Consider all right triangles with integer side lengths that form an arithmetic sequence. Compute the $2014$th smallest perimeter of all such right triangles.
1. We are given a right triangle with integer side lengths that form an arithmetic sequence. Let the side lengths be \(a\), \(b\), and \(c\) such that \(a < b < c\). Since they form an arithmetic sequence, we can write:
   \[
   b = a + d \quad \text{and} \quad c = a + 2d
   \]
   for some positive integer \(d\).

2. Since the triangle is a right triangle, it must satisfy the Pythagorean theorem:
   \[
   a^2 + b^2 = c^2
   \]
   Substituting \(b\) and \(c\) in terms of \(a\) and \(d\), we get:
   \[
   a^2 + (a + d)^2 = (a + 2d)^2
   \]

3. Expanding and simplifying the equation:
   \[
   a^2 + (a^2 + 2ad + d^2) = a^2 + 4ad + 4d^2
   \]
   \[
   2a^2 + 2ad + d^2 = a^2 + 4ad + 4d^2
   \]
   \[
   2a^2 + 2ad + d^2 - a^2 - 4ad - 4d^2 = 0
   \]
   \[
   a^2 - 2ad - 3d^2 = 0
   \]

4. This is a quadratic equation in \(a\). Solving for \(a\):
   \[
   a^2 - 2ad - 3d^2 = 0
   \]
   Using the quadratic formula \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -2d\), and \(c = -3d^2\):
   \[
   a = \frac{2d \pm \sqrt{(2d)^2 - 4 \cdot 1 \cdot (-3d^2)}}{2 \cdot 1}
   \]
   \[
   a = \frac{2d \pm \sqrt{4d^2 + 12d^2}}{2}
   \]
   \[
   a = \frac{2d \pm \sqrt{16d^2}}{2}
   \]
   \[
   a = \frac{2d \pm 4d}{2}
   \]
   \[
   a = d \quad \text{or} \quad a = -3d
   \]
   Since \(a\) must be positive, we have \(a = 3d\).

5. Therefore, the side lengths of the triangle are:
   \[
   a = 3d, \quad b = a + d = 4d, \quad c = a + 2d = 5d
   \]
   The side lengths form the triple \((3d, 4d, 5d)\).

6. The perimeter of the triangle is:
   \[
   3d + 4d + 5d = 12d
   \]

7. To find the 2014th smallest perimeter, we set \(d = 2014\):
   \[
   \text{Perimeter} = 12 \times 2014 = 24168
   \]

The final answer is \(\boxed{24168}\).