Return to Revisions

I'm really surprised if this question isn't a duplicate, but since I failed to find one that asked about the Fibonacci sequence rather than someone using it as an example, I'll answer.

The most natural approach, besides using the built-in Fibonacci function, recursion:

f[0] = 0; f[1] = 1;
f[n_] := f[n] = f[n - 1] + f[n - 2]  (* note memoization *)

Array[f, 10]

{1, 1, 2, 3, 5, 8, 13, 21, 34, 55}

Better performing may be Nest and NestList:

fibonacciList[n_] := Module[{x = 0}, NestList[x + (x = #) &, 1, n - 1]]

fibonacciList[10]

{1, 1, 2, 3, 5, 8, 13, 21, 34, 55}

Another useful way uses LinearRecurrence:

LinearRecurrence[{1, 1}, {1, 1}, 10]

{1, 1, 2, 3, 5, 8, 13, 21, 34, 55}

Hopefully these examples inspire you.

I now note that you request the sequence starting from zero. Most of these are easy to adapt or modify. The first one is simply:

Array[f, 10, 0]

{0, 1, 1, 2, 3, 5, 8, 13, 21, 34}

For the second you may instead write:

fibonacciList2[n_] := Module[{x = 1}, NestList[x + (x = #) &, 0, n - 1]]

fibonacciList2[10]

{0, 1, 1, 2, 3, 5, 8, 13, 21, 34}

The last one merely needs the proper seed:

LinearRecurrence[{1, 1}, {0, 1}, 10]

{0, 1, 1, 2, 3, 5, 8, 13, 21, 34}

Finally, taking the question at face value you can modify your code to return fPrev rather than fNext to start from zero:

fibonacciSequence[n_] := 
 Module[{fPrev = 0, fNext = 1, i = 0}, 
  While[i++ < n, {fPrev, fNext} = {fNext, fPrev + fNext}];
  fPrev
 ]

Array[fibonacciSequence, 10, 0]

{0, 1, 1, 2, 3, 5, 8, 13, 21, 34}

Addendum for rcollyer:

$fibList = {0, 1};
fibonacciList[n_] /; n <= Length@$fibList := Take[$fibList, n]
fibonacciList[n_] := $fibList =
  $fibList ~Join~ 
   Module[{x = $fibList[[-2]]}, 
    Rest@NestList[x + (x = #) &, $fibList[[-1]], n - Length@$fibList]]