Patience, young "Padovan" Everyone knows the Fibonacci sequence:
You take a square, attach an equal square to it, then repeatedly attach a square whose side length is equal to the largest side length of the resulting rectangle.
The result is a beautiful spiral of squares whose sequence of numbers is the Fibonacci sequence:



But, what if we didn't want to use squares?

If we use equilateral triangles—instead of squares—in a similar fashion, we get an equally beautiful spiral of triangles and a new sequence:  the Padovan sequence, aka A000931:



Task:

Given a positive integer, \$N\$, output \$a_N\$, the \$N\$th term in the Padovan sequence OR the first \$N\$ terms.

Assume that the first three terms of the sequence are all \$1\$.  Thus, the sequence will start as follows:
$$
1,1,1,2,2,3,...
$$

Input:


Any positive integer \$N\ge0\$
Invalid input does not have to be taken into account


Output:


The \$N\$th term in the Padovan sequence OR the first \$N\$ terms of the Padovan sequence.
If the first \$N\$ terms are printed out, the output can be whatever is convenient (list/array, multi-line string, etc.)
Can be either \$0\$-indexed or \$1\$-indexed


Test Cases:
(0-indexed, \$N\$th term)

Input | Output
--------------
0     | 1
1     | 1
2     | 1
4     | 2
6     | 4
14    | 37
20    | 200
33    | 7739


(1-indexed, first \$N\$ terms)

Input | Output
--------------
1     | 1
3     | 1,1,1
4     | 1,1,1,2
7     | 1,1,1,2,2,3,4
10    | 1,1,1,2,2,3,4,5,7,9
12    | 1,1,1,2,2,3,4,5,7,9,12,16


Rules:


This is code-golf: the fewer bytes, the better!
Standard loopholes are forbidden.
Jelly, 10 bytes

9s3’Ẓæ*³FṀ


Try it online!

1-indexed. Computes the largest element of: $$\begin{bmatrix}0&amp;0&amp;1 \\ 1&amp;0&amp;1 \\ 0&amp;1&amp;0\end{bmatrix}^n$$
where the binary matrix is conveniently computed as: $$\begin{bmatrix}\mathsf{isprime}(0)&amp;\mathsf{isprime}(1)&amp;\mathsf{isprime}(2) \\ \mathsf{isprime}(3)&amp;\mathsf{isprime}(4)&amp;\mathsf{isprime}(5) \\ \mathsf{isprime}(6)&amp;\mathsf{isprime}(7)&amp;\mathsf{isprime}(8)\end{bmatrix}$$

(this is a total coincidence.)

9s3         [[1,2,3],[4,5,6],[7,8,9]]    9 split 3
   ’        [[0,1,2],[3,4,5],[6,7,8]]    decrease
    Ẓ       [[0,0,1],[1,0,1],[0,1,0]]    isprime
     æ*³    [[0,0,1],[1,0,1],[0,1,0]]^n  matrix power by input
        FṀ                               flatten, maximum