\begin{theorem}\label{Thm: Hypercontractivity on tower of n}
  For $n=3\cdot 2^k$ and $n=2^k$ with $k\ge 1$, we have
  \[
    \norm{P_{t}f}_q\le \norm{f}_p\quad \Leftrightarrow \quad t\ge \frac{1}{2}\log(\frac{q-1}{p-1})
    \]
    for $1<p\le q<\infty$.
\end{theorem}

The hypercontractivity of the Poisson-like semigroup on the cyclic group $\mathbb{Z}_n$ is a long-standing open problem for all $n$ except $n=2$ and $n=4$. More precisely, by equipping $\mathbb{Z}_n$ with the normalized counting measure $\mu_n$, the Poisson-like semigroup $(P_t)_{t\in \mathbb{R}_+}$ is defined as the family of maps $P_t:L_\infty(\mathbb{Z}_n,\mu_n)\to L_\infty(\mathbb{Z}_n,\mu_n)$, which acts on Fourier series by
\[
 P_t:\sum_{k=0}^{n-1} a_k\chi_k(x) \mapsto \sum_{k=0}^{n-1} e^{-t\psi_n(k)}a_k\chi_k(x),
\]
where $\psi_n(k)=\min(k,n-k)$ is the word-length function on $\mathbb{Z}_n$, and $\chi_k(x)=e^{\frac{2\pi ikx}{n}}\in L_\infty(\mathbb{Z}_n)$. The hypercontractivity problem asks for the optimal time $t_{p,q}$ with $1< p\leq q<\infty$ such that
\[
 \norm{P_t f}_{q}\le \norm{f}_{p}\qquad \text{for all } t\ge t_{p,q}.
\]
For the case $n=2$, the optimal time $t_{p,q}=\tfrac{1}{2}\log\big(\tfrac{q-1}{p-1}\big)$ follows by applying the classical two-point inequality. That inequality was first proved by Bonami~\cite{MR283496}, later rediscovered by Gross~\cite{MR420249}, and was also used by Beckner~\cite{MR385456} to obtain the best constants for the Hausdorff--Young inequality. For $n=4$, Beckner, Janson, and Jerison~\cite{MR730056} obtained the same optimal time $t_{p,q}=\tfrac{1}{2}\log\big(\tfrac{q-1}{p-1}\big)$ by a clever reduction from $\mathbb{Z}_2$ to $\mathbb{Z}_4$, though the approach does not extend to other $\mathbb{Z}_{m\times n}$. For $n=3$, by Wolff's reduction \cite[Corollary 3.1]{MR2314078}, the optimal time $t_{2,q}$ of $P_t$ on $\mathbb{Z}_3$ coincides the optimal time
\[
t_{2,q}=\frac{1}{2}\log(\frac{\frac{2}{3}(\frac{1}{3})^{\frac{2}{q}-1}-\frac{1}{3}(\frac{2}{3})^{\frac{2}{q}-1}}{(\frac{2}{3})^{\frac{2}{q}}-(\frac{1}{3})^{\frac{2}{q}}})
\]
of the simple semigroup $T_t=e^{-t(\mathrm{id}-\mathbb{E}_{\delta})}$ on the weighted two-point space $L_\infty(\{-1,1\},\tfrac{1}{3}\delta_{-1}+\tfrac{2}{3}\delta_{1})$, which was computed by Oleszkiewicz and Lata\l a \cite{MR1796718,MR2073432}. For the general optimal time $t_{p,q}$ of $(P_t)_{t\in \mathbb{R}_+}$ on $\mathbb{Z}_3$, only asymptotic information is available; see \cite[Theorem 2.1]{MR2314078}. For $n=5$, Andersson~\cite{MR1883499} determined $t_{2,q}=\tfrac{1}{2}\log(q-1)$ for $q\in 2\mathbb{Z}_+$. For $n\ge 6$, Junge, Palazuelos, Parcet, and Perrin~\cite{MR3709719} proved partial results: for even $n$, $t_{2,q}=\tfrac{1}{2}\log(q-1)$ for $q\in 2\mathbb{Z}_+$; for odd $n$, the same holds when $n\ge q$, via a rather involved combinatorial method. Combining Stein's interpolation method \cite{MR82586} with Gross's extrapolation technique \cite{MR420249,MR372613,MR2325763}, they further obtained $t_{p,q}\le \tfrac{\log 3}{2}\log\big(\frac{q-1}{p-1}\big)$ in those regimes.\par

Hypercontractivity is also widely studied for semigroups on manifolds. Among the most classical examples, Weissler~\cite{MR578933} proved optimal hypercontractivity for the heat and Poisson semigroups on the circle $\mathbb{S}^1$, while Rothaus~\cite{MR593787} independently treated the heat case. For the sphere $\mathbb{S}^n$ ($n\ge 2$), Mueller and Weissler~\cite{MR674060} established optimal hypercontractivity for the heat semigroup. For Poisson-type semigroups on $\mathbb{S}^n$, see Janson~\cite{MR706641}, Beckner~\cite{MR1164616}, and Frank--Ivanisvili~\cite{MR4278128}.\par

Now we present our main result for the Poisson-like semigroup $(P_t)_{t\in\mathbb{R}_+}$ on $\mathbb{Z}_n$. Note that the standard argument, based on a simple series expansion of $\|P_t(1+\epsilon f)\|_q$ and $\|1+\epsilon f\|_p$ at $\epsilon=0$, shows the universal lower bound $t_{p,q}\ge \tfrac{1}{2}\log\big(\tfrac{q-1}{p-1}\big)$. The theorem below records the optimal times $t_{p,q}$ along a dyadic tower of $n$.

Without loss of generality, suppose $\lambda_{j}<\lambda_{j+3}$. The function $F$ strictly increases on $(0,1)$ and strictly decreases on $(1,\infty)$, so from \eqref{eq: opposite F} with $\nu_{j}=\nu_{j+3}=0$ proved in Step 1, we get
\begin{equation}
 F(\lambda_{j}) = F(\lambda_{j+3}).
\end{equation}
And hence
\[
 \lambda_{j}<1<\lambda_{j+3}.
\]
Define $\Theta(x)\coloneqq F(x)-F(2-x)$. Then $\Theta(1)=0$ and $\Theta'(x)=-\frac{2}{3}\log\big(x(2-x)\big)>0$ on $(0,1)$, so $\Theta(x)<0$ on $(0,1)$. Applying this to $x=\lambda_{j}$ gives $F(\lambda_{j})<F(2-\lambda_{j})$. Since $F$ decreases on $(1,\infty)$ and $F(\lambda_{j+3})=F(\lambda_{j})$, we deduce
\begin{equation}\label{Ineq: opposite sum lower bound}
 \lambda_{j+3}>2-\lambda_{j}
 \quad\Rightarrow\quad
 \lambda_{j}+\lambda_{j+3}>2.
\end{equation}\par
\medskip
\textbf{Step 3.} We show that for $\ell,\ell'\in\{0,1,2\}\setminus\{j_0\}$, we have $\lambda_{\ell}=\lambda_{\ell+3}=\lambda_{\ell'}=\lambda_{\ell'+3}$.\par
By the assumption $\lambda_{j_0}\neq \lambda_{j_0+3}$, we have $\lambda_{j_0}+\lambda_{j_0+3}>2$ from Step 2. Hence
\[
 \lambda_{j_0}^2+\lambda_{j_0+3}^2 \ge \frac{(\lambda_{j_0}+\lambda_{j_0+3})^2}{2} > 2.
\]
Consequently, if $\lambda_{j}\neq \lambda_{j+3}$ for all $0\leq j\le 2$, then $\sum_{k=0}^{5}\lambda_k^2>6$, contradicting $\sum_{k=0}^{5}\lambda_k^2<6$ in \eqref{Eqns: Lagrange for n=6, (0,1,2,1,2,1)}. Therefore, there exists at least one $\ell$ with $\lambda_{\ell}=\lambda_{\ell+3}$. Let $\ell$ be an index in $\{0,1,2\}\setminus\{j_0\}$ such that $\lambda_\ell=\lambda_{\ell+3}$ and $\lambda_\ell$ is minimal among all pairs $\{\lambda_j=\lambda_{j+3}\}$ with $j\in\{0,1,2\}\setminus\{j_0\}$. If $\lambda_\ell=\lambda_{\ell+3}\ge 1$, then $\sum_{k=0}^{5}\lambda_k^2>6$, which again contradicts $\sum_{k=0}^{5}\lambda_k^2<6$. So there at least one $\ell$ with $\lambda_{\ell}=\lambda_{\ell+3}<1$. Denote by $\ell'$ the remaining index in $\{0,1,2\}\setminus\{j_0,\ell\}$. \par

\begin{proof}
Recall the case $n=2$ for Theorem~\ref{Thm: Log Sobolev inequality n=6 times 2^k and n=8 times 2^k} is known in \cite{MR420249}, i.e.,
\begin{equation}\label{Ineq: 2-LSI}
 \frac{1}{4}\left( x^2\log\frac{2x^2}{x^2+y^2}+y^2\log\frac{2y^2}{x^2+y^2}\right)
\le \left( \frac{x-y}{2} \right)^2\qquad x,y\geq 0.
\end{equation}
Given $\lambda=(a_0,b_0,a_1,b_1,\ldots,a_{n-1},b_{n-1})$ with $a=(a_0,\ldots,a_{n-1})$ and $b=(b_0,\ldots,b_{n-1})$, the $2n$--entropy splits as
\[
\begin{aligned}
\mathrm{H}_{2n}[\lambda]&=\frac{1}{2n}\sum_{i=0}^{n-1} a_i^2 \log(\frac{2na_i^2}{\sum_{i=0}^{n-1}(a_i^2+b_i^2)})+\frac{1}{2n}\sum_{i=0}^{n-1} b_i^2 \log(\frac{2nb_i^2}{\sum_{i=0}^{n-1}(a_i^2+b_i^2)})\\
&=\frac{1}{2n}\sum_{i=0}^{n-1} a_i^2\log(\frac{n a_i^2}{\sum_{i=0}^{n-1}a_i^2})+\frac{1}{2n}\sum_{i=0}^{n-1} b_i^2\log(\frac{n b_i^2}{\sum_{i=0}^{n-1}b_i^2})\\
 &\quad +\frac{1}{2n}\left(\sum_{i=0}^{n-1} a_i^2\log(\frac{2 \sum_{i=0}^{n-1} a_i^2}{\sum_{i=0}^{n-1}(a_i^2+b_i^2)})+\sum_{i=0}^{n-1} b_i^2\log(\frac{2 \sum_{i=0}^{n-1} b_i^2}{\sum_{i=0}^{n-1}(a_i^2+b_i^2)})\right).
\end{aligned}
\]
Applying the $n$--LSI to the first two terms, and $2$--LSI \eqref{Ineq: 2-LSI} with $x=\norm{a}_2$ and $y=\norm{b}_2$ to the last term, yields
\[
\mathrm{H}_{2n}[\lambda]\le \inner{a,\Gamma(n)a}+\inner{b,\Gamma(n)b}
+\frac{1}{2n}\big(\norm{a}_2-\norm{b}_2\big)^2.
\]
Finally, by Proposition~\ref{Prop: Compare Dirichlet form for n and 2n},
\begin{equation}\label{Eqn: induction from n-LSI to 2n-LSI}
 \mathrm{H}_{2n}[\lambda]\le \inner{a,\Gamma(n)a}+\inner{b,\Gamma(n)b}
+\frac{1}{2n}\big(\norm{a}_2-\norm{b}_2\big)^2
\le 2\inner{\lambda,\Gamma(2n)\lambda}.
\end{equation}\par

For odd $n_0$, define the weight function on $\mathbb{Z}_n$ for even $n\geq n_0$
\begin{equation}\label{Eqn: definition of induced gamma_n for odd base case}
 \gamma_n(k)=
 \begin{cases}
 \psi_n(k), & k\ne \tfrac{n}{2},\\
 1, & k=\tfrac{n}{2}.
 \end{cases}
\end{equation}
The pair $(\psi_{n_0},\gamma_{2n_0})$ satisfies \eqref{Cond: gamma pair condition} and \eqref{Ineq: Quadratic inequality from n to 2n}. For even $n$, the pair $(\gamma_n,\gamma_{2n})$ also satisfies \eqref{Cond: gamma pair condition}, and the desired inequality \eqref{Ineq: Quadratic inequality from n to 2n} becomes
\begin{equation}\label{Ineq: quadratic for odd n}
 f(x)\coloneqq r_b (n-3) x^2+x \left(2 \sqrt{r_a+1} \sqrt{r_b+1}-2\right)+r_a (n-3)\geq 0
\end{equation}
for all $x\ge 0$ and $0\le r_a,r_b\le 1$. Since $n-3>0$, the minimum of $f$ is attained at $x=-\frac{2 \sqrt{r_a+1} \sqrt{r_b+1}-2}{2r_b(n-3)}\leq 0$ and the value of the minimum of $f$ on $[0,\infty)$ is $f(0)=r_a(n-3)\geq 0$. Hence \eqref{Ineq: quadratic for odd n} holds. Therefore Proposition~\ref{Prop: Compare Dirichlet form for n and 2n} applies to the pairs $(\gamma_n,\gamma_{2n})$ defined by \eqref{Eqn: definition of induced gamma_n for odd base case}. Iterating \eqref{Eqn: induction from n-LSI to 2n-LSI} along $\big((\psi_{n_0},\gamma_{2n_0}),(\gamma_{2n_0},\gamma_{4n_0}),\ldots\big)$, we obtain for all $m\ge 0$,
\[
 \mathrm{H}_{n_0\cdot 2^{m+1}}[\lambda]
 \le 2\inner{\lambda,\Gamma(n_0\cdot 2^{m+1})\lambda}.
\]
Let $\Psi(n)=\frac{1}{n}F_n\diag(\psi_n)F_n^{-1}$. Since $\gamma_n\leq \psi_n$ pointwise, we have
\[
 \inner{\lambda,\Gamma(n_0\cdot 2^{m+1})\lambda}
 \le \inner{\lambda,\Psi(n_0\cdot 2^{m+1})\lambda}.
\]
Therefore,
\[
 \mathrm{H}_{n_0\cdot 2^{m+1}}[\lambda]
 \le 2\inner{\lambda,\Psi(n_0\cdot 2^{m+1})\lambda}.
\]
By Lemma~\ref{Lem: Explicit form of Log Sobolev ineq}, the \(n_0\cdot 2^{m+1}\)--LSI holds for
\(\psi_{n_0\cdot 2^{m+1}}\) for all \(m\ge 0\), which proves the claim.
\end{proof}

\textbf{(2) Case $n=2^m$ with $m\geq 1$.} The case $n=2$ is classical and the case $n=4$ is due to the work \cite{MR730056} and Gross's extrapolation technique~\cite{MR420249}. Note that the pair $(\phi_4,\gamma_8)$ satisfies \eqref{Cond: gamma pair condition}, and \eqref{Ineq: Quadratic inequality from n to 2n} becomes
\[
 f(x)\coloneqq \left(2-\frac{r_b}{5}\right)x^2+2x\left(\sqrt{r_a+1}\sqrt{r_b+1}-3\right)
 -\frac{r_a}{5}+2\ \ge 0,
\]
with $0\le r_a,r_b\le 1$ and $x\ge0$. As $2-\frac{r_b}{5}>0$, the minimum of $f$ on $[0,\infty)$ is attained at $x=-\tfrac{2 \left(\sqrt{r_a^2+1} \sqrt{r_b^2+1}-3\right)}{2\left( 2-r_b^2/5 \right)}$ and the value of the minimum is
\[
 \frac{r_a(24r_b+35)+5\left(-30\sqrt{r_a+1}\sqrt{r_b+1}+7r_b+30\right)}
 {5(r_b-10)}
 =\frac{h(r_a,r_b)}{5(r_b-10)}.
\]
Since $r_b\in[0,1]$, the denominator is negative. In order to show \eqref{Ineq: Quadratic inequality from n to 2n} hold, it suffices to show $h(r_a,r_b)\le 0$ on $[0,1]^2$. A direct computation gives
\[
 \pdv[2]{h}{r_a}(r_a,r_b)=\frac{75\sqrt{r_b+1}}{2(r_a+1)^{3/2}}>0,
 \qquad
 \pdv[2]{h}{r_b}(r_a,r_b)=\frac{75\sqrt{r_a+1}}{2(r_b+1)^{3/2}}>0.
\]
Thus $h$ is convex in each variable separately. So the maximum of $h$ on $[0,1]^2$ is attained at a corner. Evaluating,
\[
 h(0,0)=0,\quad h(0,1)=h(1,0)=5\big(37-30\sqrt{2}\big)<0,\quad h(1,1)=-56<0.
\]
Hence $\max_{[0,1]^2} h\le 0$, and since $5(r_b-10)<0$ we conclude
\[
 \frac{r_a(24r_b+35)+5\left(-30\sqrt{r_a+1}\sqrt{r_b+1}+7r_b+30\right)}
 {5(r_b-10)} \ge 0.
\]
Therefore $(\phi_4,\gamma_8)$ satisfies \eqref{Ineq: Quadratic inequality from n to 2n}. Recall that $(\gamma_{8\cdot 2^m},\gamma_{8\cdot 2^{m+1}})$ also satisfies \eqref{Ineq: Quadratic inequality from n to 2n}. Hence by Theorem~\ref{Thm: Log Sobolev inequality n=4} and Theorem~\ref{Thm: induction from n-LSI to 2n-LSI} the $8\cdot 2^m$--LSI holds for $\gamma_{8\cdot 2^m}$ for all $m\ge 0$, and therefore for $\psi_{8\cdot 2^m}$.
\end{proof}

A standard route to hypercontractivity proceeds through Log--Sobolev inequalities (LSI) by Gross's celebrated work~\cite{MR420249}: $\norm{P_{t}f}_q\le \norm{f}_p$ holds whenever $t\geq \frac{C}{4}\log\big(\frac{q-1}{p-1}\big)$ if and only if the corresponding LSI holds with constant $C$. Thus, to prove Theorem~\ref{Thm: Hypercontractivity on tower of n}, it suffices to establish the following $n$--LSI with the optimal constant $2$ along the above dyadic tower of $n$. Denote by $A_{\psi_n}$ the generator of semigroup $(P_t)_{t\in\mathbb{R}_+}$, that is
\begin{equation}
  A_{\psi_n}: \sum_{k=0}^{n-1} a_k\chi_k(x)\mapsto \sum_{k=0}^{n-1} \psi_n(k)a_k\chi_k(x).
\end{equation}
 The case $n=4$ in the following theorem is due to the work \cite{MR730056} and Gross's extrapolation technique~\cite{MR420249}.
\begin{theorem}\label{Thm: Log Sobolev inequality n=6 times 2^k and n=8 times 2^k}
For $n=3\cdot 2^k$ and $n=2^k$ with $k\ge 1$, we have the following LSI with the optimal constant $2$:
\[
  \int_{\mathbb{Z}_n} f^2\log f^2\dd\mu_n-\norm{f}_2^2\log\norm{f}_2^2\le 2\inner{f,A_{\psi_n}f}_{L_2(\mathbb{Z}_n,\mu_n)},\qquad f\in L_2^+(\mathbb{Z}_n,\mu_n).
\]
\end{theorem}

Our proof of Theorem~\ref{Thm: Log Sobolev inequality n=6 times 2^k and n=8 times 2^k} is based on a new induction scheme with three key ingredients:
\begin{enumerate}
  \item Auxiliary weights $\phi_4$ on $\mathbb{Z}_4$ and $\phi_6$ on $\mathbb{Z}_6$ together with their corresponding LSIs. These LSIs are tighter than those for the word-lengths $\psi_n$; this refinement is new even when $n=4$ and plays a key role in the analysis of LSIs.
  \item Karush--Kuhn--Tucker (KKT) analysis for the $4$-- and $6$--LSI.
We develop an efficient way to handle LSIs on $\mathbb{Z}_n$ via KKT analysis, combined with the aforementioned manipulation of the length functions. The specific structure of $\phi_6$ introduces a symmetry in the KKT system that makes the analysis tractable.
  \item An induction from the $n$--LSI to the $2n$--LSI under a crucial compatibility condition. We use a Cooley--Tukey factorization of the $2n$-point discrete Fourier transform (DFT), which expresses a large DFT as a combination of smaller DFTs and yields a comparison of Dirichlet forms at the scales $n$ and $2n$. The choice of the new weights $\phi_4$ and $\phi_6$ mentioned in (1) is crucial for the base steps $4\to 8$ and $6\to 12$.
\end{enumerate}
Finally, we remark that the above ideas are also useful for studying LSIs along other towers of the form $m\cdot n^k$. The KKT analysis and the compatibility condition vary in their technical details, depending on the specific value of $n$ and on the particular choice of weights on $\mathbb{Z}_n$, and we will not carry out this analysis in this paper.\par

The article is organized as follows. After a brief introduction to the LSI formulation and the $n$-dimensional KKT framework in Section~\ref{Sec: DFT and KKT analysis}, we analyze the KKT systems associated with the LSIs for $n=6$ and $n=4$ in Section~\ref{Sec: Log Sobolev inequality n=4 and n=6}. Section~\ref{Sec: Induction from n to 2n} presents a comparison criterion for a pair of weights that allows us to compare the Dirichlet forms at the scales $n$ and $2n$. We then apply it to the transition $n\to 2n$ of LSIs and establish Theorem~\ref{Thm: Log Sobolev inequality n=6 times 2^k and n=8 times 2^k}.

To prove Theorem~\ref{Thm: Hypercontractivity on tower of n}, the text follows the standard route via Log--Sobolev inequalities (LSI): by Gross's equivalence, $\|P_t f\|_q\le \|f\|_p$ holds for $t\ge \frac{C}{4}\log\big(\frac{q-1}{p-1}\big)$ iff the corresponding LSI holds with constant $C$. Hence, it “suffices to establish” an $n$--LSI “with the optimal constant $2$” along the dyadic towers $n=2^k$ and $n=3\cdot 2^k$, namely Theorem~\ref{Thm: Log Sobolev inequality n=6 times 2^k and n=8 times 2^k}.

The proof of Theorem~\ref{Thm: Log Sobolev inequality n=6 times 2^k and n=8 times 2^k} is described as a “new induction scheme” with three ingredients: (1) introduce auxiliary weights $\phi_4$ on $\mathbb{Z}_4$ and $\phi_6$ on $\mathbb{Z}_6$ and prove their LSIs, which are “tighter than those for the word-lengths $\psi_n$,” and are “crucial for the base steps $4\to 8$ and $6\to 12$;” (2) perform “Karush--Kuhn--Tucker (KKT) analysis for the $4$-- and $6$--LSI,” using an “efficient way” via KKT combined with “manipulation of the length functions,” where “the specific structure of $\phi_6$ introduces a symmetry in the KKT system” that makes it tractable; (3) carry out an “induction from the $n$--LSI to the $2n$--LSI under a crucial compatibility condition,” using a “Cooley--Tukey factorization of the $2n$-point discrete Fourier transform (DFT)” to express the large DFT via smaller ones and obtain a “comparison of Dirichlet forms at the scales $n$ and $2n$.” Together these steps yield the LSI theorem with constant $2$, and by the Gross equivalence this gives the stated hypercontractivity threshold for Theorem~\ref{Thm: Hypercontractivity on tower of n}.

Biconditional or Equivalence
Inequality or Bound

Question:
Let \(\mathbb Z_n\) be the cyclic group with normalized counting measure \(\mu_n\), and let the Poisson-like semigroup \((P_t)_{t\ge 0}\) act on Fourier series by
\[
P_t\!:\sum_{k=0}^{n-1} a_k\chi_k(x)\mapsto \sum_{k=0}^{n-1} e^{-t\psi_n(k)}a_k\chi_k(x),
\]
where \(\psi_n(k)=\min(k,n-k)\) and \(\chi_k(x)=e^{2\pi i kx/n}\). For \(n\) of the form \(2^k\) or \(3\cdot 2^k\) with integer \(k\ge 1\), and for \(1<p\le q<\infty\), which quantitative estimate gives exactly when the \(L_p(\mathbb Z_n,\mu_n)\to L_q(\mathbb Z_n,\mu_n)\) hypercontractive bound holds?

For every \(f\in L_p(\mathbb Z_n,\mu_n)\),
\[
\|P_t f\|_{L_q(\mathbb Z_n,\mu_n)}\le \|f\|_{L_p(\mathbb Z_n,\mu_n)}
\]
holds if and only if
\[
t\ge \frac12\log\!\left(\frac{q-1}{p-1}\right).
\]

Choices:
B.
For every \(f\in L_p(\mathbb Z_n,\mu_n)\),
\[
\|P_t f\|_{L_q(\mathbb Z_n,\mu_n)}\le \|f\|_{L_p(\mathbb Z_n,\mu_n)}
\]
holds if and only if
\[
t\ge \log\!\left(\frac{q-1}{p-1}\right).
\]
C.
For every \(f\in L_p(\mathbb Z_n,\mu_n)\), if
\[
t\ge \frac12\log\!\left(\frac{q-1}{p-1}\right),
\]
then
\[
\|P_t f\|_{L_q(\mathbb Z_n,\mu_n)}\le \|f\|_{L_p(\mathbb Z_n,\mu_n)}.
\]
D.
For every \(f\in L_p(\mathbb Z_n,\mu_n)\),
\[
\|P_t f\|_{L_q(\mathbb Z_n,\mu_n)}\le \|f\|_{L_p(\mathbb Z_n,\mu_n)}
\]
holds if and only if there exists a constant \(C_n>0\), depending only on \(n\), such that
\[
t\ge C_n\log\!\left(\frac{q-1}{p-1}\right).
\]
E.
For every \(f\in L_p(\mathbb Z_n,\mu_n)\),
\[
\|P_t f\|_{L_q(\mathbb Z_n,\mu_n)}\le \|f\|_{L_p(\mathbb Z_n,\mu_n)}
\]
holds if and only if
\[
t\ge \frac12\log\!\left(\frac{q}{p}\right).
\]

dropped the 'if and only if' optimality/necessity direction

uniform exact threshold replaced by an unspecified n-dependent constant

replaced the Gross-type ratio (q-1)/(p-1) by q/p

The stem does not explicitly reveal the threshold; it only states the operator and asks for an equivalent condition. There is no direct answer leakage beyond the general topic of hypercontractivity.

This is essentially a direct recall of the sharp hypercontractive threshold for the given semigroup: the correct option restates the theorem in equivalent form rather than asking for a derived or conceptually competing conclusion.

There is some reasoning pressure because one must distinguish an exact equivalence from merely sufficient bounds, especially against option C and nearby constants. However, the item mainly tests theorem recall rather than generating a conclusion from provided information.

The distractors are plausible and mathematically targeted: B alters the sharp factor, C is a weaker true statement but not equivalent, D swaps in an incorrect ratio, and E uses a plausible but nonoptimal constant. These reflect realistic failure modes.

A solid MCQ in terms of distractor design and lack of leakage, but it is largely a theorem-recall question and only weakly tests genuine generative reasoning.


Question:
Let \(\mathbb Z_n\) be the cyclic group with normalized counting measure \(\mu_n\), and let the Poisson-like semigroup \((P_t)_{t\ge 0}\) act on Fourier series by
\[
P_t\!:\sum_{k=0}^{n-1} a_k\chi_k(x)\mapsto \sum_{k=0}^{n-1} e^{-t\psi_n(k)}a_k\chi_k(x),
\]
where \(\psi_n(k)=\min(k,n-k)\) and \(\chi_k(x)=e^{2\pi i kx/n}\). For \(n\) of the form \(2^k\) or \(3\cdot 2^k\) with integer \(k\ge 1\), and for \(1<p\le q<\infty\), which quantitative estimate gives exactly when the \(L_p(\mathbb Z_n,\mu_n)\to L_q(\mathbb Z_n,\mu_n)\) hypercontractive bound holds?

Choices:
B.
For every \(f\in L_p(\mathbb Z_n,\mu_n)\),
\[
\|P_t f\|_{L_q(\mathbb Z_n,\mu_n)}\le \|f\|_{L_p(\mathbb Z_n,\mu_n)}
\]
holds if and only if
\[
t\ge \log\!\left(\frac{q-1}{p-1}\right).
\]
C.
For every \(f\in L_p(\mathbb Z_n,\mu_n)\), if
\[
t\ge \frac12\log\!\left(\frac{q-1}{p-1}\right),
\]
then
\[
\|P_t f\|_{L_q(\mathbb Z_n,\mu_n)}\le \|f\|_{L_p(\mathbb Z_n,\mu_n)}.
\]
D.
For every \(f\in L_p(\mathbb Z_n,\mu_n)\),
\[
\|P_t f\|_{L_q(\mathbb Z_n,\mu_n)}\le \|f\|_{L_p(\mathbb Z_n,\mu_n)}
\]
holds if and only if there exists a constant \(C_n>0\), depending only on \(n\), such that
\[
t\ge C_n\log\!\left(\frac{q-1}{p-1}\right).
\]
E.
For every \(f\in L_p(\mathbb Z_n,\mu_n)\),
\[
\|P_t f\|_{L_q(\mathbb Z_n,\mu_n)}\le \|f\|_{L_p(\mathbb Z_n,\mu_n)}
\]
holds if and only if
\[
t\ge \frac12\log\!\left(\frac{q}{p}\right).
\]

The stem does not state the threshold or the sharp constant. It asks for the exact hypercontractive estimate, so there is no explicit answer leakage beyond indicating that an exact characterization exists.

This is essentially a direct recall of a specific theorem: the sharp 'if and only if' hypercontractive threshold for the given semigroup and class of n. It does not substantially reframe the result or require deriving a new consequence.

Some reasoning is needed to distinguish the exact statement from nearby variants (wrong constant, only one direction, n-dependent constant, wrong ratio), but the item mainly tests recognition of the known theorem rather than genuine generative reasoning.

The distractors are plausible and mathematically meaningful: a factor-of-2 error, a weaker true sufficiency statement, loss of uniform sharpness via an unspecified constant, and an incorrect q/p-type threshold. These reflect common failure modes and are clearly distinct.

A solid theorem-recall MCQ with strong distractors and no direct leakage, but it is largely tautological and only mildly tests reasoning.


\label{thm-main}
Let $a$ and $b$ be integers with $b>a>1$. Suppose that there exist positive integers $N$ and $D,$ and nonnegative integers $x_i, y_i$ such that
$$
N+iD = a^{x_i}+b^{y_i}, \; \; \mbox{ for } \; i \in \{ 0, 1, 2, 3, 4 \}.
$$
Then we have either
$$
(a,b,N,D)=(2,2^k+1,2^k+1,2^k)  \; \mbox{ or } \;  (3,4 \cdot 3^{k-1}+1,3^{k-1}+1,2 \cdot 3^{k-1}) \; \mbox{ for } \; k \in \mathbb{N},
$$
or
$$
\begin{array}{c}
(a,b,N,D) \in \left\{ (2,3,5,2),  (2,3,7,6), (2,3,9,8),  (2,3,17,24),  \right. \\
\left. (2,3,41,24), (2,5,5,8),  (2,9,17,24),  (2,9,41,24), (3,4,7,6) \right\}.
\end{array}
$$

Let $a$ and $b$ be integers with $b>a>1$, and define the set $S_{a,b}$ to be the sumset of the geometric progressions
$$
1, a, a^2, a^3, \ldots \; \mbox{ and } \; 1, b, b^2, b^3, \ldots,
$$
i.e.
$$
S_{a,b} = \{ n \in \mathbb{N} \; : \; n = a^x+b^y, \; \mbox{ for } x, y \in \mathbb{Z}, \; x, y \geq 0 \}.
$$
The main goal of this paper is to study the arithmetic progressions in the sets $S_{a,b}$. These sets are sufficiently thin that, were they random, our expectation for a fixed pair $(a,b)$ would be that $S_{a,b}$ should contain at most finitely many three-term arithmetic progressions, and indeed this is the case ``most'' of the time. By way of example, $S_{2,7}$ contains precisely $22$ such progressions, while, in contrast, $S_{2,9}$ contains infinitely many and $S_{2,8}$ even contains infinitely many four-term arithmetic progressions. It is likely that $S_{a,b}$ contains infinitely many
three-term arithmetic progressions precisely when either $a=2$ and $b=2^k+1$, for $k$ a positive integer, or when $a$ and $b$ are multiplicatively dependent, i.e. there exist positive integers $c$ and $d$ such that $a^c=b^d$. We will discuss this further in Section \ref{Sec9}.

In this paper, we will focus our attention on the problem of characterizing longer arithmetic progressions in the sets $S_{a,b}$.
Our starting point is recent work of Chen, Huang and Zhang \cite{ChHuZh} who showed, via a completely elementary argument, that the longest arithmetic progression in the set
$S_{2,3}$
has length $6$. Regarding such maximal progressions, we prove the following broad generalization of this, characterizing all $5$-term progressions in the sets $S_{a,b}$.

\noindent An almost immediate corollary of this is the following.
\begin{cor} \label{cor-main}
If $a$ and $b$ are integers with $b>a>1$, then the only $6$-term arithmetic progressions in $S_{a,b}$ are
$$
3,5,7,9,11,13 \; \; \mbox{ and } \; \; 17, 41, 65, 89, 113, 137,
$$
in $S_{2,3}$ and in both $S_{2,3}$ and $S_{2,9}$, respectively.
\end{cor}

Suppose that $a$ and $b$ are integers with $b>a \geq 2$ and that we have a $5$-term arithmetic progression in integers of the shape $a^x+b^y$, i.e. that there exist positive integers $N$ and $D,$ and nonnegative integers $x_i, y_i$ such that
$$
N+iD = a^{x_i}+b^{y_i}, \; \; \mbox{ for } 0 \leq i \leq 4.
$$
We will write
$$
n = \left[ \frac{\log (N+4D)}{\log a} \right] \; \mbox{ and } \; m = \left[ \frac{\log (N+4D)}{\log b} \right],
$$
where by $[x]$ we mean the greatest integer $\leq x$.
Let us call a term $N+iD$ in our arithmetic progression {\it $a$-large} if there is some representation of $N+iD = a^{x_i}+b^{y_i}$ with $x_i=n$, {\it $b$-large} if there is a representation of $N+iD = a^{x_i}+b^{y_i}$ with $y_i=m$, and {\it large}, if $N+iD$ is either $a$-large or $b$-large.

\begin{prop} \label{equality}
Let $a$ and $b$ be integers with $b>a>1$. Suppose that there exist positive integers $N$ and $D,$ and nonnegative integers $x_i, y_i$ such that
$$
N+iD = a^{x_i}+b^{y_i}, \; \; \mbox{ for } \; i \in \{ 0, 1, 2, 3, 4 \}.
$$
Suppose further that here exist integers $i, j$ and $k$ with $0 \leq i < j < k \leq 4$ and either
\begin{equation} \label{x}
x_i=x_j=x_k,
\end{equation}
or
\begin{equation} \label{y}
y_i=y_j=y_k.
\end{equation}
Then either
$$
(a,b,N,D)=(2,3,5,2), (2,3,7,6), (2,3,9,8), (3,4,7,6),
$$
\begin{equation} \label{family}
(a,b,N,D)=(2,2^{x_0}+1,2^{x_0}+1,2^{x_0})
\end{equation}
or
\begin{equation} \label{family2}
(a,b,N,D)=(3,4 \cdot 3^{x_0}+1,3^{x_0}+1,2 \cdot 3^{x_0}).
\end{equation}
\end{prop}

Suppose next that precisely three of $N+iD$, $i \in \{ 0, 1, 2, 3, 4 \}$ are large. From our previous work, we may assume that either
\begin{equation} \label{Case1}
\mbox{there exist } 0 \leq i < j \leq 4 \mbox{ and } k \not\in \{ i, j \} \mbox{ with } x_i=x_j=n, \; y_k=m,
\end{equation}
or
\begin{equation} \label{Case2}
\mbox{there exist } 0 \leq i < j \leq 4 \mbox{ and } k \not\in \{ i, j \} \mbox{ with } y_i=y_j=m, \; x_k=n.
\end{equation}
Since two of the terms $N+iD$, $i \in \{ 0, 1, 2, 3, 4 \}$ are not large, in all cases we have inequality (\ref{starty}) with $\kappa=1$.
In particular,
$$
N+D \leq a^{n-1}+b^{m-1} < \left( \frac{1}{a} + \frac{1}{b} \right) (N+4D)
$$
and hence, from $4(N+D)>N+4D$, $\frac{1}{a} + \frac{1}{b} > \frac{1}{4}$,
i.e.
\begin{equation} \label{special2}
a \in \{ 2, 3, 4 \}, \; a=5, \; 6 \leq b \leq 19, \; \; a=6, \; 7 \leq b \leq 11 \; \mbox{ or } \; a=7, \; 8 \leq b \leq 9.
\end{equation}

We therefore may suppose that $a=2$. Our goal is to show, in this case, that we have only $5$-term arithmetic progressions corresponding to
 $$
(a,b,N,D) \in \{ (2,3,17,24), (2,3,41,24), (2,9,17,24), (2,9,41,24) \}.
$$
 Let us assume first that $\delta_1=1$ and that $x_j=n-1$, whence (\ref{awesome}) and (\ref{spooky2}) imply that
 $$
 (i,j,k,\kappa) \in
 \{ (2,3,4,1), (2,4,3,1), (3,4,2,1), (1,3,4,2) \}.
 $$
 We will treat each of these four cases in turn. Suppose first that $(i,j,k,\kappa)=(2,3,4,1)$. Then
 $$
 \left( b^{\delta_2} -1 \right) b^{m-\delta_2} = 2^n-2^{x_2+1}
 $$
 and
 $$
2^n = 3 \cdot b^m - 2 b^{y_4} - b^{m-\delta_2},
 $$
 from (\ref{gaewA}) and (\ref{gaewB}), respectively. The second of these implies that
 $$
 2^n \geq 3 (b^m-b^{m-1}) > b^m
 $$
 and hence we can apply Theorem 1.6 of \cite{Be-Pill} to the first equation to conclude that $b=3$ and that
 $$
 (n,m,x_2,\delta_2) \in \{ (5,3,2,2), (8,5,3,4) \}.
 $$
 Neither coincides with a $5$-term progression (in each case $N+4D \not\in S_{2,3}$). If $(i,j,k,\kappa)=(2,4,3,1)$ or $(3,4,2,1)$, from the identity $(N+D)+(N+4D)=(N+2D)+(N+3D)$, we find that
 $$
 b^{m-\delta_2} - b^{y_k} = 2^{x_i},
 $$
 and so Lemma \ref{tech-lem} implies that
 we have one of
 \begin{equation} \label{ann1}
 b=2^{x_i}+1, \; m-\delta_2=1, \; y_k=0,
\end{equation}
\begin{equation} \label{ann3}
b=3, \; m-\delta_2=2, \; y_k=0, \; x_i=3,
\end{equation}
or
\begin{equation} \label{ann4}
b=9, \; m-\delta_2=1, \; y_k=0, \; x_i=3.
\end{equation}
In the first case, if $(i,j,k,\kappa)=(2,4,3,1)$, we have
$$
D = 2^{n-2}-2^{x_2-1} = \frac{1}{3} 2^{x_2} ( 2^{x_2}+1)^{m-1},
$$
a contradiction modulo $2^{x_2}$. If, on the other hand, we have (\ref{ann1}) and $(i,j,k,\kappa)=(3,4,2,1)$,
$$
D = 2^{n-1}-2^{x_3} = \frac{1}{3} 2^{x_3} ( 2^{x_3}+1)^{m-1},
$$
so that
$$
2^{n-1-x_3}-1 = \frac{1}{3} ( 2^{x_3}+1)^{m-1}.
$$
Modulo $3$, $x_3$ is odd and $m \geq 2$. Thus, modulo $8$, we find that either $x_3=1$, $m=2$, $n=3$, or $x_3=1$, $m=3$, $n=4$, or $x_3=3$, $m=2$, $n=4$. Only the first of these corresponds to a $5$-term arithmetic progression in $S_{a,b}$, namely $(a,b,N,D)=(2,3,5,2)$, which we have previously encountered. If $(i,j,k,\kappa)=(2,4,3,1)$, (\ref{gaewA}) becomes
$$
\left( b^{\delta_2} -1 \right) b^{m-\delta_2} = 3 \cdot (2^{n-2}-2^{x_2-1}),
$$
while $(i,j,k,\kappa)=(3,4,2,1)$ yields
$$
\left( b^{\delta_2} -1 \right) b^{m-\delta_2} = 3 \cdot (2^{n-1} - 2^{x_3}).
$$
In cases (\ref{ann3}) and (\ref{ann4}), we may thus apply Proposition \ref{DT} to these equations, concluding after a short computation with the $5$-term arithmetic progressions corresponding to
$$
(a,b,N,D) = (2,3,17,24) \; \mbox{ and } \; (2,9,17,24).
$$
Finally,
 if $(i,j,k,\kappa)=(1,3,4,2)$, then, from $(N+D)+(N+4D)=(N+2D)+(N+3D)$, we have
 $$
 b^{m-\delta_2}-b^{y_4}=2^{x_1},
 $$
 while (\ref{gaewB}) gives
 $$
 2^{n-1} = 2 b^m -b^{y_4} - b^{m-\delta_2}.
 $$
 Lemma \ref{tech-lem} thus implies that $y_4=0$ and so combining the preceding two equations,
 $$
 b^m =2^{n-2} +2^{x_1-1}+1.
 $$
 Since necessarily $m \geq 2$ (else we have three equal $y_i$), we may appeal to Theorem 1 of \cite{BeBuMi} to conclude that
 $$
 b^m \in \{ 7^2, 23^2, 3^4 \},
 $$
 so that
 $$
 (b,m,n,x_1) \in \{ (7,2,7,5), (23,2,11,5), (3,4,8,5), (9,2,8,5) \};
 $$
 none of these correspond to a $5$-term arithmetic progression in $S_{a,b}$.

If $S_{a,b}$ contains a $6$-term arithmetic progression, then, from Theorem \ref{thm-main}, we require $N+5D \in S_{a,b}$ for
$$
(a,b,N,D)=(2,2^k+1,2^k+1,2^k) \; \mbox{ or } \; (3,4 \cdot 3^{k-1}+1,3^{k-1}+1,2 \cdot 3^{k-1}) \; \mbox{ for } \; k \in \mathbb{N},
$$
or
$$
\begin{array}{c}
(a,b,N,D) \in \left\{ (2,3,5,2), (2,3,7,6), (2,3,9,8), (2,3,17,24), \right. \\
\left. (2,3,41,24), (2,5,5,8), (2,9,17,24), (2,9,41,24), (3,4,7,6) \right\}.
\end{array}
$$
A short check of the latter $9$ sporadic cases, reveals the $6$-term progressions corresponding to
$$
(a,b,N,D) = (2,3,17,24) \; \mbox{ and } \; (2,9,17,24),
$$
and no $7$-term progressions. Suppose that
$(a,b,N,D)=(2,2^k+1,2^k+1,2^k)$, for $k$ a positive integer, and that
$$
N+5D = 2^k+1 + 5 \cdot 2^k =3 \cdot 2^{k+1}+1= 2^{x_5} + (2^k+1)^{y_5},
$$
for $x_5$ and $y_5$ nonnegative integers. If $k=1$, then $x_5=y_5=2$, so that $(a,b,N,D) = (2,3,3,2)$. If $k \geq 2$, then
$$
(2^k+1)^2 \geq 3 \cdot 2^{k+1}+1,
$$
whence necessarily $y_5 \in \{ 0, 1 \}$. Thus
$$
3 \cdot 2^{k+1} = 2^{x_5} \; \mbox{ or } \; 2^{k+2}+1= 2^{x_5},
$$
each an immediate contradiction.
Finally, suppose that $(a,b,N,D)=(3,4 \cdot 3^{k-1}+1,3^k+1,2 \cdot 3^k)$ for $k$ a positive integer, and that
$$
N+5D = 3^{k-1}+1+ 10 \cdot 3^{k-1} = 11 \cdot 3^{k-1}+1=3^{x_5}+(4 \cdot 3^{k-1}+1)^{y_5},
$$
with $x_5$ and $y_5$ nonnegative integers. For each positive integer $k$, we have
$$
(4 \cdot 3^{k-1}+1)^2 > 11 \cdot 3^{k-1}+1
$$
and so $y_5 \in \{ 0, 1 \}$, corresponding to
$$
11 \cdot 3^{k-1}=3^{x_5} \; \mbox{ and } \;
7 \cdot 3^{k-1}=3^{x_5},
$$
respectively. These contradictions complete the proof of
Corollary \ref{cor-main}.

\noindent An almost immediate corollary of this is the following.
\begin{cor} \label{cor-main}
If $a$ and $b$ are integers with $b>a>1$,  then the only $6$-term arithmetic progressions in $S_{a,b}$ are
$$
3,5,7,9,11,13 \; \; \mbox{ and } \; \; 17, 41, 65, 89, 113, 137,
$$
in $S_{2,3}$ and in both $S_{2,3}$ and $S_{2,9}$, respectively.
\end{cor}

Our arguments, in contrast to those of  \cite{ChHuZh}, rely, implicitly or explicitly, upon various results from Diophantine approximation, including bounds for linear forms in logarithms, both complex and $p$-adic, and upon various Diophantine consequences of the modularity of Frey-Hellegouarch curves. The outline of this paper is the following.
In Section \ref{Sec2}, we present the various results we will need for explicitly solving certain polynomial-exponential and $3, 4$ and $5$-term $S$-unit equations. Sections \ref{Sec3}--\ref{Sec7} are devoted to dealing with $5$-term arithmetic progressions in $S_{a,b}$ with fixed numbers of ``large'' terms, proving Theorem \ref{thm-main}. In Section \ref{Sec8}, we prove Corollary \ref{cor-main}. Finally, in Section \ref{Sec9}, we discuss various families of pairs $(a,b)$ for which the $S_{a,b}$ are known to have $3$ or $4$-term progressions.

The post-theorem introduction gives only a high-level outline for proving Theorem~\ref{thm-main}: the arguments (unlike \cite{ChHuZh}) "rely, implicitly or explicitly, upon various results from Diophantine approximation, including bounds for linear forms in logarithms, both complex and $p$-adic," and "upon various Diophantine consequences of the modularity of Frey-Hellegouarch curves." In Section~\ref{Sec2} the paper "present[s] the various results" needed to explicitly solve certain polynomial-exponential and $3,4,5$-term $S$-unit equations, and Sections~\ref{Sec3}--\ref{Sec7} "deal[] with $5$-term arithmetic progressions in $S_{a,b}$ with fixed numbers of \`\`large\'\' terms," thereby proving Theorem~\ref{thm-main}.

The post-theorem introduction gives only a high-level outline for proving the main theorem: the arguments (unlike \cite{ChHuZh}) "rely, implicitly or explicitly, upon various results from Diophantine approximation, including bounds for linear forms in logarithms, both complex and $p$-adic," and "upon various Diophantine consequences of the modularity of Frey-Hellegouarch curves." Next the paper "present[s] the various results" needed to explicitly solve certain polynomial-exponential and $3,4,5$-term $S$-unit equations, and later sections "deal[] with $5$-term arithmetic progressions in $S_{a,b}$ with fixed numbers of ``large'' terms," thereby establishing the main theorem.

Question:
Let $a$ and $b$ be integers with $b>a>1$. Suppose there exist positive integers $N$ and $D$, and nonnegative integers $x_i,y_i$ for $i=0,1,2,3,4$, such that
$$
N+iD=a^{x_i}+b^{y_i}
$$
for each $i\in\{0,1,2,3,4\}$. In other words, the five terms $N,N+D,N+2D,N+3D,N+4D$ of an arithmetic progression are all representable in the form $a^x+b^y$ with $x,y\ge 0$. Which of the following conclusions about the quadruple $(a,b,N,D)$ holds?

One must have either
$$
(a,b,N,D)=(2,2^k+1,2^k+1,2^k)
\quad\text{or}\quad
(a,b,N,D)=(3,4\cdot 3^{k-1}+1,3^{k-1}+1,2\cdot 3^{k-1})
$$
for some positive integer $k$, or else
$$
(a,b,N,D)\in\{(2,3,5,2),(2,3,7,6),(2,3,9,8),(2,3,17,24),(2,3,41,24),(2,5,5,8),(2,9,17,24),(2,9,41,24),(3,4,7,6)\}.
$$

Choices:
B.
One must have either
$$
(a,b,N,D)=(2,2^k+1,2^k+1,2^k)
\quad\text{or}\quad
(a,b,N,D)=(3,4\cdot 3^{k-1}+1,3^{k-1}+1,2\cdot 3^{k-1})
$$
for some nonnegative integer $k$, or else
$$
(a,b,N,D)\in\{(2,3,5,2),(2,3,7,6),(2,3,9,8),(2,3,17,24),(2,3,41,24),(2,5,5,8),(2,9,17,24),(2,9,41,24),(3,4,7,6)\}.
$$
C.
One must have either
$$
(a,b,N,D)=(2,2^k+1,2^k+1,2^k)
$$
for some positive integer $k$, or else
$$
(a,b,N,D)\in\{(2,3,5,2),(2,3,7,6),(2,3,9,8),(2,3,17,24),(2,3,41,24),(2,5,5,8),(2,9,17,24),(2,9,41,24),(3,4,7,6),(3,4\cdot 3^{k-1}+1,3^{k-1}+1,2\cdot 3^{k-1})\text{ for some }k\in\mathbb N\}.
$$
D.
One must have either
$$
(a,b,N,D)=(2,2^k+1,2^k+1,2^k)
\quad\text{or}\quad
(a,b,N,D)=(3,4\cdot 3^{k-1}+1,3^{k}+1,2\cdot 3^{k-1})
$$
for some positive integer $k$, or else
$$
(a,b,N,D)\in\{(2,3,5,2),(2,3,7,6),(2,3,9,8),(2,3,17,24),(2,3,41,24),(2,5,5,8),(2,9,17,24),(2,9,41,24),(3,4,7,6)\}.
$$
E.
One must have either
$$
(a,b,N,D)=(2,2^k+1,2^k+1,2^k)
\quad\text{or}\quad
(a,b,N,D)=(3,4\cdot 3^{k-1}+1,3^{k-1}+1,2\cdot 3^{k-1})
$$
for some positive integer $k$, and in addition one necessarily has $a\in\{2,3\}$; equivalently, outside these two infinite families one must have
$$
(a,b,N,D)\in\{(2,3,5,2),(2,3,7,6),(2,3,9,8),(2,3,17,24),(2,3,41,24),(2,5,5,8),(2,9,17,24),(2,9,41,24)\}.
$$

The stem does not state or strongly hint at the exact classification. It only poses the hypothesis and asks for the full description of all admissible quadruples.

This is essentially a direct theorem-recall item: the stem asks for the complete classification, and the correct option is the theorem's conclusion almost verbatim.

There is some pressure to compare exactness, parameter ranges, and sporadic exceptions across choices, especially against the weaker-true option. However, the item mainly tests recall/recognition of a known classification rather than generating a conclusion from mathematical reasoning.

The distractors are plausible and well targeted: one weakens the claim, one alters a boundary condition, one perturbs a formula in an infinite family, and one adds a plausible extra sporadic case. These reflect realistic mathematical failure modes.

Well-constructed distractors and no answer leakage, but the question is largely a theorem-statement recognition task rather than a genuinely generative reasoning problem.


A surjective endomorphism of a projective algebraic variety $f\colon X
\to X$ is said to be \textit{int-amplified} if there exists an ample
Cartier divisor $A$ on $X$ such that $f^*A - A$ is ample. Equivalently,
$f$ is int-amplified if all eigenvalues of the operator $f^* \colon
N^1(X)_{\mathbb{C}} \to N^1(X)_{\mathbb{C}}$ have magnitude greater than $1$
\cite[Theorem 1.1]{Meng-Building-Blocks}, where $N^1(X)$ is the group
of divisors on $X$ up to numerical equivalence. Roughly speaking, an
endomorphism is int-amplified if it expands the algebraic variety ``in
all directions.''

Whether a given variety admits any int-amplified endomorphism is an
interesting question of algebraic dynamics and seems to be a rare
property. Abelian varieties admit int-amplified endomorphisms (e.g. the
multiplication-by-$m$ endomorphism) and normal toric varieties admit
int-amplified endomorphisms (e.g. the $m$-th power Frobenius morphism).
Meng and Zhong conjecture \cite[Question 1.2]{Meng-Rigidity} that a
smooth projective rationally connected complex variety admits an
int-amplified endomorphism if and only if it is a toric variety (see
also \cite[Question 4.4]{Fakhruddin}). This has been proved for smooth
surfaces \cite[Theorem 3]{Nakayama02} and smooth Fano threefolds
\cite[Theorem 1.4]{Meng-Zhang-Zhong}, \cite[Theorem
6.1]{Totaro-Log-Bott}.

When we consider singular Fano varieties, we find examples that admit
int-amplified endomorphisms which are not toric, but are finite
quotients of toric varieties.
\begin{example}\label{quaternion-example}
Consider the two-dimensional representation of the quaternion group
given by
\[
 i\mapsto \begin{pmatrix} i&0\\0&-i\end{pmatrix},\quad
 j\mapsto \begin{pmatrix}0&1\\-1&0\end{pmatrix},\quad
 k=ij\mapsto \begin{pmatrix}0&i\\i&0\end{pmatrix}.
\]
Extending this to a projective representation in $\PGL(3)$, we can
consider it as an action of $Q_8$ on $\mathbb{P}^2$. Let $X$ be the quotient of
$\mathbb{P}^2$ by this action. Then the image of $[0:0:1]$ is a $D_4$
singularity: $D_4$ singularities are not toric so $X$ cannot be a toric
variety. Nevertheless we can easily see that the ``Frobenius
endomorphism'' $[x:y:z] \mapsto [x^5:y^5:z^5]$ of $\mathbb{P}^2$ commutes with
the action of $Q_8$ and thus descends to an int-amplified endomorphism
of $X$.
\end{example}

In view of such examples, it is a folklore conjecture that a klt Fano
variety admits an int-amplified endomorphism if and only if it a finite
quotient of a toric variety. Nakayama proved \cite[Lemma
2.6]{NakayamaStruct} that a quotient of a toric surface by a finite
group which acts freely in codimension one and preserves the open torus
inherits an int-amplified endomorphism from the toric variety. Nakayama
also proved that for a klt del Pezzo surface $X$ of Picard rank $>1$,
$X$ admits an int-amplified endomorphism iff $X$ is a quotient of a
toric variety by a finite group acting freely in codimension $1$ such
that the action preserves the open torus \cite[Theorem 1.3]{Nakayama}.

In this article, I study the case of Picard rank $1$ and consider del
Pezzo surfaces with Gorenstein singularities. Note that an
endomorphism of a projective variety of Picard rank $1$ is int-amplified
if and only if it has degree greater than one, since in this case
$N^1(X)_{\mathbb{C}} \cong \mathbb{C}$ and $f^*: N^1(X)_{\mathbb{C}} \to N^1(X)_{\mathbb{C}}$ is just
multiplication by $(\deg f)^{\dim X}$. Define $S'(E_8)$ as the weighted
sextic surface $\{ X^5Y + X^4Z + Z^3+ W^2 = 0 \} \subset \mathbb{P}(1, 1, 2, 3)$
\cite{Gurjar}: it is a Gorenstein del Pezzo surface of Picard rank 1.
Then the main theorem of this paper is the following.

\begin{abstract}
We prove that, in all except one case, a Gorenstein del Pezzo surface of
Picard rank $1$ admits an int-amplified endomorphism if and only if it
is a quotient of a toric variety by a finite group which acts freely in
codimension one and preserves the open torus. We classify all such
quotients.
\end{abstract}
\maketitle

When we consider singular Fano varieties, we find examples that admit
int-amplified endomorphisms which are not toric, but are finite
quotients of toric varieties.
\begin{example}\label{quaternion-example}
Consider the two-dimensional representation of the quaternion group
given by
\[
 i\mapsto \begin{pmatrix} i&0\\0&-i\end{pmatrix},\quad
 j\mapsto \begin{pmatrix}0&1\\-1&0\end{pmatrix},\quad
 k=ij\mapsto \begin{pmatrix}0&i\\i&0\end{pmatrix}.
\]
Extending this to a projective representation in $\PGL(3)$, we can
consider it as an action of $Q_8$ on $\P^2$. Let $X$ be the quotient of
$\P^2$ by this action. Then the image of $[0:0:1]$ is a $D_4$
singularity: $D_4$ singularities are not toric so $X$ cannot be a toric
variety. Nevertheless we can easily see that the ``Frobenius
endomorphism'' $[x:y:z] \mapsto [x^5:y^5:z^5]$ of $\P^2$ commutes with
the action of $Q_8$ and thus descends to an int-amplified endomorphism
of $X$.
\end{example}

In view of such examples, it is a folklore conjecture that a klt Fano
variety admits an int-amplified endomorphism if and only if it a finite
quotient of a toric variety. Nakayama proved \cite[Lemma
2.6]{NakayamaStruct} that a quotient of a toric surface by a finite
group which acts freely in codimension one and preserves the open torus
inherits an int-amplified endomorphism from the toric variety. Nakayama
also proved that for a klt del Pezzo surface $X$ of Picard rank $>1$,
$X$ admits an int-amplified endomorphism iff $X$ is a quotient of a
toric variety by a finite group acting freely in codimension $1$ such
that the action preserves the open torus \cite[Theorem 1.3]{Nakayama}.

In this article, I study the case of Picard rank $1$ and consider del
Pezzo surfaces with Gorenstein singularities. Note that an
endomorphism of a projective variety of Picard rank $1$ is int-amplified
if and only if it has degree greater than one, since in this case
$N^1(X)_{\C} \isom \C$ and $f^*: N^1(X)_{\C} \to N^1(X)_{\C}$ is just
multiplication by $(\deg f)^{\dim X}$. Define $S'(E_8)$ as the weighted
sextic surface $\{ X^5Y + X^4Z + Z^3+ W^2 = 0 \} \subset \P(1, 1, 2, 3)$
\cite{Gurjar}: it is a Gorenstein del Pezzo surface of Picard rank 1.
Then the main theorem of this paper is the following.

We give an overview of rank $1$ Gorenstein del Pezzo surfaces below and
their classification by singularity type. To prove Theorem
\ref{main_theorem}, we first identify which rank $1$ Gorenstein del
Pezzo surfaces are quotients of toric surfaces. The ``easy direction''
(that is, the statement that if $X$ is such a quotient of a toric
variety, it admits an int-amplified endomorphism) follows from
\cite[Lemma 2.6]{NakayamaStruct}. For the rest, following Kawakami and
Totaro \cite{Kawakami-Totaro}, we use the tool of Bott vanishing,
combined with lifting to universal covers, to deduce that they do not
admit int-amplified endomorphisms. To find failures of Bott vanishing,
we use a version of the Riemann-Roch formula for normal surfaces. This
formula is automated in a Python program which is in the accompanying
code repository \cite{CodeRepo}.

We work in characteristic zero, but everything in this paper should
hold over an algebraically closed field of characteristic $\neq 2, 3$,
as long as we everywhere replace ``int-amplified endomorphism'' with
``int-amplified endomorphism of degree invertible in the base field''.
The classification of rank $1$ Gorenstein del Pezzo surfaces is the same
over algebraically closed fields of characteristic $\neq 2, 3$
\cite[Theorem B.6]{Lacini}, and all the endomorphisms described in this
paper have degree coprime to $2$ and $3$.

\begin{proposition}\label{toric_quotient} The five surfaces
$S(3A_1+D_4)$, $S(A_1+2A_3)$, $S(A_1+A_2+A_5)$, $S(4A_2)$ and
$S(2A_1+2A_3)$ are quotients of toric surfaces by a finite group that
acts freely in codimension one and preserves the open torus.
\end{proposition}
\begin{proof}
We describe each of these surfaces as a quotient of a toric variety, and
find an int-amplified endomorphism of the toric variety that descends to
the quotient (we can also apply \cite[Lemma 2.6]{NakayamaStruct} to show
that the quotients admit int-amplified endomorphisms).
\begin{itemize}
\item $S(3A_1+D_4)$ is the surface constructed in Example
\ref{quaternion-example}.

The quasi-universal covers of $S(A_2+E_6)$ and $S'(A_2+E_6)$ are
Gorenstein del Pezzo surfaces of rank $3$ with a single $D_4$
singularity. The quasi-universal covers of $S(A_1+E_7)$ and
$S'(A_1+E_7)$ are Gorenstein del Pezzo surfaces of rank $2$ with a
single $E_6$ singularity. Let $Y$ be one of the four surfaces
$S(A_2+E_6)$, $S'(A_2+E_6)$, $S(A_1+E_7)$ or $S'(A_1+E_7)$ and let $\pi:
X \to Y$ be the quasi-universal cover. Since $D_4$ and $E_6$
singularities are not toric, $X$ is not a toric variety. Furthermore, we
know $f^{-1}(Y^0)$ is simply-connected and is $X^0$ minus a finite set
of points: since $X^0$ is smooth, $X^0$ is also simply-connected. Thus
$X$ cannot be a quotient of a toric variety by a finite group acting
freely in codimension one. However, by \cite[Theorem 1.3]{Nakayama}, if
a klt del Pezzo surface of Picard rank $>1$ admits a endomorphism of
degree $>1$, it is a quotient of a toric variety by a finite group
acting freely in codimension one. If $Y$ admitted an int-amplified
endomorphism, it would lift to endomorphism of $X$ of degree $>1$, which
is impossible. Thus the surfaces $S(A_2+E_6)$, $S'(A_2+E_6)$,
$S(A_1+E_7)$ and $S'(A_1+E_7)$ do not admit int-amplified endomorphisms.

\begin{theorem}\label{main_theorem} Let $X$ be a Gorenstein del Pezzo
surface of Picard rank $1$ over an algebraically closed field of
characteric zero which is not isomorphic to the surface $S'(E_8)$. Then
$X$ admits an int-amplified endomorphism (i.e. an endomorphism of degree
$>1$) if and only if $X$ is a quotient of a toric variety by a finite
group acting freely in codimension one that preserves the open torus.
\end{theorem}

We give an overview of rank $1$ Gorenstein del Pezzo surfaces below and
their classification by singularity type. To prove Theorem
\ref{main_theorem}, we first identify which rank $1$ Gorenstein del
Pezzo surfaces are quotients of toric surfaces. The ``easy direction''
(that is, the statement that if $X$ is such a quotient of a toric
variety, it admits an int-amplified endomorphism) follows from
\cite[Lemma 2.6]{NakayamaStruct}. For the rest, following Kawakami and
Totaro \cite{Kawakami-Totaro}, we use the tool of Bott vanishing,
combined with lifting to universal covers, to deduce that they do not
admit int-amplified endomorphisms. To find failures of Bott vanishing,
we use a version of the Riemann-Roch formula for normal surfaces. This
formula is automated in a Python program which is in the accompanying
code repository \cite{CodeRepo}.

We work in characteristic zero, but everything in this paper should
hold over an algebraically closed field of characteristic $\neq 2, 3$,
as long as we everywhere replace ``int-amplified endomorphism'' with
``int-amplified endomorphism of degree invertible in the base field''.
The classification of rank $1$ Gorenstein del Pezzo surfaces is the same
over algebraically closed fields of characteristic $\neq 2, 3$
\cite[Theorem B.6]{Lacini}, and all the endomorphisms described in this
paper have degree coprime to $2$ and $3$.

\subsection*{Acknowledgments} I would like to thank Burt Totaro for many
helpful conversations. The author is supported by NSF Graduate Research
Fellowship Grant No. DGE-2034835.

To prove Theorem~\ref{main_theorem}, we first identify which rank $1$ Gorenstein del Pezzo surfaces are quotients of toric surfaces. The ``easy direction'' (if $X$ is such a quotient of a toric variety, then it admits an int-amplified endomorphism) follows from \cite[Lemma 2.6]{NakayamaStruct}. For the remaining direction, following Kawakami and Totaro \cite{Kawakami-Totaro}, we use Bott vanishing, combined with lifting to universal covers, to deduce that the other surfaces do not admit int-amplified endomorphisms. To find failures of Bott vanishing, we use a version of the Riemann--Roch formula for normal surfaces (implemented in a Python program in \cite{CodeRepo}).

Biconditional or Equivalence
Implication

Question:
Let $X$ be a Gorenstein del Pezzo surface of Picard rank $1$ over an algebraically closed field of characteristic $0$, and assume that $X$ is not isomorphic to the weighted sextic surface
$$S'(E_8)=\{X^5Y+X^4Z+Z^3+W^2=0\}\subset \mathbb{P}(1,1,2,3).$$
Recall that a surjective endomorphism $f\colon X\to X$ is called int-amplified if there exists an ample Cartier divisor $A$ on $X$ such that $f^*A-A$ is ample; for Picard-rank-$1$ surfaces, this is equivalent to $\deg f>1$. A toric variety has a dense open torus, called its open torus. Under these assumptions, which statement about $X$ holds?

$X$ admits an int-amplified endomorphism (equivalently, an endomorphism of degree $>1$) if and only if $X$ is a quotient of a toric variety by a finite group that acts freely in codimension one and preserves the open torus.

Choices:
B. $X$ admits an int-amplified endomorphism (equivalently, an endomorphism of degree $>1$) if and only if $X$ is itself a toric variety.
C. If $X$ is a quotient of a toric variety by a finite group that acts freely in codimension one and preserves the open torus, then $X$ admits an int-amplified endomorphism (equivalently, an endomorphism of degree $>1$).
D. $X$ admits an int-amplified endomorphism (equivalently, an endomorphism of degree $>1$) if and only if $X$ is a quotient of a toric variety by a finite group that preserves the open torus.
E. $X$ admits an int-amplified endomorphism (equivalently, an endomorphism of degree $>1$) whenever $X$ is a quotient of a toric variety by a finite group that acts freely in codimension one and preserves the open torus, and conversely every such endomorphism exists only after replacing $X$ by a finite cover that is toric.

existence on $X$ versus only after passing to a toric finite cover

The stem gives background definitions and hypotheses but does not explicitly reveal the correct equivalence. It does not single out the quotient-of-toric characterization in a way that makes the answer trivial.

The item is close to a theorem-recall question: under the stated hypotheses, it asks which exact characterization holds. However, it is not a pure restatement because the options include weaker, stronger, and subtly altered variants.

Some reasoning is needed to compare the logical strength of the choices, especially distinguishing the exact iff statement from a weaker one-way implication and from overstrong or hypothesis-dropped alternatives. Still, it mainly tests precise theorem recognition rather than deeper derivation.

The distractors are mathematically plausible and well-targeted: one is too strong (toric instead of quotient-of-toric), one is a weaker true statement, one drops a crucial codimension-one freeness condition, and one introduces a misleading finite-cover reformulation.

A solid MCQ with little answer leakage and strong distractors, though it leans more toward exact theorem recall than genuinely generative mathematical reasoning.


\begin{theorem} \label{thm:main}
    For any sequence of binary alphabet $\A_1,\A_2,\dots$, there exists an infinite cube-free word in $\A_1\times\A_2\times\dots$.
\end{theorem}

\begin{lemma}\label{lem:eigenvector}
    Let $p=12$. There exist nonnegative weights $(\lambda_s)_{s\in \minsuff}$ with $\lambda_{\varepsilon}>0$ such that for every word $w$ over the alphabet $\mathbb{N}$
    and every $\A\in \binom{\mathbb{N}}{2}$,
    \begin{equation}\label{ineq:eigen}
        \alpha \lambda_{\minsuff(w)} \le \sum_{\substack{ c\in\A\\wc\in \widetilde{\mathcal{C}}^{(p)}}} \lambda_{\minsuff(wc)},
    \end{equation}
    where $\alpha = 1.457$.
\end{lemma}

A \emph{square} (\textit{resp.} a \emph{cube}) is a word $uu$ (\textit{resp.} $uuu$) for some nonempty word $u$. A \emph{cube-free word} (resp. \emph{square-free word}) is a word that contains no cube (resp. square) as a factor.
Thue proved that there exists an infinite cube-free word over the binary alphabet and an infinite square-free word over the ternary alphabet \cite{Thue1}. These results are regarded as the first results in combinatorics on words, and many generalizations of these problems have been considered.

One such generalization is the notion of nonrepetitive coloring of graphs introduced by Alon et al. \cite{Alon2002}. A graph coloring (of the edges or of the vertices) is said to be \emph{nonrepetitive} if the sequence of colors along any path avoids squares (see \cite{WoodThueChoiceNumber} for a recent survey on this topic). This notion naturally led to the notion of nonrepetitive list-coloring, where instead of having one fixed set of colors every vertex has to choose a color from a list of colors specific to this vertex. The \emph{Thue-list number} of a graph is the smallest integer $k$ such that if all the lists have size at least $k$ then the graph can be nonrepetitively colored in such a way that each vertex receives a color from its list. In this article, we will consider a variant of the following challenging conjecture.
\begin{conjecture}[{\cite[\ldots]{rosenfeldThueList, CzerwinskiThueChoiceNumber, GagolThueChoiceNumber, GrytczukSurveyThueChoiceNumber, GrytczukThueChoiceNumber, WoodThueChoiceNumber,ZhaoThueChoiceNumber,Grytczuk2011Jan,Fiorenzi2011Oct,Skrabulakova2015Aug}}]\label{conj:ThueListe}
 The Thue list number of the infinite path is $3$.
\end{conjecture}

We provide a formulation of this conjecture in terms of combinatorics on words. By an \emph{alphabet}, we mean any finite set. Given a sequence $(\A_i)_{i\ge1}$ of alphabets, we say that a word $w=w_1\ldots w_n$ \emph{respects} $(\A_i)_{i\ge 1}$, if for all $i$, $w_i\in\A_i$ (the definition naturally extends to infinite words). Conjecture \ref{conj:ThueListe} can be rephrased as follows.
\begin{conjecture}
Given a sequence of alphabets $(\A_i)_{i\ge1}$ with $|A_i|\ge3$ for all $i$, there exists an infinite square-free word that respects $(\A_i)_{i\ge 1}$.
\end{conjecture}

Without loss of generality, we can assume that all the $\A_i$ are sets of integers, so we adopt this convention for the sake of notation in the rest of the article. It was first proven by Grytczuk, Przyby{\l}o, and Zhu that the conjecture is true if the condition $|\A_i|\ge3$ is replaced by $|\A_i|\ge4$ \cite{Grytczuk2011Jan}.
The second author recently proved that if $|\bigcup_{i\ge1} \A_i|=4$ and $|\A_i|\ge3$ for every $i$ then there exists an infinite square-free word that respects $(\A_i)_{i\ge 1}$ \cite{rosenfeld2022avoiding}. This result is far from proving the conjecture, but the author argues that if it were not for the computational limitations, the technique used in this could certainly prove the full conjecture. Indeed, this technique requires one to verify by a computer-assisted proof some growth property of a very large automaton associated with square-free words.

In the current article, we consider a natural variant of the question. Since cubes are avoidable over the binary alphabet, is it possible to avoid cubes if all of the lists are of size $2$? We give a positive answer to this question in the following theorem.

\begin{conjecture}[{\cite[\ldots]{rosenfeldThueList, CzerwinskiThueChoiceNumber, GagolThueChoiceNumber, GrytczukSurveyThueChoiceNumber, GrytczukThueChoiceNumber, WoodThueChoiceNumber,ZhaoThueChoiceNumber,Grytczuk2011Jan,Fiorenzi2011Oct,Skrabulakova2015Aug}}]\label{conj:ThueListe}
  The Thue list number of the infinite path is $3$.
\end{conjecture}

\section{Introduction}
A \emph{square} (\textit{resp.} a \emph{cube}) is a word $uu$ (\textit{resp.} $uuu$) for some nonempty word $u$. A \emph{cube-free word} (resp. \emph{square-free word}) is a word that contains no cube (resp. square) as a factor.
Thue proved that there exists an infinite cube-free word over the binary alphabet and an infinite square-free word over the ternary alphabet \cite{Thue1}. These results are regarded as the first results in combinatorics on words, and many generalizations of these problems have been considered.

We provide a formulation of this conjecture in terms of combinatorics on words. By an \emph{alphabet}, we mean any finite set. Given a sequence $(\A_i)_{i\ge1}$ of alphabets, we say that a word $w=w_1\ldots w_n$ \emph{respects} $(\A_i)_{i\ge 1}$, if for all $i$, $w_i\in\A_i$ (the definition naturally extends to infinite words). Conjecture \ref{conj:ThueListe} can be rephrased as follows.
\begin{conjecture}
Given a sequence of alphabets $(\A_i)_{i\ge1}$ with $|A_i|\ge3$ for all $i$, there exists an infinite square-free word that respects $(\A_i)_{i\ge 1}$.
\end{conjecture}

In the current article, we consider a natural variant of the question. Since cubes are avoidable over the binary alphabet, is it possible to avoid cubes if all of the lists are of size $2$? We give a positive answer to this question in the following theorem.

The idea behind the first proof is to show that the growth rate of the language is not far from the growth rate of the regular approximation of the language which yields the following result.
\begin{theorem} \label{thm:growth}
For any sequence of binary alphabet $\A_1,\A_2,\dots$, and for every $n$, there exist at least $1.35^n$ cube-free words in $\A_1\times\A_2\times\dots\times\A_n$.
\end{theorem}
The core idea allowing us to deduce bounds on the growth of a power-free language $\mathcal L$ from its approximation $\mathcal L$ was first used by Kolpakov to estimate the growth rate of power-free languages \cite{kopakov2007efficient,Kolpakov2007Dec}. Shur later improved upon this technique \cite{shur2010Jul,Shur2012Nov}. This technique was later adapted by the second author to provide a partial answer to Conjecture \ref{conj:ThueListe} in \cite{rosenfeld2022avoiding}.
The proof of this result applies the technique developed in \cite{rosenfeld2022avoiding} for squares, but provides a complete answer in the case of cubes.
The second proof of our main result does not imply anything about the number of solutions, but it contains algorithmic implications.
\begin{theorem} \label{thm:computable}
Let $\A_1,\A_2,\dots$ be a computable sequence of binary alphabet. Then there exists a computable infinite cube-free word $w=w_1w_2w_3\ldots$ in $\A_1\times\A_2\times\dots$. Moreover, there is an algorithm that given $w_1\ldots w_n$ and\ $\A_{n+1}$ computes $w_{n+1}$ in time polynomial in $n$.
\end{theorem}
This proof also relies on computing properties of the same automaton, but then we use a different argument for adding the large cubes back.
Intuitively, the idea is to introduce a weight function that measures how difficult it is to extend a word, and to prove that under some condition if the weight is not too large then at least one of the extensions also has a small weight (which we can then use inductively to construct a word). This idea was originally due to Miller \cite{miller2012two}, and was recently generalized in \cite{Rosenfeld2025Mar}. However, the weight functions in \cite{miller2012two} and \cite{Rosenfeld2025Mar} are much simpler and in our article the precise definition of the function depends on the automaton. Our result is in fact even stronger, as everything still holds if for all $i$ an opponent is allowed to choose the alphabet $\A_{i+1}$ after we choose $w_i$.

\section{Approximation by a regular language}\label{sec:notation}
Our approach is to bound the difference between the number of cube-free words and the number of words that avoids all cubes up to a given length $p$.
We let $\mathcal{C}$ be the set of all cube-free words.
For a given integer $p$, we let $\widetilde{\mathcal{C}}^{(p)}$ be the set of words that avoid cubes of period at most $p$.
Given a sequence of alphabets $(\A_i)_{i\ge1}$, we let $\mathcal{C}[(\A_i)_{i\ge1}]$ (resp. $\widetilde{\mathcal{C}}^{(p)}[(\A_i)_{i\ge1}]$) be the elements of $\mathcal{C}$ (resp. $\widetilde{\mathcal{C}}^{(p)}$) that respect $(\A_i)_{i\ge1}$.

This implies
\begin{align*}
 \sum_{\substack{c\in\A\\uc\in \widetilde{\mathcal{C}}^{(p)}}} \omega(uc)
 &\le \beta\omega(u)+\sum_{i>p} \lambda_{\minsuff(u)}\beta^{1-i}\\
 &<\lambda_{\minsuff(u)}\left(\beta+\frac{\beta^{1-p}}{\beta-1}\right)\\
 &\le\lambda_{\minsuff(u)}\alpha\\
 &\le \sum_{\substack{c\in\A\\uc\in \widetilde{\mathcal{C}}^{(p)}}} \lambda_{\minsuff(uc)}\,,
\end{align*}
where the third inequality is by our theorem hypothesis, and the last inequality was proven earlier in this proof.
This implies that there exists $c\in \A$ such that $uc\in \widetilde{\mathcal{C}}^{(p)}$ and $\omega(uc)<\lambda_{\minsuff(uc)}$ as desired.
\end{proof}
\begin{proof}[Proof of Theorem \ref{thm:computable}]
 In particular, with $p=12$ and $\alpha=1.457$ we can set $\beta=1.35$ to satisfy the condition of Lemma \ref{lem:betacond}.
 So we inductively define $u_0=\varepsilon$, and for all $n$, $u_{n+1} = u_n a_{n+1}$ where $a_{n+1}\in\A_{n+1}$ is chosen such that $\omega(u_{n+1})<\lambda_{\minsuff(u_{n+1})}$. By Lemma \ref{lem:smallweighisfine}, each of the $u_i$ is cube-free, and so is the infinite word $a_1a_2a_3\ldots$ and by construction it belongs to $\A_1\times\A_2\times\A_3\ldots$ as desired.

\begin{lemma}\label{lem:counting_argument}
For all $n\in\mathbb{N}$ and $t\in\minsuff$, we have the following inequality
 \begin{equation}\label{eq:exclusion}
 |C_{n+1}^t| \ge \sum_{v\in C_n}\sum_{\substack{c\in\A_{n+1}\\ vc\in\widetilde{\mathcal{C}}^{(p)}\\\minsuff(vc)=t}} 1 - \sum_{i=p+1}^{n+1} |C_{n+1-i}^t|.
 \end{equation}
\end{lemma}
\begin{proof}
 We let $G^t$ be the set of words of length $n+1$ in $A_1\times\ldots\times A_{n+1}$ so that
 \begin{itemize}
 \item the prefix of size $n$ is cube-free,
 \item the word does not contain any cubes of period at most $p$,
 \item the word ends in state $t$.
 \end{itemize}
 In other words,
 \[
 G^t= \left\{va\in \widetilde{\mathcal{C}}^{(p)}: v\in C_{n}, a\in A_{n+1}, \minsuff(va)=t\right\}\,.
 \]
 Hence,
 \[
 |G^t|= \sum_{v\in C_n}\sum_{\substack{c\in\A_{n+1}\\ vc\in\widetilde{\mathcal{C}}^{(p)}\\\minsuff(vc)=t}} 1\,.
 \]
 Let $F_i$ be the set of words from $G_t$ that contain a cube of period $i$ as a suffix. By definition, a word from $G_t$ can only contain a cube as a suffix and this cube must be of period more than $p$ (and at most $n+1$).\footnote{We could easily replace this $n+1$ by $\lfloor(n+1)/3\rfloor$, but it does not make any difference in the latter part of the proof.} We get
 \[
 C_{n+1}^t= G^t \setminus \bigcup_{i=p+1}^{n+1} F_i\,,
 \]
 which implies
 \[
 |C_{n+1}^t|\ge |G^t| - \sum_{i=p+1}^{n+1} |F_i|\,.
 \]
 It remains to prove that $|F_i|\le |C_{n+1-i}^t|$ for every $i$ to conclude the proof.
 Let $va\in F_i$, then by definition $va=uyyy$ with $|y|=i>p$. Since $v\in C_n$ and $va\in \widetilde{\mathcal{C}}^{(p)}$, we can apply Corollary \ref{cor:easingOnePeriodPreservesState} to deduce that $\minsuff(uyy)=\minsuff(vc)=t$, that is, $uyy\in C_{n+1-i}^t$. Given the word $uyy$ and the period $i=|y|$, we can uniquely reconstruct $vc=uyyy$. This implies $|F_i|\le |C_{n+1-i}^t|$, which concludes our proof.
\end{proof}

In fact, we do not prove this result directly, but we prove two different stronger results, each implying Theorem \ref{thm:main}.
These two proofs share a key idea: we can use an automaton to study the variant of the problem where we only forbid cubes of length at most $p$, for some arbitrary choice of $p$. Intuitively, short cubes are more difficult to avoid than long cubes, so by taking $p$ large enough, we should obtain a good approximation of the problem. Adapting standard tools from combinatorics on words and automata theory, studying this approximation is quite simple to achieve.
The delicate part is then to prove that if $p$ is large enough this approximation is good enough to deduce something about the original problem where all cubes are forbidden.

The idea behind the first proof is to show that the growth rate of the language is not far from the growth rate of the regular approximation of the language which yields the following result.
\begin{theorem} \label{thm:growth}
For any sequence of binary alphabet $\A_1,\A_2,\dots$, and for every $n$, there exist at least $1.35^n$ cube-free words in $\A_1\times\A_2\times\dots\times\A_n$.
\end{theorem}
The core idea allowing us to deduce bounds on the growth of a power-free language $\mathcal L$ from its approximation $\mathcal L$ was first used by Kolpakov to estimate the growth rate of power-free languages \cite{kopakov2007efficient,Kolpakov2007Dec}. Shur later improved upon this technique \cite{shur2010Jul,Shur2012Nov}. This technique was later adapted by the second author to provide a partial answer to Conjecture \ref{conj:ThueListe} in  \cite{rosenfeld2022avoiding}.
The proof of this result applies the technique developed in \cite{rosenfeld2022avoiding} for squares, but provides a complete answer in the case of cubes.
The second proof of our main result does not imply anything about the number of solutions, but it contains algorithmic implications.
\begin{theorem} \label{thm:computable}
Let $\A_1,\A_2,\dots$ be a computable sequence of binary alphabet. Then there exists a computable infinite cube-free word $w=w_1w_2w_3\ldots$ in $\A_1\times\A_2\times\dots$. Moreover, there is an algorithm that given $w_1\ldots w_n$ and\ $\A_{n+1}$ computes $w_{n+1}$ in time polynomial in $n$.
\end{theorem}
This proof also relies on computing properties of the same automaton, but then we use a different argument for adding the large cubes back.
Intuitively, the idea is to introduce a weight function that measures how difficult it is to extend a word, and to prove that under some condition if the weight is not too large then at least one of the extensions also has a small weight (which we can then use inductively to construct a word). This idea was originally due to Miller \cite{miller2012two}, and was recently generalized in \cite{Rosenfeld2025Mar}. However, the weight functions in \cite{miller2012two} and \cite{Rosenfeld2025Mar} are much simpler and in our article the precise definition of the function depends on the automaton. Our result is in fact even stronger, as everything still holds if for all $i$ an opponent is allowed to choose the alphabet $\A_{i+1}$ after we choose $w_i$.

Note that Theorem \ref{thm:growth} and Theorem \ref{thm:computable} both independently imply Theorem \ref{thm:main}, but none of them seem to imply the other. We discuss the possible connection between this result in Section \ref{sec:discussion}.

The article is organized as follows. In Section \ref{sec:notation}, we introduce some common notations. In particular, we define the language $\widetilde{\mathcal{C}}^{(p)}$ that approximate the language of cube-free words, and we give the statement of Lemma \ref{lem:eigenvector} that is later crucial to our two proofs. In Section \ref{sec:potential}, we prove Theorem \ref{thm:computable}. In Section \ref{sec:growth}, we prove Theorem \ref{thm:growth}. Finally, we provide in Section \ref{sec:computations} a short description of the computation used to verify Lemma \ref{lem:eigenvector}.

We "do not prove [Theorem \ref{thm:main}] directly", but instead prove "two different stronger results, each implying Theorem \ref{thm:main}" (Theorem~\ref{thm:growth} and Theorem~\ref{thm:computable}). Both proofs share a key idea: "use an automaton" to study an approximation where one "only forbid[s] cubes of length at most $p$". The motivation is that "short cubes are more difficult to avoid than long cubes", so with $p$ large this gives "a good approximation"; the "delicate part" is to show that for $p$ large enough, this approximation lets one conclude results for the original language where "all cubes are forbidden".

For the route via Theorem~\ref{thm:growth}, the idea is "to show that the growth rate of the language is not far from the growth rate of the regular approximation of the language" and then apply the technique (Kolpakov; improved by Shur; adapted in \cite{rosenfeld2022avoiding}) to "deduce bounds on the growth of a power-free language" from its approximation.

For the route via Theorem~\ref{thm:computable}, one again uses "computing properties of the same automaton", but with "a different argument for adding the large cubes back": "introduce a weight function that measures how difficult it is to extend a word" and prove that "under some condition if the weight is not too large then at least one of the extensions also has a small weight", enabling an inductive construction of an infinite cube-free word; the weight function here "depends on the automaton" (in contrast to simpler weight functions in \cite{miller2012two,Rosenfeld2025Mar}).

Question: Let \((\mathcal A_i)_{i\ge 1}\) be any sequence of alphabets, where each \(\mathcal A_i\) is binary (that is, \(|\mathcal A_i|=2\)). A cube is a word of the form \(uuu\) for some nonempty word \(u\), and an infinite word is cube-free if none of its finite factors is a cube. Which statement holds about infinite words \(w=a_1a_2a_3\cdots\) with \(a_i\in \mathcal A_i\) for every \(i\)?

There exists an infinite cube-free word \(w=a_1a_2a_3\cdots\) such that \(a_i\in \mathcal A_i\) for every \(i\); equivalently, for any sequence of binary alphabets \(\mathcal A_1,\mathcal A_2,\dots\), there exists an infinite cube-free word in \(\mathcal A_1\times \mathcal A_2\times \cdots\).

Choices:
B. There exists an infinite word \(w=a_1a_2a_3\cdots\) such that \(a_i\in \mathcal A_i\) for every \(i\), and \(w\) is square-free; equivalently, for any sequence of binary alphabets \(\mathcal A_1,\mathcal A_2,\dots\), there exists an infinite square-free word in \(\mathcal A_1\times \mathcal A_2\times \cdots\).
C. There exists an infinite word \(w=a_1a_2a_3\cdots\) such that \(a_i\in \mathcal A_i\) for every \(i\), and every finite prefix of \(w\) is cube-free; in particular, \(w\) has arbitrarily long cube-free finite prefixes respecting \((\mathcal A_i)_{i\ge 1}\).
D. For every sequence of binary alphabets \(\mathcal A_1,\mathcal A_2,\dots\), there exists an integer \(N\) such that for every \(n\ge N\) one can choose letters \(a_1\in\mathcal A_1,\dots,a_n\in\mathcal A_n\) so that \(a_1\cdots a_n\) is cube-free; equivalently, every such sequence admits cube-free words of all sufficiently large lengths.
E. There exists a single infinite cube-free binary word \(w=b_1b_2b_3\cdots\) with the following universal property: for every sequence of binary alphabets \(\mathcal A_1,\mathcal A_2,\dots\), one has \(b_i\in \mathcal A_i\) for every \(i\); equivalently, the same infinite cube-free word belongs to \(\mathcal A_1\times\mathcal A_2\times\cdots\) for all choices of binary alphabets.

The stem only defines binary alphabets and cube-freeness, then asks which existence statement is true. It does not explicitly or implicitly reveal that the infinite cube-free conclusion in choice A is correct.

The item is close to theorem recognition: one option is essentially the target existence theorem, while the others are weakened, strengthened, or quantifier-tampered variants. So it is not fully tautological, but it is still a mild reformulation of a theorem statement rather than a derived application.

Solving requires some reasoning about logical strength, quantifiers, and the distinction between cube-free and square-free. However, it mainly tests recognition/comparison of candidate statements rather than substantial generative mathematical reasoning.

The distractors are strong and mathematically meaningful: B alters infinite extension, C is a weaker true statement, D confuses cube-free with square-free, and E introduces a subtle quantifier-dependence error. These are plausible and distinct failure modes.

A solid MCQ with no answer leakage and high-quality distractors, but it is somewhat theorem-recognition-based and only moderately pressures genuine generative reasoning.


\begin{thm}
\label{thm:main}
Let $\beta(n)\colonequals \max_{w\in S_n}|\supp(\mathfrak S_w)|$. Then
\[
\lim_{n\to\infty}\frac{\ln(\beta(n))}{n\ln(n)} = 1.
\]
More precisely,
\[
-\ln(4)-1 \leq \liminf_{n\to\infty}\frac{\ln(\beta(n)) - n\ln(n)}n \leq \limsup_{n\to\infty}\frac{\ln(\beta(n)) - n\ln(n)}n \leq -1.
\]
\end{thm}

Schubert polynomials $\mathfrak S_w(x_1, \dots, x_n)$, indexed by permutations $w\in S_n$, are lifts of Schubert cycles in the cohomology of the flag variety \cite{ls82}. The specialization $\mathfrak S_w(1,\dots,1)$ is equal to the number of reduced pipe dreams of $w$, and has a geometric interpretation as the degree of the matrix Schubert variety of $w$. Writing $u(n)\colonequals \max_{w\in S_n}\mathfrak S_w(1, \dots,1)$, Stanley \cite{stanley17} observed that
\[
\frac14 \leq \liminf_{n\to\infty} \frac{\log_2(u(n))}{n^2} \leq \limsup_{n\to\infty}\frac{\log_2(u(n))}{n^2} \leq \frac12
\]
and asked whether $\lim_{n\to\infty}\frac{\log_2(u(n))}{n^2}$ exists, and, if so, what the value of this limit is. His question remains open, but see \cite{mpp19,gao21,gl24,mppy25,zhang25} for recent progress on this problem and its variants.

We study the growth rate of the maximal sizes of \emph{supports} of Schubert polynomials. Write $\mathrm{supp}(\mathfrak S_w)\colonequals \{\alpha \in \ZZ_{\geq 0}^n\colon \mathbf x^\alpha\textup{ appears in } \mathfrak S_w\}$.

The key ingredient in our proof of Theorem~\ref{thm:main} is the observation that a certain \emph{layered} permutation has support of size asymptotically at least $n!/4^n$. Layered permutations have previously appeared in Merzon--Smirnov's conjecture \cite{ms16} that $\mathfrak S_w(1,\dots,1)$ is maximized at a layered permutation. Guo--Lin conjectured \cite[Prob 5.3]{gl24} that there exist permutations simultaneously maximizing $\mathfrak S_w(1, \dots,1)$ and $|\supp(\mathfrak S_w)|$.

We will use the following (crude) estimates for the factorial function, whose proofs we include for completeness.
\begin{lem}
\label{lem:factorial-estimate}
The inequalities
\[
e\left(\frac ne\right)^n \leq n! \leq e\left(\frac{n+1}e\right)^{n+1}
\]
hold for all positive integers $n$.
\end{lem}
\begin{proof}
Estimate $\ln(n!) = \sum_{k=1}^n\ln(k)$ by the integrals
\[
\int_1^n \ln(x)\,dx \leq \sum_{k=1}^n \ln(k) \leq \int_1^{n+1}\ln(x)\,dx.
\]
The definite integrals evaluate to $n\ln(n) - n + 1$ and $(n+1)\ln(n+1) - (n+1) + 1$. The claim follows.
\end{proof}
\begin{lem}
\label{lem:floor-fact-est}
The inequality
\[
\lfloor k\rfloor! \geq \frac1{2k}\left(\frac ke\right)^k
\]
holds for all real numbers $k>1$.
\end{lem}
\begin{proof}
Using Lemma~\ref{lem:factorial-estimate}, compute that
\[
\lfloor k\rfloor! \geq \frac1{k+1} \lceil k\rceil! \geq \frac1{k+1} \cdot \frac{\lceil k\rceil^{\lceil k\rceil}}{e^{\lceil k\rceil-1}} \geq \frac1{k+1}\cdot \frac{k^k}{e^k}.
\]
As $k > 1$, the claim follows.
\end{proof}
\section{A layered permutation with large Schubitope}
Our main technical result is as follows.
\begin{thm}
\label{thm:tech-main}
Let $w = w(b_1, \dots, b_m)$ be a layered permutation in $S_n$. If $b_m \leq n/2$, then
\[
|\supp(\mathfrak S_w)| \geq b_m!\cdot |\supp(\mathfrak S_{w'})|, \qquad \textup{ where } w'\colonequals w(b_1, \dots, b_{m-1})\in S_{n-b_m}.
\]
\end{thm}
Our proof of Theorem~\ref{thm:tech-main} will use the following lemma.
\begin{lem}
\label{lem:last-block}
Let $w\in S_n$ be a layered permutation of the form $w = w(1, \dots, 1, b_m)$ for $b_m \leq n/2$. For any subset $S\subseteq D(w)$, there exists a diagram $C$ satisfying $C\leq D(w)$ and $C\cap D(w) = S$.
\end{lem}
\begin{proof}
Each column $D_j$ of $D(w)$ is of the form $\{n-b_m + 1, \dots, n-c\}$ for some $c \geq 1$. As $|D_j| \leq b_m-1 < n-b_m$, it follows that for any subset $S_j$ of $D_j$ there exists a set $C_j\leq D_j$ with $C_j \cap D_j = S_j$. Applying this column by column to $D(w)$, the claim follows.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:tech-main}]
For each of the $b_m!$ many vectors $\alpha\in\ZZ_{\geq0}^{b_m}$ satisfying $(0,\dots,0) \leq \alpha \leq (b_m - 1, b_m - 2, \dots, 1, 0)$, we construct an embedding $i_\alpha\colon \mathcal S_{D(w')} \hookrightarrow \mathcal S_{D(w)}$ so that $\img(i_\alpha)\cap\img(i_{\alpha'}) = \emptyset$ for all $\alpha\neq\alpha'$.

For each $\alpha$ satisfying $(0,\dots,0)\leq\alpha\leq (b_m - 1, b_m - 2, \dots, 1)$, use Lemma~\ref{lem:last-block} to choose a diagram $C_\alpha \leq D(w_{\mathrm{LB}})$ whose weight satisfies $\wt(C_\alpha)_{n-b_i + j} = \alpha_j$. Then define the map
\begin{align*}
i_\alpha\colon \mathcal S_{D(w')}&\to \mathcal S_{D(w)}\\
\wt(C')&\mapsto\wt((C', C_\alpha));
\end{align*}
the map $i_\alpha$ is well-defined and injective because it is in fact the translation map $\gamma \mapsto \gamma +\wt(C_\alpha)$. Furthermore, any $\gamma' \in \mathcal S_{D(w')}$ satisfies $\gamma'_{n-b_m+j} = 0$ for all $j\geq 0$, so any vector $\gamma\in\img(i_\alpha)$ satisfies $\gamma_{n-b_m+j} = \alpha_j$. In particular, the $\img(i_\alpha)$ are disjoint subsets of $\mathcal S_{D(w)}$.
\end{proof}
\begin{cor}
\label{cor:upshot}
Fix an integer $n\geq 3$ and let $c\colonequals \max\{k \colon \lfloor n/2^k\rfloor \geq 1\}$ and $d \colonequals n - \sum_{k=1}^c \lfloor n/2^k\rfloor$. Let $w$ be the layered permutation with blocks of size
\[
\underbrace{1,\dots,1}_{d\textup{ many}}, \left\lfloor \frac n{2^c}\right\rfloor, \left\lfloor \frac n{2^{c-1}}\right\rfloor, \dots, \left\lfloor\frac n4\right\rfloor,\left\lfloor \frac n2\right\rfloor.
\]
Then
\[
|\supp(\mathfrak S_w)| \geq \prod_{k=1}^c\left\lfloor \frac n{2^k}\right\rfloor!.
\]
In particular,
\[
\ln(|\supp(\mathfrak S_w)|) \geq \left(n - \frac1{\ln(2)}\ln(n) - 2\right)\ln(n) - n\ln(4) - (n-2).
\]
\end{cor}
\begin{proof}
From the inequality
\[
\left\lfloor \frac n{2^k}\right\rfloor \leq \frac n{2^k} = \frac12\left(n - \sum_{i=1}^{k-1} \frac n{2^i}\right) \leq \frac12\left(n - \sum_{i=1}^{k-1} \left\lfloor\frac n{2^i}\right\rfloor\right),
\]
Theorem~\ref{thm:tech-main} may be repeatedly applied to the layered permutation $w$ to obtain
\[
|\supp(\mathfrak S_w)| \geq \prod_{k=1}^c\left\lfloor \frac n{2^k}\right\rfloor!.
\]
Lemma~\ref{lem:floor-fact-est} implies
\begin{align*}
\prod_{k=1}^c\left \lfloor \frac n{2^k}\right\rfloor! &\geq \prod_{k=1}^c\left(\frac{2^{k-1}}n \left(\frac n{2^k e}\right)^{\frac n{2^k}}\right) \\&\geq \frac 1{n^c} \cdot \underbrace{\prod_{k=1}^c \left(\frac1{2^k}\right)^{\frac n{2^k}}}_{\geq 2^{-2n}}\cdot \underbrace{\prod_{k=1}^c \left(\frac ne\right)^{\frac n{2^k}}}_{\geq \left(\frac ne\right)^{n-2}}\label{eqn:estimate}\tag{$*$}\\
&\geq \frac{n^{n-c-2}}{4^n\cdot e^{n-2}}\\&\geq \frac{n^{n - \log_2(n) - 2}}{4^n \cdot e^{n-2}}
\end{align*}
where the estimates in~\eqref{eqn:estimate} follow from
\[
\sum_{k=1}^c k\cdot \frac n{2^k} \leq \sum_{k=1}^\infty k \cdot \frac n{2^k} = 2n
\]
and
\[
\frac n2 + \dots + \frac n{2^c} = n - \frac n{2^c} \geq n-2.
\]
\end{proof}

\newtheorem*{thm:main}{Theorem~\ref{thm:main}}
\begin{thm:main}
Let $\beta(n)\colonequals \max_{w\in S_n}|\supp(\mathfrak S_w)|$. Then
\[
\lim_{n\to\infty}\frac{\ln(\beta(n))}{n\ln(n)} = 1.
\]
More precisely,
\[
-\ln(4)-1 \leq \liminf_{n\to\infty}\frac{\ln(\beta(n)) - n\ln(n)}n \leq \limsup_{n\to\infty}\frac{\ln(\beta(n)) - n\ln(n)}n \leq -1.
\]
\end{thm:main}
\begin{proof}[Proof of Theorem~\ref{thm:main}]
By Lemma~\ref{lem:upper-bound}, the inequality
\[
\ln(|\supp(\mathfrak S_w)|) \leq \ln(n!) \leq (n+1)\ln(n+1) - n
\]
holds for all $w\in S_n$. By Corollary~\ref{cor:upshot}, there exists $w\in S_n$ with
\[
\ln(|\supp(\mathfrak S_w)|) \geq \left(n-\frac1{\ln(2)}\ln(n) - 2\right)\ln(n) - n\ln(4) - (n-2).
\]
The inequalities
\[
-\ln(4)-1 \leq \liminf_{n\to\infty}\frac{\ln(\beta(n)) - n\ln(n)}n \leq \limsup_{n\to\infty}\frac{\ln(\beta(n)) - n\ln(n)}n \leq -1
\]
follow.
\end{proof}

\begin{prop}
\label{prop:groth-precise}
Fix an integer $n$, let $k$ be the unique integer so that $\binom{k+1}2\leq n<\binom{k+2}2$, and let $b \colonequals n - \sum_{i=1}^k i$. Let $w$ be the layered permutation with blocks of size $1, 2, \dots, k, b$. Then
\[
|\supp(\mathfrak G_w)| \geq \frac{n!}{n^{k+1}}.
\]
In particular,
\[
\ln(|\supp(\mathfrak G_w)|) \geq n\ln(n) - n - (\sqrt{2n}+1)\ln(n)
\]
\end{prop}
\begin{proof}
Let $\mathbf c = \mathrm{wt}(D(w))$ denote the Lehmer code of $w$, and let $\mathbf d = \mathrm{wt}(\overline{D(w)})$. Proposition~\ref{prop:fireworks-grothendieck} implies that $[\mathbf c, \mathbf d] \subseteq \supp(\mathfrak G_w)$, so
\begin{equation}
\label{eqn:crude-grothendieck}
|\supp(\mathfrak G_w)| \geq \prod_{i=1}^n (d_i - c_i + 1).
\end{equation}

\newtheorem*{thm:groth}{Theorem~\ref{thm:groth}}
\begin{thm:groth}
Let $\beta^{\mathfrak G}(n)\colonequals\max_{w\in S_n}|\supp(\mathfrak G_w)|$. Then
\[
\lim_{n\to\infty}\frac{\ln(\beta^{\mathfrak G}(n)) - n\ln(n)}n = -1.
\]
\end{thm:groth}
\begin{proof}[Proof of Theorem~\ref{thm:groth}]
By Lemma~\ref{lem:upper-bound}, the inequality
\[
\ln(|\supp(\mathfrak G_w)|)\leq \ln(n!) \leq (n+1)\ln(n+1)-n
\]
holds for all $w\in S_n$. By Proposition~\ref{prop:groth-precise}, there exists $w\in S_n$ with
\[
\ln(|\supp(\mathfrak G_w)|) \geq n\ln(n) - n - (\sqrt{2n}+1)\ln(n).
\]
The claim follows.
\end{proof}

Theorem~\ref{thm:main} answers a problem \cite[Prob 5.5]{gl24} posed by Guo--Lin. We do not know if the limit $\lim_{n\to\infty}\frac{\ln(\beta(n)) - n\ln(n)}n$ exists, nor do we have a conjecture for its value.

The key ingredient in our proof of Theorem~\ref{thm:main} is the observation that a certain \emph{layered} permutation has support of size asymptotically at least $n!/4^n$. Layered permutations have previously appeared in Merzon--Smirnov's conjecture \cite{ms16} that $\mathfrak S_w(1,\dots,1)$ is maximized at a layered permutation. Guo--Lin conjectured \cite[Prob 5.3]{gl24} that there exist permutations simultaneously maximizing $\mathfrak S_w(1, \dots,1)$ and $|\mathrm{supp}(\mathfrak S_w)|$.

Grothendieck polynomials $\mathfrak G_w$ are inhomogeneous deformations of Schubert polynomials; they are generating functions for (possibly nonreduced) pipe dreams of $w$. We are able to produce a layered permutation with support of size asymptotically at least $n!/e^{\sqrt{2n} \cdot \ln(n)}$, i.e., of size $n!$ up to a subexponential factor. In particular, the maximal sizes of supports of Grothendieck polynomials satisfy the following more precise asymptotics.
\begin{thm}
\label{thm:groth}
Let $\beta^{\mathfrak G}(n)\colonequals\max_{w\in S_n}|\mathrm{supp}(\mathfrak G_w)|$. Then
\[
\lim_{n\to\infty}\frac{\ln(\beta^{\mathfrak G}(n)) - n\ln(n)}n = -1.
\]
\end{thm}

The post-theorem introduction indicates that “the key ingredient in our proof of Theorem~\ref{thm:main} is the observation that a certain \emph{layered} permutation has support of size asymptotically at least $n!/4^n$.”

\label{thm:main}
Let $\beta(n)\colonequals \max_{w\in S_n}|\mathrm{supp}(\mathfrak S_w)|$. Then
\[
\lim_{n\to\infty}\frac{\ln(\beta(n))}{n\ln(n)} = 1.
\]
More precisely,
\[
-\ln(4)-1 \leq \liminf_{n\to\infty}\frac{\ln(\beta(n)) - n\ln(n)}n \leq \limsup_{n\to\infty}\frac{\ln(\beta(n)) - n\ln(n)}n \leq -1.
\],

Question:
For each positive integer $n$, let $S_n$ be the symmetric group on $n$ letters. For a permutation $w\in S_n$, let $\mathfrak S_w(x_1,\dots,x_n)$ denote its Schubert polynomial, and define its support by
\[
\operatorname{supp}(\mathfrak S_w)\coloneqq \{\alpha\in \mathbb Z_{\ge 0}^n : \mathbf x^\alpha \text{ appears in } \mathfrak S_w\}.
\]
Set
\[
\beta(n)\coloneqq \max_{w\in S_n} |\operatorname{supp}(\mathfrak S_w)|.
\]
Which asymptotic estimate holds for $\beta(n)$?

One has
\[
\lim_{n\to\infty}\frac{\ln(\beta(n))}{n\ln(n)}=1.
\]
More precisely,
\[
-\ln(4)-1\le \liminf_{n\to\infty}\frac{\ln(\beta(n))-n\ln(n)}{n}
\le \limsup_{n\to\infty}\frac{\ln(\beta(n))-n\ln(n)}{n}\le -1.
\]

Choices:
B.
One has
\[
\lim_{n\to\infty}\frac{\ln(\beta(n))}{n\ln(n)}=1.
\]
More precisely,
\[
-\ln(4)-1\le \liminf_{n\to\infty}\frac{\ln(\beta(n))-n\ln(n)}{n}
\le \limsup_{n\to\infty}\frac{\ln(\beta(n))-n\ln(n)}{n}\le 0.
\]
C.
One has
\[
\limsup_{n\to\infty}\frac{\ln(\beta(n))}{n\ln(n)}=1.
\]
D.
One has
\[
\lim_{n\to\infty}\frac{\ln(\beta(n))}{n^2}=0.
\]
More precisely,
\[
\limsup_{n\to\infty}\frac{\ln(\beta(n))-n\ln(n)}{n}= -1.
\]
E.
One has
\[
\lim_{n\to\infty}\frac{\ln(\beta(n))}{n\ln(n)}=1.
\]
More precisely,
\[
\lim_{n\to\infty}\frac{\ln(\beta(n))-n\ln(n)}{n}= -1.
\]

The stem gives only definitions of Schubert polynomials, support, and β(n), then asks for the asymptotic estimate. It does not reveal the correct asymptotic form or its refined constants.

The item is close to a theorem-recall question: it essentially asks for the correct asymptotic statement of β(n). However, it is not a pure tautology, since the options vary in strength, precision, and limiting form, so the student must distinguish among competing formulations.

Some reasoning is required to compare nearby asymptotic claims, especially the weaker true statement (C), the overstrengthened statement (E), and the slightly altered bound in (B). Still, the question mainly tests recognition of the precise theorem statement rather than generating a conclusion from intermediate facts.

The distractors are mathematically plausible and well targeted: one weakens the conclusion, one slightly relaxes a bound, one mismatches normalization, and one improperly upgrades liminf/limsup bounds to an exact limit. These reflect realistic asymptotic reasoning errors.

A solid asymptotic-discrimination MCQ with no answer leakage and strong distractors, but it leans toward precise theorem recognition rather than deeper generative reasoning.