100 points. Python 3

n=int(input().strip())
numerators=[1]
denominators=[2]
i=2
while i<=n:
    numerators.append(denominators[-1])
    denominators.append(denominators[-1]*2+numerators[-2])
    i+=1
for i in range(n):
    n=denominators[i]+numerators[i]
    d=denominators[i]
    if len(str(n))>len(str(d)):
        print(i+1)

JAva Code

import java.io.*;
import java.util.*;
import java.math.*;

public class Solution {

    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();
        scanner.close();

        List<Integer> iterations = new ArrayList<>();
        BigInteger numerator = BigInteger.valueOf(3);
        BigInteger denominator = BigInteger.valueOf(2);

        for (int i = 2; i <= N; i++) {
            BigInteger newNumerator = numerator.add(denominator.multiply(BigInteger.valueOf(2)));
            BigInteger newDenominator = numerator.add(denominator);

            if (newNumerator.toString().length() > newDenominator.toString().length()) {
                iterations.add(i);
            }

            numerator = newNumerator;
            denominator = newDenominator;
        }

        for (int iteration : iterations) {
            System.out.println(iteration);
        }
    }
}

Paythan

[def continued_fraction_expansion_sqrt2(N): numerators = [3] denominators = [2] iterations = []

for i in range(1, N):
    numerator = numerators[-1] + 2 * denominators[-1]
    denominator = numerators[-1] + denominators[-1]
    numerators.append(numerator)
    denominators.append(denominator)

    # Check if the numerator has more digits than the denominator
    if len(str(numerator)) > len(str(denominator)):
        iterations.append(i + 1)  # Add 1 to make it 1-based index

return iterations

Input

N = int(input())

Get the iterations where the numerator has more digits than the denominator

result = continued_fraction_expansion_sqrt2(N)

Output the result

for iteration in result: print(iteration)](https://)

include

using namespace std;

deque Add(deque a, deque b) { deque res;

if(a.size() < b.size()) swap(a, b);

short carry = 0;

while(!a.empty())
{
    short sum = a.back() + ((!b.empty()) ? b.back() : 0) + carry;

    carry = sum / 10;
    res.push_front(sum % 10);
    a.pop_back();

    if(!b.empty()) b.pop_back();
}
while(carry) 
{
    res.push_front(carry);
    carry /= 10;
}
return res;

}

bool DB = false;

int main() { int n; cin >> n;

deque<short> N = {1}, D = {2};

for(int i=1; i<=n; i++)
{        
    if(Add(N, D).size() > D.size())
    {
        cout << i << endl;
    }

    deque<short> next_n = Add(D, D);        
    deque<short> next_d = D;

    next_n = Add(next_n, N);

    N = next_n;
    D = next_d;        
    swap(N, D);
}        
return 0;

}

HINTS:

This is like an IQ test: Find similarities of series (3/2, 7/5, 17/2, 41/29,...)

For example with 7/5:

Numerator (N) 7 = 3 + 2 x 2

Denominator (D) 5 = 3 + 2

Then we got:

N[i] = N[i-1] + 2*D[i-1]

D[i] = N[i-1] + D[i-1]

Then you can solve yourself