Fibonacci Modified : Challenge | Dynamic Programming | Algorithms | HackerRank

Fibonacci Modified : Challenge | Dynamic Programming | Algorithms | HackerRank
A series is defined in the following manner:
Given the n^th and (n+1)^th terms, the (n+2)^th can be computed by the following relation
T_n+2 = (T_n+1)² + T_n
So, if the first two terms of the series are 0 and 1:
the third term = 1² + 0 = 1
fourth term = 1² + 1 = 2
fifth term = 2² + 1 = 5
... And so on.
Given three integers A, B and N, such that the first two terms of the series (1^st and 2^nd terms) are A and B respectively, compute the N^th term of the series.
http://blog.csdn.net/jmspan/article/details/51740378

        Scanner scanner = new Scanner(System.in);
        int a = scanner.nextInt();
        int b = scanner.nextInt();
        int n = scanner.nextInt();
        BigInteger[] nums = new BigInteger[n];
        nums[0] = new BigInteger(Integer.toString(a), 10);
        nums[1] = new BigInteger(Integer.toString(b), 10);
        for(int i = 0; i < n - 2; i ++) {
            nums[i + 2] = nums[i].add(nums[i + 1].multiply(nums[i + 1]));
        }
        System.out.println(nums[n - 1].toString());
    }

https://github.com/bhajunsingh/programming-challanges/blob/master/hackerrank/algorithms/fibonacci-modified/fibonacci-modified.py
if __name__ == '__main__':
    A, B, N = list(map(int, sys.stdin.readline().split()))
 
    for _ in range(N - 2):
        A, B = B, A + B * B
 
    print(B)
https://github.com/17patelumang/HackerRank/blob/master/fibonacci-modified.py
    if n==1:
        print a
    elif n==2:
        print b
    else:
     
        while count<n:
            count=count+1
            c=(b*b)+a
            a=b
            b=c
     
        print b
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