Can fold be used to create infinite lists? -- by pigworker Sep 6 '12 at 11:37 -- http://stackoverflow.com/a/12299223/849891

gistfile1.hs

"A cheeky person can thus express any infinitary recursion as a non-recursive foldr 
on an infinite list. E.g.,

foldr (\_ fibs -> 1 : 1 : zipWith (+) fibs (tail fibs)) undefined [1..]

(Edit: or, for that matter

foldr (\_ fib a b -> a : fib b (a + b)) undefined [1..] 1 1

which is closer to the definition in the question.)"

------

Prelude> let fact n = foldr (\k fact -> if k > n then 1 else k*fact) undefined [1..]
(0.00 secs, 528100 bytes)
Prelude> fact 5
120
(0.01 secs, 525032 bytes)
Prelude> let fact n = foldr (\_ fact k a -> if k > n then a else fact (k+1) (k*a) ) undefined [1..] 1 1
(0.00 secs, 525232 bytes)
Prelude> fact 5
120
(0.01 secs, 525168 bytes)
Prelude>