For diversity reasons:
Table[a[i] /.
RSolve[{a[n] == a[n - 1] + a[n - 2], a[1] == 1, a[2] == 1}, a, n][[1]],
{i, 0, 15}]
{0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610}
or as @J.M. ironically remarked:
Array[DifferenceRoot[
Function[{a, n}, {a[n] == a[n - 1] + a[n - 2], a[1] == 1, a[2] == 1}]], 16, 0]