Let's divide this problem up into cases:
Case 1 (2 1's, 2 2's, 3 4's) - There are \({7! \over 2! 2! 3!} = 210\) cases.
Case 2 (1 1's, 3 2's, 3 4's) - There are \({7! \over 3! 3!} = 140\) cases.
Case 3 (1 1's, 2 2's, 4 4's) - There are \({7! \over 2! 4!} = 105\) cases.
Case 3 (1 1's, 2 2's, 3 4's, 1 3, 5, or 6) - There are \({7! \over 2! 3!} \times 3 = 1260\) cases.
So, there are \(210 + 140 + 105 + 1260 = \color{brown}\boxed{1,715}\) sequences.