arXiv subjects

Aiken Kazin

Publications and source records attributed to Aiken Kazin.

2 recordsLinked to original sources

arXiv2026

Bounded continued fractions and quadratic orders in $\Q(\sqrt{127})$

We study the least periods and multiplier orders in two constructions of Mercat for continued fractions with bounded partial quotients. Our main explicit result concerns a family in $\Q(\sqrt{127})$ with partial quotients at most $4$. For every integer $j\geq1$, we prove that its primitive period has length $92j+2$ and its multiplier order has conductor $y_{4j}/4$, where $(4730624+419775\sqrt{127})^k=x_k+y_k\sqrt{127}$. The totally positive unit group of this order has index $4j$ in that of the maximal order. These exact formulas give divisor-sum lower bounds at fixed traces and a lower bound of order $(\log X)^2$ for the number of bounded continued fractions with discriminant at most $X$. We also prove a primitive-period refinement of Mercat's palindromic construction. Combining this refinement with Hensley's counting theorem gives, for every $m\geq2$ and $η>0$, at least $X^{δ_m/2-η}$ quadratic fields, up to a positive constant, that admit infinitely many purely periodic continued fractions with partial quotients at most $m$. Here $δ_m$ is the Hausdorff dimension of the corresponding continued fraction Cantor set.

math.NT↗

arXiv2026

Infinitely Many Off-Critical-Line Zeros of the Tempered Xi Function: A Disproof of Yang's Conjecture

Yang introduced a tempered xi function $\widehatξ(s)$ by replacing the hyperbolic cosine in a classical integral representation of the Riemann xi function by a hyperbolic sine, and conjectured that every zero of $\widehatξ$ lies on the critical line $\Re s=1/2$. We disprove this conjecture. After the change of variables $x=e^{2t}$, one has \[ \widehatξ\!\left(\tfrac12+iz\right)=iS(z), \qquad S(z)=\int_0^\infty K(t)\sin(zt)\,dt, \] where $K$ is positive, smooth, and doubly exponentially decaying. Two integrations by parts give $S(y)=K(0)/y+O(y^{-2})$ for real $|y|\to\infty$, with $K(0)>0$; hence $S$ has only finitely many real zeros. On the other hand $S$ is an entire function of order at most one. If it had only finitely many zeros in the complex plane, Hadamard factorization would force $S(z)=P(z)e^{az+b}$, which is incompatible with $S(y)\to0$ as $y\to\pm\infty$. Thus $S$ has infinitely many nonreal zeros, and consequently $\widehatξ$ has infinitely many zeros off the critical line. The proof is unconditional and does not use the Riemann Hypothesis or numerical zero finding. A short numerical illustration is included.

math.NT↗