Finding taxicab Numbers

algorithm, c, hardy-ramanujan, numbers

Solution

I figured out the answer could be obtained this way:

#include<stdio.h>

int main() {
    int n, i, count=0, j, k, int_count;
    printf("Enter the number of values needed: ");
    scanf("%d", &n);
    i = 1;
    while(count < n) {
       int_count = 0;
       for (j=1; j<=pow(i, 1.0/3); j++) {
          for(k=j+1; k<=pow(i,1.0/3); k++) {
              if(j*j*j+k*k*k == i)
              int_count++;
          }
       }
       if(int_count == 2) {
          count++;
          printf("\nGot %d Hardy-Ramanujan numbers %d", count, i);  
       }
       i++;
    }
}

Since `a^3+b^3 = n`, `a` should be less than `n^(1/3)`.

Problem

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 ```

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