But wait — no: \(F_{n+2}\) gives the count for \(k=1\) to \(k=\lfloor (n+1)/2 \rfloor\), but we want **exactly 4 H’s**? No — the problem does not fix number of H’s. It allows any number, as long as no two are adjacent.

Thu, 30 Apr 2026 02:11:35 +0000

Content under review