For u transcendental over a subfield K of ℂ, span_K{u^s : s ∈ {0, ±1} ∪ {±4^j : j ≥ 1}} carries no 2×3 configuration

A $p \times q$ configuration over a field $K$ in a $K$-subspace $V \subseteq \mathbb{C}$ is a pair $x_1, \dots, x_p$ and $y_1, \dots, y_q$ of complex numbers, each family linearly independent over $K$, with every product $x_iy_j$ in $V$; it is the input of Roy's strong six exponentials theorem when $p = 2$, $q = 3$, $K = \overline{\mathbb{Q}}$ and $V = \widetilde{\mathcal{L}}$.

Let $K$ be a subfield of $\mathbb{C}$ and $u$ transcendental over $K$. The span of the powers $u^s$ with $s \in \{0, \pm1\} \cup \{\pm 4^j : j \ge 1\}$ carries no $2 \times 3$ configuration over $K$.

At a candidate, the hypothesis "$u^{4^j} \in \widetilde{\mathcal{L}}$ for every $j \ge 1$" puts this whole hull inside $\widetilde{\mathcal{L}}$ (by conjugation). So Roy's strong six exponentials theorem, used through Laurent hulls, cannot exclude it, although it excludes many pairs of powers (DiazModulus.power_pair_hull_two_by_three_iff). Nothing is claimed about other configurations in $\widetilde{\mathcal{L}}$.

Proof. The six products lie in the span of finitely many of the powers, so by DiazModulus.laurent_hull_config_iff and DiazModulus.two_sumset_iff_difference_count some $d > 0$ would have three $s$ with $s, s + d$ in the exponent set $T$. The non-zero absolute values in $T$ are powers of $4$, so two of them are equal or differ by a factor at least $4$. For a pair $s < s + d$ in $T$ with $M = \max(|s|, |s + d|)$, either $s = -M$ and $d = 2M$, or $|d - M| \le M/4$ and $s \in \{M - d, -M\}$; the factor-$4$ gap forces the same $M$ and the same case for all pairs with difference $d$, so $s$ takes at most two values.

Novelty. Not found in the sources read (not in Diaz 2007, whose p. 390 summary only says the six exponentials route will not go far, nor in Fischler 2001); it is short.