We are given sufficiently many stones of the forms of a rectangle $2\times 1$ and square $1\times 1$. Let $n > 3$ be a natural number. In how many ways can one tile a rectangle $3 \times n$ using these stones, so that no two $2 \times 1$ rectangles have a common point, and each of them has the longer side parallel to the shorter side of the big rectangle?
To solve the problem, we need to find the number of ways to tile a \(3 \times n\) rectangle using \(2 \times 1\) and \(1 \times 1\) tiles, with the condition that no two \(2 \times 1\) tiles share a common point and each \(2 \times 1\) tile is placed vertically.

1. **Define the recurrence relation:**
   Let \(a_n\) be the number of ways to tile a \(3 \times n\) rectangle under the given conditions. We can consider the following cases for the last column of the rectangle:
   - If the last column is filled with three \(1 \times 1\) tiles, then the remaining \(3 \times (n-1)\) rectangle can be tiled in \(a_{n-1}\) ways.
   - If the last column contains one \(2 \times 1\) tile, it can be placed in two ways (either occupying the top two squares or the bottom two squares). In either case, the next-to-last column must be filled with three \(1 \times 1\) tiles, leaving a \(3 \times (n-2)\) rectangle to be tiled. This gives \(2a_{n-2}\) ways.

   Therefore, the recurrence relation is:
   \[
   a_n = a_{n-1} + 2a_{n-2} \quad \text{for all} \quad n \geq 3
   \]

2. **Initial conditions:**
   We need to determine the initial values \(a_1\) and \(a_2\):
   - For \(n = 1\), the \(3 \times 1\) rectangle can only be tiled with three \(1 \times 1\) tiles. Thus, \(a_1 = 3\).
   - For \(n = 2\), the \(3 \times 2\) rectangle can be tiled in the following ways:
     - All six squares are filled with \(1 \times 1\) tiles.
     - One \(2 \times 1\) tile is placed vertically in the first column, and the remaining four squares are filled with \(1 \times 1\) tiles. This can be done in 2 ways (top two or bottom two squares).
     - Two \(2 \times 1\) tiles are placed vertically in both columns. This is not possible since it would violate the condition that no two \(2 \times 1\) tiles share a common point.

     Therefore, \(a_2 = 5\).

3. **Solve the recurrence relation:**
   We solve the recurrence relation \(a_n = a_{n-1} + 2a_{n-2}\) with initial conditions \(a_1 = 3\) and \(a_2 = 5\). The characteristic equation of the recurrence relation is:
   \[
   x^2 - x - 2 = 0
   \]
   Solving this quadratic equation, we get:
   \[
   x = 2 \quad \text{or} \quad x = -1
   \]
   Therefore, the general solution to the recurrence relation is:
   \[
   a_n = A \cdot 2^n + B \cdot (-1)^n
   \]
   Using the initial conditions to find \(A\) and \(B\):
   \[
   a_1 = 3 \implies 2A - B = 3
   \]
   \[
   a_2 = 5 \implies 4A + B = 5
   \]
   Solving these simultaneous equations:
   \[
   2A - B = 3 \quad \text{(1)}
   \]
   \[
   4A + B = 5 \quad \text{(2)}
   \]
   Adding equations (1) and (2):
   \[
   6A = 8 \implies A = \frac{4}{3}
   \]
   Substituting \(A = \frac{4}{3}\) into equation (1):
   \[
   2 \cdot \frac{4}{3} - B = 3 \implies \frac{8}{3} - B = 3 \implies B = \frac{8}{3} - 3 = \frac{8}{3} - \frac{9}{3} = -\frac{1}{3}
   \]
   Therefore, the solution to the recurrence relation is:
   \[
   a_n = \frac{4}{3} \cdot 2^n - \frac{1}{3} \cdot (-1)^n = \frac{4 \cdot 2^n + (-1)^{n+1}}{3}
   \]

4. **Conclusion:**
   The number of ways to tile a \(3 \times n\) rectangle using \(2 \times 1\) and \(1 \times 1\) tiles, with the given conditions, is:
   \[
   a_n = \frac{2^{n+2} + (-1)^{n+1}}{3}
   \]

The final answer is \(\boxed{\frac{2^{n+2} + (-1)^{n+1}}{3}}\).