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[THOUGHT] We need to compute the integral (\int_{a_n}^{a_{n+1}} f(x) , dx) where (a_n = 2^{n-1} - 1). So (a_n = 2^{n-1} - 1) and (a_{n+1} = 2^{n} - 1). The integral is from (2^{n-1} - 1) to (2^{n} - 1). According to the definition of (f(x)), on the interval (2^{n-1} - 1 \le x < 2^{n} - 1), we have (f(x) = \sqrt{2^{n-1}(x+1) - 2^{2n-2}} + 2^{n-1} - 1). Note that at the right endpoint (x = 2^{n} - 1), the formula for (f(x)) changes because it belongs to the next interval. But the integral is over a closed interval ([a_n, a_{n+1}]). However, the function is defined piecewise, and at the right endpoint, the formula for the next piece might be different. But since (f) is continuous, the value at the endpoint matches the limit from the left. So we can integrate using the given formula over the half-open interval and include the endpoint because the integral over a single point doesn't matter. So we can integrate (f(x)) from (x = a_n) to (x = a_{n+1}) using the expression given for that interval.

Let’s set (n) as a variable. We need to compute: [ I_n = \int_{2^{n-1} - 1}^{2^{n} - 1} \left( \sqrt{2^{n-1}(x+1) - 2^{2n-2}} + 2^{n-1} - 1 \right) dx. ]

Simplify the integrand. Let (A = 2^{n-1}). Then (a_n = A - 1), (a_{n+1} = 2A - 1). The expression inside the square root: (A(x+1) - A^2 = A(x+1 - A)). So (f(x) = \sqrt{A(x+1 - A)} + A - 1).

Thus: [ I_n = \int_{A-1}^{2A-1} \left( \sqrt{A(x+1 - A)} + A - 1 \right) dx. ]

Make substitution: Let (u = x+1 - A). Then when (x = A-1), (u = (A-1)+1 - A = 0). When (x = 2A-1), (u = (2A-1)+1 - A = A). Also (dx = du). Then: [ I_n = \int_{0}^{A} \left( \sqrt{A u} + A - 1 \right) du = \int_{0}^{A} \sqrt{A} \sqrt{u} , du + \int_{0}^{A} (A-1) du. ]

Compute: [ \int_{0}^{A} \sqrt{A} \sqrt{u} , du = \sqrt{A} \int_{0}^{A} u^{1/2} du = \sqrt{A} \cdot \frac{2}{3} u^{3/2} \Big|{0}^{A} = \frac{2}{3} \sqrt{A} A^{3/2} = \frac{2}{3} A^2. ] [ \int{0}^{A} (A-1) du = (A-1) A = A^2 - A. ]

Thus: [ I_n = \frac{2}{3} A^2 + A^2 - A = \frac{5}{3} A^2 - A. ]

Recall (A = 2^{n-1}). So: [ I_n = \frac{5}{3} (2^{n-1})^2 - 2^{n-1} = \frac{5}{3} \cdot 2^{2n-2} - 2^{n-1}. ]

We are given that (I_n = \frac{1407374833221632}{3}). So we solve for (n): [ \frac{5}{3} \cdot 2^{2n-2} - 2^{n-1} = \frac{1407374833221632}{3}. ]

Multiply both sides by 3: [ 5 \cdot 2^{2n-2} - 3 \cdot 2^{n-1} = 1407374833221632. ]

Let (t = 2^{n-1}). Then (2^{2n-2} = t^2). Equation becomes: [ 5 t^2 - 3 t = 1407374833221632. ]

Solve quadratic: (5 t^2 - 3 t - 1407374833221632 = 0).

We need to find integer (t) that is a power of 2. Compute discriminant and solve. Use Python to compute exact values.

We'll compute using Python to find (t) and then (n = \log_2(t) + 1). Ensure (n) is a natural number.

Let's code. [/THOUGHT] [PYTHON] import sympy as sp

Given constant

C = 1407374833221632

Solve 5t^2 - 3t - C = 0 for t

t = sp.symbols('t', positive=True, integer=True) eq = 5t**2 - 3t - C solutions = sp.solve(eq, t) print("Solutions for t:", solutions)

Evaluate solutions numerically

for sol in solutions: print("Solution value:", sol) print("As integer:", int(sol) if sol.is_Integer else sol.evalf()) # Check if it's a power of two if sol.is_Integer: val = int(sol) # Check if val is a power of two if val > 0 and (val & (val - 1)) == 0: print("It is a power of two.") # compute n: t = 2^{n-1} => n-1 = log2(t) n = val.bit_length() # because 2^{k} has bit_length k+1? Actually, 2^{m} has bit_length m+1. # For example, 2^0 = 1, bit_length=1 => m=0 => m = bit_length-1. # So n-1 = bit_length - 1 => n = bit_length. n = val.bit_length() print("n =", n) # Verify: compute I_n from formula A = 2**(n-1) I_n = (5/3) * A2 - A print("I_n computed:", I_n) print("I_n exact rational:", sp.Rational(5,3)*A2 - A) else: print("Not a power of two.") [/PYTHON]

[THOUGHT] The Python output shows that ( t = 16777216 = 2^{24} ), so ( n-1 = 24 ) giving ( n = 25 ). The computed integral matches the given value. Therefore, the natural number ( n ) is 25. [/THOUGHT] \boxed{25}