Finding taxicab Numbers

Question

Find the first n taxicab numbers. Given a value n. I would like to find the first n taxicab numbers. A taxicab being a number that can be expressed as the sum of two perfect cubes in more than one way.

(Note that there are two related but different sets referred to as 'taxicab numbers': the sums of 2 cubes in more than 1 way, and the smallest numbers that are the sum of 2 positive integral cubes in n ways. This question is about the former set, since the latter set has only the first six members known)

For example:

1^3 + 12^3 = 1729 = 9^3 + 10^3

I would like a rough overview of the algorithm or the C snippet of how to approach the problem.

The first five of these are:

   I    J      K    L      Number 
---------------------------------
   1   12      9   10      1729       
   2   16      9   15      4104      
   2   24     18   20     13832       
  10   27     19   24     20683      
   4   32     18   30     32832    

Answer 1

Below is the even better java code for printing N ramanujan numbers as it has even less time complexity. Because, it has only one for loop.

import java.util.*;
public class SolutionRamanujan 
{
    public static void main(String args[] ) throws Exception 
    {
        SolutionRamanujan s=new SolutionRamanujan();
        Scanner scan=new Scanner(System.in);
        int n=scan.nextInt();
        int i=0,k=1;
        while(i<n){
            if(s.checkRamanujan(k))
            {
                i=i+1;
                System.out.println(i+" th ramanujan number is "+k);
            }
            k++;
        }
        scan.close();
    }
    //checking whether a number is ramanujan number
    public boolean checkRamanujan(int a)
    {   
        int count=0;
        int cbrt=(int)Math.cbrt(a);
        //numbers only below and equal to cube th root of number
        for(int i=1;i<=cbrt;i++)
        {
            int difference=a-(i*i*i);
            int a1=(int) Math.cbrt(difference);                //checking whether the difference is perfect cube 

            if(a1==Math.cbrt(difference)){
                count=count+1;
            }
            if(count>2){            //checking if two such pairs exists i.e. (a*a*a)+(b*b*b)=(c*c*c)+(d*d*d)=number
                return true;
            }
        }
        return false;
    }
}