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1 + [Summation of i(i+1)/2 where i runs from 1 to n]

Also, did a bit of math and found a closed form formula for this:

1 + [n(n+1)(2n+1) + 3n(n+1)]/12

Another way of looking at this.

You want the number of triples (i,j,k) such that n > i >= j >= k >= 0

Now if you consider the triple (i+2, j+1, k), then we have that n+1 >= i+2 > j+1 >k >= 0. Thus all we need to do is select 3 numbers among 0,1,2,..., n+2 and sort them, subtract 1 from the middle and 2 from the max to get our possible i,j,k.

Thus, the number is

n+1 choose 3.

Since temp starts out as 1, the answer is

1 + (n+1 choose 3)

I initially thought it to be n^2 + (n-1), but later realized this does not look to have a generic formula at least to me... for e.g. if n=2, temp will be 5, if n=3, temp will be 11, the above formula only works for n=2 and 3, but if n=4, temp = 21 (n^2 + (n+1)) and if n=5, temp = 36. So this does not look to have a generic formula.. Correct me if I'm wrong.

So, if n=2,3,4,5,6 ==> then temp = 5,11,21,36,57. Let me know if you folks see a formula here, it does not come to my head at this point :-).

Well my Math Says so.

1> Loop 1 runs upto (n-1)
2> Loop 2 runs upto (n-1)*n(Square)/2 , coz its summation where n is infact (n-1) here.
3> Loop 3 this is the most complex one, which my calculation shows whopping
((n-1)*n(Square))/4*((n-1)*n(Square)/2+1)

Now Final Formula is the Product of Loop 1 * Loop 2 * Loop 3 . Which is Strange ..

F[n] = F[n-1] + n*(n+1)/2

F(n)=2*F(n-1)+n+1;

this one gives correct answer
C(n-1+r,r)

here, n is no elements and r is no of loops...

one more solution to the same and somewhat explainable....

n*(n+1)(n+1)/r!

where n is no of elements and r is no loops....

1 + the number of combinations with repetition allowed (n choose r).
The formula is (n+r-1)! / r!(n-1)! and r = 3 (the number of nested loops)

1 + (n+2)! / 6(n-1)!

answer is temp=(((n*(n+1)*(2n+1))+(3n*(n+1)))/12)+1

The answer is 1 + Summation ( Sum (i) ) where i = 1 to n

Adding 1 because, temp is initialized to 1.
So it will be 1 + Summation (i(i+1)/2) = 1 + (Summation(i^2) + Summation(i))/2 for i =1 to n
= 1+ [ ( n(n+1)(2n+1)/6 ) + ( n(n+1)/2 ) ] /2
And when simplified, gets to 1+ ( n(n+1)(n+2)/6 )

it's ((n-1)^i)^k

Algorithm for the above problem

if n=1 value is 1
n=2 value is 4
if n>=3, use below
Formula is (n*n)+ValueOf(n-2)

Java Code for the above algorithm.

public class CountForLoopUsingRecursion {

	/**
	 * @param args
	 */
	//what is the value of temp for N
	//int temp=0;
	//for (int i = 0; i <N; i++) {
	//	for (int j = 0; j <=i; j++) {
	//		for (int k = 0; k <=j; k++) {
	//			temp++;
	//		}
	//	}
		
	public static void main(String[] args) {
		// TODO Auto-generated method stub
			System.out.println(findLoopCount(1));
			System.out.println(findLoopCount(2));
			System.out.println(findLoopCount(3));
			System.out.println(findLoopCount(4));
			System.out.println(findLoopCount(5));
			System.out.println(findLoopCount(6));
			System.out.println(findLoopCount(7));
			System.out.println(findLoopCount(8));
	}
	
	static int findLoopCount(int n)
	{
		if(n==1||n<=0)
		{
			return 1;
		}
		if(n==2)
		{
			return 4;
		}
		
			return (n*n)+findLoopCount(n-2);
		
	}

}

n * (n+1)/2 * (n+1)/4

Program will throw compilation error as the variable 'n' is not declared.