[THOUGHT] I will compute the integral over [a_n,a_{n+1}] in closed form in terms of n, set it equal to the given value, and solve for n by searching for the power-of-two parameter m=2^{n-1} using Python verification. [/THOUGHT] [PYTHON] import sympy as sp

N = sp.Integer(1407374833221632) m = sp.symbols('m', integer=True, positive=True)

Solve quadratic 5m^2 - 3m - N = 0

sol = sp.solve(sp.Eq(5m**2 - 3m - N, 0), m) print(sol)

Check which solution is a power of two

def is_power_of_two(k): return k > 0 and (k & (k-1) == 0)

for s in sol: if s.is_integer and s > 0: si = int(s) print("candidate m:", si, "power_of_two:", is_power_of_two(si))

If one is power of two, compute n

for s in sol: if s.is_integer and s > 0: si = int(s) if is_power_of_two(si): n = sp.log(si, 2) + 1 print("n =", n)

Verify integral formula directly for that n

def I(n): # I_n = (52^(2n-2) - 32^(n-1))/3 return (5sp.Integer(2)**(2n-2) - 3*sp.Integer(2)**(n-1)) / 3

for n in range(1, 80): if sp.simplify(I(n) - N/3) == 0: print("Verified n:", n, "I(n):", I(n)) break [/PYTHON]

[THOUGHT] The verified solution from Python gives (m=2^{n-1}=16777216=2^{24}), hence (n=25). [ \boxed{25} ]