For a finite set A⊂N and k∈N, let ωk(A)=∑i∈A,i=k1. For each n∈N, define ak,n = ∣{E⊂N:E=∅orωk(E)<minE⩽maxE⩽n}∣. First, we prove that ak,k+ℓ = 2Fk+ℓ,forallℓ⩾0andk⩾ℓ+2, where Fn is the nth Fibonacci number. Second, we show that ∣{E⊂N:maxE=n+1,minE>ω2,3(E),and∣E∣=2}∣ = Fn, where ω2,3(E)=∑i∈E,i=2,31.