[THOUGHT] Let me analyze this problem step by step.

First, let me understand the function f(x):

• For $2^{n-1} - 1 \le x < 2^n - 1$, we have $f(x) = \sqrt{2^{n-1}(x+1) - 2^{2n-2}} + 2^{n-1} - 1$

The sequence $a_n = 2^{n-1} - 1$, so:

• $a_1 = 2^0 - 1 = 0$
• $a_2 = 2^1 - 1 = 1$
• $a_3 = 2^2 - 1 = 3$
• $a_n = 2^{n-1} - 1$
• $a_{n+1} = 2^n - 1$

So the interval $[a_n, a_{n+1}]$ is exactly $[2^{n-1} - 1, 2^n - 1]$, which matches the domain where the formula for f(x) applies with parameter n.

Let me compute the integral $\int_{a_n}^{a_{n+1}} f(x) dx$.

For $x \in [2^{n-1} - 1, 2^n - 1]$: $f(x) = \sqrt{2^{n-1}(x+1) - 2^{2n-2}} + 2^{n-1} - 1$

Let me substitute $u = x + 1$, so when $x = 2^{n-1} - 1$, $u = 2^{n-1}$, and when $x = 2^n - 1$, $u = 2^n$.

$f(x) = \sqrt{2^{n-1} \cdot u - 2^{2n-2}} + 2^{n-1} - 1 = \sqrt{2^{n-1}(u - 2^{n-1})} + 2^{n-1} - 1$

Let me compute this integral using Python. [/THOUGHT] [PYTHON] from sympy import symbols, sqrt, integrate, simplify

n = symbols('n', positive=True, integer=True) x = symbols('x')

For the interval [2^(n-1) - 1, 2^n - 1]

f(x) = sqrt(2^(n-1)*(x+1) - 2^(2n-2)) + 2^(n-1) - 1

Let's compute the integral symbolically first for a specific n

def compute_integral(n_val): # f(x) = sqrt(2^(n-1)*(x+1) - 2^(2n-2)) + 2^(n-1) - 1 a = 2**(n_val - 1) # 2^(n-1) lower = 2**(n_val - 1) - 1 upper = 2**n_val - 1

# f(x) = sqrt(a*(x+1) - a^2) + a - 1
# = sqrt(a*(x+1-a)) + a - 1
# = sqrt(a) * sqrt(x+1-a) + a - 1

f_x = sqrt(a * (x + 1) - a**2) + a - 1

result = integrate(f_x, (x, lower, upper))
return simplify(result)

Test for small values of n

for n_val in range(1, 10): result = compute_integral(n_val) print(f"n = {n_val}: integral = {result}") [/PYTHON]

[THOUGHT] I can see a pattern in the integrals. Let me find which n gives the value 1407374833221632/3.

Looking at the pattern:

• n=1: 2/3
• n=2: 14/3
• n=3: 68/3
• n=4: 296/3
• n=5: 1232/3
• n=6: 5024/3
• n=7: 20288/3
• n=8: 81536/3
• n=9: 326912/3

The numerators are: 2, 14, 68, 296, 1232, 5024, 20288, 81536, 326912...

I need to find n such that the numerator equals 1407374833221632.

Let me check: 1407374833221632 is a very large number. Let me compute more values or find the pattern.

Looking at ratios: 14/2 = 7, 68/14 ≈ 4.86, 296/68 ≈ 4.35, 1232/296 ≈ 4.16, 5024/1232 ≈ 4.08...

The pattern seems to approach 4. Let me check if there's a formula. The numerator appears to follow a pattern like $2^{2n-1} + 2^n - 2$ or similar.

Actually, let me verify: for n=1: 2^1 + 2^1 - 2 = 2 ✓ n=2: 2^3 + 2^2 - 2 = 8 + 4 - 2 = 10 ✗

Let me try another pattern. Looking at 2, 14, 68, 296, 1232:

• 2 = 2
• 14 = 2 + 12 = 2 + 4·3
• 68 = 14 + 54 = 14 + 4·13.5...

Actually, the formula might be $\frac{2^{2n} + 2^{n+1} - 4}{3} \cdot 2 = \frac{2^{2n+1} + 2^{n+2} - 8}{3}$

Let me just compute: 1407374833221632 = 2^{2n+1} + 2^{n+2} - 8 for some n.

1407374833221632 ≈ 1.4 × 10^15 ≈ 2^50

So 2n+1 ≈ 50, meaning n ≈ 25.

Let me verify: 2^51 = 2251799813685248, 2^50 = 1125899906842624

1407374833221632 - 2^50 = 281474926379008 2^48 = 281474976710656, close!

Actually 1407374833221632 = 2^50 + 2^48 - 8 = 1125899906842624 + 281474976710656 - 8 = 1407374883553272 ≠ 1407374833221632

Let me recalculate. The target numerator is 1407374833221632.

2^50 = 1125899906842624 1407374833221632 - 1125899906842624 = 281474926379008

2^48 = 281474976710656 281474926379008 is close to 2^48 but not exact.

Let me try n=25: 2^51 + 2^27 - 8 = 2251799813685248 + 134217728 - 8 = 2251799947902968 ≠ target

I need to run more Python to find the exact n. [/THOUGHT]

Actually, I need to run more code to find the answer. Let me continue.

[PYTHON] target = 1407374833221632

def compute_integral_numerator(n_val): from sympy import symbols, sqrt, integrate, simplify, Rational x = symbols('x') a = 2**(n_val - 1) lower = 2**(n_val - 1) - 1 upper = 2n_val - 1 f_x = sqrt(a * (x + 1) - a2) + a - 1 result = integrate(f_x, (x, lower, upper)) return simplify(result)

Binary search or direct computation

for n_val in range(20, 60): result = compute_integral_numerator(n_val) numerator = result * 3 print(f"n = {n_val}: numerator = {numerator}") if numerator == target: print(f"FOUND: n = {n_val}") break [/PYTHON]

[THOUGHT] The Python computation confirms that when n = 25, the integral equals 1407374833221632/3, which matches the target value exactly. [/THOUGHT]

\boxed{25}