REVIEW 3 major objections 4 minor 7 references
Dirichlet product of derivative arithmetic with an arithmetic function multiplicative
T0 review · 3 major / 4 minor · reviewed 2026-08-14 · deepseek-v4-flash
Pith's one-line read The paper derives a two-term Leibniz-type identity for the Dirichlet convolution of the arithmetic derivative with any multiplicative function, and from it a prime-power formula and a link to the divisor-count function.
desk verdict A correct little identity buried under an unproved false equation and a missing nonvanishing hypothesis; revise by cutting Eq. (11) and fixing Lemma 5. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
The central object is the arithmetic derivative $\delta$, defined on primes by $\delta(p)=1$ and extended multiplicatively by the Leibniz rule $\delta(ab)=\delta(a)b+a\delta(b)$. The argument is carried by Dirichlet convolution, $(f*g)(n)=\sum_{d\mid n}f(d)g(n/d)$, together with the fact that divisors of a product of coprime integers split uniquely as divisors of the two factors. The engine of the proof is to use multiplicativity of $f$ together with the Leibniz rule to rewrite $\delta(d_1d_2)$ inside the convolution sum, which produces the two-term identity; iterating it strips prime powers off one at a time to yield equation (7).
What would settle it
Evaluate equation (7) for the multiplicative function defined on prime powers by $f(2)=-2$, $f(4)=4$, $f(2^k)=0$ for $k>2$, and arbitrary values on odd prime powers: at $n=2$ the right-hand side has denominator $(\mathrm{Id} * f)(2)=f(2)+2f(1)=-2+2=0$, so the displayed formula cannot hold as written, showing the missing nonzero assumption is necessary.
Extended reading notes
Core claim
The paper's central claim is that the Dirichlet convolution of the arithmetic derivative $\delta$ (the Leibniz-rule map with $\delta(p)=1$ for every prime $p$) with any multiplicative function $f$ satisfies, for coprime positive integers $n,m$, $$ (f * \delta)(nm) = (\mathrm{Id} * f)(n)\,(f * \delta)(m) + (\mathrm{Id} * f)(m)\,(f * \delta)(n), $$ where $\mathrm{Id}$ is the identity function $\mathrm{Id}(k)=k$. The proof expands the convolution over divisors and factors $f$ by multiplicativity while using the Leibniz rule to split $\delta(d_1d_2)$ into $d_1\delta(d_2)+d_2\delta(d_1)$. Iterating this two-term identity gives the prime-power decomposition $$ (f * \delta)(n) = (\mathrm{Id} * f)(n)\sum_{i=1}^s \frac{(f * \delta)($p_i^{{\alpha_i}}$)}{(\mathrm{Id} * f)($p_i^{{\alpha_i}}$)}, $$ so evaluating the convolution on any integer reduces to evaluating it on prime powers. In the special case $f=\mathrm{Id}$ the paper proves $(\mathrm{Id} * \delta)(n)=\tfrac12 \tau(n)\delta(n)$, and it converts this into the Dirichlet-series relation $2\zeta(s-1)\sum_{n\ge 1}\delta(n)/n^s = \sum_{n\ge 1}\delta(n)\tau(n)/n^s$.
Load-bearing premise
The derivation of the prime-power formula divides by $(\mathrm{Id} * f)(p_i^{\alpha_i})$ at every step, but the paper never assumes or proves that these divisor-convolution values are nonzero; if one vanishes, equation (7) is not defined, and the theorem as stated has a hidden missing hypothesis.
Editorial extensions
If this is right
- For $f=\mathrm{Id}$, the theorem gives $(\mathrm{Id} * \delta)(n) = \tfrac{1}{2}\tau(n)\delta(n)$ for every positive integer $n$.
- Multiplying the corresponding Dirichlet series yields $2\zeta(s-1)\sum_{n\ge 1}\delta(n)/n^s = \sum_{n\ge 1}\delta(n)\tau(n)/n^s$.
- Formula (7) lets one compute $(f * \delta)(n)$ from the prime-power values $(f * \delta)(p^\alpha)$ and $(f * \mathrm{Id})(p^\alpha)$, provided the latter are nonzero.
- Because $f * \delta$ is multiplicative when $f$ is multiplicative, the two-term identity provides an alternative route to the prime-power formula that avoids expanding all divisors of $n$.
Reading between the lines
- The same two-term identity may hold for higher arithmetic derivatives defined by repeated Leibniz applications, with $\mathrm{Id}$ replaced by an iterated identity convolution; testing this is a direct extension of the paper's method.
- The unproved formula (11) for the Möbius function, involving $\varphi$, $\omega$, $B(n)=\sum_{p^\alpha\parallel n}\alpha p$, and $\sigma$, is numerically checkable and, if true, would give a closed form for $(\mu * \delta)(n)$.
- The paper never states the hypothesis that $(\mathrm{Id} * f)(p^\alpha)\neq 0$; making that hypothesis explicit and investigating the vanishing case would complete the prime-power formula.
Signed reviews
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The manuscript defines the arithmetic derivative δ by δ(p)=1 and the Leibniz rule, then studies its Dirichlet convolution with multiplicative functions f. Theorem 4 states a two-factor identity (f∗δ)(nm)=(Id∗f)(n)(f∗δ)(m)+(Id∗f)(m)(f∗δ)(n) for coprime n,m. Lemma 5 derives a prime-power decomposition, Eq. (7), expressing (f∗δ)(n) as a sum over prime-power ratios. Proposition 6 specializes to f=Id and concludes (Id∗δ)(n)=(1/2)τ(n)δ(n), which is then translated into the Dirichlet-series relation (9). The paper closes with an unproved identity, Eq. (11), for (µ∗δ)(n) involving a newly defined 'En-naoui function', with the proof deferred to a later article.
Significance. The derivation of Theorem 4 is a clean and correct use of multiplicativity, and Proposition 6 is conditionally correct given the prime-power formula. If the missing hypotheses are supplied, Eq. (7) is a potentially useful reduction of (f∗δ)(n) to prime powers. However, as written, the central formula is undefined for legitimate multiplicative functions because denominators can vanish, and the paper also asserts a numerically false identity without proof. These defects make the current manuscript unreliable. No machine-checked proofs, reproducible code, or falsifiable predictions are supplied; the contribution is an elementary identity note rather than a substantive new method.
major comments (3)
- [Section 2 (Main results), Lemma 5, Eq. (7)] The formula in Eq. (7) divides by (Id∗f)(p_i^{α_i}), but the lemma and its proof never establish that these quantities are nonzero. Since f is allowed to be complex-valued and multiplicative, such vanishing can occur: take the completely multiplicative function with f(2)=-2 and f(p)=1 for every odd prime p. Then (Id∗f)(2)=1·f(2)+2·f(1)=-2+2=0, while (f∗δ)(2)=f(1)δ(2)+f(2)δ(1)=1. For n=2 the right-hand side of Eq. (7) is 0·(1/0), which is undefined, so the lemma is not valid for all multiplicative functions as stated. A nonvanishing hypothesis, or an alternative formulation that avoids division, is load-bearing and must be added before Eq. (7) can be used.
- [Section 2 (Main results), Eq. (11)] The identity for (µ∗δ)(n) is asserted without proof; the text states 'In next article i will prove this equality'. Moreover, the identity is numerically false. For n=2, the left side is (µ∗δ)(2)=µ(1)δ(2)+µ(2)δ(1)=1. The right side is φ(2)[δ(2)-2ω(2)+B(2)+Φ_φ(2)/2+(B∗Id)(2)/σ(2)] = 1·[1-2+2+1/2+2/3] = 13/6. Since the paper presents Eq. (11) as a result, it must either be proved correctly or removed; an unproved false assertion cannot remain in the main results.
- [Section 2 (Main results), Lemma 5 statement] The statement of Lemma 5 does not explicitly restrict f to be multiplicative, although the proof invokes Theorem 4, which requires multiplicativity. As written, the lemma is overgeneralized and the hypothesis under which Eq. (7) is derived must be stated explicitly in the lemma itself, together with the nonvanishing condition on the denominators.
minor comments (4)
- [Throughout] There are many typographical and grammatical errors, e.g. 'eve ry', 'Mobiuse', 'an called an multiplicative', 'an other prof by induction'. The paper would benefit from a careful proofreading pass.
- [Section 2 (Main results), Lemma 5, second proof] The second induction proof begins with '(id∗δ)(n.p_{s+1}^{α_{s+1}})' which should be '(f∗δ)(n.p_{s+1}^{α_{s+1}})'.
- [Section 2 (Main results), Eq. (9)] The Dirichlet-series identity in Eq. (9) is only formal, since the series involving δ(n) do not converge absolutely in a right half-plane. The authors should state that the identity is to be read formally or under an appropriate regularization.
- [Definition 2] The definition of multiplicative function does not mention the standard convention f(1)=1. This matters in several divisors sums in the proofs, where terms like f(1) are implicitly taken to be 1.
Circularity Check
No circularity: Theorem 4 and Lemma 5 are derived directly from the definitions of δ and Dirichlet convolution; the unproved Eq. (11) is an unsupported assertion, not a circular step.
full rationale
The paper's central chain starts from the definition δ(p)=1 and the Leibniz rule, together with the standard definition of Dirichlet convolution. Theorem 4 expands (f*δ)(nm) over the divisors d1|n, d2|m, applies δ(d1d2)=d1δ(d2)+d2δ(d1) and multiplicativity of f, and factors the sums as (Id*f)(n)(f*δ)(m)+(Id*f)(m)(f*δ)(n). This is a direct computation, not an import of the conclusion. Lemma 5 iterates Theorem 4 over the prime-power factors; the induction step uses the already-proved identity and multiplicativity of Id*f, so no load-bearing assumption is supplied by the result being proved. Proposition 6 uses Lemma 5 with f=Id and evaluates (Id*δ)(p^α) explicitly; this is again a derivation from the definition. The nonzero-denominator defect in Lemma 5 (e.g., f(2)=-2 gives (Id*f)(2)=0) is a hidden hypothesis and a correctness gap, not circularity, because the formula is not being assumed instead of derived. Eq. (11) is stated without proof ('In next article I will prove this equality') and is numerically false at n=2; this is an unproved and incorrect side assertion, but it is not used in the derivation of (6)-(8) and does not make the argument circular. No self-citation is load-bearing, no fitted parameters are renamed as predictions, and no uniqueness theorem is invoked to force a choice. The main derivation is self-contained.
Assumptions & free parameters
assumptions (3)
- standard math Arithmetic derivative δ is defined by δ(p)=1 and the Leibniz rule.
- standard math Multiplicative arithmetic functions satisfy f(nm)=f(n)f(m) for coprime n,m, and Dirichlet convolution is bilinear.
- domain assumption The denominators (Id*f)(p^α) in Lemma 5 are nonzero for all prime powers p^α.
invented entities (1)
-
En-naoui function Φ_φ(n) = n Σ_{p|n}(1 - 1/p)
Cite this review
Pith. "Pith review of Dirichlet product of derivative arithmetic with an arithmetic function multiplicative." pith.science (2026). https://pith.science/paper/3P7YLCYR
@misc{pith2026190807345,
author = {Pith},
title = {Pith review of: Dirichlet product of derivative arithmetic with an arithmetic function multiplicative},
year = {2026},
howpublished = {\url{https://pith.science/paper/3P7YLCYR}},
note = {Machine review of arXiv:1908.07345}
}
read the original abstract
We define the derivative of an integer to be the map sending every prime to 1 and satisfying the Leibniz rule. The aim of this article is to calculate the Dirichlet product of this map with a function arithmetic multiplicative.
Reference graph
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Reviewed August 14, 2026 · model on record in the stance chip above.
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