Separation exponent at least 2.127294

For every real number $\varepsilon>0$ and every natural number $S\ge 2$, there exist a natural number $n$ and a Boolean function $f:\{0,1\}^{n}\to\{0,1\}$ such that $S\le s(f)$ and $s(f)^{\,2.127294-\varepsilon}\le \operatorname{bs}(f)$, where the latter inequality uses real numbers and real exponentiation, and $2.127294$ is the exact rational number $1063647/500000$. For an input $x$ and a subset $B$ of its $n$ coordinates, let $x^{B}$ denote the input obtained by complementing exactly the coordinates in $B$. Here $s(f)$ is the maximum, over all inputs $x$, of the number of coordinates $i$ for which $f(x^{\{i\}})\ne f(x)$; and $\operatorname{bs}(f)$ is the maximum, over all inputs $x$, of the largest cardinality of a finite family of pairwise disjoint, nonempty coordinate subsets $B$ such that $f(x^{B})\ne f(x)$ for every member $B$ of that family. The empty family is allowed and has cardinality zero; in the defining maximum, any family failing the stated requirements also contributes zero. The witnesses $n$ and $f$ may depend on both $\varepsilon$ and $S$. Although the quantifier over $n$ includes zero, $n=0$ has sensitivity zero and therefore cannot satisfy $S\le s(f)$ under $S\ge2$. There is no upper bound on $\varepsilon$, so the exponent may be zero or negative; the required base satisfies $s(f)\ge2$.