fails because the ordinary boundary carries are two apart and the old digits force another discrepancy at $2k+2$ when the row continues that far. A valid replacement must retain both mixed carries, the common forcing, those boundary offsets and the pending mass. The displayed joint constraints supply no all-future invariant, so uniform collapse remains unknown, not refuted. The direct full-weight construction retains two candidate terminal carries determined by the actual pending mass. Its endpoint cases always pass; its interior case leaves the explicit phase interval on the multiplier page. No successor invariant excludes that interval. Restarting every block at carry $1$ is false, since $c_{32}\ge2$ for $\frac12$. Always choosing the smaller of the two candidate carries is also false: $E_{32}<1$ but $c_{64}=2$. Exact renewals $\frac{w_d}2=S_G+\frac{w_L}2$ cannot replace it: Lemma 14 classifies all such identities and proves that they stop at endpoint $6$. A finite set of fixed rational multiples of $w_L$ also cannot provide unbounded renewal positions. The variable-carry selection theorem remains unknown, not refuted by these failures. Finally, the finite census of all multiplier sets inside $\{1,\ldots,12\}$ suggests that every finite set has a half seed, but integer multiplicities and index-dependent sets can lose the half target, so an argument blind to the fixed Boolean multiplier structure cannot prove it.
• Standing of the input. Unproved here and not refuted by the finite checks. The uniform-position model is no premise about either rational. No weaker sufficient input is established here: selection at the full-weight doubling boundaries is equivalent to it, while seeds for unbounded initial segments, the window rule and whole-word collapse imply it.

Computations

The evidence program certifies the following finite statements; none proves membership in the full Mersenne achievement set or an irrationality.

1. Census. Of the 19654 reduced fractions $p/Q\in(0,E)$ with $Q\le200$, 41 are finite subsums, each ending at a step allowed by Lemma 9, 7395 meet a certified gap, every one at a step at most 12, and the remaining 12218 pass step 1000 without terminating, so by Lemma 9 any of them that lies in $\Sigma$ has infinite support. The fraction passing among those that are not finite subsums is $12218/19613=0.6230$, against $1/E=0.6224$. The step-$n$ gap counts, as fractions of the fractions that met no gap before step $n$, follow the gap fractions $g_n/T_{n-1}$ of Lemma 3:

   step $n$	1	2	3	4	5	6	7
   gaps	4809	1470	600	268	132	66	32
   observed fraction	0.2447	0.0990	0.0449	0.0210	0.0106	0.0053	0.0026
   $g_n/T_{n-1}$	0.2448	0.0988	0.0452	0.0217	0.0106	0.0053	0.0026

   The later gaps are 8, 6 and 4 at steps 8, 9 and 12.

2. Long orbits. $\frac12$ and $\frac1{21}$ pass every step up to 20000, selecting 10028 and 10006 of the first 20000 integers. The least value of $|2^nr_{n-1}-1|$ over $15\le n\le20000$ is about $2^{-13.3}$ for $\frac12$ and $2^{-13.5}$ for $\frac1{21}$, at steps 15999 and 2215 respectively. Both minima lie strictly between $2^{-14}$ and $2^{-13}$, as the integer enclosures certify. A gap at step $n$ requires $\sigma_n-1<2^nr_{n-1}-1<w_n$, with $w_n=2^{-n}/(1-2^{-n})$.

3. Near misses. Among the 6830 reduced fractions with $Q\le150$ that pass step 400 without terminating, the steps $15\le n\le400$ with $n>t+2$ and $|2^nr_{n-1}-1|<2^{t-n}$ number 2, 10, 22, 120 and 438 for $t=2,4,6,8,10$, against 1.7, 6.7, 26.7, 106.7 and 426.9 if each $2^nr_{n-1}-1$ were uniform on $[-1,1]$. Distinct fractions can have equal residuals at the same step, meaning $q-S_F=q'-S_{F'}$ for their selected prefixes. Their later greedy choices then coincide. The events are not independent: they come from 1, 3, 9, 36 and 139 distinct pairs $(n,r_{n-1})$, respectively. The comparison is descriptive and makes no asymptotic prediction for these dependent orbits.

4. Periodic supports. For each of the 1965 nonempty residue sets $R$ modulo $M\le10$ that are not periodic with a smaller period, the certified enclosure of $\sum_{n\ge1,\,n\bmod M\in R}w_n$ contains no fraction with denominator below $2^{1000}$. So no nonempty support periodic with period at most 10 has a rational sum with denominator below $2^{1000}$. Changing finitely many members adds to the sum a rational $\rho$ with odd denominator, so an eventually periodic infinite support with eventual period at most 10 and rational sum $a/b$ is excluded when the least common multiple of $b$ and the denominator of $\rho$ is below $2^{1000}$. The empty support is excluded from this census; its sum is zero.

5. Finite arithmetic witnesses. The reduced dyadic rational $q=4203/32768$ first meets a gap at step 17, after selecting exactly $F=\{4,5,6,7,8,10,11,13,14,16\}$. Its residual satisfies $L_{17,2}+E_{17,2}<q-S_F<w_{17}$, with the exact bounds of Lemma 8. Thus a dyadic denominator $2^J$ alone does not confine gaps to steps below $J$: here $J=15$. The same check reproduces the failed two-state extension and the actual truncated greedy carries at steps 106--108 in the multiplier construction.

6. Window rule. The rule of Lemma 12 gives bits through step 20000 and reproduces there the greedy digits of $\frac12$ certified by check 2; at every step $4\le n\le20000$ its digit equals the closed form of Lemma 12(b). With $f_m$ restricted to the multipliers of $C$, it reproduces the exact truncated greedy digits for $C=\{1,2,3,4,6\}$ and $C=\{1,\ldots,8\}$ through step 400, with the carries $1,2,3,1$ at steps 105 to 108 for the first set.

7. Finite seeds. The integer equation of Lemma 13 is checked for $C=\{1\}$, $\{1,2\}$, $\{1,2,3\}$, $\{1,2,3,4,6\}$, $\{1,2,3,4,6,12\}$ and $\{1,\ldots,8\}$ at $M=\max(2,\operatorname{lcm}(C))$. Four doubling blocks are constructed; every bit, carry, earlier divisor contribution and boundary carry is checked through $16M$. Exact greedy comparison runs through $\min(400,16M)$, with index 1 omitted. The stated seed at $M=12$ and its carry-$3$ transition are pinned, and an altered seed is rejected. These finite checks support the implementation of the extension; the extension proof is in Lemma 13 and the linked multiplier page.

8. Signed seed repair. The displayed seeds at $M=60$ for $C_4$ and $C_5$ have exact row sum $2^{59}-1$. Their finite carries obey (1) on the multiplier page and lie in $[1,|C|]$, ending at $1$, but at position $29$ the carry changes from $2$ to $1$. Check 8 recomputes the row identities, all finite carries, the transition at positions $28$--$30$ and the signed repair equation. The seed theorem supplies their infinite extensions. This excludes a nonnegative-carry repair invariant; it does not construct seeds for unbounded initial segments.

9. Multiplier sets. Every set $C$ with $1\in C\subseteq\{1,\ldots,12\}$, 2048 sets, has its half seed at $M=\max(2,\operatorname{lcm}(C))$, and for the 1542 with $M\le2520$ the whole-word iteration of the multiplier page collapses to it. For $C=\{1,2,3,4,6\}$ and $M=12$, the stable-pair proposition is tested at all 2048 targets below $2^{11}$, 427 of which give a nontrivial pair. Two multiplicity examples and one index-dependent family are checked to lose the half target in exact arithmetic. These finite checks construct no seed for an unbounded family of sets. The 169 nontrivial different-target pairs with first disagreement $k\ge3$ also check the mixed-carry rejoin, the ordinary-carry offsets and the subsequent forcing. Their terminal carries need not be $1$, so they do not test the half-target bound or prove collapse there.

10. Full-weight doubling blocks. Along the certified prefix of $\frac12$, the block identity and its two-candidate window of the multiplier page hold exactly at $H=2,4,\ldots,64$, and its gap separation holds for every word with $H\le7$. The value of $E_{16}$, the carry $c_{32}=2$, and the two explicit terminal-carry candidates and their residuals are checked, including $E_{32}<1$, $c_{64}=2$, which requires the larger candidate. A different-target control has both candidates fail. The exceptional endpoints $L=2,6$ in Lemma 14 give exactly the three displayed renewal identities by exhaustive subset checking. Bang–Zsigmondy and the written argument, not this finite check, exclude all other endpoints. No later doubling block is decided.

Limits of proposed full-weight certificates

These are scoped obstructions to certificates of the kinds tried here, not barriers to every route.

• Raw carries. By Lemma 6 the carries of any infinite support are unbounded, so no finite list of carry values, and no eventually periodic carry sequence, can certify membership.
• Dyadic denominators. Check 5 gives a gap at step 17 for a rational of denominator $2^{15}$. Parity excludes finite representations of nonintegral dyadics but does not exclude their later gaps. This witness refutes that proposed uniform cutoff, not the half target or Lemma 10's horizon, which also depends on the denominator of the finite subsum.
• Periodic supports. Check 4 above excludes the stated period and denominator ranges. It does not exclude larger periods, larger denominators, or periodic normalized carry rules for nonperiodic supports. The computed prefixes of $\frac12$ and $\frac1{21}$ each select about half of their indices; this proves neither an asymptotic density nor a structural rule.
• Averaging over many rationals. For $n\ge\log_2Q+2$ a level-$(n-1)$ cylinder has length $T_{n-1}\le\sigma_1 2^{1-n}<1/Q$ and contains at most one point of $\frac1Q\mathbb Z$, so the number of fractions $p/Q$ with a gap at step $n$ is a sum of single-orbit events. A bound by interval length plus one per occupied cylinder then permits every remaining fraction to fail. This is a limitation of that bound, not of every counting or Fourier argument. Poisson summation expresses the number of fractions $p/Q$ in the $2^{n-1}$ gaps of step $n$ through the transforms at the frequencies $hQ$, whose moduli are $2^{n-1}\prod_{k<n}|\cos(\pi hQw_k)|$ times the transform of one gap. A bound computed from these moduli alone holds equally for every translate of the union of gaps, since translation changes only the phases, and some translate puts a fraction $p/Q$ in a gap of step $n$. Such a bound is therefore at least $1$ at every step, and its sum over $n$ diverges; a counting argument for a fixed $Q$ needs the phases, which encode the positions of the individual orbits. Measure-class information alone (Lebesgue density, $L^p$ norms, Fourier coefficients of $\mathbf 1_\Sigma$) is unchanged on removing the countable set $\mathbb Q$ from $\Sigma$. Without additional pointwise information it cannot decide rational membership. The modified set need not be closed, so this observation does not rule out using the closed-set structure of $\Sigma$ together with analytic estimates.
• Fixed points. Each binary expansion $(\epsilon_n)$ of $t=\sum_n\epsilon_n2^{-n}\in[0,1]$ gives the point $\sum_n\epsilon_nw_n$ of $\Sigma$, and the order is preserved. The two expansions of an interior dyadic point of depth $n$ give two points $g_n$ apart, the ends of a gap, so either convention gives a map from $[0,1]$ to $\Sigma$ that jumps at every interior dyadic point. Writing the target as a fixed-point equation for $t$ and invoking a continuity theorem is therefore invalid; an invariant must avoid every jump. The finite whole-word iteration on the multiplier page has a separate obstruction: nested integer bounds can stabilize in a two-cycle around a valid seed. Its explicit witness changes the target from $2^{M-1}-1$. By the stable-pair proposition there, such a two-cycle coexisting with a seed forces a long run of equal digits beside a dyadic point, and without a seed a two-cycle always occurs. These are restrictions on a possible proof, not an exclusion of this route. At the half target, positivity and the terminal value $1$ give the additional mixed-carry bounds of that page, but no uniform contradiction is proved.
• Exact renewals. Lemma 14 classifies every identity $\frac{w_d}2=S_G+\frac{w_L}2$ with $d<L$ and $G\subseteq\{d+1,\ldots,L\}$. The half orbit can restart in this form only at $2$ and $6$, so this proposed infinite renewal chain is impossible. More generally, at a surviving half-prefix the reduced denominator of $r_L/w_L$ is at least $\lceil(L+2)/2\rceil$ for $L\ne6$. Finitely many fixed rational multiples of $w_L$ therefore do not give an alternative renewal chain. These conclusions concern exact terminal forms, not arbitrary half membership, approximate shadows, or the variable terminal carries of the multiplier page.
• Base two is critical. For an integer $t\ge2$, the sums of distinct $1/(t^n-1)$ form a Cantor set of measure $\lim_n2^nT^{(t)}_n$, where $T^{(t)}_n=\sum_{k>n}\frac1{t^k-1}$; the limit is $1$ for $t=2$ and $0$ for every $t\ge3$. The proof of Kovač and Tao's Theorem 2.3 shows that merged sums over several bases contain a nondegenerate interval under their hypothesis $\sum_k\frac1{t_k-1}>1$; for the single base $2$ that sum equals $1$, the boundary case, and the route cannot borrow that flexibility.
• Deforming the base. For $\frac12<x<1$, $\frac{x^n}{1-x^n}\le\sum_{k>n}\frac{x^k}{1-x^k}$ once $x^{n+1}\le2x-1$, so by Kakeya's criterion the sums of distinct $\frac{x^n}{1-x^n}$ have gaps at only finitely many levels. The pairs (base, sum) for bases in $[\frac12,x_0]$ with $x_0<1$ form a compact set, so $q\in\Sigma$ would follow from $q$ being such a sum along a sequence of bases tending to $\frac12$. But a gap of $\Sigma$ at step $n$ that contains $q$ stays a gap containing $q$ at every base close enough to $\frac12$, because the inequalities $w_k>T_k$ for $k\le n$ and the gap's defining inequalities are strict at $\frac12$. The number of levels with gaps grows without bound as the base tends to $\frac12$, so certifying membership this way means excluding gaps at more and more levels, the same infinite family of tests.

What composes

• Borwein's Theorem 1 and the other classes on the problem page give irrational values for particular supports; they constrain where a rational sum can come from but supply no membership.
• Kovač and Tao's Theorems 2.3 and 2.5 produce rational values with several bases or with shifted denominators. Their constructions use exactly the freedom that Problem 257 removes.
• Cook's note and his weighted-support claim are pending partial claims, used here as no premise. If the first holds, a support with rational sum has divergent reciprocal sum. The computed prefixes select about half of their indices, but no limiting density or reciprocal-sum divergence for their infinite continuations is proved here.
• The binary attacks on Problem 249 and Problem 251 concern other coefficient sequences; their comparison classes do not preserve $\theta_A$ for an arbitrary zero-one support, and nothing from them is used here.

Next steps and reopening

The route is blocked at Step 4. It reopens on any of the following.

• A structural invariant of the orbit of $\frac12$, or of another explicit rational, forcing $\{s_{n-1}\}\notin W_n$ for all large $n$. A renewal rule for the rational shadows of Lemma 10 counts only if it carries structure beyond Dirichlet's theorem: its argument does not guarantee renewal at near-dyadic states, including those necessarily preceding a gap. The dyadic roundings of the corollary supply an exact first shadow at step $J$.
• A rational whose infinite greedy support has a finite description (eventually periodic, automatic, or otherwise) together with a proof that the sum over it is that rational. Check 4 excludes the periodic supports with period at most 10 for denominators below $2^{1000}$.
• An argument that certifies a rational point of $\Sigma$ outside the finite subsums without following one orbit, using additional structure beyond the measure calculation or finite counts alone.

For truncated weights $\sum_{a=1}^J2^{-an}$, Lemma 13 gives a finite construction task: produce the seed at $M=\operatorname{lcm}(1,\ldots,J)$ for unbounded $J$. The doubling extension then supplies the entire infinite support for each such $J$, and Lemma 11's transfer proves Step 4. The common-multiple adjoining rule alone does not produce these sets. A proposed seed construction must solve the exact integer equation, not just a longer prefix or a bounded list of values of $J$. The necessary two-prefix interface restricts an induction adjoining the next multiplier; its continuation must solve the signed repair equation with the Boolean digit bounds. A sufficient invariant must permit negative carry changes and control their effects at later multiples. No such invariant is supplied. The exact residual-capacity bounds on the multiplier page give another test: exclude its first-failure interval at every position through $M/2$, for unbounded initial segments, while retaining every pending divisor contribution. Collapse of the whole-word iteration is another sufficient construction. Its stable-pair proposition and mixed-carry bounds leave a concrete test: exclude the coupled continuation after the first borrow block for unbounded $b$, or repair it to an ordinary seed while paying both boundary offsets and every later divisor contribution. Equality of the mixed carries at $2k$ does not discharge this test. The census of all multiplier sets inside $\{1,\ldots,12\}$ suggests proving the seed for every finite multiplier set at once, which would permit adjoining multipliers in any order. Its proof must use the bit coefficients of one fixed set: integer multiplicities and index-dependent sets can lose the half target. These are proposed constructions, not additional completed steps of the full proof.

Each of these interfaces implies gap avoidance along one orbit, the full orbit of $\frac12$ or its truncated orbit through $M/2$, and none is weaker than that avoidance. Selection at the full-weight doubling boundaries and exclusion of the first-failure interval are equivalent to it; the window rule, whole-word collapse and exclusion of the mixed-carry continuation imply it and are not known to be equivalent to it. These implication directions do not rule out proving one of the construction theorems. In particular, the multiplier page's explicit replacement for a fixed terminal carry keeps $E_H$ and chooses between $\lfloor E_H\rfloor+1$ and $\lfloor E_H\rfloor+2$. Its endpoint cases are proved, and its interior phase avoidance is unproved. The fixed carry-$1$ restart and the exact renewals of Lemma 14 cannot supply that missing argument. A successful invariant must derive the needed bounds from the half-rooted states and the divisor contributions; those bounds cannot be assumed merely because the variables describe the orbit.

The next discriminating test for the direct orbit route requires a proposed invariant. State its exact state variables, prove it contains a certified prefix of one rational, and prove that every successor both remains in the invariant and avoids $W_n$. A normalized state must also imply the original carry bound or residual tail bound and retain the divisor contributions of all earlier choices. Such a proof would discharge Step 4; failure would refute that particular invariant. For a renewal rule the test is concrete: express the shadow at $m_{j+1}$ through the shadow at $m_j$ and the digits selected in between, and bound its denominator and error well enough that Lemma 10 reaches $m_{j+2}$, without assuming that the intermediate states avoid near-dyadic positions. No invariant or renewal rule with these properties is supplied here. An invariant may instead be stated through Lemma 12; it must then show that the integers $N_{n-1}$ never come within the bound of Lemma 12(b) below a multiple of $2^{n-2}$. Extending the finite prefix or inspecting additional digits alone does not perform this test.