Alternating dyadic intervals contain disjoint IP sets ($\mathrm{FS}(4^j),\mathrm{FS}(2\cdot 4^j)$)

Let $E$ be the union of integer intervals $[2^{2j},2^{2j+1})$ for $j\ge 0$. Every nonempty sum of distinct $4^j$ has largest summand $4^J$ and lies in $[4^J,2\cdot 4^J)$, since the sum of all smaller summands is $(4^J-1)/3$. Thus $\mathrm{FS}(4^j)\subseteq E$. Doubling gives $\mathrm{FS}(2\cdot 4^j)\subseteq E^c$. Both sides are IP sets. Each can be adjoined to the IP-star filter while preserving the finite intersection property, yielding two distinct IP ultrafilters.