/* 
 * Solution to Project Euler problem 28
 * Copyright (c) Project Nayuki. All rights reserved.
 * 
 * https://www.nayuki.io/page/project-euler-solutions
 * https://github.com/nayuki/Project-Euler-solutions
 */


public final class p028 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p028().run());
	}
	
	
	/* 
	 * From the diagram, let's observe the four corners of an n * n square (where n is odd).
	 * It's not hard to convince yourself that the top right corner always has the value n^2.
	 * Working counterclockwise (backwards), the top left corner has the value n^2 - (n - 1),
	 * the bottom left corner has the value n^2 - 2(n - 1), and the bottom right is n^2 - 3(n - 1).
	 * Putting it all together, this outermost ring contributes 4n^2 - 6(n - 1) to the final sum.
	 * 
	 * Incidentally, the closed form of this sum is (4m^3 + 3m^2 + 8m - 9) / 6, where m = size.
	 */
	private static final int SIZE = 1001;  // Must be odd
	
	public String run() {
		long sum = 1;  // Special case for size 1
		for (int n = 3; n <= SIZE; n += 2)
			sum += 4 * n * n - 6 * (n - 1);
		return Long.toString(sum);
	}
	
}