[THOUGHT] To solve the problem, we start by analyzing the integral of the piecewise function $f(x)$ over the interval $[a_n, a_{n+1})$. The function $f(x)$ is defined as:

$$f(x) = \sqrt{2^{n-1}(x+1) - 2^{2n-2}} + 2^{n-1} - 1 \quad \text{for} \quad 2^{n-1} - 1 \le x < 2^n - 1.$$

We need to compute the integral:

$$\int_{a_n}^{a_{n+1}} f(x) \, dx = \int_{2^{n-1} - 1}^{2^n - 1} f(x) \, dx.$$

First, we perform a substitution to simplify the integral. Let $u = x + 1$. Then $du = dx$, and the limits of integration change as follows:

• When $x = 2^{n-1} - 1$, $u = 2^{n-1}$.
• When $x = 2^n - 1$, $u = 2^n$.

Thus, the integral becomes:

$$\int_{2^{n-1}}^{2^n} \left( \sqrt{2^{n-1} u - 2^{2n-2}} + 2^{n-1} - 1 \right) \, du.$$

Next, we simplify the expression inside the square root:

$$2^{n-1} u - 2^{2n-2} = 2^{n-1} (u - 2^{n-1}).$$

Let $v = u - 2^{n-1}$. Then $dv = du$, and the limits of integration change as follows:

• When $u = 2^{n-1}$, $v = 0$.
• When $u = 2^n$, $v = 2^{n-1}$.

The integral now becomes:

$$\int_{0}^{2^{n-1}} \left( \sqrt{2^{n-1} v} + 2^{n-1} - 1 \right) \, dv.$$

We split this into two integrals:

$$\int_{0}^{2^{n-1}} \sqrt{2^{n-1} v} \, dv + \int_{0}^{2^{n-1}} (2^{n-1} - 1) \, dv.$$

First, we evaluate the integral of the square root term:

$$\int_{0}^{2^{n-1}} \sqrt{2^{n-1} v} \, dv = \sqrt{2^{n-1}} \int_{0}^{2^{n-1}} \sqrt{v} \, dv.$$

The integral of $\sqrt{v}$ is:

$$\int \sqrt{v} \, dv = \frac{2}{3} v^{3/2}.$$

Evaluating this from 0 to $2^{n-1}$:

$$\sqrt{2^{n-1}} \left[ \frac{2}{3} v^{3/2} \right]_{0}^{2^{n-1}} = \sqrt{2^{n-1}} \cdot \frac{2}{3} \left( (2^{n-1})^{3/2} - 0 \right) = \sqrt{2^{n-1}} \cdot \frac{2}{3} \cdot 2^{3(n-1)/2} = \frac{2}{3} \cdot 2^{(n-1)/2} \cdot 2^{3(n-1)/2} = \frac{2}{3} \cdot 2^{2(n-1)} = \frac{2}{3} \cdot 4^{n-1}.$$

Next, we evaluate the integral of the constant term:

$$\int_{0}^{2^{n-1}} (2^{n-1} - 1) \, dv = (2^{n-1} - 1) \int_{0}^{2^{n-1}} 1 \, dv = (2^{n-1} - 1) \cdot 2^{n-1} = 4^{n-1} - 2^{n-1}.$$

Combining both results, the total integral is:

$$\frac{2}{3} \cdot 4^{n-1} + 4^{n-1} - 2^{n-1} = \left( \frac{2}{3} + 1 \right) 4^{n-1} - 2^{n-1} = \frac{5}{3} \cdot 4^{n-1} - 2^{n-1}.$$

We are given that this integral equals:

$$\frac{1407374833221632}{3}.$$

Thus, we set up the equation:

$$\frac{5}{3} \cdot 4^{n-1} - 2^{n-1} = \frac{1407374833221632}{3}.$$

Multiplying through by 3 to clear the denominator:

$$5 \cdot 4^{n-1} - 3 \cdot 2^{n-1} = 1407374833221632.$$

Let $k = 2^{n-1}$. Then $4^{n-1} = k^2$, and the equation becomes:

$$5k^2 - 3k = 1407374833221632.$$

We solve this quadratic equation for $k$:

$$5k^2 - 3k - 1407374833221632 = 0.$$

Using the quadratic formula $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 5$, $b = -3$, and $c = -1407374833221632$:

$$k = \frac{3 \pm \sqrt{9 + 4 \cdot 5 \cdot 1407374833221632}}{10}.$$

Calculating the discriminant:

$$9 + 4 \cdot 5 \cdot 1407374833221632 = 9 + 28147496664432640 = 28147496664432649.$$

Taking the square root of the discriminant:

$$\sqrt{28147496664432649} = 16777216.$$

Thus, the positive root is:

$$k = \frac{3 + 16777216}{10} = \frac{16777219}{10} = 1677721.6.$$

Since $k$ must be a power of 2, we check $k = 2^{24} = 16777216$:

$$5 \cdot (2^{24})^2 - 3 \cdot 2^{24} = 5 \cdot 2^{48} - 3 \cdot 2^{24} = 5 \cdot 281474976710656 - 3 \cdot 16777216 = 1407374883553280 - 50331648 = 1407374833221632.$$

This matches the given value. Therefore, $k = 2^{24}$, so $n-1 = 24$ and $n = 25$.

[PYTHON]

Verification of the solution

n = 25 k = 2 ** (n - 1) result = 5 * (k ** 2) - 3 * k print(result == 1407374833221632) [/PYTHON]

[THOUGHT] The Python code confirms that $n = 25$ satisfies the equation. Therefore, the natural number $n$ is:

$$\boxed{25}$$

[THOUGHT] The Python output confirms that $n = 25$ satisfies the equation. Therefore, the natural number $n$ is:

$$\boxed{25}$$