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The equality cases $P_t(\mathbb{N})=\tfrac12$ for the deconvolved sum-of-digits measures

T0 review · 0 major / 4 minor · reviewed 2026-08-28 · deepseek-v4-flash

Pith's one-line read The deconvolved sum-of-digits measure assigns exactly half of its mass to the nonnegative integers precisely when the inner binary word is saturated.

desk verdict Settles Cheng's saturation problem with a clean characterization; the proof checks out, with only a couple of presentational gaps. read the letter →

arxiv 2608.24948 v1 pith:3YJIIU2Q submitted 2026-08-24 math.NT math.COmath.PR

classification math.NTmath.COmath.PR MSC 11A6360G4005A0505A20
keywords sumofdigitsmedianpropertybinarytreesstoppedrandomwalkssubwordordersaturatedwordsasymptoticdensitiesexpansion
verification ladder T0 review T1 audit T2 compute T3 formal

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The reading

This paper solves the saturation problem for the deconvolved sum-of-digits measures: it gives a complete, finite characterization of the odd integers $t$ for which the probability measure $P_t$ assigns exactly mass $1/2$ to the nonnegative integers. Writing the odd integer as $t=(1w1)_2$, the paper proves that $P_t(\mathbb{N})=1/2$ if and only if the inner binary word $w$ is saturated, meaning that in the block decomposition $w=1^{a_0}01^{a_1}0\cdots01^{a_k}$ with $k$ zeros, every block of ones, including the first and last, has length at least $k$. It also proves that every non-saturated $w$ gives a strict excess, with $P_w(\mathbb{N})\ge 1/2+2^{-2\min(\ell_0(w),\ell_1(w)+1)-1}$. The interest is that the median inequality $P_t(\mathbb{N})\ge1/2$ lies behind the long-standing sum-of-digits median conjecture; identifying the equality cases turns that inequality into a precise, finitely checkable combinatorial statement and shows that equality is rare.

What carries the argument

The load-bearing object is the tree representation of the measures: $P_w$ is the law of a fair random walk stopped when it first leaves the internal nodes of a recursively grown binary tree $T_w$, and the internal node language of $T_w$ is the set of scattered subwords, obtained by deleting some digits, of the reversed word $\overleftarrow{w}$. The paper writes the excess $P_w(\mathbb{N})-1/2$ as a finite sum $\sum_{m\ge1}2^{-2m-1}\Delta_m^-(w)$, where $\Delta_m^-(w)=2|B_{2m-1}^{-1}(w)|-|B_{2m}^0(w)|$ and $B_n^x(w)$ is the number of subwords of $w$ of length $n$ and balance $x$. An insertion-counting identity then yields the key equivalence: $P_w(\mathbb{N})=1/2$ iff every subword $v$ of $w$ with $\ell_0(v)=\ell_1(v)+1$ has the property that inserting a single digit $1$ into any slot still gives a subword of $w$.

What would settle it

Evaluate the finite leaf sum of Corollary 4.8 for a saturated word with two zeros, for instance $w=11011011$: the theorem predicts $P_t(\mathbb{N})=1/2$ exactly, so any deviation falsifies it. Conversely, evaluate the same finite sum for the non-saturated word $w=1001$, whose middle block of ones between the two zeros has length $0<2$; the theorem predicts a value strictly larger than $1/2$.

Watch

Extended reading notes

Core claim

Theorem 1.2 is the paper's central claim: for $w\in\{0,1\}^*$ and $t=(1w1)_2$, $P_t(\mathbb{N})=1/2$ if and only if $w$ is saturated. Saturation means that when $w$ is written as $1^{a_0}01^{a_1}0\cdots01^{a_k}$ with exactly $k$ zeros, every $a_i$ satisfies $a_i\ge k$; equivalently, every zero is isolated and every maximal run of ones, including the first and last runs, has length at least the number of zeros. The proof establishes both directions: for saturated $w$, every subword $v$ with one more zero than ones remains a subword after inserting a single digit $1$ in any slot, which forces all correction terms in the formula for $P_w(\mathbb{N})$ to vanish; for non-saturated $w$, the paper constructs a short deficient subword whose forbidden insertion contributes a positive correction, giving the explicit lower bound.

Load-bearing premise

The characterization rests on the previously established identification of $P_w$ with the law of a stopped random walk on a recursively generated binary tree, together with an unproved combinatorial observation about block profiles that is used in both directions of the proof; if either of these failed, the saturation criterion would not control the measures.

Editorial extensions

If this is right

  • The equality set $\{t:P_t(\mathbb{N})=1/2\}$ has natural density zero, since the saturated words are exponentially sparse; the shortest saturated word with $k$ zeros has length $k(k+2)$.
  • For every non-saturated $w$, the strict inequality $P_w(\mathbb{N})>1/2$ holds with the explicit gap $2^{-2\min(\ell_0(w),\ell_1(w)+1)-1}$, extending the earlier lower bound beyond the $0$-initial case.
  • Corollary 1.3 gives an exact count of saturated words of each length, $E(\ell)=\sum_k\binom{\ell-k(k+1)}{k}$, so the equality cases are fully enumerated, not merely described qualitatively.
  • The saturation criterion is checkable in linear time from the run-length profile of $w$, so deciding whether a given odd integer is an equality case requires no numerical evaluation of the measure.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

  • The paper does not state the negative-side analogue, but the complement symmetry of the subword counts implies that $P_w(-\mathbb{N})=1/2$ should hold exactly when the complemented word $\overline{w}$ is saturated; this is a direct corollary of the same machinery.
  • The insertion-counting proof suggests a general rule: for any tree-defined stopped random-walk measure, the median defect is controlled by the shortest subword with one excess zero, so the exponential constant in lower bounds is set by a minimal deficient subword rather than by the whole tree geometry.
  • Because saturated words are exponentially rare, the strict median inequality $P_t(\mathbb{N})>1/2$ holds for a set of parameters of density one; the paper notes the zero-density conclusion but does not phrase it this way.
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Formalized claims in Lean

  1. Claim #1: Theorem 1.2 is the paper's central claim: for $w\in\{0,1\}^*$ and $t=(1w1)_2$, $P_t(\mathbb{N})=1/2$ if and only if $w$ is saturated. Saturation means that when $w$ is written as $1^{a_0}01^{a_1}0\cdots01^{a_k}$ with exactly $k$ zeros, every $a_i$ satisfies $a_i\ge k$; equivalently, every zero is isolated and every maximal run of ones, including the first and last runs, has length at least the num

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Referee Report

0 major / 4 minor

Summary. The paper studies the family of finitely supported measures P_t obtained by deconvolving the sum-of-digits correlation measures μ_t. For an odd integer t=(1w1)_2, the main result (Theorem 1.2) is that P_t(N)=1/2 if and only if the inner word w is saturated, i.e. in the block decomposition w=1^{a_0}0...01^{a_k} with k zeros every block length satisfies a_i≥k. The proof passes through the tree representation of [9] and the subword description of node languages from [2], reduces P_w(N) to an insertion-counting series, and shows that the deficit terms vanish exactly in the saturated case. The paper also extends Cheng's lower bound to all non-saturated w and gives an explicit enumeration formula for saturated words.

Significance. The result resolves completely the saturation problem left open in [2] and previously studied only numerically. The characterization is clean and testable, and the proof is elementary and mostly self-contained after Section 4, with explicit verifiable identities (Proposition 5.1, Corollary 5.3, Theorem 5.8). The strengthened lower bound for non-saturated words is a quantitative bonus. I found no load-bearing mathematical error; the main theorem is sound, assuming the cited tree representation of [9] is correct. The only in-paper gap is the unproved Observation 6.1, which I verified; supplying its short proof is a minor revision.

minor comments (4)
  1. [Section 6, Observation 6.1] This observation is asserted with the remark that the proof is an easy exercise, but it is used in both directions of Theorem 1.2: part (1) in the sufficiency argument and part (2) in Lemma 6.2. I verified the statement: part (1) follows by greedily embedding the blocks of u into the blocks of w, each of length at least k, and part (2) follows because an embedding of a word with exactly k zeros must send the j-th zero of u to the j-th zero of w. Please include this short proof (or at least a detailed sketch) so that the main theorem is fully supported in the manuscript itself.
  2. [Section 5, proof of Theorem 5.2] In the proof of the second inequality in (5.4), the complement map sends B^x_n(w) to B^{-x}_n(\bar w), so the displayed chain should involve the complement of w rather than w itself; as written, the notation is confusing. Please correct this.
  3. [Section 6, first line] There is a typo: 'We start with twe following' should be 'We start with the following'.
  4. [Section 5, Corollary 5.9] The statement of Corollary 5.9 uses P_t(N) although the hypothesis is expressed for w; for consistency with the rest of the paper, either write P_w(N) or recall the identification t=(1w1)_2.

Circularity Check

0 steps flagged · score 0.0 of 10

No significant circularity: the saturation characterization is derived from a self-contained insertion-counting argument; the cited tree framework is scaffolding, not a forced input.

full rationale

The derivation of Theorem 1.2 is self-contained after the imported tree/subword setup. The paper does not fit a parameter and then call it a prediction: P_w(N) is expressed by the telescoping identity (3.5)/(4.12), with Delta^-_m(w) a genuine count of subwords, and Corollary 5.5 reduces the equality P_w(N)=1/2 to an insertion property. Theorem 1.2 then follows from Observation 6.1, an elementary subword-block lemma that is asserted without proof (Section 6, 'the proof is an easy exercise') but is independent of P_t and is easily verified: saturated block profiles permit embedding of any word with at most k zeros and k ones, and with equal zero counts the block-wise comparison c_j <= a_j is necessary and sufficient. Lemma 6.2 and Corollary 6.3 are derived from this lemma and Corollary 5.9, not assumed. The only same-author citations are [9] for the recurrences (2.1), the tree growth (2.3), and the identification P_w = L(S_{tau_w}) (2.8); these are prior structural results with stated assumptions that do not include the saturation criterion, and the combinatorial Proposition 4.4 from [2] is re-proved in Section 4. Thus the claimed characterization is not equivalent by construction to any input; no circular step is present.

Assumptions & free parameters 0 free parameters · 3 assumptions · 0 invented entities

The paper introduces no free parameters and no invented entities. The central claim depends on the prior structural encoding of P_t via trees, imported from [9], plus standard subword combinatorics that the paper proves or cites.

assumptions (3)
  • domain assumption P_1=delta_0, P_{2t}=P_t, P_{2t+1}=Phi(P_{t+1},P_t), and mu_t=mu_1*P_t (eqs. (2.1), (2.2)).
    Imported from [9, Corollary 7]; these recurrences define the measures whose median property is studied.
  • domain assumption For the tree family defined by (2.3), the law of the stopped fair random walk equals P_w (eq. (2.8), [9, Theorem 11]).
    Imported from [9]; the entire saturation theorem is stated for these tree measures.
  • standard math Cylinder probabilities and fair increments: P([u])=2^{-ell(u)} and S_n=b(u) on [u].
    Standard product-measure facts used in Sections 2 and 3.

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Pith. "Pith review of The equality cases $P_t(\mathbb{N})=\tfrac12$ for the deconvolved sum-of-digits measures." pith.science (2026). https://pith.science/paper/3YJIIU2Q

@misc{pith2026260824948,
  author       = {Pith},
  title        = {Pith review of: The equality cases $P_t(\mathbbN)=\tfrac12$ for the deconvolved sum-of-digits measures},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/3YJIIU2Q}},
  note         = {Machine review of arXiv:2608.24948}
}
abstract

Let $s(n)$ denote the number of ones in the binary expansion of an integer $n\in\mathbb{N}$, and let $\mu_t$ be the probability measure on $\mathbb{Z}$ defined by the asymptotic densities of the level sets of the function $\mathbb{N}\ni n\mapsto s(n+t)-s(n)\in\mathbb{Z}$. Let $P_t$ be the family of finitely supported measures defined by the convolution $\mu_t=\mu_1*P_t$. Recently, Tarlowski (2026) has shown that the family $P_t$ may be represented as a recursively grown binary tree $T_t$, and that the Cusick's conjecture - $\mu_t(\mathbb{N})>\frac12$, $t\in\mathbb{N}$, - follows from the asymmetry property of the family $T_t$, which was posed there as an open problem. Next, Cheng (2026) has provided the combinatorial description of the family $T_t$ in the language of principal subsequence ideals, and proved both conjectures. Both of these problems are directly related to the problem of determining the zeros of the function $\mathbb{N}\ni t \mapsto P_t(\mathbb{N})-\frac12\in[0,\tfrac12]$, a problem left open by Cheng (2026) as a saturation problem, and previously analyzed only numerically. In this paper we solve this problem completely. Writing an odd integer $t\ge3$ as $t=(1\,w\,1)_2$ with $w\in\{0,1\}^{\star}$, we show that $P_t(\mathbb{N})=\frac12$ if and only if $w$ is \emph{saturated} in the following sense: in the block decomposition $w=1^{a_0}\,0\,1^{a_1}\,0\cdots0\,1^{a_k}$ with exactly $k$ zeros, every block of "1" satisfies $a_i\ge k$. Additionally, we show that the lower bound for $P_t(\mathbb{N})$ established by Cheng for $0$-initial words holds true for all non-saturated words.

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Works this paper leans on

9 extracted references · 8 canonical work pages

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