Which starting number, under one million, produces the longest chain?

The following iterative sequence is defined for the set of positive integers:

n → n / 2 ( n is even)
n → 3 n + 1 ( n is odd)

Using the rule above and starting with 13, we generate the following sequence:

13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.

Which starting number, under one million, produces the longest chain?

Train of thought

• Using 1 million loop exhaustively calculate the length of the chain, and obtain the longest chain by comparison
• When the end of the sequence is not 1, it is calculated according to the specified odd and even iterative sequence

Solutions

public static void findLongestSeq() {
    // record start number
    long startingNumber = 0;
    // record max sequence length
    long maxLength = 0;

    long n;
    for (int i = 1; i < 1000000; i++) {
        long length = 0;
        n = i;
        // according to specific iterative way
        while (n != 1) {
            length++;
            if (n % 2 == 0) {
                n = n / 2;
            } else {
                n = n * 3 + 1;
            }
        }
        // record maxLength & startingNumber
        if (length > maxLength) {
            maxLength = length;
            startingNumber = i;
        }
    }
    System.out.println(startingNumber);
    System.out.println(maxLength);
}

The complexity

Time complexity: O(n²)