RosettaCodeData/Task/Fibonacci-sequence/Haskell/fibonacci-sequence-18.hs
2023-07-01 13:44:08 -04:00

1 line
78 B
Haskell

f (n,(a,b)) = (2*n,(a*a+b*b,2*a*b+b*b)) -- iterate f (1,(0,1)) ; b is nth