A Positive Proportion of the Reduced D'Arcais Polynomials is not Hurwitz

Abstract: Heretofore, the second and third author conjectured that the D'Arcais polynomials, related to the coefficients of the powers of the Dedekind $η$-function, are Hurwitz polynomials except for a root at the origin. We show that this does in fact not hold for a positive proportion of all natural numbers.

• The paper proves that a positive proportion of reduced D’Arcais polynomials are not Hurwitz, meaning they have at least one zero with positive real part.
• The authors apply the Hurwitz–Routh criterion to reciprocal polynomials and show that a negative second determinant occurs whenever the divisor sum satisfies σ₋₁(n) > 215.264, including all multiples of an explicit 82-digit integer’s factorial.
• The result establishes positive density but gives an astronomically weak effective bound, leaving the first counterexample, the true density, and the behavior of right-half-plane roots open.

The D'Arcais polynomials $P_n^\sigma(X)$, defined by the generating product

$$\sum_{n=0}^\infty P_n^\sigma(X)q^n = \prod_{m=1}^\infty (1-q^m)^{-X} = \exp\Bigg( X \sum_{j=1}^\infty \sigma(j)\frac{q^j}{j}\Bigg),$$

encode the Fourier coefficients of powers of the Dedekind $\eta$-function and coincide with the Nekrasov–Okounkov polynomials of combinatorics. Heim and Neuhauser conjectured that the reduced polynomials $R_n^\sigma(X) := P_n^\sigma(X)/X$ are Hurwitz — that is, all their zeros lie in the open left half-plane $\{z : \Re(z)<0\}$ — a property relevant to stability theory via the Routh–Hurwitz criterion. The conjecture was supported by computation: $R_n^\sigma(X)$ is Hurwitz for all $n \leq 1\,000$. This paper disproves the conjecture in a strong quantitative sense (2608.18842).

Main result

Let $\mathcal{A}(n)$ denote the number of $k \leq n$ for which $R_k^\sigma(X)$ fails to be Hurwitz. The main theorem states

$$\sum_{n=0}^\infty P_n^\sigma(X)q^n = \prod_{m=1}^\infty (1-q^m)^{-X} = \exp\Bigg( X \sum_{j=1}^\infty \sigma(j)\frac{q^j}{j}\Bigg),$$0

so a positive proportion of all reduced D'Arcais polynomials are not Hurwitz. The proof is constructive in principle: writing $$\sum_{n=0}^\infty P_n^\sigma(X)q^n = \prod_{m=1}^\infty (1-q^m)^{-X} = \exp\Bigg( X \sum_{j=1}^\infty \sigma(j)\frac{q^j}{j}\Bigg),$$1, an integer with 82 digits, every multiple of $$\sum_{n=0}^\infty P_n^\sigma(X)q^n = \prod_{m=1}^\infty (1-q^m)^{-X} = \exp\Bigg( X \sum_{j=1}^\infty \sigma(j)\frac{q^j}{j}\Bigg),$$2 yields a counterexample. Since $$\sum_{n=0}^\infty P_n^\sigma(X)q^n = \prod_{m=1}^\infty (1-q^m)^{-X} = \exp\Bigg( X \sum_{j=1}^\infty \sigma(j)\frac{q^j}{j}\Bigg),$$3 has roughly $$\sum_{n=0}^\infty P_n^\sigma(X)q^n = \prod_{m=1}^\infty (1-q^m)^{-X} = \exp\Bigg( X \sum_{j=1}^\infty \sigma(j)\frac{q^j}{j}\Bigg),$$4 digits, this explicit bound is astronomically far beyond any feasible computation.

Method of proof

The argument rests on the Hurwitz–Routh determinant criterion: a degree-$$\sum_{n=0}^\infty P_n^\sigma(X)q^n = \prod_{m=1}^\infty (1-q^m)^{-X} = \exp\Bigg( X \sum_{j=1}^\infty \sigma(j)\frac{q^j}{j}\Bigg),$$5 polynomial is Hurwitz if and only if all minors $$\sum_{n=0}^\infty P_n^\sigma(X)q^n = \prod_{m=1}^\infty (1-q^m)^{-X} = \exp\Bigg( X \sum_{j=1}^\infty \sigma(j)\frac{q^j}{j}\Bigg),$$6, and a strictly negative minor forces a root in the open right half-plane. Applying the criterion to the reciprocal polynomial $$\sum_{n=0}^\infty P_n^\sigma(X)q^n = \prod_{m=1}^\infty (1-q^m)^{-X} = \exp\Bigg( X \sum_{j=1}^\infty \sigma(j)\frac{q^j}{j}\Bigg),$$7 (which is Hurwitz exactly when $$\sum_{n=0}^\infty P_n^\sigma(X)q^n = \prod_{m=1}^\infty (1-q^m)^{-X} = \exp\Bigg( X \sum_{j=1}^\infty \sigma(j)\frac{q^j}{j}\Bigg),$$8 is), the second minor is

$$\sum_{n=0}^\infty P_n^\sigma(X)q^n = \prod_{m=1}^\infty (1-q^m)^{-X} = \exp\Bigg( X \sum_{j=1}^\infty \sigma(j)\frac{q^j}{j}\Bigg),$$9

Thus it suffices to show $\eta$0 for infinitely many $\eta$1 of positive density.

The required estimates combine three ingredients:

• Lower bound on $\eta$2: if $\eta$3 divides $\eta$4, then $\eta$5, where $\eta$6 is the $\eta$7-th harmonic number.
• Lower bound on $\eta$8: a combinatorial count of compositions gives $\eta$9 for $R_n^\sigma(X) := P_n^\sigma(X)/X$0.
• Upper bounds on $R_n^\sigma(X) := P_n^\sigma(X)/X$1 and $R_n^\sigma(X) := P_n^\sigma(X)/X$2: using the bounds from prior work by Charlton, Heim, and Stumpenhusen together with $R_n^\sigma(X) := P_n^\sigma(X)/X$3, one obtains $R_n^\sigma(X) := P_n^\sigma(X)/X$4 and $R_n^\sigma(X) := P_n^\sigma(X)/X$5 for $R_n^\sigma(X) := P_n^\sigma(X)/X$6.

Combining these yields, for $R_n^\sigma(X) := P_n^\sigma(X)/X$7,

$R_n^\sigma(X) := P_n^\sigma(X)/X$8

so $R_n^\sigma(X) := P_n^\sigma(X)/X$9 is not Hurwitz whenever

$\{z : \Re(z)<0\}$0

Since $\{z : \Re(z)<0\}$1 is unbounded along multiples of $\{z : \Re(z)<0\}$2 (by multiplicativity, $\{z : \Re(z)<0\}$3), the inequality holds for all multiples of $\{z : \Re(z)<0\}$4, giving density at least $\{z : \Re(z)<0\}$5.

Sharpening the threshold

The paper's remarks substantially improve the picture. Starr's asymptotic formula for the $\{z : \Re(z)<0\}$6-fold convolution of $\{z : \Re(z)<0\}$7,

$\{z : \Re(z)<0\}$8

suggests asymptotically sharp bounds under which failure of the Hurwitz criterion occurs once

$\{z : \Re(z)<0\}$9

Remarkably, the $R_n^\sigma(X)$0rd superabundant number already satisfies $R_n^\sigma(X)$1, exceeding this threshold — so $R_n^\sigma(X)$2 serves as a realistic benchmark for where an explicit counterexample should lie, though the paper's method cannot certify it because $R_n^\sigma(X)$3 is sufficient but not necessary for non-Hurwitz behavior.

A striking connection to the Riemann hypothesis emerges: Robin's equivalence states that RH holds if and only if $R_n^\sigma(X)$4 for all $R_n^\sigma(X)$5. Under RH, the threshold $R_n^\sigma(X)$6 cannot be met below roughly $R_n^\sigma(X)$7, several orders of magnitude beyond the superabundant benchmark — underscoring how loose the proven error terms are relative to the expected truth.

Limitations and open questions

The authors concede explicitly that their approach does not identify the smallest counterexample: negativity of $R_n^\sigma(X)$8 is only a sufficient condition, so the true first failure could occur far earlier than $R_n^\sigma(X)$9. Moreover, the proven density lower bound $n \leq 1\,000$0 is vanishingly small compared to what the asymptotic analysis suggests. Several questions remain open:

• Whether the apparent families of non-real root trajectories visible for $n \leq 1\,000$1 eventually cross into the right half-plane, and whether their slopes relate to $n \leq 1\,000$2;
• Whether there exist infinitely many pairwise distinct roots with positive real part, or whether any compact set of positive measure eventually contains a root of some $n \leq 1\,000$3;
• Whether $n \leq 1\,000$4 is Hurwitz for all primes $n \leq 1\,000$5;
• Whether all zeros of $n \leq 1\,000$6 are simple; the dominant root is known to be real and simple, and simplicity of the rest is supported numerically but unproven.

Conclusion

This paper refutes the Heim–Neuhauser Hurwitz conjecture for reduced D'Arcais polynomials by proving that a positive proportion of them possess zeros in the open right half-plane. The proof combines elementary divisor-sum estimates with the Hurwitz–Routh criterion, and its quantitative content — a counterexample guaranteed only above $n \leq 1\,000$7, against a heuristic benchmark near the $n \leq 1\,000$8rd superabundant number — leaves a wide gap between what is proven and what is expected. The result reframes the analytic study of D'Arcais polynomials around locating the first genuine counterexample and describing the right-half-plane roots whose existence it establishes.