Does the set of squares $S = \{n^2: n\in\omega\}$ adhere to Benford's law for the first digit in every base $b\geq 2$?
Precise formulation of what it means for a set $T\subseteq \omega$ to "adhere to Benford's law". Let $b \geq 2$ be an integer. For $x\in\omega$, let $f^1_b(x)$ be the first digit of $x$ in $b$-ary representation. We say that a set $T\subseteq \omega$ adheres to Benford's law in base $b$ if for every $d\in\{0,\ldots , b-1\} = b$ we have $$\lim \sup_{n\to\infty} \frac{|\{t\in (T\cap n): f^1_b(t) = d\}|}{|T\cap n| + 1} \; = \; \log_b\Big(1 + \frac{1}{d}\Big).$$
Best Answer
No. Benford's law works well for sequences that grow exponentially, and the squares grow too slowly.
In particular, fix a base $b \geq 3$, consider the case of $d = 1$, and choose $n = 2 \cdot b^{2k}$. For this $n$, we have $$ |\{ t \in (S \cap n) : f_{b}^{1}(t) = 1 \}| = \sum_{r=1}^{2k} \lfloor b^{r/2} \sqrt{2} \rfloor - \lfloor b^{r/2} \rfloor + 1, $$ while the size of $|T \cap n| + 1 = \lfloor b^{k} \sqrt{2} \rfloor + 1$.
This leads to a value for the ratio which tends to $$ \frac{(\sqrt{2} - 1) \left(1 + \frac{1}{\sqrt{b}} + \frac{1}{b} + \cdots\right)}{\sqrt{2}} = \frac{\frac{\sqrt{2} - 1}{\sqrt{2}}}{\frac{\sqrt{b} - 1}{\sqrt{b}}} $$ as $k \to \infty$. We have that $$ \frac{\frac{\sqrt{2} - 1}{\sqrt{2}}}{\frac{\sqrt{b} - 1}{\sqrt{b}}} > \log_{b}(2) $$ for all $b \geq 3$.