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}`