p028.mat.txt

(* 
 * Solution to Project Euler problem 28
 * By Nayuki Minase
 * 
 * http://nayuki.eigenstate.org/page/project-euler-solutions
 * https://github.com/nayuki/Project-Euler-solutions
 *)


(* Use this in Wolfram 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