 Solution

Use a programming closure whose private variable holds the next Fibonacci number. Use arbitrary-precision integers to enumerate all terms exactly:At the cell containing $F_i$, the private state immediately before computing its successor is following = F_(i+1). The successor returns that value and sets following to $F_i+F_{i+1}=F_{i+2}$, preserving the recursion invariant. The resulting sequence startsEach successor is computed once by the memoized implementation in part (c); interleaved copies see the same already generated successor instead of independently advancing shared programming closure state. Replacing BigInt by Int and the two initial values by 0,1 gives exactly the hinted Int => Int implementation, but eventually incurs integer overflow, so it does not enumerate all mathematical Fibonacci numbers.

An alternative avoiding mutable programming closure state represents the current and next Fibonacci numbers by a pair $(a,b)$ and uses the pure function $(a,b)\mapsto (b,a+b)$ with initial state $(0,1)$; a projecting iterator returns the first component. The closure construction above meets the suggested one-number interface directly.