Problem

Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:

21 22 23 24 25
20  7  8  9 10
19  6  1  2 11
18  5  4  3 12
17 16 15 14 13

It can be verified that the sum of the numbers on the diagonals is 101.

What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?

Answer

669171001

Python

#!/usr/bin/env python
def sum_diagonals_of_spiral(size):
    n = 1
    step = 2
    total = 0
    since_last = 0
    while n <= size**2:
        total += n
        n += step
        since_last += 1
        if since_last == 4:
            step += 2
            since_last = 0
    return total

print(sum_diagonals_of_spiral(1001))

Ruby

#!/usr/bin/env ruby
i = 1
sum = i
step = 2
until i >= 1001**2
  4.times { sum += i += step }
  step += 2
end
puts sum

Haskell

main :: IO ()
main = print $ foldr (+) 1 [4*n^2 - 6*n + 6 | n <- [3,5..1001]]

Mathematica

(* 
 * 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.
 *)
size = 1001;  (* Must be odd *)
Sum[Block[{n = 2k + 1}, 4 * n^2 - 6 * (n - 1)], {k, (size - 1) / 2}] + 1

Java

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);
	}
	
}