Piltz Divisor Problem for k-Fold Divisor Error Terms

Piltz Divisor Problem for k-Fold Divisor Error Terms

Definitions and mathematical statement

For $k\ge3$, let $d_k(n)$ count ordered $k$-tuples of positive integers with product $n$. Define the error in the summatory function by \[ \Delta_k(x)=\sum_{n\le x}d_k(n) -\operatorname*{Res}_{s=1}\frac{\zeta(s)^k x^s}{s},\qquad x\ge1. \] The residue is the complete polynomial-in-$\log x$ main term multiplied by $x$, not just its highest-degree term. Define \[ \alpha_k=\inf\{a:\ \text{for every }{\varepsilon}>0, \ \Delta_k(x)=O_{k,{\varepsilon}}(x^{a+{\varepsilon}})\}. \] Determine $\alpha_k$ for every $k\ge3$. Constants may depend on $k$ and ${\varepsilon}$, but not on $x$. This is a pointwise error-exponent problem rather than a mean-square or short-interval average problem.

Formulation and concept references: [S1].

Short English statement

What is the best possible pointwise error exponent when counting ordered factorizations into a fixed number of positive factors?

Research attempt

Process and precise target.

This attempt was generated by GPT-6 Astra Ultra on 27 September 2026. We derive an exact multidimensional hyperbola identity, carry it through completely for $k=3$, and examine the cancellation lost in this elementary approach. The resulting exponent $2/3$ is only a baseline upper bound for $\alpha_3$, far from a determination of the exponent. No claim of a new record is made.

The pointwise exponent in the question must be distinguished from exponents defined by mean-square integrals. The introduction of [S1] records the conjectural pointwise value $(k-1)/(2k)$ and its known omega obstruction, as well as the separate mean-square problem. The sign-change results in [S2] and number-field resonance results in [S3] have their own conclusions; they do not supply the missing pointwise upper estimate here. We retain the complete residue polynomial throughout, since subtracting only its highest-degree term changes the error by an amount of order $x$ times a logarithmic power.

An exact identity before any estimation.

Let $D_r(x)=\sum_{n\le x}d_r(n)$ for $r\ge1$, with $D_0(t)=1$ for $t\ge1$. Fix $k\ge2$ and $y=\lfloor x^{1/k}\rfloor$. Every ordered $k$-tuple of positive integers with product at most $x$ has at least one coordinate at most $y$: otherwise its product is at least $(y+1)^k>x$. Inclusion-exclusion on these $k$ coordinate events gives \[ D_k(x)=\sum_{j=1}^k(-1)^{j-1}\binom kj \sum_{1\le a_1,\ldots,a_j\le y} D_{k-j}\!\left(\frac{x}{a_1\cdots a_j}\right). \tag{438.1} \] When $j=k$, every selected tuple is admissible because its product is at most $y^k\le x$, and that term equals $(-1)^{k-1}y^k$. The identity counts ordered tuples, as required by $d_k$; it would not be valid with unordered factorizations and the same binomial coefficients.

For $k=3$ this becomes \[ D_3(x)=3\sum_{a\le y}D_2(x/a) -3\sum_{a,b\le y}\left\lfloor\frac{x}{ab}\right\rfloor+y^3. \tag{438.2} \] This exact relation is also useful for finite verification because its two sides can be computed by different counting procedures.

The two-dimensional input with its error accounted for.

For $t\ge1$, set $z=\lfloor\sqrt t\rfloor$. Splitting lattice points below a hyperbola yields \[ D_2(t)=2\sum_{a\le z}\lfloor t/a\rfloor-z^2 =2tH_z-z^2+O(z), \] where $H_z=\sum_{a\le z}1/a$. Using $H_z=\log z+\gamma+O(1/z)$, $z^2=t+O(\sqrt t)$, and $\log z=\tfrac12\log t+O(t^{-1/2})$ gives \[ D_2(t)=t\log t+(2\gamma-1)t+O(\sqrt t). \tag{438.3} \] The estimate is uniform for $t\ge1$ after enlarging the absolute constant over a bounded initial interval. Inserting $438.3$ in the first sum of $438.2$ costs \[ O\!\left(\sqrt x\sum_{a\le y}a^{-1/2}\right) =O(\sqrt{xy})=O(x^{2/3}). \] Replacing each floor in its second sum by its argument costs $O(y^2)=O(x^{2/3})$. Thus, with $L=\log x$ and $J_y=\sum_{a\le y}(\log a)/a$, \[ D_3(x)=3x\{(L+2\gamma-1)H_y-J_y-H_y^2\}+y^3+O(x^{2/3}). \tag{438.4} \] No cancellation between the floor errors has been used at this point.

Recovering the complete main polynomial.

Fix the convention \[ \zeta(1+u)=u^{-1}+\gamma-\gamma_1u+O(u^2). \] In this convention $\gamma_1=\lim_{y\to\infty}(J_y-\tfrac12\log^2y)$. The first Euler summation corrections give, for integral $y\to\infty$ and $\ell=\log y$, \[ H_y=\ell+\gamma+\frac1{2y}+O(y^{-2}),\qquad J_y=\frac{\ell^2}{2}+\gamma_1+\frac{\ell}{2y} +O((1+\ell)y^{-2}). \tag{438.5} \] These estimates follow by applying the trapezoidal summation formula to $1/t$ and $(\log t)/t$ on the tail; the derivatives have integrable variation of the asserted orders. The limits specify the constants and prevent a sign change in $\gamma_1$.

Here $\ell=L/3+O(1/y)$. The expression in braces in $438.4$, omitting the $1/y$ corrections, is \[ (L+2\gamma-1)(\ell+\gamma)-\tfrac12\ell^2-\gamma_1-(\ell+\gamma)^2. \] Its derivative in $\ell$ is $L-3\ell-1$, bounded when $\ell=L/3+O(1/y)$. Replacing $\ell$ by $L/3$ therefore changes it by $O(1/y)$, without an extra factor $L$. Likewise, the combined coefficient of $1/(2y)$ in $438.5$ is $L-3\ell-1$, again bounded. The remaining error contributes $O(x(1+L)/y^2)$, which is $O(x^{2/3})$. Finally $y^3=x+O(x^{2/3})$. Simplification yields \[ \begin{split} D_3(x)&=xP_2(\log x)+O(x^{2/3}),\\ P_2(L)&=\tfrac12L^2+(3\gamma-1)L +1-3\gamma+3\gamma^2-3\gamma_1. \end{split} \tag{438.6} \] The derivation was for sufficiently large $x$; a change of constant covers $x\ge1$.

To verify that this is the main term specified in the problem, expand \[ \zeta(1+u)^3=u^{-3}+3\gamma u^{-2} +(3\gamma^2-3\gamma_1)u^{-1}+O(1), \] and $x^{1+u}/(1+u)=x[1+(L-1)u+(L^2/2-L+1)u^2+O(u^3)]$. The coefficient of $u^{-1}$ in their product is exactly $xP_2(L)$. Consequently $438.6$ proves $\alpha_3\le2/3$ with the catalogue's normalization, not merely a bound after an approximate subtraction.

Where the attempted improvement fails.

One might try replacing the absolute bounds on the two error sums by square-root cancellation. The second is a sum of fractional parts evaluated at $x/(ab)$ over a highly dependent multiplication table. The values are neither independent random variables nor uniformly distributed by any argument above. The first contains errors $\Delta_2(x/a)$ at correlated arguments. An upper bound for each term gives the exponent already obtained; asserting random signs would be an additional unproved hypothesis.

The target suggested by the literature for $k=3$ is $1/3$, so saving a small power from a single error estimate would still not establish the required endpoint exponent. Moreover $\alpha_3$ is an infimum with an $\varepsilon$ allowance. Determining it does not require a bound with exactly $x^{1/3}$ and no logarithms, but it does require estimates for every positive $\varepsilon$, with constants independent of $x$.

Nor can a mean-square estimate simply be read as a pointwise one. If $\int_X^{2X}|E(t)|^2dt\le X^{1+2b}$, Chebyshev's inequality controls the measure of the set where $|E(t)|>X^{b+\eta}$ by $X^{1-2\eta}$. It does not prove that this exceptional set is empty. Narrow spikes are compatible with such an integral bound. The actual divisor error has arithmetic structure, but a separate argument exploiting that structure is necessary to eliminate every large exceptional value.

For larger $k$, $438.1$ remains available. Its usefulness is organizational: it records intersections exactly and makes the family of correlated lower-dimensional errors explicit. It does not automatically determine their joint cancellation. Treating those errors as independent would import the conclusion rather than derive it.

Verification and outcome.

The finite checks compare $438.1$ with divisor counts computed by convolution for several $k$ and all integer $x$ in a fixed initial range. Symbolic coefficient arithmetic separately checks the residue polynomial and the cancellations of the apparent $x\log x/y$ terms. These calculations test identities, not the asymptotic optimal exponent.

UNRESOLVED. The exact hyperbola decomposition and the full $k=3$ calculation give a checked elementary upper bound. No optimal pointwise exponent is established for any of the general $k\ge3$ targets, and the missing cancellation estimate is explicitly unproved.

Sources

• S1 Bellotti and Yang, On the generalised Dirichlet divisor problem (2023 preprint; BLMS 2024). https://arxiv.org/html/2303.05028. Formulation source.
• S2 Status source for Piltz Divisor Problem for k-Fold Divisor Error Terms. https://academic.oup.com/imrn/article/2024/20/13191/7755133. Further reference; not a proof endorsement.
• S3 Status source for Piltz Divisor Problem for k-Fold Divisor Error Terms. https://arxiv.org/html/2506.22587v1. Further reference; not a proof endorsement.