Sum of Diagonals in 1001x1001 Spiral

[AntonVrba]

Consider a spiral starting with the number 1 and moving to the right in a clockwise direction, below is an example of a 5 by 5 spiral and the diagonal is highlighted:


21 22 23 24 25
20 07 08 09 10
19 06 01 02 11
18 05 04 03 12
17 16 15 14 13

The sum of both diagonals is 101.

if the spiral formed in the same way to 1001 by 1001, then what is the sum of both diagonals?

 

[AntonVrba]

HINT:

the number in the top right corner is n^2, the other corners are given by: n^2-n+1, n^2-2n+2, and n^2-3n+3.

 

[Galileo]

Answer:


Argh, my stupid computer crashed while I finished typing out the solution here.
Luckily my closed form formula was still copied to mthe memory:
Sum of diagonals for the nxn spiral (n is odd): (2/3)n^3+(1/2)n^2+(4/3)n-3/2

Which gives for n=1001: 669171001

 

[BicycleTree]

I wrote a program and came up with the same answer.

class Diagonals{
	public static void main(String args[])
	{
		int x = 0, y = 0, sum = 0;
		for(int i = 1; i <= 1001*1001; i++)
		{
			if(Math.abs(x) == Math.abs(y))
				sum+=i;
			if(y >= Math.abs(x))
				x++;
			else if((y <= -x && x > 0) || (y < x && x <= 0))
				x--;
			else if(x > 0)
				y--;
			else if(x < 0)
				y++;
		}
		System.out.println(sum);
	}
}

How can I prove my program equivalent to your formula?

 

[Andrex]

Answer:


Argh, my stupid computer crashed while I finished typing out the solution here.
Luckily my closed form formula was still copied to mthe memory:
Sum of diagonals for the nxn spiral (n is odd): (2/3)n^3+(1/2)n^2+(4/3)n-3/2

Which gives for n=1001: 669171001

You were once the only author on the net to publish this formula. How do you come to this result? I published a similar formula much later. Sorry for not quoting you.

My contribution in APL Quote Quad (December 20027) fell in the category "APL as a tool of thought".

Regards

 

[mathal]

Tribute and addendum to Galileo's solution .
Perhaps someone can make use of the iteration math I am contributing to find a way to Galileo's formula.

Spoiler

Galileo's solution formula
(2/3)n^3+(1/2)n^2+(4/3)n-3/2The algorithm I made up used n^2 +((n-1)^2)+1)*3 for each +2 iteration starting at 1.
Starting at 2 it yields the even diagonal sums.
Your formula for even diagonals would leave out the last -3/2. Yours is a straightforward answer. Good one Galileo.

mathal

 

[Andrex]

You were once the only author on the net to publish this formula. How do you come to this result? I published a similar formula much later. Sorry for not quoting you.

My contribution in APL Quote Quad (December 20027) fell in the category "APL as a tool of thought".

Regards

Erratum: I mean "December 2007" (not "December 20027").