 Left scan

Return the successive accumulator values of a left fold, including the initial accumulator: scanl f e [] = [e] and scanl f e (x:xs) = e : scanl f (f e x) xs. The $k$th result is the fold of the first $k$ elements, proved by the fold equation. Thus the scan equals map (foldl f e) . inits, but computes $n$ updates instead of $n(n+1)/2$ on a finite input of length $n$. Accumulator strictness and the cost of an update remain separate from this operation count.