[THOUGHT] We need to understand the function f(x). The condition says: for n = 1,2,3,..., when x ∈ [2^{n-1} - 1, 2^{n} - 1), we have: f(x) = sqrt(2^{n-1}(x+1) - 2^{2n-2}) + 2^{n-1} - 1.

Let's simplify the expression inside the square root: 2^{n-1}(x+1) - 2^{2n-2} = 2^{n-1}x + 2^{n-1} - 2^{2n-2}.

But note 2^{2n-2} = (2^{n-1})^2. So: = 2^{n-1}x + 2^{n-1} - (2^{n-1})^2 = 2^{n-1}(x + 1 - 2^{n-1}).

Thus f(x) = sqrt(2^{n-1}(x + 1 - 2^{n-1})) + 2^{n-1} - 1.

Check domain: x ∈ [2^{n-1} - 1, 2^{n} - 1). Then x+1 ∈ [2^{n-1}, 2^{n}]. So x+1 - 2^{n-1} ∈ [0, 2^{n-1}]. So inside sqrt: 2^{n-1}*(x+1 - 2^{n-1}) ∈ [0, 2^{2n-2}], non-negative, good.

Now define a_n = 2^{n-1} - 1. Then a_n = 2^{n-1} - 1, a_{n+1} = 2^{n} - 1.

We need ∫{a_n}^{a{n+1}} f(x) dx = given value: 1407374833221632 / 3.

We can compute the integral analytically for each n. Since f(x) is defined piecewise over intervals [a_n, a_{n+1}) for n (but careful: the definition uses same n? Let's check: For x in [2^{n-1} - 1, 2^{n} - 1), f(x) = ... with same n. So indeed for interval from a_n to a_{n+1}, the n in definition is exactly the n we are using. Because a_n = 2^{n-1} - 1, a_{n+1} = 2^{n} - 1, which matches the interval for that n.

Thus for n, the integral I_n = ∫{a_n}^{a{n+1}} f(x) dx = ∫_{2^{n-1}-1}^{2^{n}-1} [sqrt(2^{n-1}(x+1 - 2^{n-1})) + 2^{n-1} - 1] dx.

Let's compute I_n analytically.

Let t = x+1 - 2^{n-1}. Then x = t + 2^{n-1} - 1. When x = a_n = 2^{n-1} - 1, t = (2^{n-1} - 1 + 1) - 2^{n-1} = 0. When x = a_{n+1} = 2^{n} - 1, t = (2^{n} - 1 + 1) - 2^{n-1} = 2^{n} - 2^{n-1} = 2^{n-1}. So t ∈ [0, 2^{n-1}].

Also dx = dt.

f(x) = sqrt(2^{n-1} * t) + 2^{n-1} - 1 = sqrt(2^{n-1}) * sqrt(t) + 2^{n-1} - 1.

Thus I_n = ∫_{t=0}^{2^{n-1}} [sqrt(2^{n-1}) * sqrt(t) + 2^{n-1} - 1] dt.

Compute: sqrt(2^{n-1}) = 2^{(n-1)/2}.

∫ sqrt(t) dt = (2/3) t^{3/2}.

So ∫ sqrt(2^{n-1}) sqrt(t) dt = 2^{(n-1)/2} * (2/3) t^{3/2}.

∫ (2^{n-1} - 1) dt = (2^{n-1} - 1) t.

Evaluate from 0 to 2^{n-1}:

I_n = 2^{(n-1)/2} * (2/3) * (2^{n-1})^{3/2} + (2^{n-1} - 1) * 2^{n-1}.

Simplify: (2^{n-1})^{3/2} = 2^{(n-1)*3/2} = 2^{3(n-1)/2}.

Multiply by 2^{(n-1)/2}: 2^{(n-1)/2} * 2^{3(n-1)/2} = 2^{(n-1)/2 + 3(n-1)/2} = 2^{4(n-1)/2} = 2^{2(n-1)} = 4^{n-1}.

Thus first term: (2/3) * 4^{n-1}.

Second term: (2^{n-1} - 1) * 2^{n-1} = 2^{2n-2} - 2^{n-1}.

So I_n = (2/3) * 4^{n-1} + 2^{2n-2} - 2^{n-1}.

But 4^{n-1} = 2^{2n-2}. So (2/3)*4^{n-1} = (2/3)*2^{2n-2}.

Thus I_n = (2/3)*2^{2n-2} + 2^{2n-2} - 2^{n-1} = (2/3 + 1)*2^{2n-2} - 2^{n-1} = (5/3)*2^{2n-2} - 2^{n-1}.

Simplify: 2^{2n-2} = 4^{n-1}. So I_n = (5/3)*4^{n-1} - 2^{n-1}.

Check: For n=1: a1 = 0, a2 = 1. f(x) for n=1: interval [0,1). f(x) = sqrt(2^{0}(x+1 - 1)) + 2^{0} - 1 = sqrt(x) + 0. So I_1 = ∫_0^1 sqrt(x) dx = 2/3. Our formula: (5/3)*4^{0} - 2^{0} = (5/3)*1 - 1 = 5/3 - 1 = 2/3. Good.

Now we need I_n = 1407374833221632 / 3.

Thus (5/3)*4^{n-1} - 2^{n-1} = 1407374833221632 / 3.

Multiply both sides by 3: 54^{n-1} - 32^{n-1} = 1407374833221632.

Let m = n-1, then m ≥ 0 integer. Then equation: 54^{m} - 32^{m} = 1407374833221632.

Let y = 2^m. Then 4^m = (2^m)^2 = y^2. So equation: 5 y^2 - 3 y - 1407374833221632 = 0.

Solve quadratic: y = [3 ± sqrt(9 + 451407374833221632)]/(25) = [3 ± sqrt(9 + 201407374833221632)]/10.

Compute discriminant D = 9 + 20*1407374833221632 = 9 + 28147496664432640 = 28147496664432649.

We need y positive integer, so y = (3 + sqrt(D))/10.

We need sqrt(D) to be integer. Let's compute D exactly.

We'll compute using Python to see if sqrt(D) is integer and find m.

Then n = m+1.

We'll write code to solve.

Also note: the given number is large: 1407374833221632. That's 1.407374833221632e15. It resembles 2^something? Let's see: 2^50 = 1125899906842624, 2^51 = 2251799813685248. So our number is between 2^50 and 2^51. 1407374833221632 is close to 2^50? Actually 2^50 = 1125899906842624, difference ~2814749666443264. Wait that's exactly 2^51? No.

Better to compute directly.

We'll compute D and sqrt.

Let's code.

[/THOUGHT] [PYTHON] import math

Given constant

const = 1407374833221632

Equation: 5y^2 - 3y - const = 0

y = (3 + sqrt(9 + 20*const)) / 10

D = 9 + 20 * const sqrtD = math.isqrt(D) # integer sqrt if perfect square if sqrtD * sqrtD != D: print("D is not a perfect square") else: y = (3 + sqrtD) // 10 # Check if y is integer and satisfies equation if 5yy - 3*y == const: print(f"y = {y}") # y = 2^m, m = log2(y) # Check if y is power of 2 if (y & (y - 1)) == 0 and y > 0: m = int(math.log2(y)) n = m + 1 print(f"m = {m}, n = {n}") else: print("y is not a power of 2") else: print("No integer solution found") [/PYTHON]

[THOUGHT] The Python output shows that n = 25 satisfies the condition. The final answer is n = 25.

\boxed{25}