{"id":"adb73e24-b749-423f-96f6-31784836c31d","arxiv_id":"2412.02337","paper_version":2,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":7.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":0,"one_line_summary":"A positive algebraic irrational number is normal to base b if and only if a certain averaged sum of Riemann zeta values over vertical arithmetic progressions vanishes for every integer k≥0.","lead":"This paper links the normal digits of algebraic numbers, such as the square root of 2, to averages of the Riemann zeta function. It proves that an algebraic irrational is normal to a base exactly when a zeta-based sum tends to zero.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"No load-bearing defect found; the Ridout step is correctly applied and the ψ1 endpoint in Corollary 1.2 is a harmless presentational slip.","rationale":"The central claim is an exact equivalence between normality of algebraic α and vanishing of the stated zeta averages. The proof hangs on Theorem 1.1, which separates cleanly into k>0 and k=0 cases. For k>0, Theorem 2.2 has error O(log log N), so transferring the vanishing of zeta averages to vanishing of ψ_{k+1}(b^h α) averages is immediate. For k=0, the boundary term G_N must be o(log N); the paper uses Ridout's theorem exactly where needed, and the uniformity in h is supplied by the theorem's fixed-prime-set denominator statement. I checked the exponent in the derived lower bound: ||b^h α|| ≫_γ b^{−γh} follows by multiplying the rational-approximation lower bound by q=b^h, and the subsequent dyadic-style h-splitting proves G_N = o(log N) correctly. The remaining mathematical steps in the Euler-Maclaurin and stationary-phase portions are consistent with the stated error terms, including the optimized (log N)^{2/3}(log log N)^{5/9} error. The only real defect I see is in the wording of Corollary 1.2's proof: ψ1 has a discontinuity/non-polynomial value at integers, so the phrase that each ψ_k is a polynomial is inaccurate, and the passage from Weierstrass polynomial averages to individual ψ-averages needs a one-line Riemann-integrability justification. This does not affect the theorem because α is algebraic irrational and b^h α is never an integer. The reader's conditional verdict is therefore understandable but the concern is not load-bearing; the paper would benefit from a small presentational patch, not a change to the central argument.","tokens_in":26665,"tokens_out":34279,"duration_ms":377650,"concrete_test":"Re-derive the final step of Theorem 2.1 in Section 9: multiply the un-normalized identity ∑_{1≤|n|≤N} ζ(2πidn)e(θn)/n = ∑_{0<h<d log N}∑_{1≤|m|≤de^{−(θ+h)/d}N} e(me^{((θ+h)/d)})/m + O(...) by (2πi)^{-1}, then apply Lemma 6.4 and the bound |sin πx| ≥ 2‖x‖. If the resulting expression is exactly (2.2), the factor and ψ1 endpoint are confirmed; if a factor of 2πi persists, Theorem 2.1's statement would need correction.","verdict_should_be":"UNCHANGED","load_bearing_attack":"Scrutiny of the chain Corollary 1.2 ⇐ Theorem 1.1 ⇐ Theorems 2.1/2.2 shows that the only point at which the equivalence can fail is the k=0 boundary term G_N in Theorem 2.1. The proof controls G_N through Ridout's theorem (Theorem 2.3), and the application is legitimate: with q=b^h, Ridout gives |α−a/q| ≫_γ q^{−1−γ} for all h, so ||b^h α|| = q·|α−a/q| ≫_γ b^{−γh} uniformly in h. The h-range split then gives G_N = O(1)+O(γ log N), hence G_N = o(log N). I found no circular or omitted step in the stationary-phase argument strong enough to threaten the claim. The one genuine inaccuracy is in Corollary 1.2: ψ1 is not a polynomial on [0,1] because of its special value at integers, and the reduction from polynomial averages to ψ-averages should cite Riemann integrability of the periodic Bernoulli functions. Since b^h α is never an integer for algebraic irrational α, the argument survives by working with B_1({x}) on the sequence. This is presentational, not load-bearing.","agreement_with_reader":"partial"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper proves an asymptotic identity (Theorem 1.1) relating weighted partial sums of the Riemann zeta function at the arithmetic progression s = -k + 2πin/log b to averages of periodic Bernoulli polynomials at b^h α. For positive algebraic irrational α the error is o(log N), and the authors deduce Corollary 1.2: α is normal to base b if and only if all the displayed zeta averages vanish. The proof is built on two general results (Theorems 2.1 and 2.2) established through the functional equation, Euler-Maclaurin summation, van der Corput estimates, stationary phase, and Ridout's theorem for the k=0 boundary term.","tokens_in":26942,"tokens_out":43236,"duration_ms":422175,"significance":"If correct, this is a striking and substantial contribution: it converts the notoriously intractable property of digit normality for algebraic numbers into concrete analytic vanishing conditions for zeta values on vertical arithmetic progressions. The paper is largely self-contained, the error terms are optimized explicitly, and the application of Ridout's theorem to control the k=0 boundary term is legitimate. The results also give clean corollaries for rational powers and for integer α. However, one step in the proof of the general identities uses an inequality that is false as stated, so the main theorems are not yet fully established.","major_comments":[{"comment":"The uniform bound |∑_{a≤|m|≤b} e(mx)/m| ≪ 1 is false. For example, with a=1, b=N, and x=1/N, the sum equals 2∑_{m=1}^N cos(2πm/N)/m, which is asymptotic to 2 log N. Lemma 6.4 bounds the sine partial sums, not the symmetric exponential sums. This inequality is used immediately after (9.4) to discard the interval H2(1)-2d ≤ h < dλ+θ at a cost of O(log log N), and that step is needed for the proofs of both Theorems 2.1 and 2.2. A replacement estimate is required; for k=0 the partial sums can grow logarithmically, and Theorem 2.1 makes no Diophantine assumption on θ or d.","section":"Section 9, Eq. (9.3)"},{"comment":"The proof states that every real polynomial on [0,1] can be represented as a finite linear combination of the ψ_k because 'each ψ_k is a polynomial of degree k'. This is false for k=1: ψ1 is defined to vanish at integers, so it is not identical to the polynomial B1(x) on all of [0,1]. The argument can be repaired by noting that for algebraic irrational α, the numbers b^h α are never integers, so ψ1(b^h α) = B1({b^h α}) on the sequence in question; the proof should state this explicitly.","section":"Corollary 1.2, proof"}],"minor_comments":[{"comment":"The paper uses e(x) for e^{2πix} but also writes e^{(h+θ)/d} for the natural exponential. This double use is very confusing in formulas such as e(me(θ-h)/d) in Section 9; exp(...) should be used for the natural exponential.","section":"Throughout"},{"comment":"The heuristic sketch knowingly chooses η=0 in Lemma 3.2 although the lemma requires η>0. The eventual rigorous argument does not use this choice, but the sketch would be clearer if it noted that the rigorous proof supplies the missing η parameter.","section":"Section 3, heuristic"},{"comment":"The two displayed forms of the theorem in (2.1) and (2.2) would be easier to read if the notation for the natural exponential in the summation limits were defined explicitly at that point.","section":"Theorem 2.1, statement"}],"recommendation":"major_revision","confidential_remarks":"The main issue is not the overall strategy or the k=0 Diophantine step, but the false inequality (9.3) in Section 9. If the authors can replace it with a correct bound for the range in question, the paper may be salvageable; as written, the proof of the central theorems is incomplete. The ψ1 issue in Corollary 1.2 is minor and easily fixed."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Dear Colleague,\n\nThe thing to know: this paper proves an exact equivalence between normality of a positive algebraic irrational to base b and the vanishing of certain zeta averages for every k ≥ 0. It does not settle Borel's conjecture and does not produce a new normal number, but it is the first general analytic criterion of this kind, and the proof is substantial.\n\nWhat's new: [KS23] handled only simple normality of 2^{p/q} to base 2. Here they extend to all algebraic irrationals, all bases, and all k ≥ 0. Corollary 1.2 is new. The proof proceeds through Theorem 1.1, derived from Theorems 2.1/2.2 via the functional equation, Euler-Maclaurin, stationary phase, and Ridout's theorem. The error terms are handled carefully; the k=0 case optimizes to O((log N)^{2/3} (log log N)^{5/9}). The exposition is honest — the sketch in Section 3 genuinely helps, and the authors say they don't fully understand why the transformation occurs, which is a nice touch.\n\nSoft spots, in proportion. The one real inaccuracy is in the proof of Corollary 1.2: 'each ψ_k is a polynomial' is false for k=1 because of the special value at integers. It's presentational, not load-bearing: for algebraic irrational α, b^h α is never an integer, so the argument works with B_1({x}) on the sequence. The stress-test note is correct that the Ridout step is legitimate: with q=b^h, Ridout gives ||b^h α|| ≫_γ b^{-γ h} uniformly, and the h-range split yields G_N = o(log N). No circular step or omitted argument threatens the main claim.\n\nThere are inherent limitations, not flaws: the zeta-average condition is probably as hard to verify as normality itself, so this is a structural characterization rather than a decision procedure. The citation pattern is fine; the self-reference to [KS23] is appropriate.\n\nWho it's for: number theorists working on normality, uniform distribution mod 1, or the Riemann zeta function. It deserves serious refereeing. I'd ask the authors to fix the ψ1 sentence and add a short remark that b^h α is never an integer. Recommend conditional acceptance.\n\nBest,","headline":"A genuine and largely clean equivalence between normality of algebraic numbers and vanishing of zeta averages; the small ψ1 slip is presentational, not load-bearing.","tokens_in":27458,"tokens_out":3471,"would_cite":true,"duration_ms":32807,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["11K16","11M06"],"pacs":[],"model":"deepseek-v4-flash","headline":"The paper claims that an algebraic irrational number is normal to a base exactly when a family of Riemann zeta averages vanishes.","keywords":["normality","algebraic numbers","Riemann zeta function","periodic Bernoulli polynomials","uniform distribution modulo one","Ridout's theorem","stationary phase","base-b expansion"],"falsifier":"For a fixed algebraic irrational $\\alpha$ and base $b$ (for instance $\\alpha=\\sqrt[3]{2}$, $b=2$), compute the boundary sum $G_N=\\sum_{0<h<\\log_b N}\\min(1,b^h/(\\|b^h\\alpha\\|N))$ at $N=10^5,10^6,\\ldots$; the proof requires $G_N=o(\\log N)$. If these numbers grow like a positive multiple of $\\log N$, the $k=0$ case of Theorem 1.1 fails. Alternatively, evaluate the zeta average in Corollary 1.2 at $k=0$; if it tends to zero while $b^h\\alpha$ is visibly not uniformly distributed, the claimed equivalence fails.","tokens_in":26473,"feed_emoji":"🔢","tokens_out":12248,"duration_ms":113351,"temperature":0.7,"pith_summary":"Normality to a base is the statement that every block of digits occurs with the expected frequency; for algebraic irrationals this is an open problem, and no explicit algebraic normal number is known. The paper establishes a sharp criterion: a positive algebraic irrational $\\alpha$ is normal to base $b$ if and only if a certain logarithmic average of Riemann zeta values at $-k+2\\pi i n/\\log b$, weighted by $e^{2\\pi i n\\log\\alpha/\\log b}/n^{k+1}$, tends to zero for every integer $k\\ge 0$. The proof works by deriving an asymptotic identity that converts these zeta averages into averages of periodic Bernoulli polynomials over the fractional parts of $\\alpha b^h$. If the criterion is correct, the old conjecture that all algebraic irrational numbers are normal in every base becomes exactly the statement that all such zeta averages vanish for every algebraic irrational in every base.","feed_headline":"Algebraic normality pinned down by zeta sums","feed_subtitle":"A new criterion ties the normality of algebraic numbers to averages of the Riemann zeta function.","key_machinery":"The load-bearing mechanism is a two-sided transform. The functional equation of $\\zeta$ converts $\\zeta(-k+2\\pi i d n)$ into a sum over integers $m$ of $e(d n \\log(me/(dn)) + \\theta n)$, and the stationary phase method then evaluates the exponential sum over $n$. The Fourier expansion of the periodic Bernoulli polynomial $\\psi_{k+1}$ turns the result into the sum $-\\frac{1}{\\log^k b}\\sum_{0<h<\\log_b N}\\psi_{k+1}(b^h\\alpha)$. For $k=0$, Ridout's theorem supplies the lower bound $\\|b^h\\alpha\\|\\gg_\\gamma b^{-\\gamma h}$ that makes the boundary term $o(\\log N)$.","core_discovery":"The central claim is that a positive algebraic irrational $\\alpha$ is normal to base $b$ if and only if, for every integer $k\\ge 0$, $$\\lim_{N\\to\\infty}\\frac{1}{\\log N}\\sum_{1\\le |n|\\le N} \\zeta\\left(-k+\\frac{2\\pi i n}{\\log b}\\right)\\frac{$e^{{2\\pi i n \\log \\alpha/\\log b}}$}{$n^{{k+1}}$}=0.$$ This is Corollary 1.2. It is proved through Theorem 1.1, an asymptotic identity expressing the left-hand side as $-\\frac{1}{\\log^k b}\\sum_{0<h<\\log_b N}\\psi_{k+1}(b^h\\alpha)+o(\\log N)$, where $\\psi_{k+1}$ is the $(k+1)$-st periodic Bernoulli polynomial. Since $\\alpha$ is normal to base $b$ exactly when the sequence $b^h\\alpha$ is uniformly distributed modulo one, and uniform distribution is equivalent to the averages of all periodic Bernoulli polynomials tending to zero, the identity turns a digit-counting property into a statement about the Riemann zeta function on vertical arithmetic progressions.","pith_inferences":["Because the asymptotic identity in Theorem 2.1 is stated for arbitrary real $\\theta$, the zeta-average to Bernoulli-average connection holds for transcendental numbers as well; only the boundary term, which for algebraic $\\alpha$ is controlled by Ridout's theorem, would need a different estimate.","An effective version of the criterion would follow from replacing the qualitative lower bound on $\\|\\alpha b^h\\|$ with an explicit one, turning the equivalence into a finitary test for normality up to a given number of digits.","If the equivalence is correct, it relocates the normality problem entirely: one would need to prove that the displayed averages vanish for every algebraic irrational, a statement about $\\zeta$ that does not mention digits."],"forward_implications":["If Corollary 1.2 is correct, proving the classical conjecture for a given algebraic $\\alpha$ is equivalent to checking that all displayed zeta averages vanish; no digit-counting is required.","Theorem 1.1 gives a quantitative form: digit-counting averages and zeta averages agree up to $o(\\log N)$, so a non-normal algebraic number would force at least one zeta average to fail to vanish.","For $\\alpha=b^{p/q}$ with $\\gcd(p,q)=1$ and $b$ not a $q$-th power of an integer, normality to base $b$ is equivalent to vanishing of the same zeta averages with $e^{2\\pi i n p/q}$ in place of the $\\alpha$-dependent exponential.","For integer $\\alpha$, the zeta averages converge to explicit constants: nonzero only when $k$ is odd, in which case the value is proportional to $\\zeta(k+1)/\\log^{k+1} b$; the even-$k$ case corresponds to trivial zeros of $\\zeta$.","The $k=0$ case of the theorem reproduces the earlier characterization of simple normality of $2^{p/q}$, so the new result contains that earlier one as a special case."],"supporting_citations":[{"why":"Ridout's theorem supplies the diophantine lower bound on the distance from $\\alpha b^h$ to the nearest integer, which is used to show the boundary term in the $k=0$ case is $o(\\log N)$.","marker":"[Rid57]"},{"why":"This reference gives the theorem that normality to base $b$ is equivalent to uniform distribution of the sequence $\\alpha b^h$ modulo one, the first step in proving Corollary 1.2.","marker":"[Bug12]"},{"why":"The authors' earlier work connects simple normality of $2^{p/q}$ to the Riemann zeta function, and the present paper extends that method to all algebraic numbers.","marker":"[KS23]"},{"why":"This reference supplies the functional equation of the Riemann zeta function and the exponential-sum tools used in Sections 4 and 6.","marker":"[Tit86]"},{"why":"This reference contains the stationary phase integral estimate used to convert oscillatory integrals into sums involving Bernoulli polynomials.","marker":"[Hux94]"},{"why":"This reference provides the Fourier expansion of periodic Bernoulli polynomials and the Euler-Maclaurin formula that underlie the proof.","marker":"[AIK14]"},{"why":"This reference provides the bound on partial sums of $\\sin(2\\pi kx)/(\\pi k)$ that controls boundary sums in the digit-counting argument.","marker":"[MV07]"}],"fun_headline_variants":["Zeta sums reveal normality of algebraic numbers","Normality of algebraic numbers tied to zeta averages","Algebraic normality equivalent to zeta mean vanishing","Riemann zeta criterion for normal algebraic numbers","Zeta average test pins down algebraic normality"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The proof for the $k=0$ case rests on the deep fact that for an algebraic irrational $\\alpha$, the fractional part of $\\alpha b^h$ stays away from $0$ by at least a small exponential amount for every $h$; if that uniformity fails, the boundary term is no longer $o(\\log N)$ and the equivalence for $k=0$ collapses.","fun_headline_variants_meta":{"raw":{"variants":["Zeta sums reveal normality of algebraic numbers","Normality of algebraic numbers tied to zeta averages","Algebraic normality equivalent to zeta mean vanishing","Riemann zeta criterion for normal algebraic numbers","Zeta average test pins down algebraic normality"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000494,"raw_usage":{"total_tokens":2432,"prompt_tokens":962,"completion_tokens":1470,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":578,"completion_tokens_details":{"reasoning_tokens":1414}},"tokens_in":578,"tokens_out":1470,"duration_ms":9834,"temperature":1.0,"reasoning_tokens":1414,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-11T23:34:35.973888+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"For a fixed algebraic irrational $\\alpha$ and base $b$ (for instance $\\alpha=\\sqrt[3]{2}$, $b=2$), compute the boundary sum $G_N=\\sum_{0<h<\\log_b N}\\min(1,b^h/(\\|b^h\\alpha\\|N))$ at $N=10^5,10^6,\\ldots$; the proof requires $G_N=o(\\log N)$. If these numbers grow like a positive multiple of $\\log N$, the $k=0$ case of Theorem 1.1 fails. Alternatively, evaluate the zeta average in Corollary 1.2 at $k=0$; if it tends to zero while $b^h\\alpha$ is visibly not uniformly distributed, the claimed equivalence fails.","supporting_citations":[],"review_version":1}