[THOUGHT] We have derived that the integral ( I_n = \int_{a_n}^{a_{n+1}} f(x) , dx = \frac{5}{3} \cdot 2^{2n-2} - 2^{n-1} ). Setting this equal to ( \frac{1407374833221632}{3} ) leads to the equation ( 5 \cdot 2^{2n-2} - 3 \cdot 2^{n-1} = 1407374833221632 ). Let ( t = 2^{n-1} ), then ( 5t^2 - 3t = 1407374833221632 ). Solving this quadratic equation for ( t ) and checking if ( t ) is a power of 2 will determine ( n ). We'll compute the discriminant ( D = 9 + 20 \cdot 1407374833221632 ) and verify if ( \sqrt{D} ) is an integer such that ( t = \frac{3 + \sqrt{D}}{10} ) is a power of 2. [/THOUGHT] [PYTHON] constant = 1407374833221632 D = 9 + 20 * constant import math s = math.isqrt(D) print("D =", D) print("s =", s) print("s^2 =", ss) print("D - s^2 =", D - ss)

Check if D is a perfect square

if s*s == D: print("D is a perfect square") else: print("D is not a perfect square")

Compute t = (3 + s) / 10

t_val = (3 + s) / 10 print("t_val =", t_val)

Check if t_val is integer

if t_val.is_integer(): t_int = int(t_val) print("t_int =", t_int) # Check if t_int is a power of 2 if t_int > 0 and (t_int & (t_int - 1)) == 0: n = t_int.bit_length() print("n =", n) else: print("t_int is not a power of 2") else: print("t_val is not integer") [/PYTHON]

[THOUGHT] The Python output confirms that ( D = 28147496664432649 ) is a perfect square with ( s = 167772157 ), and solving ( t = \frac{3 + s}{10} ) yields ( t = 16777216 ), which is ( 2^{24} ). Thus, ( 2^{n-1} = 2^{24} ) implies ( n = 25 ). Verification shows this satisfies the integral equation exactly. [/THOUGHT] \boxed{25}