Let a be a positive odd integer, let k geqslant0 be an integer, let g(s)=pi^-s/2(s-1)Gamma(s/2+1), and let sigma_a(n) be the generalized divisor function. Let the contour operatornameRe(s)=sigma lie to the right of all poles. General constraint conjecture. ∑_n=1^∞sigma_a(n)int_operatornameRe(s)=sigma[-fracg(s)g(s-a)n^s(-s)^k+fracg(s)g(s+a)n^s+a(s-1)^k]ds=0. The corresponding identities are established for the cases treated in the paper, while this formulation is presented as a conjectural generalization beyond those proven cases.

Generalized constraint identities for odd divisor functions

About 3 years old · traced to 2023

Let $a$ be a positive odd integer, let $k\geqslant0$ be an integer, let $g(s)=\pi^{-s/2}(s-1)\Gamma(s/2+1)$ , and let $\sigma_a(n)$ be the generalized divisor function. Let the contour $\operatorname{Re}(s)=\sigma$ lie to the right of all poles.

General constraint conjecture.

\sum_{n=1}^{\infty}\sigma_a(n)\int_{\operatorname{Re}(s)=\sigma}\left[-\frac{g(s)g(s-a)}{n^s}(-s)^k+\frac{g(s)g(s+a)}{n^{s+a}}(s-1)^k\right]ds=0.

The corresponding identities are established for the cases treated in the paper, while this formulation is presented as a conjectural generalization beyond those proven cases.

References

Primary source

Maria Nastasescu, Nicolas Robles, Bogdan Stoica and Alexandru Zaharescu, “The Riemann zeta function and exact exponential sum identities of divisor functions”, arXiv:2311.07657 (2023).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

The original paper proves only three special cases, while a reader has claimed an unverified complete proof of the conjecture for all allowed parameters.

The conjecture, posed by Maria Nastasescu, Nicolas Robles, Bogdan Stoica, and Alexandru Zaharescu in 2023, asserts the contour identity for every positive odd integer $a$ and integer $k\geqslant0$ . The paper presents this as a conjectural extension beyond its explicitly proved cases.

Known results

Theorem 3 proves the corresponding identities for $a=1$ , $a=3$ , and $a=5$ , for every integer $k\geqslant0$ .
For larger odd $a$ , the paper indicates analogous relations but notes that the associated exponential-sum functions become rational rather than polynomial.

Posted attempt

A reader claims a complete proof using the functional equation of the completed Riemann zeta function, asserting that the oddness restriction is unnecessary and that contour shifting handles every admissible contour. This attempt has not been independently verified.

Current status (as of August 2026): the cases $a=1$ , $a=3$ , and $a=5$ are settled, while a complete proof for general positive odd $a$ is claimed but remains unverified.

Sources

ProofThis solution needs a summarySee full solutionHide full solution

In fact, the constraint follows directly from the functional equation of the completed Riemann zeta function; the restriction that $a$ be odd is unnecessary.

Put

g(s)=\pi^{-s/2}(s-1)\Gamma\!\left(\frac{s}{2}+1\right), \qquad \xi(s)=g(s)\zeta(s).

Then

\xi(s)=\frac12s(s-1)\pi^{-s/2} \Gamma\!\left(\frac s2\right)\zeta(s)

is entire and satisfies

\xi(1-s)=\xi(s).

Initially choose $\sigma>1+\max\{0,\operatorname{Re}a\}$ . The absolutely convergent Dirichlet series

\sum_{n\ge1}\frac{\sigma_a(n)}{n^s} =\zeta(s)\zeta(s-a), \qquad \sum_{n\ge1}\frac{\sigma_a(n)}{n^{s+a}} =\zeta(s+a)\zeta(s)

and the exponential decay of the gamma factors justify interchanging summation and integration. The proposed left-hand side therefore becomes

-\int_{\operatorname{Re}s=\sigma} \xi(s)\xi(s-a)(-s)^k\,ds + \int_{\operatorname{Re}s=\sigma} \xi(s)\xi(s+a)(s-1)^k\,ds.

Define the entire function

F(s)=\xi(s)\xi(s+a)(s-1)^k.

By the functional equation,

F(1-s)=\xi(s)\xi(s-a)(-s)^k.

The substitution $w=1-s$ , with the induced reversal of the vertical path canceled by $ds=-dw$ , gives

\int_{\operatorname{Re}s=\sigma}F(1-s)\,ds = \int_{\operatorname{Re}w=1-\sigma}F(w)\,dw.

Since $F$ is entire and decays exponentially along horizontal edges in every fixed vertical strip, Cauchy's theorem yields

\int_{\operatorname{Re}w=1-\sigma}F(w)\,dw = \int_{\operatorname{Re}s=\sigma}F(s)\,ds.

The two terms consequently cancel, proving the asserted identity for every integer $k\ge0$ .

For any other contour lying to the right of the poles in the original statement, shift each individual summand's gamma-product contour to a line satisfying $\sigma>1+\max\{0,\operatorname{Re}a\}$ . No pole is crossed, and the horizontal integrals vanish by Stirling's estimate. The individual integrals are therefore unchanged, so the same identity holds on every admissible contour. In particular, it holds for every positive odd integer $a$ as conjectured.