arXiv subjects

Yaoran Yang

Publications and source records attributed to Yaoran Yang.

23 records · Page 2Linked to original sources

arXiv2026

A Local Valuation Criterion for Quadratic-Permutation Interleaved Zadoff--Chu Sequences

Berggren and Popović introduced quadratic-permutation-polynomial interleaved Zadoff--Chu sequences and, from exhaustive data, conjectured that all normalized QPP-interleaved Zadoff--Chu sequences are inequivalent to ordinary Zadoff--Chu sequences precisely for prime-power lengths $N=p^n$ with $p>3$ and $n>1$. We give an exact local arithmetic criterion. For a normalized QPP $π_{a,b}(k)=ak^2+bk\pmod N$, the interleaved sequence is equivalent, under the standard five CAZAC-preserving operations, to a Zadoff--Chu sequence if and only if, for every prime power $p^α\Vert N$, the valuation of $a$ satisfies \[ ν_p(a)\ge \begin{cases} 0, & p=2,\ α=1,\\ α-1, & p=2,\ α\ge2,\\ α-1, & p=3,\\ α, & p>3. \end{cases} \] The proof is based on a third finite-difference invariant of the lifted Zadoff--Chu phase, namely \[ Δ^3\bigl((ak^2+bk+\varepsilon_N+2q)(ak^2+bk)\bigr) =12a(2ak+3a+b). \] As a consequence, the conjectured prime-power boundary is not correct: the exact non-vacuous condition for all nonzero normalized QPPs to be inequivalent to Zadoff--Chu sequences is that $N$ is odd, $9\nmid N$, and $p^2\mid N$ for at least one prime $p\ge5$. In particular, $N=75=3\cdot5^2$ is the smallest non-prime-power counterexample to the conjectured ``only if'' direction. A second corollary records the corresponding statement for irreducible QPPs.

math.NT↗

arXiv2026

Finite factorization is detected by undermonoids

Let $M$ be a cancellative commutative monoid and call a submonoid $S$ of $M$ an undermonoid if $\G(S)=\G(M)$ inside the Grothendieck group of $M$. Gotti and Li asked whether the finite factorization property is hereditary once it is known on all undermonoids: if every undermonoid of $M$ is a finite factorization monoid, must every submonoid of $M$ be a finite factorization monoid? We give an affirmative answer. Equivalently, for every cancellative commutative monoid $M$, the following two conditions coincide: every submonoid of $M$ is an FFM, and every undermonoid of $M$ is an FFM. The proof isolates a fixed length $\ell$ and an infinite set of length-$\ell$ factorizations of one element $b$. In the non-group case, a divisor-complement ideal $I=\{m\in M:m\nmid_M b\}$ enlarges the bad submonoid to a bad undermonoid while preserving the chosen length-$\ell$ factorizations. In the group case, a maximality argument over submonoids for which these factorizations survive is combined with a two-sided perturbation $S\mapsto S+\Nzero(2b+u)$. The key point is that the perturbation creates no new units and does not split any atom occurring in the fixed factorizations. This yields an undermonoid with infinitely many factorizations of $b$, contradicting the hypothesis.

math.GR↗

arXiv2026

A compensation theorem for the Sylow-integral invariant and counterexamples to an \texorpdfstring{$A_5$}{A5}-characterization conjecture

Let $ν_p(G)$ be the number of Sylow $p$-subgroups of a finite group $G$, let $σ_p(G)$ be their common order, and set \[ γ(G)=\int_0^1\sum_{p\inπ(G)}ν_p(G)x^{σ_p(G)}\,dx =\sum_{p\inπ(G)}\frac{ν_p(G)}{σ_p(G)+1}. \] A recent conjectural extension of the simple-group theorem for this invariant asserted that a nonsolvable finite group has $γ(G)=9/2$ precisely when $G\cong A_5$. We disprove this assertion by a direct and verifiable construction. More generally, we prove an exact direct-product compensation formula for $A_5$ with an arbitrary nilpotent factor. The formula reduces the equality $γ(A_5\times N)=9/2$ to a finite Egyptian-fraction equation in the orders of the Sylow subgroups of $N$. Taking $N=\C_2\times\C_7\times\C_{11}\times\C_{13}\times\C_{17}\times\C_{19}\times\C_{29}\times\C_{71}\times\C_{83}$, the loss in the $2$-Sylow contribution is exactly compensated by the new normal Sylow subgroups. Consequently $G=A_5\times N$ is nonsolvable, is not isomorphic to $A_5$, has solvable radical $N$, and nevertheless satisfies $γ(G)=9/2$. Several further explicit compensation certificates are also recorded.

math.GR↗

arXiv2026

Weighted Uniform Endpoint Majorants for Integrals Involving Modified Bessel Functions

We give an affirmative full-range solution to Gaunt's 2019 Open Problem~2.10. The problem asks whether, for every $ν>-1/2$ and $0<γ<1$, the reciprocal-power integral $\int_0^x e^{-γt}I_ν(t)t^{-ν}\,\dd t$ is bounded by a constant multiple of $e^{-γx}I_{ν+1}(x)x^{-ν}$, uniformly for all $x>0$. Earlier exponential-tilt estimates proved such endpoint majorants only under an additional smallness condition on $γ$. We prove the estimate throughout the natural range $0<γ<1$, with an explicit admissible constant. More generally, if $μ>-1$, $q>-1$, $0<γ<1$, and $w(x)x^{-q}$ is nondecreasing on $(0,\infty)$, then for every $θ\in(γ,1)$, $\int_0^x e^{-γt}w(t)t^{-μ}I_μ(t)\,\dd t$ is controlled by an explicit multiple of $e^{-γx}w(x)x^{-μ}I_{μ+1}(x)$. The case $w\equiv1$, $q=0$, and $μ=ν$ resolves Gaunt's problem. The weighted theorem also yields shifted-order and moment estimates, applies to approximate power weights and monotone regularly varying amplitudes, and provides two-sided estimates under a reversed comparison. We further analyze the sharp power-weighted quotient via endpoint expansions, a stationary equation, and parameter monotonicity.

math.CA↗

arXiv2026

A Uniform Random-Lattice Tail Bound for the SVP Kissing-Profile Parameter

A recent SICOMP paper on classical and quantum algorithms for the shortest vector problem introduced a lattice-dependent parameter $γ(L)$, bounded universally in the exponential sense by $2^{0.402n+o(n)}$, and conjectured that this parameter is $2^{o(n)}$ for most lattices. We prove the Haar--Siegel random-lattice version in a stronger, dimension-uniform form. Let $X_n=\operatorname{SL}_n(\R)/\operatorname{SL}_n(\Z)$, let $μ_n$ be its invariant probability measure, and let $γ(L)=\sup_{r\ge1} N_L(rλ_1(L))/r^n$, where $N_L(R)$ counts nonzero vectors of $L$ of Euclidean norm at most $R$. For every $n\ge3$ and every $T>0$, \[ μ_n\{L\in X_n:γ(L)>T\}\le C T^{-1} \] with an absolute constant $C$. Consequently, for every sequence $a_n\to\infty$, $γ(L_n)\le a_n$ with $μ_n$-probability tending to one; in particular $γ(L_n)=2^{o(n)}$ with high probability. In the product model of independent Haar--Siegel lattices, $γ(L_n)\le \exp(\sqrt n)$ eventually almost surely. The proof uses Rogers's second-moment estimate only through a dyadic self-normalization argument around the random scale $λ_1(L)$.

math.NT↗