Yes. You can say a = zipWith (+) (0 : b) (tail c ++ [0]) and that would make a[i] = b[i-1] (where b[-1] is 0) + c[i+1] (EDIT: where the last element of c is

Yes. You can say a = zipWith (+) (0 : b) (tail c ++ [0]) and that would make a[i] = b[i-1] (where b[-1] is 0) + c[i+1] (EDIT: where the last element of c is 0 (thanks tkahn6!)). Also, interestingly, if you have lazy evaluation, you can use a zip to produce the fibonacci numbers: just zip the list up with itself. fibs = 0 : 1 : zipWith (+) fibs (tail fibs) so you cons 0 to 1 to the zip, and the zip is combining 0 with 1, and then 1 with 1, and then 1 with 2. fibs[0] = 0 fibs[1] = 1 fibs[i, i>=2] = fibs[i-2] + fibs[i-1] and this looks like the math behind it.