math.NT — Number Theory

On Problem 5 of Vitorino and Vukusic: $\sup_{k\ge 0} B(2^k\alpha)\ge 12$

Contributed by David Niedbala Giraudin

For an irrational $\alpha$ let $B(\alpha)=\limsup_n a_n(\alpha)$ be the upper limit of its partial quotients. Vitorino and Vukusic proved $\sup_{k\ge 0}B(2^k\alpha)\ge 5$ for every irrational $\alpha$ and asked, as their Problem 5, for an improvement. We prove $\sup_{k\ge 0}B(2^k\alpha)\ge 12$. The proof is a finite computation organised around an invariant $N(C)$: the least number of doubling levels at which the recurrent kernel of a synchronous product of Raney transducers becomes a finite union of pure cycles. We show that $N(C)$ is intrinsic, that $N(C)<\infty$ for every $C$ would imply the $2$-adic Littlewood conjecture, and that $N(C)=\infty$ is equivalent to a rigidity failure; we compute $N(C)=1,1,3,5,5,8,15,21,29,30,31$ for $C=1,\dots,11$. In the language of Badziahin's set LMad this settles a level-bounded form of his Problem B for $C\le 11$. The cycles which survive are ideal classes of discriminant $D\equiv 1 \pmod 8$ whose whole $\mathfrak{p}_2$-orbit has partial quotients at most $C$; we list them, and check the lists by an independent arithmetic route, in both directions for $D\le 10^5$. The result concerns the upper limit $B$ only. On the side of the supremum $M$ it gives $\sup_k M(2^k\alpha)\ge 12$, which is weaker than what is known: Vitorino and Vukusic prove $15$, and Badziahin's bound $1/25$ yields $24$. Two self-contained Python files, without external dependency, reproduce the table and the lists of discriminants; the table up to $C=8$ takes about fourteen seconds.