Brute forces perhaps

This can be solved just by number theory methods. The last Fibonacci prime before 200 has index 137 OEIS .

Closed-form expression can be used to calculate the logarithm. Since for large numbers the second term of the closed form expression can be ignored, it becomes: $log{_1}{_3}{_7} ( \varphi ^{137} / \sqrt{5}) = 13.23608812$

where $\varphi = \frac{1 + \sqrt{5}}{2} \approx 1.61803\,39887\cdots$

I have Fast Fibonacci algorithm to calculate n-th Fibonacci Number but no Prime Testing Algorithm for so big numbers. So I calculated $F(200)$ and then checked out the list of prime Fibonacci on oeis .

I too checked out OEIS to find that the last fibonacci prime before $n=200$ is $k=137$ . Rest was computed using Binet's formula.

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import math

def fib(n):
    if(n==0):
        return 0
    else:
        #Binet's formula
        return math.floor((1/math.sqrt(5)*((1+math.sqrt(5))**n - (1-math.sqrt(5))**n)/2**n))

print("ans : ",math.floor(math.log(fib(137))/math.log(137)))

Since log (x)=x^n/1-n then n would have to be a prime number no larger than 13. At least, this is what made sense in my head, I just don't know exactly how to write the notation.