Denominator Periods, Rational-Value Constraints and Achievement-Set Geometry

The finite-support theorem.

For a finite nonempty $F\subseteq\mathbb{N}_{>0}$ and an integer $b\ge2$, write $$x_F(b)=\sum_{n\in F}\frac1{b^n-1}=\frac{N_F}{D_F} \qquad (\gcd(N_F,D_F)=1,\ D_F>0).$$ We use the standard trivial-modulus convention $\operatorname{ord}_1(b)=1$.

The order statement is , its reduced-denominator form is , coprimality is , and the growth clause is .

Structure.

Section  develops the arithmetic of Theorem . Section  collects what rationality would force on an arbitrary infinite support, which is the part of this note that quantifies over every support rather than sampling. Section  gives representative supports already known to give irrational values, grouped by the argument that reaches them. Section  treats the squarefree support, whose values are known at every power-of-two base and which two of the arguments used here provably cannot reach at any even base. Section  gives the unrestricted and support-restricted topology and measure classifications, followed by the exact $1/2$ and $1/21$ frontiers. Section  states what remains.

Integral scaled tails, and unboundedness.

Binary long division turns a hypothetical rational value into an integer recurrence. Fix an integer $v\ge1$ and a sequence $f\colon\mathbb{N}_{>0}\to\mathbb{N}$ of coefficients. Call an integer sequence $u\colon\mathbb{N}\to\mathbb{Z}$ a for $f$ with multiplier $v$ if $$u(n+1)=2u(n)-v\,f(n+1)\quad\text{for every }n\ge0, \qquad\text{and}\qquad u(n)=o(2^{n}).$$ The recurrence is one step of long division in base $2$: double the remainder, then pay out the next coefficient. The growth condition is what pins the solution down, since two solutions of the recurrence differ by $C\,2^{n}$ for a constant $C$, and only $C=0$ survives $u(n)=o(2^{n})$. For any coefficient sequence with $f(n)\le n$ — and $\operatorname{sc}_A(n)\le d(n)\le n$ qualifies, $d$ being the divisor function — the series $\sum f(n)2^{-n}$ is rational exactly when a tempered scaled-tail sequence exists for some $v\ge1$, and every such sequence is then the scaled tail $u(n)=v\sum_{j\ge1}f(n+j)2^{-j}$, so the orbit is unique rather than merely available (). The next theorem exhibits such an orbit for $X_A$ itself, with the coefficient sequence shifted past the power of $2$ in the denominator.

Checked as . The useful clause is exact: these scaled-tail states cannot remain bounded. In particular they cannot eventually cycle through a finite set of states.

Divisor coverage cannot have long gaps.

Say that $\operatorname{sc}_A$ has a of length $h$ at $N$ if $\operatorname{sc}_A(N+1)=\dots=\operatorname{sc}_A(N+h)=0$, that is, if no element of $A$ divides any of $h$ consecutive integers. At $A=\{2,3\}$, for instance, $\operatorname{sc}_A$ vanishes exactly on the integers prime to $6$, so every zero window has length at most $1$: one of any two consecutive integers is even.

Checked as . Read as a constraint on counterexamples, this rules out zero windows of any fixed positive proportion of $\log_2N$ once $N$ is large. The proof compares an exponential lower bound forced on the scaled tail with an upper bound for divisor sums that grows more slowly than any fixed power of $N$.

A reciprocal-sum lower bound.

Let $\operatorname{ord}_v(2)$ denote the multiplicative order of $2$ modulo an odd $v>1$, and write the reciprocal mass of $A$ for $\sum_{a\in A}1/a$.

Checked as . A finite support already supplies an instance: $A=\{2,3\}$ has $X_A=10/21$, so $c=0$, $v=21$, $\gcd(p,v)=1$ and $\operatorname{ord}_{21}(2)=6$, and the bound reads $\tfrac12+\tfrac13\ge\tfrac16$. The summability hypothesis is doing real work: a rational value with fixed odd denominator part $v$ and a convergent reciprocal sum requires mass at least $1/\operatorname{ord}_v(2)$. The statement does not apply when the reciprocal sum diverges, and it gives no positive lower bound uniform in $v$; it constrains the pair (support, denominator) jointly.

The critical dyadic case has a different endpoint.

Checked as . The two alternatives are an exact necessary consequence of a dyadic rational value. They do not contradict rationality by themselves: a contradiction requires separate proofs that the reciprocal terms are summable and that the mass is at most $1$. The formal nonsummability alternative also does not, by itself, assert a particular asymptotic law for the partial sums.

A Boolean–Möbius characterisation.

The preceding theorems give necessary conditions. The next statement is an equivalence, and is the sharpest description of rational-valued supports the development has. It removes the support from the description altogether, which is possible because the support can always be read back off the incidence sequence: with $\mu$ the Möbius function, $*$ Dirichlet convolution $(g*h)(n)=\sum_{d\mid n}g(d)h(n/d)$ and $\mathbf 1_A$ the indicator of $A$, the identity $\operatorname{sc}_A=\mathbf 1_A*1$ inverts to $\mu*\operatorname{sc}_A=\mathbf 1_A$.

For $p\in\mathbb{Z}$, $q\ge1$ and $U\colon\mathbb{N}\to\mathbb{Z}$, set $$Q_U(0)=0,\qquad Q_U(n)=\frac{2U(n-1)-U(n)}q\quad(n\ge1).$$ Call $U$ an for $(p,q)$ if $$\begin{gathered} U(0)=p,\quad U(N)>0,\quad U(N)\le q\bigl(2\sqrt N+4\bigr)\quad(N\ge0),\\ q\mid 2U(N)-U(N+1)\quad(N\ge0),\qquad (\mu*Q_U)(n)\in\{0,1\}\quad(n\ge1). \end{gathered} \tag{2}\label{eq:bmc}$$ The divisibility condition makes $Q_U$ integral; the last condition says that Möbius inversion of the quotient is the indicator of a set.

Checked as . The finite example $A=\{2,3\}$ makes the definition concrete. Here $p/q=10/21$, and $$Q_U(1),\ldots,Q_U(6)=0,1,1,1,0,2,\qquad U(0),\ldots,U(6)=10,20,19,17,13,26,10.$$ Möbius inversion gives $(\mu*Q_U)(1),\ldots,(\mu*Q_U)(6)=0,1,1,0,0,0$, the indicator of $\{2,3\}$ through that range. This finite calculation illustrates the correspondence; the admissible scaled-tail sequence itself is an infinite object, and the finite support is not a counterexample to Problem .

One boundary on the equivalence. The support it produces is not required to be infinite, and a finite support supplies an admissible scaled-tail sequence for its own value, so existence of such a sequence is not by itself a counterexample to Problem . What the equivalence changes is the search space, not the difficulty.

Combined constraints on a rational counterexample.

Taken together: a counterexample to Problem  would have an unbounded scaled-tail sequence obeying an exact linear recurrence and sublogarithmic divisor-coverage gaps. If its reciprocal sum converged and its reduced odd denominator part were greater than $1$, it would also satisfy Theorem ; at a dyadic rational value it would instead satisfy Theorem . Theorem  reconstructs every rational value, but by itself does not force the reconstructed support to be infinite. These qualifications matter: the statements have different hypotheses, and no jointly contradictory combination has been proved.

The values are known at every power-of-two base

Duverney and Tachiya’s Corollary 1.2, applied with $E$ the primes, $s=2$ and $\ell=1$, gives $L=\operatorname{lcm}(1)=1$ and the admissibility condition $|q|\le2$. The exponent $j$ in their displayed family remains arbitrary. Thus, for every $h\ge1$, their Example 1.1 says that the numbers $$1,\qquad \sum_{n\ge1}\frac{|\mu(n)|}{2^{jn}-1}\quad(1\le j\le h)$$ are linearly independent over $\mathbb{Q}$  [2]. Since $|\mu|$ is the indicator of the squarefree integers including $n=1$, and $A_{\mathrm{sf}}$ omits only $n=1$, whose weight at base $2^j$ is $(2^j-1)^{-1}\in\mathbb{Q}$, $$X_{A_{\mathrm{sf}}}(2^j) =\sum_{n\ge1}\frac{|\mu(n)|}{2^{jn}-1}-\frac1{2^j-1},$$ and rational translation preserves the joint linear independence with $1$.

Two boundaries on that. It is a citation, not a formalisation: nothing in this development proves it. The admissibility condition $|q|^L\le s$ at $s=2$ allows only $|q|=2$, so this citation covers the bases $2^j$ and does not cover bases that are not powers of $2$; this note proves no result about those remaining values.

Two block-certificate hypotheses have no instance

Fix $b\ge2$ and a nonnegative coefficient sequence $f$. The two hypotheses used below are best printed rather than named. For every precision $q\ge1$, they ask for $N,K,L,C\in\mathbb{N}$ with $K\le L$ satisfying the common conditions $$\begin{gathered} \sum_{r=K+1}^{L}f(N+r)b^{L-r}\le C,\qquad \exists t\in\mathbb{N}:\ f(N+L+1+t)>0,\\ q(C+N+L+2)<b^L, \end{gathered} \tag{3}\label{eq:block-common}$$ together with one of the two first-block conditions $$\begin{array}{ll} \text{digitwise:}& b^r\mid f(N+r)\quad(1\le r\le K),\\[2mm] \text{weighted:}& b^K\mid\displaystyle\sum_{r=1}^{K}f(N+r)b^{K-r}. \end{array} \tag{4}\label{eq:block-first}$$ Thus the data $N,K,L,C$ may depend on $q$. The parity of Theorem  refutes both alternatives for the squarefree incidence sequence at every even base.

The two engine-facing nonexistence statements are checked directly as and . The displayed proof is retained to expose the empty- and one-position-block boundary cases rather than leaving the scope hidden behind the interfaces.

The obstruction is a normalisation, not the value

Corollary  is a statement about two arguments failing on a value that Corollary  shows to be irrational. The question it raises is therefore not whether the value can be reached, but what the failure is a property . It is a property of where the support starts.

Adjoin $1$ to the support and write $A_{\mathrm{sf}}^{+}=A_{\mathrm{sf}}\cup\{1\}$, the full squarefree support. Then $$X_{A_{\mathrm{sf}}^{+}}(b)-X_{A_{\mathrm{sf}}}(b)=\frac1{b-1}\in\mathbb{Q} ,$$ so the two supports pose the same irrationality question at every base, while the divisor incidence changes from $2^{\omega(n)}-1$ to $2^{\omega(n)}$ — from odd to even at every $n\ge2$. The parity obstruction of Corollary  evaporates under a shift that provably cannot change the answer. The shifted coefficient identity and the exact equivalence of the two irrationality questions are checked as and . More is true: the shifted first-block condition at base $2$ asks for $2^{r}\mid 2^{\omega(N+r)}$, that is $\omega(N+r)\ge r$ for $1\le r\le K$, and a Chinese-remainder construction reserving $r$ fresh primes for each shift $r$ supplies such an $N$ for every $K$. This is checked both as the arithmetic block theorem and in the engine-facing form .

It does follow that either argument certifies the shifted support: the opening block is one of the conditions listed above, and the middle bound and the arithmetic inequality are untouched by this observation. What does follow is a methodological point: an obstruction stated against a coefficient sequence can be an artefact of the normalisation chosen for that sequence. Before a no-go result is reported as a property of a problem, the coordinate it is stated in should be varied by a transformation the problem is known to be invariant under. Here the transformation is adding one rational number, and it removes the obstruction entirely.

The same invariance holds for every finite change, not only this one. If $A\mathbin{\triangle}B$ is finite, choose $M$ above all of its elements. The two support series then have the same tail beyond $M$, while each omitted prefix is rational. The two directions of this argument are exactly the checked prefix lemmas and . Thus $X_A(b)$ is irrational if and only if $X_B(b)$ is irrational for every integer $b\ge2$. The present shift is the smallest instance of a general checked finite-change principle.

This is the second time in this note that a boundary turns out to belong to the method rather than to the mathematics. In Remark  the terminating alternative of the signed periodic dichotomy is empty, and a theorem of Luca and Tachiya is what shows it; here a parity obstruction survives only until the support is shifted by one element. In each case the correction came from outside the development — once from the literature, once from asking what the statement was invariant under. A formalised no-go result carries exactly the authority of its hypotheses, and its hypotheses include the coordinates it was written in.

Geometry after restricting the allowed exponents

The injectivity statement has a geometric completion for every set $J\subseteq\mathbb{N}$ of allowed digit positions. Let $$\mathcal A_J= \left\{\sum_{k\in J}\varepsilon_kw_{k+1}: \varepsilon_k\in\{0,1\}\right\}.$$ Formally, the allowed strings form the , which is . Its range is the ; the range and image descriptions agree by . Consequently $\mathcal A_J$ is and . It lies inside $\mathcal A$ by and is therefore . If $J$ is infinite, the digit space is ; injectivity transfers that property to , so the closed set is . Thus every infinite allowed support gives a compact perfect nowhere-dense set with a unique selector for each point.

The metric classification is exact. The summability lemma , and changing one coordinate changes the value by precisely its signed weight by . If $k\notin J$, allowing $k$ splits the new set into $\mathcal A_J$ and its translate by $w_{k+1}$ . The two pieces are , so adjoining one coordinate . At $J=\mathbb{N}$, the restricted set is the full set $\mathcal A$ .

For finite $F$, the division-free identity and its solved form . Monotonicity under enlarging $J$ is . It traps a set with infinitely many forbidden coordinates below finite-codimension faces of arbitrarily small dyadic measure, yielding . The two cases are assembled in . This theorem classifies the size of the value set generated inside any prescribed family of exponents. It does not classify the arithmetic nature of its individual points and therefore does not settle Problem .

The value $1/2$

A rational point of $\mathcal A$ attained by an infinite support refutes Problem . The distinguished candidate is $1/2$, and Theorem  gives a self-contained pair of exact alternatives.

The greedy expansion of $1/2$ is an exact finite computation as far as one cares to take it. Through level $21$ it takes the exponents $$2,\ 3,\ 6,\ 7,\ 14,\ 20,\ 21$$ and skips the others, the skipped runs being $\{1\}$, $\{4,5\}$, $\{8,\dots,13\}$ and $\{15,\dots,19\}$; every level through $21$ is survived. The second condition below asks whether the list of skipped exponents is infinite, and no finite computation answers that.

In the statement, $u=(u_1,\dots,u_d)\in\{0,1\}^{d}$ is a finite prefix, $V(u)=\sum_{n\le d}u_nw_{n}$ is its value, and $T_n=\sum_{k>n}w_{k}$ as above.

Checked as , , , , , and .

The formal source also proves equivalent terminal-bit and unbounded skipped-rank formulations in its internal half-cylinder seam coordinate; the two middle source links above are those versions. Their definitions are not needed for the self-contained greedy and fatal-gap statement used here. The two sides are asymmetric: non-membership is witnessed by a fatal gap and is therefore semi-decidable, whereas membership is an infinite condition. Computing further can only raise a lower bound on where a fatal gap could occur; it cannot establish survival.

One natural route to $1/2$ is closed. The Boolean support selected by the negative values of the Möbius function has value exactly $1/2$ plus the positive Möbius tail, hence at least $1/2+1/63$ (, ). That closes the sign-truncation construction; it excludes no other support.

A second rational target

The target $1/21$ has a useful property that does not depend on any finite search.

This is checked by . The reduced denominator $21$ has doubling order $6$, so the exact denominator–lcm identity forces every selected exponent to divide $6$. The lower bound $n\ge2$ leaves only $2,3,6$, and the remaining eight subsets are discharged by exact rational arithmetic. Thus any representation of $1/21$ along this rank-$\ge2$ route, if one exists, must be infinite. The theorem proves nontermination only: it does not prove that $1/21$ lies in $\mathcal A$.

A separate arithmetic module isolates the primitive cone that appears in one candidate expansion route. Write $$\mathcal P_{23}(n)= \{(p,q)\in\mathbb{N}_{>0}^2:\gcd(p,q)=1,\ 2p+3q=n\}.$$ Every $n\ge11$ has a member of $\mathcal P_{23}(n)$ (), whereas $\mathcal P_{23}(10)$ is empty (). Multiplicity begins immediately at $11$, and every $10k$ with $k\ge2$ has at least two distinct primitive representations (, ). These are exact Diophantine facts, not a counterexample construction. In particular the recurring collisions show why primitive-cone coverage is not already a Boolean expansion: the missing step is a proof that the relevant integer multiplicities can be converted into zero–one reciprocal-Mersenne digits without changing the value.

The canonical $1/21$ frontier

The finite-support obstruction makes $1/21$ useful only if the infinite problem is stated in coordinates that retain the actual greedy orbit. Let $$r_N=\operatorname{greedyMersenneRemainder}(1/21,N)$$ be the real remainder after rank $N$, and let $$Q_N=\left\lfloor\frac{2^N}{21}\right\rfloor-P_N,$$ where $P_N$ is the integral numerator of the binary divisor-incidence prefix generated by that same greedy support. Thus $Q_N$ is the nonnegative denominator-$21$ defect after the six-periodic residue of $2^N\operatorname{mod}21$ has been removed. These are not independent models: the formal source proves an exact identity relating $Q_N$, $2^Nr_N$ and a finite-prefix divisor tail ().

The sharper route works directly with the denominator-$21$ greedy orbit. At even depth $2R$, write $$\begin{aligned} D_R&=\operatorname{twentyOneEvenQuotientGreedySupport}(R),\\ s_R&=\operatorname{twentyOneEvenQuotientGreedyRemainder}(R). \end{aligned}$$ Here $D_R\subseteq\{2,\ldots,R\}$ is the Boolean support obtained by descending integer-greedy selection on the exact scaled quotient weights, and $s_R$ is its terminal scalar remainder (, ). Write $\mathcal F_{21}$ for TwentyOneFatalAlignedBranch, the explicit branch in which the real greedy orbit has a fatal witness, only finitely many skipped exponents (hence cofinite eventual selection), eventual quotient/rational-greedy alignment, and eventual occupation of every doubling block.

The equivalence is ; the closed-row compactness step is ; and the eventual affine regime is . Explicitly, on $\mathcal F_{21}$ the quotient orbit is eventually strictly supercapacity and loses its last Boolean choice: if $\tau_R$ is the periodic target pulse and $\pi_R$ the divisor pulse generated by $D_R$, then for every sufficiently large $R$, $$s_R>2^R,\qquad D_{R+1}=D_R\cup\{R+1\},\qquad s_{R+1}=4s_R+\tau_R-\pi_R-(2^{R+1}+1).$$ The denominator-specific separation theorem additionally proves that every closed Boolean quotient row is exactly the canonical quotient-greedy row, including exact saturation at the boundary (). An aligned crossing from saturation into strict supercapacity forces a missing canonical ancestor at one-third or two-thirds scale and a real skipped exponent with square-root-bounded defect (, ).

These conclusions remove alternative late Boolean branches; they do not exclude the full fatal/cofinite/aligned branch. On that branch the permanent affine-supercapacity recurrence is forced, so a contradiction of the recurrence would be sufficient, but the recurrence alone is not the exact membership endpoint. It is conditional, not a contradiction: synthetic pulses can sustain such an affine recurrence without being the actual divisor pulse.