On the counting function of square-full numbers

Ayyadurai Sankaranarayanan; P. Akhilesh

REVIEW

On the counting function of square-full numbers

Number Theory · Submitted 2 Sep 2026

Pith's verdict: Nice idea, but the main conditional theorem (1.2) relies on an unstated Lindelöf hypothesis and is not proved under the strong Riemann hypothesis alone.

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arXiv ID2609.02327 (version 2)

Pith IDISRBWQNG, Pith's permanent record of this paper and its review

CitationAyyadurai Sankaranarayanan, P. Akhilesh. "On the counting function of square-full numbers." arXiv:2609.02327 (2026). https://pith.science/paper/ISRBWQNG

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The paper tries to improve the error term for square-full numbers to essentially x^ε under the strong Riemann hypothesis, and gives an explicit extra main term from zeros of ζ(6s). The unconditional formula is also new in form. The approach — contour integration with residues from ζ(6s) zeros — is a natural extension of existing techniques.

But the central result has a load-bearing flaw. The proof of Theorem 1.2 repeatedly uses the estimate ζ(1/2+it) ≪ |t|^ε, asserting that the strong Riemann hypothesis (zeros on the critical line, all simple) justifies κ = ε. That is wrong. Strong RH does not imply the Lindelöf hypothesis; the best bound that follows from RH alone is still the convexity bound ζ(1/2+it) ≪ |t|^{1/4+ε}. Without κ = ε the horizontal‑line estimates explode, and the claimed error O_ε(x^{10ε}) collapses. This is not a minor gap — it kills Theorem 1.2 as stated. The authors may have confused the strong RH with the Lindelöf hypothesis, but the paper doesn't mention LH until Remark 1.5.4 and explicitly says κ = ε is justified by the strong RH.

The unconditional Theorem 1.4 uses κ = 13/84+ε (Bourgain), which is fine, and the sketch is plausible. But that part only improves the error from about 0.1388 to 1/12 ≈ 0.0833, and the proof is incomplete (the sketch is too brief to verify).

I would reject this paper for publication in its current form unless the authors either replace the strong RH by the Lindelöf hypothesis (which would change the title and novelty) or rework the estimates with κ = 1/4 and accept a much larger error exponent. The unconditional part alone is not enough for a separate paper.

Verdict: not acceptable as is. The authors should be told plainly about the misunderstanding of the strong RH.

Referee report

Major objections (1)

[Section 3 (page 12, estimation of horizontal contributions H2–H6)] The authors state: 'Since we assume the Strong Riemann Hypothesis, we can very well take κ=ε for any ε>0.' This is incorrect. The strong Riemann hypothesis (zeros on the critical line and simple) does not imply the Lindelöf hypothesis ζ(1/2+it) ≪ t^ε. The best bound under RH is still ζ(1/2+it) ≪ t^{1/4+ε} from the convexity principle, and even under RH the exponent cannot be reduced to 0. The argument therefore implicitly assumes the Lindelöf hypothesis, which is not among the paper's stated assumptions. Without κ=ε the estimates for H2, H3, and H4 become much larger, and the claimed error terms O_ε(x^{10ε}) and O_ε(x^{1/12+10ε}) are not established. The authors must either adopt the Lindelöf hypothesis as an additional assumption or rework the estimates with a valid bound on ζ(1/2+it) and accordingly adjust the error terms in Theorem 1.2.

Minor objections (5)

[Abstract and Introduction] The abstract says 'under the assumptions of Riemann hypothesis and that the zeros of ζ(s) are simple' – this is the strong RH, but the proof actually needs the Lindelöf hypothesis. This mismatch should be clarified.
[Lemma 2.5] The lemma states a bound ζ(σ+it) ≪ |t|^{2κ(1-σ)} log|t|. The constant κ is introduced as the growth exponent on the critical line. The proof uses the maximum modulus principle, which is standard. However, the paper should explicitly note that for the proof of Theorem 1.2 the authors intend to take κ as small as possible, but the allowed value is constrained by known results.
[Proof of Theorem 1.2, choice of σ₀] The choices σ₀ = -1/6+ε and σ₀ = -3/34+ε are given without justification of why these specific values yield the claimed bounds. A brief explanation would improve readability.
[Remark 1.5 (4)] The remark notes that the optimal T depends on κ, and that the Lindelöf hypothesis corresponds to κ=ε. This shows awareness of the issue, making the oversight in the proof more puzzling.
[Section 4, proof of Theorem 4.1] The estimate 'x^{1/k} T T^{kε}' appears to contain a typographical error (likely 'x^{1/k} T^{-1} T^{kε}' or similar). Please verify the algebraic manipulation in the horizontal integral bounds.

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The paper

The paper claims

Under the strong Riemann hypothesis, the counting function of square-full numbers gains an extra sum over zeta zeros, sharpening the error term to O(x^{10ε}).

The reading

Square-full numbers are positive integers where every prime factor appears at least twice. This paper establishes a precise asymptotic formula for their counting function under the strong Riemann hypothesis (all nontrivial zeros of ζ(s) lie on the critical line and are simple). The formula contains not only the classical $x^{{1/2}}$ and $x^{{1/3}}$ main terms but also an infinite sum over the zeros of ζ(6s), indexed by ρ = 1/12 + iγ/6 with ζ(6ρ)=0. Each zero contributes a term $x^{{1/12 + iγ/6}}$ with an explicit coefficient. The error is reduced to O_ε($x^{{10ε}}$), far smaller than any previous result under RH. A second formulation pushes the error to O_ε($x^{{1/12+10ε}}$) while keeping the zero sum short, revealing the true oscillatory structure of the remainder.

What carries the argument

The proof uses Perron’s formula to write the counting sum as an integral of ζ(2s)ζ(3s)/ζ(6s)·x^s / s over a vertical line with Re(s) = 1/2 + 1/log x. The contour is shifted left to Re(s) = σ_0 with −1/2 < σ_0 < 0, collecting residues at the poles of the integrand. The main terms come from the poles at s = 1/2 and s = 1/3. The additional poles at s = 1/12 + iγ/6 arise from the zeros of ζ(6s) in the denominator; the strong Riemann hypothesis guarantees that these zeros lie on the critical line Re(s) = 1/2 and are simple, so the residues are well‑defined first‑order residues. The horizontal and vertical contour integrals are estimated using convexity bounds for ζ(s) (which under RH become essentially |t|^ε) and a lemma of Ramachandra–Sankaranarayanan that provides near‑polynomial bounds for 1/ζ on horizontal lines. The free parameter T (the height of the contour) and σ_0 are chosen to balance the various error terms, yielding the stated ε‑power savings.

Extended reading notes

Core claim

The paper’s central result is Theorem 1.2: assuming the strong Riemann hypothesis, for any ε > 0 and sufficiently large x, the sum over square‑full numbers n ≤ x equals ζ(3/2)/ζ(3) $x^{{1/2}}$ + ζ(2/3)/ζ(2) $x^{{1/3}}$ + Σ_{ρ = 1/12 + iγ/6, |γ| < c_1 $x^{{1/2}}$, ζ(6ρ)=0} c_{γ/6} $x^{{1/12 + iγ/6}}$ + O_ε($x^{{10ε}}$). The coefficients c_{γ/6} are effective complex constants determined by the residue of ζ(2s)ζ(3s)/ζ(6s)·x^s/s at the pole s = 1/12 + iγ/6. The presence of the zero sum is the ‘extra main term’ announced in the title. The error O_ε($x^{{10ε}}$) is essentially a power of log x, representing a far smaller remainder than the earlier conditional exponent θ = 0.1406… or any unconditional bound.

Load-bearing premise

The proof requires that all nontrivial zeros of the Riemann zeta function lie on the critical line Re(s)=1/2 and are simple (the strong Riemann hypothesis). If any zero lies off the line or has multiplicity greater than one, the contour‑shift argument breaks down because the poles of the integrand are no longer simple or the zero sum cannot be truncated cleanly.

Editorial extensions

If this is right

The error term O_ε(x^{10ε}) is nearly as small as x^ε, suggesting that the true order of the remainder may be a power of log x or even bounded. This is far below the conditional exponent (0.14…) that was previously known.
The explicit extra sum over ζ(6ρ)=0 reveals a connection between the distribution of square‑full numbers and the zeros of the Riemann zeta function, analogous to the explicit formula for the prime numbers.
If the strong Riemann hypothesis is true, then the counting function of square‑full numbers is determined up to an almost‑polynomial error by the first two main terms and a finite number of low‑height zeros. This makes numerical verification or heuristic modelling of square‑full numbers feasible to high precision.
The method generalises to k‑full numbers; Theorem 4.1 shows that under the Lindelöf hypothesis the error exponent for k‑full numbers (k ≥ 4) is at most (k+1)/(k(2k+3)), improving earlier bounds.

Reading between the lines

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

The assumption of simple zeros is critical. If some nontrivial zero of ζ(s) were multiple, the residue at the corresponding s would involve derivatives of the integrand, potentially introducing logarithmic factors in the main term and spoiling the ε‑power savings. Therefore, the result is almost a characterisation of the strong Riemann hypothesis: the simple‑zero condition is exactly what is neede
The error term O_ε(x^{10ε}) is so small that it likely cannot be reduced further without a fundamentally new idea: any sub‑polynomial error would require understanding cancellations in the zero sum, which is as hard as the Lindelöf hypothesis itself.
The unconditional result (Theorem 1.4) still contains an explicit sum over zeros with possible multiplicities and a worse error, showing that removing the hypothesis makes the problem substantially harder – the trade‑off between hypothesis strength and error sharpness is laid out clearly.

Original abstract

We study the distribution of square-full numbers under the assumptions of Riemann hypothesis and that the zeros of $\zeta(s)$ are simple. Thus the error term is improved to a considerable extent with an extra main term.

The compiled record

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What the abstract itself states, as probabilities: claims first 0.03 · states limitation 0.56 · human participants 0.01 · controlled comparison 0.04 · randomized assignment 0.01 · states quantitative result 0.15 · compares with prior results 0.51 · mentions code or data availability 0.01.

The abstract, sentence by sentence

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We study the distribution of square-full numbers under the assumptions of Riemann hypothesis and that the zeros of $\zeta(s)$ are simple. Thus the error term is improved to a considerable extent with an extra main term.

background contribution method result interpretation main claim

What the classifier reports for each sentence. The number is the probability it gave the label.
#	sentence	role	the authors' own work	states the main conclusion	strength
s0	We study the distribution of square-full numbers under the assumptions of Riemann hypothesis and that the zeros of $\zeta(s)$ are simple.	contribution 0.96	0.80	0.09	asserted 0.82
s1	Thus the error term is improved to a considerable extent with an extra main term.	result 0.96	0.79	0.61	asserted 1.00

What kind of paper the classifier reports this is: theoretical result 1.00 · evidence from proof or derivation 1.00.

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