I know we can perform memoization in Haskell. The well known recursive Fibonacci example works v. well. f(10000) returns a virtually instant answer which would not be possible otherwise. My (probably naive) function to give the number of partitions of a number :- p = ((map p' [0 .. ]) !!) where p' 0 = 1 p' 1 = 1 p' n = sum [(((-1) ^ (k+1)) * ( p' (n-((k*(3*k-1)) `div` 2)) + p' (n-((k*(3*k+1)) `div` 2)))) | k <- [1 .. n]] It is an attempt to apply the Euler recurrence formula (no 11 in http://mathworld.wolfram.com/PartitionFunctionP.html ) It works but it is shockingly slow. It is orders of magnitude slower than the Python memoized version which runs very fast. parts = {0:1, 1:1} def P(n): if not n in parts: parts[n] = sum ([( ((-1) ** (k+1)) * ( P(n-((k*(3*k-1))//2)) + P(n-((k*(3*k+1))//2)) ) ) for k in xrange (1, n+1)]) return parts[n] Why? Its as if memoization is being ignored in the haskell version. How to fix?