Taxicab numbers: Difference between revisions

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Content added Content deleted
m (→‎{{header|REXX}}: changed/added comments, replaced pivot sort with a (shorter) bubble sort.)
m (→‎{{header|REXX}}: eliminated the need for some variables to store literals.)
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mx=mx+mx%10; ww=length(mx)*3 /*cushion; compensate for the triples.*/
mx=mx+mx%10; ww=length(mx)*3 /*cushion; compensate for the triples.*/
w=ww%2; numeric digits max(9,ww) /*prepare to use some larger numbers. */
w=ww%2; numeric digits max(9,ww) /*prepare to use some larger numbers. */
@.=.; #=0; @@.=0; $.= /*initialize some REXX variables. */
@.=.; #=0; @@.=0; @and=" ──and── "; $.= /*set some REXX vars and handy literals*/
a=right('+',4); @and=" ──and── " /* " " " literals (vars).*/
/* [↓] generate extra taxicab numbers.*/
/* [↓] generate extra taxicab numbers.*/
do j=1 until #>=mx; C=j**3 /*taxicab numbers may not be in order. */
do j=1 until #>=mx; C=j**3 /*taxicab numbers may not be in order. */
Line 1,885: Line 1,884:
end /* [↑] define one cube sum at a time. */
end /* [↑] define one cube sum at a time. */
has=@@.s /*has this number been defined before? */
has=@@.s /*has this number been defined before? */
if \has then $.s=right(s,ww) '───►' r(@.s,a)r(b.s) @and r(j,a)r(k)
if \has then $.s=right(s,ww) '───►' r(@.s," +")r(b.s) @and r(j,' +')r(k)
else $.s=$.s @and r(j,a)r(k) /*build a string for display.*/
else $.s=$.s @and r(j,' +')r(k) /*◄─ build a display string. [↑] */
@@.s=1 /*mark taxicab number as a sum of cubes*/
@@.s=1 /*mark taxicab number as a sum of cubes*/
if has then iterate /*S is a triple (or sometimes better).*/
if has then iterate /*S is a triple (or sometimes better).*/
#=#+1 /*bump the taxicab number counter. */
#=#+1; #.#=s /*bump taxicab counter; define taxicab#*/
#.#=s /*define a #. taxicab number. */
end /*k*/ /* [↑] build the cubes one─at─a─time. */
end /*k*/ /* [↑] build the cubes one─at─a─time. */
end /*j*/ /* [↑] complete with overage numbers. */
end /*j*/ /* [↑] complete with overage numbers. */

Revision as of 19:35, 15 March 2016

Taxicab numbers is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

A   taxicab number   (the definition that is being used here)  is a positive integer that can be expressed as the sum of two positive cubes in more than one way.

The first taxicab number is   1729:

1729     =     13 + 123     =     93 + 103
Task requirements
  • Compute and display the lowest 25 taxicab numbers (in numeric order, and in a human-readable format).
  • For each of the taxicab numbers, show the number as well as it's constituent cubes.
  • Show the 2,000th taxicab number and a half dozen more   (extra credit)
See also

C

Using a priority queue to emit sum of two cubs in order. It's reasonably fast and doesn't use excessive amount of memory (the heap is only at 245 length upon the 2006th taxi). <lang c>#include <stdio.h>

  1. include <stdlib.h>

typedef unsigned long long xint; typedef unsigned uint; typedef struct { uint x, y; // x > y always xint value; } sum_t;

xint *cube; uint n_cubes;

sum_t *pq; uint pq_len, pq_cap;

void add_cube(void) { uint x = n_cubes++; cube = realloc(cube, sizeof(xint) * (n_cubes + 1)); cube[n_cubes] = (xint) n_cubes*n_cubes*n_cubes; if (x < 2) return; // x = 0 or 1 is useless

if (++pq_len >= pq_cap) { if (!(pq_cap *= 2)) pq_cap = 2; pq = realloc(pq, sizeof(*pq) * pq_cap); }

sum_t tmp = (sum_t) { x, 1, cube[x] + 1 }; // upheap uint i, j; for (i = pq_len; i >= 1 && pq[j = i>>1].value > tmp.value; i = j) pq[i] = pq[j];

pq[i] = tmp; }

void next_sum(void) { redo: while (!pq_len || pq[1].value >= cube[n_cubes]) add_cube();

sum_t tmp = pq[0] = pq[1]; // pq[0] always stores last seen value if (++tmp.y >= tmp.x) { // done with this x; throw it away tmp = pq[pq_len--]; if (!pq_len) goto redo; // refill empty heap } else tmp.value += cube[tmp.y] - cube[tmp.y-1];

uint i, j; // downheap for (i = 1; (j = i<<1) <= pq_len; pq[i] = pq[j], i = j) { if (j < pq_len && pq[j+1].value < pq[j].value) ++j; if (pq[j].value >= tmp.value) break; } pq[i] = tmp; }

uint next_taxi(sum_t *hist) { do next_sum(); while (pq[0].value != pq[1].value);

uint len = 1; hist[0] = pq[0]; do { hist[len++] = pq[1]; next_sum(); } while (pq[0].value == pq[1].value);

return len; }

int main(void) { uint i, l; sum_t x[10]; for (i = 1; i <= 2006; i++) { l = next_taxi(x); if (25 < i && i < 2000) continue; printf("%4u:%10llu", i, x[0].value); while (l--) printf(" = %4u^3 + %4u^3", x[l].x, x[l].y); putchar('\n'); } return 0; }</lang>

Output:
   1:      1729 =   12^3 +    1^3 =   10^3 +    9^3
   2:      4104 =   15^3 +    9^3 =   16^3 +    2^3
   3:     13832 =   20^3 +   18^3 =   24^3 +    2^3
   4:     20683 =   27^3 +   10^3 =   24^3 +   19^3
   5:     32832 =   30^3 +   18^3 =   32^3 +    4^3
   6:     39312 =   33^3 +   15^3 =   34^3 +    2^3
   7:     40033 =   33^3 +   16^3 =   34^3 +    9^3
   8:     46683 =   30^3 +   27^3 =   36^3 +    3^3
   9:     64232 =   36^3 +   26^3 =   39^3 +   17^3
  10:     65728 =   33^3 +   31^3 =   40^3 +   12^3
  11:    110656 =   40^3 +   36^3 =   48^3 +    4^3
  12:    110808 =   45^3 +   27^3 =   48^3 +    6^3
  13:    134379 =   43^3 +   38^3 =   51^3 +   12^3
  14:    149389 =   50^3 +   29^3 =   53^3 +    8^3
  15:    165464 =   48^3 +   38^3 =   54^3 +   20^3
  16:    171288 =   54^3 +   24^3 =   55^3 +   17^3
  17:    195841 =   57^3 +   22^3 =   58^3 +    9^3
  18:    216027 =   59^3 +   22^3 =   60^3 +    3^3
  19:    216125 =   50^3 +   45^3 =   60^3 +    5^3
  20:    262656 =   60^3 +   36^3 =   64^3 +    8^3
  21:    314496 =   66^3 +   30^3 =   68^3 +    4^3
  22:    320264 =   66^3 +   32^3 =   68^3 +   18^3
  23:    327763 =   58^3 +   51^3 =   67^3 +   30^3
  24:    373464 =   60^3 +   54^3 =   72^3 +    6^3
  25:    402597 =   61^3 +   56^3 =   69^3 +   42^3
2000:1671816384 = 1168^3 +  428^3 =  944^3 +  940^3
2001:1672470592 = 1124^3 +  632^3 = 1187^3 +   29^3
2002:1673170856 = 1034^3 +  828^3 = 1164^3 +  458^3
2003:1675045225 = 1153^3 +  522^3 = 1081^3 +  744^3
2004:1675958167 = 1096^3 +  711^3 = 1159^3 +  492^3
2005:1676926719 = 1188^3 +   63^3 = 1095^3 +  714^3
2006:1677646971 =  990^3 +  891^3 = 1188^3 +   99^3

C#

<lang csharp>using System; using System.Collections.Generic; using System.Linq; using System.Text;

namespace TaxicabNumber {

   class Program
   {
       static void Main(string[] args)
       {
           IDictionary<long, IList<Tuple<int, int>>> taxicabNumbers = GetTaxicabNumbers(2006);
           PrintTaxicabNumbers(taxicabNumbers);
           Console.ReadKey();
       }
       private static IDictionary<long, IList<Tuple<int, int>>> GetTaxicabNumbers(int length)
       {
           SortedList<long, IList<Tuple<int, int>>> sumsOfTwoCubes = new SortedList<long, IList<Tuple<int, int>>>();
           for (int i = 1; i < int.MaxValue; i++)
           {
               for (int j = 1; j < int.MaxValue; j++)
               {
                   long sum = (long)(Math.Pow((double)i, 3) + Math.Pow((double)j, 3));
                   if (!sumsOfTwoCubes.ContainsKey(sum))
                   {
                       sumsOfTwoCubes.Add(sum, new List<Tuple<int, int>>());
                   }
                   sumsOfTwoCubes[sum].Add(new Tuple<int, int>(i, j));
                   if (j >= i)
                   {
                       break;
                   }
               }
               // Found that you need to keep going for a while after the length, because higher i values fill in gaps
               if (sumsOfTwoCubes.Count(t => t.Value.Count >= 2) >= length * 1.1)
               {
                   break;
               }
           }
           IDictionary<long, IList<Tuple<int, int>>> values = (from t in sumsOfTwoCubes where t.Value.Count >= 2 select t)
               .Take(2006)
               .ToDictionary(u => u.Key, u => u.Value);
           return values;
       }
       private static void PrintTaxicabNumbers(IDictionary<long, IList<Tuple<int, int>>> values)
       {
           int i = 1;
           foreach (long taxicabNumber in values.Keys)
           {
               StringBuilder output = new StringBuilder().AppendFormat("{0,10}\t{1,4}", i, taxicabNumber);
               foreach (Tuple<int, int> numbers in values[taxicabNumber])
               {
                   output.AppendFormat("\t= {0}^3 + {1}^3", numbers.Item1, numbers.Item2);
               }
               if (i <= 25 || (i >= 2000 && i <= 2006))
               {
                   Console.WriteLine(output.ToString());
               }
               i++;
           }
       }
   }

}</lang>

D

High Level Version

Translation of: Python

<lang d>void main() /*@safe*/ {

   import std.stdio, std.range, std.algorithm, std.typecons, std.string;
   auto iCubes = iota(1u, 1201u).map!(x => tuple(x, x ^^ 3));
   bool[Tuple!(uint, uint)][uint] sum2cubes;
   foreach (i, immutable i3; iCubes)
       foreach (j, immutable j3; iCubes[i .. $])
           sum2cubes[i3 + j3][tuple(i, j)] = true;
   const taxis = sum2cubes.byKeyValue.filter!(p => p.value.length > 1)
                 .array.schwartzSort!(p => p.key).release;
   foreach (/*immutable*/ const r; [[0, 25], [2000 - 1, 2000 + 6]]) {
       foreach (immutable i, const t; taxis[r[0] .. r[1]])
           writefln("%4d: %10d =%-(%s =%)", i + r[0] + 1, t.key,
                    t.value.keys.sort().map!q{"%4d^3 + %4d^3".format(a[])});
       writeln;
   }

}</lang>

Output:
   1:       1729 =   1^3 +   12^3 =   9^3 +   10^3
   2:       4104 =   2^3 +   16^3 =   9^3 +   15^3
   3:      13832 =   2^3 +   24^3 =  18^3 +   20^3
   4:      20683 =  10^3 +   27^3 =  19^3 +   24^3
   5:      32832 =   4^3 +   32^3 =  18^3 +   30^3
   6:      39312 =   2^3 +   34^3 =  15^3 +   33^3
   7:      40033 =   9^3 +   34^3 =  16^3 +   33^3
   8:      46683 =   3^3 +   36^3 =  27^3 +   30^3
   9:      64232 =  17^3 +   39^3 =  26^3 +   36^3
  10:      65728 =  12^3 +   40^3 =  31^3 +   33^3
  11:     110656 =   4^3 +   48^3 =  36^3 +   40^3
  12:     110808 =   6^3 +   48^3 =  27^3 +   45^3
  13:     134379 =  12^3 +   51^3 =  38^3 +   43^3
  14:     149389 =   8^3 +   53^3 =  29^3 +   50^3
  15:     165464 =  20^3 +   54^3 =  38^3 +   48^3
  16:     171288 =  17^3 +   55^3 =  24^3 +   54^3
  17:     195841 =   9^3 +   58^3 =  22^3 +   57^3
  18:     216027 =   3^3 +   60^3 =  22^3 +   59^3
  19:     216125 =   5^3 +   60^3 =  45^3 +   50^3
  20:     262656 =   8^3 +   64^3 =  36^3 +   60^3
  21:     314496 =   4^3 +   68^3 =  30^3 +   66^3
  22:     320264 =  18^3 +   68^3 =  32^3 +   66^3
  23:     327763 =  30^3 +   67^3 =  51^3 +   58^3
  24:     373464 =   6^3 +   72^3 =  54^3 +   60^3
  25:     402597 =  42^3 +   69^3 =  56^3 +   61^3

2000: 1671816384 = 428^3 + 1168^3 = 940^3 +  944^3
2001: 1672470592 =  29^3 + 1187^3 = 632^3 + 1124^3
2002: 1673170856 = 458^3 + 1164^3 = 828^3 + 1034^3
2003: 1675045225 = 522^3 + 1153^3 = 744^3 + 1081^3
2004: 1675958167 = 492^3 + 1159^3 = 711^3 + 1096^3
2005: 1676926719 =  63^3 + 1188^3 = 714^3 + 1095^3
2006: 1677646971 =  99^3 + 1188^3 = 891^3 +  990^3

Run-time: about 2.9 seconds with dmd compiler.

Heap-Based Version

Translation of: Java

<lang d>import std.stdio, std.string, std.container;

struct CubeSum {

   ulong x, y, value;
   this(in ulong x_, in ulong y_) pure nothrow @safe @nogc {
       this.x = x_;
       this.y = y_;
       this.value = x_ ^^ 3 + y_ ^^ 3;
   }

}

final class Taxi {

   BinaryHeap!(Array!CubeSum, "a.value > b.value") pq;
   CubeSum last;
   ulong n = 0;
   this() {
       last = nextSum();
   }
   CubeSum nextSum() {
       while (pq.empty || pq.front.value >= n ^^ 3)
           pq.insert(CubeSum(++n, 1));
       auto s = pq.front;
       pq.removeFront;
       if (s.x > s.y + 1)
           pq.insert(CubeSum(s.x, s.y + 1));
       return s;
   }
   CubeSum[] nextTaxi() {
       CubeSum s;
       typeof(return) train;
       while ((s = nextSum).value != last.value)
           last = s;
       train ~= last;
       do {
           train ~= s;
       } while ((s = nextSum).value == last.value);
       last = s;
       return train;
   }

}

void main() {

   auto taxi = new Taxi;
   foreach (immutable i; 1 .. 2007) {
       const t = taxi.nextTaxi;
       if (i > 25 && i < 2000)
           continue;
       writef("%4d: %10d", i, t[0].value);
       foreach (const s; t)
           writef(" = %4d^3 + %4d^3", s.x, s.y);
       writeln;
   }

}</lang>

Output:
   1:       1729 =   10^3 +    9^3 =   12^3 +    1^3
   2:       4104 =   15^3 +    9^3 =   16^3 +    2^3
   3:      13832 =   20^3 +   18^3 =   24^3 +    2^3
   4:      20683 =   24^3 +   19^3 =   27^3 +   10^3
   5:      32832 =   30^3 +   18^3 =   32^3 +    4^3
   6:      39312 =   33^3 +   15^3 =   34^3 +    2^3
   7:      40033 =   33^3 +   16^3 =   34^3 +    9^3
   8:      46683 =   30^3 +   27^3 =   36^3 +    3^3
   9:      64232 =   39^3 +   17^3 =   36^3 +   26^3
  10:      65728 =   40^3 +   12^3 =   33^3 +   31^3
  11:     110656 =   40^3 +   36^3 =   48^3 +    4^3
  12:     110808 =   45^3 +   27^3 =   48^3 +    6^3
  13:     134379 =   51^3 +   12^3 =   43^3 +   38^3
  14:     149389 =   50^3 +   29^3 =   53^3 +    8^3
  15:     165464 =   48^3 +   38^3 =   54^3 +   20^3
  16:     171288 =   54^3 +   24^3 =   55^3 +   17^3
  17:     195841 =   57^3 +   22^3 =   58^3 +    9^3
  18:     216027 =   59^3 +   22^3 =   60^3 +    3^3
  19:     216125 =   50^3 +   45^3 =   60^3 +    5^3
  20:     262656 =   60^3 +   36^3 =   64^3 +    8^3
  21:     314496 =   66^3 +   30^3 =   68^3 +    4^3
  22:     320264 =   68^3 +   18^3 =   66^3 +   32^3
  23:     327763 =   67^3 +   30^3 =   58^3 +   51^3
  24:     373464 =   60^3 +   54^3 =   72^3 +    6^3
  25:     402597 =   69^3 +   42^3 =   61^3 +   56^3
2000: 1671816384 = 1168^3 +  428^3 =  944^3 +  940^3
2001: 1672470592 = 1124^3 +  632^3 = 1187^3 +   29^3
2002: 1673170856 = 1164^3 +  458^3 = 1034^3 +  828^3
2003: 1675045225 = 1153^3 +  522^3 = 1081^3 +  744^3
2004: 1675958167 = 1159^3 +  492^3 = 1096^3 +  711^3
2005: 1676926719 = 1095^3 +  714^3 = 1188^3 +   63^3
2006: 1677646971 =  990^3 +  891^3 = 1188^3 +   99^3

Run-time: about 0.31 seconds with ldc2 compiler. It's faster than the Java solution.

Low Level Heap-Based Version

Translation of: C

<lang d>struct Taxicabs {

   alias CubesSumT = uint; // Or ulong.
   static struct Sum {
       CubesSumT value;
       uint x, y;
   }
   // The cubes can be pre-computed if CubesSumT is a BigInt.
   private uint nCubes;
   private Sum[] pq;
   private uint pq_len;
   private void addCube() pure nothrow @safe {
       nCubes = nCubes ? nCubes + 1 : 2;
       if (nCubes < 2)
           return; // 0 or 1 is useless.
       pq_len++;
       if (pq_len >= pq.length)
           pq.length = (pq.length == 0) ? 2 : (pq.length * 2);
       immutable tmp = Sum(CubesSumT(nCubes - 2) ^^ 3 + 1,
                           nCubes - 2, 1);
       // Upheap.
       uint i = pq_len;
       for (; i >= 1 && pq[i >> 1].value > tmp.value; i >>= 1)
           pq[i] = pq[i >> 1];
       pq[i] = tmp;
   }


   private void nextSum() pure nothrow @safe {
       while (!pq_len || pq[1].value >= (nCubes - 1) ^^ 3)
           addCube();
       Sum tmp = pq[0] = pq[1]; //pq[0] always stores last seen value.
       tmp.y++;
       if (tmp.y >= tmp.x) { // Done with this x; throw it away.
           tmp = pq[pq_len];
           pq_len--;
           if (!pq_len)
               return nextSum(); // Refill empty heap.
       } else
           tmp.value += tmp.y ^^ 3 - (tmp.y - 1) ^^ 3;
       // Downheap.
       uint i = 1;
       while (true) {
           uint j = i << 1;
           if (j > pq_len)
               break;
           if (j < pq_len && pq[j + 1].value < pq[j].value)
               j++;
           if (pq[j].value >= tmp.value)
               break;
           pq[i] = pq[j];
           i = j;
       }
       pq[i] = tmp;
   }


   Sum[] nextTaxi(size_t N)(ref Sum[N] hist)
   pure nothrow @safe {
       do {
           nextSum();
       } while (pq[0].value != pq[1].value);
       uint len = 1;
       hist[0] = pq[0];
       do {
           hist[len] = pq[1];
           len++;
           nextSum();
       } while (pq[0].value == pq[1].value);
       return hist[0 .. len];
   }

}


void main() nothrow {

   import core.stdc.stdio;
   Taxicabs t;
   Taxicabs.Sum[3] x;
   foreach (immutable uint i; 1 .. 2007) {
       const triples = t.nextTaxi(x);
       if (i > 25 && i < 2000)
           continue;
       printf("%4u: %10lu", i, triples[0].value);
       foreach_reverse (const s; triples)
           printf(" = %4u^3 + %4u^3", s.x, s.y);
       '\n'.putchar;
   }

}</lang>

Output:
   1:       1729 =   12^3 +    1^3 =   10^3 +    9^3
   2:       4104 =   15^3 +    9^3 =   16^3 +    2^3
   3:      13832 =   20^3 +   18^3 =   24^3 +    2^3
   4:      20683 =   27^3 +   10^3 =   24^3 +   19^3
   5:      32832 =   30^3 +   18^3 =   32^3 +    4^3
   6:      39312 =   33^3 +   15^3 =   34^3 +    2^3
   7:      40033 =   33^3 +   16^3 =   34^3 +    9^3
   8:      46683 =   30^3 +   27^3 =   36^3 +    3^3
   9:      64232 =   36^3 +   26^3 =   39^3 +   17^3
  10:      65728 =   33^3 +   31^3 =   40^3 +   12^3
  11:     110656 =   40^3 +   36^3 =   48^3 +    4^3
  12:     110808 =   45^3 +   27^3 =   48^3 +    6^3
  13:     134379 =   43^3 +   38^3 =   51^3 +   12^3
  14:     149389 =   50^3 +   29^3 =   53^3 +    8^3
  15:     165464 =   48^3 +   38^3 =   54^3 +   20^3
  16:     171288 =   54^3 +   24^3 =   55^3 +   17^3
  17:     195841 =   57^3 +   22^3 =   58^3 +    9^3
  18:     216027 =   59^3 +   22^3 =   60^3 +    3^3
  19:     216125 =   50^3 +   45^3 =   60^3 +    5^3
  20:     262656 =   60^3 +   36^3 =   64^3 +    8^3
  21:     314496 =   66^3 +   30^3 =   68^3 +    4^3
  22:     320264 =   66^3 +   32^3 =   68^3 +   18^3
  23:     327763 =   58^3 +   51^3 =   67^3 +   30^3
  24:     373464 =   60^3 +   54^3 =   72^3 +    6^3
  25:     402597 =   61^3 +   56^3 =   69^3 +   42^3
2000: 1671816384 = 1168^3 +  428^3 =  944^3 +  940^3
2001: 1672470592 = 1124^3 +  632^3 = 1187^3 +   29^3
2002: 1673170856 = 1034^3 +  828^3 = 1164^3 +  458^3
2003: 1675045225 = 1153^3 +  522^3 = 1081^3 +  744^3
2004: 1675958167 = 1096^3 +  711^3 = 1159^3 +  492^3
2005: 1676926719 = 1188^3 +   63^3 = 1095^3 +  714^3
2006: 1677646971 =  990^3 +  891^3 = 1188^3 +   99^3

Run-time: about 0.08 seconds with ldc2 compiler.

DCL

We invoke external utility SORT which I suppose technically speaking is not a formal part of the language but is darn handy at times; <lang DCL>$ close /nolog sums_of_cubes $ on control_y then $ goto clean $ open /write sums_of_cubes sums_of_cubes.txt $ i = 1 $ loop1: $ write sys$output i $ j = 1 $ loop2: $ sum = i * i * i + j * j * j $ if sum .lt. 0 $ then $ write sys$output "overflow at ", j $ goto next_i $ endif $ write sums_of_cubes f$fao( "!10SL,!10SL,!10SL", sum, i, j ) $ j = j + 1 $ if j .le. i then $ goto loop2 $ next_i: $ i = i + 1 $ if i .le. 1289 then $ goto loop1  ! cube_root of 2^31-1 $ close sums_of_cubes $ sort sums_of_cubes.txt sorted_sums_of_cubes.txt $ close /nolog sorted_sums_of_cubes $ open sorted_sums_of_cubes sorted_sums_of_cubes.txt $ count = 0 $ read sorted_sums_of_cubes prev_prev_line  ! need to detect when there are more than just 2 different sums, e.g. 456 $ prev_prev_sum = f$element( 0, ",", f$edit( prev_prev_line, "collapse" )) $ read sorted_sums_of_cubes prev_line $ prev_sum = f$element( 0,",", f$edit( prev_line, "collapse" )) $ loop3: $ read /end_of_file = done sorted_sums_of_cubes line $ sum = f$element( 0, ",", f$edit( line, "collapse" )) $ if sum .eqs. prev_sum $ then $ if sum .nes. prev_prev_sum then $ count = count + 1 $ int_sum = f$integer( sum ) $ i1 = f$integer( f$element( 1, ",", prev_line )) $ j1 = f$integer( f$element( 2, ",", prev_line )) $ i2 = f$integer( f$element( 1, ",", line )) $ j2 = f$integer( f$element( 2, ",", line )) $ if count .le. 25 .or. ( count .ge. 2000 .and. count .le. 2006 ) then - $ write sys$output f$fao( "!4SL:!11SL =!5SL^3 +!5SL^3 =!5SL^3 +!5SL^3", count, int_sum, i1, j1, i2, j2 ) $ endif $ prev_prev_line = prev_line $ prev_prev_sum = prev_sum $ prev_line = line $ prev_sum = sum $ goto loop3 $ done: $ close sorted_sums_of_cubes $ exit $ $ clean: $ close /nolog sorted_sums_of_cubes $ close /nolog sums_of_cubes</lang>

Output:
$ @taxicab_numbers
   1:       1729 =   10^3 +    9^3 =   12^3 +    1^3
   2:       4104 =   15^3 +    9^3 =   16^3 +    2^3
   3:      13832 =   20^3 +   18^3 =   24^3 +    2^3
   4:      20683 =   24^3 +   19^3 =   27^3 +   10^3
   5:      32832 =   30^3 +   18^3 =   32^3 +    4^3
   6:      39312 =   33^3 +   15^3 =   34^3 +    2^3
   7:      40033 =   33^3 +   16^3 =   34^3 +    9^3
   8:      46683 =   30^3 +   27^3 =   36^3 +    3^3
   9:      64232 =   36^3 +   26^3 =   39^3 +   17^3
  10:      65728 =   33^3 +   31^3 =   40^3 +   12^3
  11:     110656 =   40^3 +   36^3 =   48^3 +    4^3
  12:     110808 =   45^3 +   27^3 =   48^3 +    6^3
  13:     134379 =   43^3 +   38^3 =   51^3 +   12^3
  14:     149389 =   50^3 +   29^3 =   53^3 +    8^3
  15:     165464 =   48^3 +   38^3 =   54^3 +   20^3
  16:     171288 =   54^3 +   24^3 =   55^3 +   17^3
  17:     195841 =   57^3 +   22^3 =   58^3 +    9^3
  18:     216027 =   59^3 +   22^3 =   60^3 +    3^3
  19:     216125 =   50^3 +   45^3 =   60^3 +    5^3
  20:     262656 =   60^3 +   36^3 =   64^3 +    8^3
  21:     314496 =   66^3 +   30^3 =   68^3 +    4^3
  22:     320264 =   66^3 +   32^3 =   68^3 +   18^3
  23:     327763 =   58^3 +   51^3 =   67^3 +   30^3
  24:     373464 =   60^3 +   54^3 =   72^3 +    6^3
  25:     402597 =   61^3 +   56^3 =   69^3 +   42^3
2000: 1671816384 =  944^3 +  940^3 = 1168^3 +  428^3
2001: 1672470592 = 1124^3 +  632^3 = 1187^3 +   29^3
2002: 1673170856 = 1034^3 +  828^3 = 1164^3 +  458^3
2003: 1675045225 = 1081^3 +  744^3 = 1153^3 +  522^3
2004: 1675958167 = 1096^3 +  711^3 = 1159^3 +  492^3
2005: 1676926719 = 1095^3 +  714^3 = 1188^3 +   63^3
2006: 1677646971 =  990^3 +  891^3 = 1188^3 +   99^3

EchoLisp

Using the heap library, and a heap to store the taxicab numbers. For taxi tuples - decomposition in more than two sums - we use the group function which transforms a list ( 3 5 5 6 8 ...) into ((3) (5 5) (6) ...). <lang scheme> (require '(heap compile))

(define (scube a b) (+ (* a a a) (* b b b))) (compile 'scube "-f") ; "-f" means : no bigint, no rational used

is n - a^3 a cube b^3?
if yes return b, else #f

(define (taxi? n a (b 0)) (set! b (cbrt (- n (* a a a)))) ;; cbrt is ∛ (when (and (< b a) (integer? b)) b)) (compile 'taxi? "-f")

  1. |-------------------

looking for taxis


|#

remove from heap until heap-top >= a
when twins are removed, it is a taxicab number
push it
at any time (top stack) = last removed

(define (clean-taxi H limit: a min-of-heap: htop) (when (and htop (> a htop)) (when (!= (stack-top S) htop) (pop S)) (push S htop) (heap-pop H) (clean-taxi H a (heap-top H)))) (compile 'clean-taxi "-f")

loop on a and b, b <=a , until n taxicabs found

(define (taxicab (n 2100)) (for ((a (in-naturals))) (clean-taxi H (* a a a) (heap-top H)) #:break (> (stack-length S) n) (for ((b a)) (heap-push H (scube a b)))))

  1. |------------------

printing taxis


|#

string of all decompositions

(define (taxi->string i n) (string-append (format "%d. %d " (1+ i) n) (for/string ((a (cbrt n))) #:when (taxi? n a) (format " = %4d^3 + %4d^3" a (taxi? n a)))))

(define (taxi-print taxis (nfrom 0) (nto 26)) (for ((i (in-naturals nfrom)) (taxi (sublist taxis nfrom nto))) (writeln (taxi->string i (first taxi))))) </lang>

Output:

<lang scheme> (define S (stack 'S)) ;; to push taxis (define H (make-heap < )) ;; make min heap of all scubes

(taxicab 2100) (define taxis (group (stack->list S))) (taxi-print taxis )

1. 1729 = 10^3 + 9^3 = 12^3 + 1^3 2. 4104 = 15^3 + 9^3 = 16^3 + 2^3 3. 13832 = 20^3 + 18^3 = 24^3 + 2^3 4. 20683 = 24^3 + 19^3 = 27^3 + 10^3

  1. | ... |#

24. 373464 = 60^3 + 54^3 = 72^3 + 6^3 25. 402597 = 61^3 + 56^3 = 69^3 + 42^3 26. 439101 = 69^3 + 48^3 = 76^3 + 5^3

(taxi-print taxis 1999 2006) 2000. 1671816384 = 944^3 + 940^3 = 1168^3 + 428^3 2001. 1672470592 = 1124^3 + 632^3 = 1187^3 + 29^3 2002. 1673170856 = 1034^3 + 828^3 = 1164^3 + 458^3 2003. 1675045225 = 1081^3 + 744^3 = 1153^3 + 522^3 2004. 1675958167 = 1096^3 + 711^3 = 1159^3 + 492^3 2005. 1676926719 = 1095^3 + 714^3 = 1188^3 + 63^3 2006. 1677646971 = 990^3 + 891^3 = 1188^3 + 99^3

extra bonus
print all taxis which are triplets

(define (taxi-tuples taxis (nfrom 0) (nto 2000)) (for ((i (in-naturals nfrom)) (taxi (sublist taxis nfrom nto))) #:when (> (length taxi) 1) ;; filter for tuples is here (writeln (taxi->string i (first taxi)))))

(taxi-tuples taxis)

455. 87539319 = 414^3 + 255^3 = 423^3 + 228^3 = 436^3 + 167^3 535. 119824488 = 428^3 + 346^3 = 492^3 + 90^3 = 493^3 + 11^3 588. 143604279 = 423^3 + 408^3 = 460^3 + 359^3 = 522^3 + 111^3 655. 175959000 = 525^3 + 315^3 = 552^3 + 198^3 = 560^3 + 70^3 888. 327763000 = 580^3 + 510^3 = 661^3 + 339^3 = 670^3 + 300^3 1299. 700314552 = 828^3 + 510^3 = 846^3 + 456^3 = 872^3 + 334^3 1398. 804360375 = 920^3 + 295^3 = 927^3 + 198^3 = 930^3 + 15^3 1515. 958595904 = 856^3 + 692^3 = 984^3 + 180^3 = 986^3 + 22^3 1660. 1148834232 = 846^3 + 816^3 = 920^3 + 718^3 = 1044^3 + 222^3 1837. 1407672000 = 1050^3 + 630^3 = 1104^3 + 396^3 = 1120^3 + 140^3 </lang>

Elixir

<lang elixir>defmodule Taxicab do

 def numbers(n \\ 1200) do
   (for i <- 1..n, j <- i..n, do: {i,j})
   |> Enum.group_by(fn {i,j} -> i*i*i + j*j*j end)
   |> Enum.filter(fn {_,v} -> length(v)>1 end)
   |> Enum.sort
 end

end

nums = Taxicab.numbers |> Enum.with_index Enum.each(nums, fn {x,i} ->

 if i in 0..24 or i in 1999..2005 do
   IO.puts "#{i+1} : #{inspect x}"
 end

end)</lang>

Output:
1 : {1729, [{9, 10}, {1, 12}]}
2 : {4104, [{9, 15}, {2, 16}]}
3 : {13832, [{18, 20}, {2, 24}]}
4 : {20683, [{19, 24}, {10, 27}]}
5 : {32832, [{18, 30}, {4, 32}]}
6 : {39312, [{15, 33}, {2, 34}]}
7 : {40033, [{16, 33}, {9, 34}]}
8 : {46683, [{27, 30}, {3, 36}]}
9 : {64232, [{26, 36}, {17, 39}]}
10 : {65728, [{31, 33}, {12, 40}]}
11 : {110656, [{36, 40}, {4, 48}]}
12 : {110808, [{27, 45}, {6, 48}]}
13 : {134379, [{38, 43}, {12, 51}]}
14 : {149389, [{29, 50}, {8, 53}]}
15 : {165464, [{38, 48}, {20, 54}]}
16 : {171288, [{24, 54}, {17, 55}]}
17 : {195841, [{22, 57}, {9, 58}]}
18 : {216027, [{22, 59}, {3, 60}]}
19 : {216125, [{45, 50}, {5, 60}]}
20 : {262656, [{36, 60}, {8, 64}]}
21 : {314496, [{30, 66}, {4, 68}]}
22 : {320264, [{32, 66}, {18, 68}]}
23 : {327763, [{51, 58}, {30, 67}]}
24 : {373464, [{54, 60}, {6, 72}]}
25 : {402597, [{56, 61}, {42, 69}]}
2000 : {1671816384, [{940, 944}, {428, 1168}]}
2001 : {1672470592, [{632, 1124}, {29, 1187}]}
2002 : {1673170856, [{828, 1034}, {458, 1164}]}
2003 : {1675045225, [{744, 1081}, {522, 1153}]}
2004 : {1675958167, [{711, 1096}, {492, 1159}]}
2005 : {1676926719, [{714, 1095}, {63, 1188}]}
2006 : {1677646971, [{891, 990}, {99, 1188}]}

Go

<lang go>package main

import ( "container/heap" "fmt" )

type CubeSum struct { x, y uint16 value uint64 }

func (c *CubeSum) fixvalue() { c.value = cubes[c.x] + cubes[c.y] }

type CubeSumHeap []*CubeSum

func (h CubeSumHeap) Len() int { return len(h) } func (h CubeSumHeap) Less(i, j int) bool { return h[i].value < h[j].value } func (h CubeSumHeap) Swap(i, j int) { h[i], h[j] = h[j], h[i] } func (h *CubeSumHeap) Push(x interface{}) { (*h) = append(*h, x.(*CubeSum)) } func (h *CubeSumHeap) Pop() interface{} { x := (*h)[len(*h)-1] *h = (*h)[:len(*h)-1] return x }

type TaxicabGen struct { n int h CubeSumHeap }

var cubes []uint64 // cubes[i] == i*i*i func cubesExtend(i int) { for n := uint64(len(cubes)); n <= uint64(i); n++ { cubes = append(cubes, n*n*n) } }

func (g *TaxicabGen) min() CubeSum { for len(g.h) == 0 || g.h[0].value > cubes[g.n] { g.n++ cubesExtend(g.n) heap.Push(&g.h, &CubeSum{uint16(g.n), 1, cubes[g.n] + 1}) } // Note, we use g.h[0] to "peek" at the min heap entry. c := *(g.h[0]) if c.y+1 <= c.x { // Instead of Pop and Push we modify in place and fix. g.h[0].y++ g.h[0].fixvalue() heap.Fix(&g.h, 0) } else { heap.Pop(&g.h) } return c }

// Originally this was just: type Taxicab [2]CubeSum // and we always returned two sums. Now we return all the sums. type Taxicab []CubeSum

func (t Taxicab) String() string { //return fmt.Sprintf("%12d =%5d³ +%5d³ =%5d³ +%5d³", // t[0].value, t[0].x, t[0].y, t[1].x, t[1].y)

// Likely more efficient to use a bytes.Buffer here. s := fmt.Sprintf("%12d", t[0].value) for _, p := range t { s = s + fmt.Sprintf(" =%5d³ +%5d³", p.x, p.y) } return s }

func (g *TaxicabGen) Next() Taxicab { a, b := g.min(), g.min() for a.value != b.value { a, b = b, g.min() } //return Taxicab{a,b}

// Originally this just returned Taxicab{a,b} and we didn't look // further into the heap. Since we start by looking at the next // pair, that is okay until the first Taxicab number with four // ways of expressing the cube, which doesn't happen until the // 97,235th Taxicab: // 6963472309248 = 16630³ + 13322³ = 18072³ + 10200³ // = 18948³ + 5436³ = 19083³ + 2421³ // Now we return all ways so we need to peek into the heap. t := Taxicab{a, b} for g.h[0].value == b.value { t = append(t, g.min()) } return t }

func main() { const ( low = 25 mid = 2e3 high = 4e4 ) var tg TaxicabGen firstn := 3 // To show the first tripple, quadruple, etc for i := 1; i <= high+6; i++ { t := tg.Next() switch { case len(t) >= firstn: firstn++ fallthrough case i <= low || (mid <= i && i <= mid+6) || i >= high: //fmt.Printf("h:%-4d ", len(tg.h)) fmt.Printf("%5d: %v\n", i, t) } } }</lang> On a modern 64 bit processor this takes ~89 msec for the first 2000 and ~9.2 sec for the first 40,000.

Output:
    1:         1729 =   12³ +    1³ =   10³ +    9³
    2:         4104 =   16³ +    2³ =   15³ +    9³
    3:        13832 =   24³ +    2³ =   20³ +   18³
    4:        20683 =   27³ +   10³ =   24³ +   19³
    5:        32832 =   32³ +    4³ =   30³ +   18³
    6:        39312 =   34³ +    2³ =   33³ +   15³
    7:        40033 =   34³ +    9³ =   33³ +   16³
    8:        46683 =   36³ +    3³ =   30³ +   27³
    9:        64232 =   36³ +   26³ =   39³ +   17³
   10:        65728 =   40³ +   12³ =   33³ +   31³
   11:       110656 =   48³ +    4³ =   40³ +   36³
   12:       110808 =   48³ +    6³ =   45³ +   27³
   13:       134379 =   51³ +   12³ =   43³ +   38³
   14:       149389 =   53³ +    8³ =   50³ +   29³
   15:       165464 =   54³ +   20³ =   48³ +   38³
   16:       171288 =   55³ +   17³ =   54³ +   24³
   17:       195841 =   58³ +    9³ =   57³ +   22³
   18:       216027 =   60³ +    3³ =   59³ +   22³
   19:       216125 =   60³ +    5³ =   50³ +   45³
   20:       262656 =   64³ +    8³ =   60³ +   36³
   21:       314496 =   68³ +    4³ =   66³ +   30³
   22:       320264 =   66³ +   32³ =   68³ +   18³
   23:       327763 =   58³ +   51³ =   67³ +   30³
   24:       373464 =   72³ +    6³ =   60³ +   54³
   25:       402597 =   69³ +   42³ =   61³ +   56³
  455:     87539319 =  436³ +  167³ =  423³ +  228³ =  414³ +  255³
 2000:   1671816384 = 1168³ +  428³ =  944³ +  940³
 2001:   1672470592 = 1187³ +   29³ = 1124³ +  632³
 2002:   1673170856 = 1164³ +  458³ = 1034³ +  828³
 2003:   1675045225 = 1081³ +  744³ = 1153³ +  522³
 2004:   1675958167 = 1096³ +  711³ = 1159³ +  492³
 2005:   1676926719 = 1188³ +   63³ = 1095³ +  714³
 2006:   1677646971 =  990³ +  891³ = 1188³ +   99³
40000: 976889700163 = 8659³ + 6894³ = 9891³ + 2098³
40001: 976942087381 = 7890³ + 7861³ = 8680³ + 6861³
40002: 976946344920 = 9476³ + 5014³ = 9798³ + 3312³
40003: 976962998375 = 9912³ + 1463³ = 8415³ + 7250³
40004: 976974757064 = 9365³ + 5379³ = 9131³ + 5997³
40005: 977025552984 = 9894³ + 2040³ = 9792³ + 3366³
40006: 977104161000 = 9465³ + 5055³ = 9920³ +  970³

J

<lang J> 25{.(,.~ <@>:@i.@#) ;({."#. <@(0&#`({.@{.(;,)<@}."1)@.(1<#))/. ])/:~~.,/(+,/:~@,)"0/~3^~1+i.100 ┌──┬──────┬────────────┬─────────────┐ │1 │1729 │1 1728 │729 1000 │ ├──┼──────┼────────────┼─────────────┤ │2 │4104 │8 4096 │729 3375 │ ├──┼──────┼────────────┼─────────────┤ │3 │13832 │8 13824 │5832 8000 │ ├──┼──────┼────────────┼─────────────┤ │4 │20683 │1000 19683 │6859 13824 │ ├──┼──────┼────────────┼─────────────┤ │5 │32832 │64 32768 │5832 27000 │ ├──┼──────┼────────────┼─────────────┤ │6 │39312 │8 39304 │3375 35937 │ ├──┼──────┼────────────┼─────────────┤ │7 │40033 │729 39304 │4096 35937 │ ├──┼──────┼────────────┼─────────────┤ │8 │46683 │27 46656 │19683 27000 │ ├──┼──────┼────────────┼─────────────┤ │9 │64232 │4913 59319 │17576 46656 │ ├──┼──────┼────────────┼─────────────┤ │10│65728 │1728 64000 │29791 35937 │ ├──┼──────┼────────────┼─────────────┤ │11│110656│64 110592 │46656 64000 │ ├──┼──────┼────────────┼─────────────┤ │12│110808│216 110592 │19683 91125 │ ├──┼──────┼────────────┼─────────────┤ │13│134379│1728 132651 │54872 79507 │ ├──┼──────┼────────────┼─────────────┤ │14│149389│512 148877 │24389 125000 │ ├──┼──────┼────────────┼─────────────┤ │15│165464│8000 157464 │54872 110592 │ ├──┼──────┼────────────┼─────────────┤ │16│171288│4913 166375 │13824 157464 │ ├──┼──────┼────────────┼─────────────┤ │17│195841│729 195112 │10648 185193 │ ├──┼──────┼────────────┼─────────────┤ │18│216027│27 216000 │10648 205379 │ ├──┼──────┼────────────┼─────────────┤ │19│216125│125 216000 │91125 125000 │ ├──┼──────┼────────────┼─────────────┤ │20│262656│512 262144 │46656 216000 │ ├──┼──────┼────────────┼─────────────┤ │21│314496│64 314432 │27000 287496 │ ├──┼──────┼────────────┼─────────────┤ │22│320264│5832 314432 │32768 287496 │ ├──┼──────┼────────────┼─────────────┤ │23│327763│27000 300763│132651 195112│ ├──┼──────┼────────────┼─────────────┤ │24│373464│216 373248 │157464 216000│ ├──┼──────┼────────────┼─────────────┤ │25│402597│74088 328509│175616 226981│ └──┴──────┴────────────┴─────────────┘</lang>

Explanation:

First, generate 100 cubes.

Then, form a 3 column table of unique rows: sum, small cube, large cube

Then, gather rows where the first entry is the same. Keep the ones with at least two such entries (sorted by ascending order of sum).

Then, place an counting index (starting from 1) in front of each row, so the columns are now: counting index, sum, small cube, large cube.

Note that the cube root of the 25th entry is slightly smaller than 74, so testing against the first 100 cubes is more than sufficient.

Note that here we have elected to show the constituent cubes as themselves rather than as expressions involving their cube roots.

Extra credit:

<lang J> x:each 7 {. 1999 }. (,.~ <@>:@i.@#) ;({."#. <@(0&#`({.@{.(;,)<@}."1)@.(1<#))/. ])/:~~.,/(+,/:~@,)"0/~3^~1+i.10000 ┌────┬──────────┬────────────────────┬────────────────────┬┐ │2000│1671816384│78402752 1593413632 │830584000 841232384 ││ ├────┼──────────┼────────────────────┼────────────────────┼┤ │2001│1672470592│24389 1672446203 │252435968 1420034624││ ├────┼──────────┼────────────────────┼────────────────────┼┤ │2002│1673170856│96071912 1577098944 │567663552 1105507304││ ├────┼──────────┼────────────────────┼────────────────────┼┤ │2003│1675045225│142236648 1532808577│411830784 1263214441││ ├────┼──────────┼────────────────────┼────────────────────┼┤ │2004│1675958167│119095488 1556862679│359425431 1316532736││ ├────┼──────────┼────────────────────┼────────────────────┼┤ │2005│1676926719│250047 1676676672 │363994344 1312932375││ ├────┼──────────┼────────────────────┼────────────────────┼┤ │2006│1677646971│970299 1676676672 │707347971 970299000 ││ └────┴──────────┴────────────────────┴────────────────────┴┘</lang>

The extra blank box at the end is because when tackling this large of a data set, some sums can be achieved by three different pairs of cubes.

Java

<lang java>import java.util.PriorityQueue; import java.util.ArrayList; import java.util.List; import java.util.Iterator;

class CubeSum implements Comparable<CubeSum> { public long x, y, value;

public CubeSum(long x, long y) { this.x = x; this.y = y; this.value = x*x*x + y*y*y; }

public String toString() { return String.format("%4d^3 + %4d^3", x, y); }

public int compareTo(CubeSum that) { return value < that.value ? -1 : value > that.value ? 1 : 0; } }

class SumIterator implements Iterator<CubeSum> { PriorityQueue<CubeSum> pq = new PriorityQueue<CubeSum>(); long n = 0;

public boolean hasNext() { return true; } public CubeSum next() { while (pq.size() == 0 || pq.peek().value >= n*n*n) pq.add(new CubeSum(++n, 1));

CubeSum s = pq.remove(); if (s.x > s.y + 1) pq.add(new CubeSum(s.x, s.y+1));

return s; } }

class TaxiIterator implements Iterator<List<CubeSum>> { Iterator<CubeSum> sumIterator = new SumIterator(); CubeSum last = sumIterator.next();

public boolean hasNext() { return true; } public List<CubeSum> next() { CubeSum s; List<CubeSum> train = new ArrayList<CubeSum>();

while ((s = sumIterator.next()).value != last.value) last = s;

train.add(last);

do { train.add(s); } while ((s = sumIterator.next()).value == last.value); last = s;

return train; } }

public class Taxi { public static final void main(String[] args) { Iterator<List<CubeSum>> taxi = new TaxiIterator();

for (int i = 1; i <= 2006; i++) { List<CubeSum> t = taxi.next(); if (i > 25 && i < 2000) continue;

System.out.printf("%4d: %10d", i, t.get(0).value); for (CubeSum s: t) System.out.print(" = " + s); System.out.println(); } } }</lang>

Output:
   1:       1729 =   10^3 +    9^3 =   12^3 +    1^3
   2:       4104 =   15^3 +    9^3 =   16^3 +    2^3
   3:      13832 =   20^3 +   18^3 =   24^3 +    2^3
   4:      20683 =   24^3 +   19^3 =   27^3 +   10^3
   5:      32832 =   30^3 +   18^3 =   32^3 +    4^3
   6:      39312 =   33^3 +   15^3 =   34^3 +    2^3
   7:      40033 =   34^3 +    9^3 =   33^3 +   16^3
   8:      46683 =   30^3 +   27^3 =   36^3 +    3^3
   9:      64232 =   36^3 +   26^3 =   39^3 +   17^3
  10:      65728 =   33^3 +   31^3 =   40^3 +   12^3
  11:     110656 =   40^3 +   36^3 =   48^3 +    4^3
  12:     110808 =   45^3 +   27^3 =   48^3 +    6^3
  13:     134379 =   43^3 +   38^3 =   51^3 +   12^3
  14:     149389 =   50^3 +   29^3 =   53^3 +    8^3
  15:     165464 =   48^3 +   38^3 =   54^3 +   20^3
  16:     171288 =   54^3 +   24^3 =   55^3 +   17^3
  17:     195841 =   57^3 +   22^3 =   58^3 +    9^3
  18:     216027 =   59^3 +   22^3 =   60^3 +    3^3
  19:     216125 =   50^3 +   45^3 =   60^3 +    5^3
  20:     262656 =   60^3 +   36^3 =   64^3 +    8^3
  21:     314496 =   66^3 +   30^3 =   68^3 +    4^3
  22:     320264 =   66^3 +   32^3 =   68^3 +   18^3
  23:     327763 =   58^3 +   51^3 =   67^3 +   30^3
  24:     373464 =   60^3 +   54^3 =   72^3 +    6^3
  25:     402597 =   61^3 +   56^3 =   69^3 +   42^3
2000: 1671816384 = 1168^3 +  428^3 =  944^3 +  940^3
2001: 1672470592 = 1124^3 +  632^3 = 1187^3 +   29^3
2002: 1673170856 = 1164^3 +  458^3 = 1034^3 +  828^3
2003: 1675045225 = 1153^3 +  522^3 = 1081^3 +  744^3
2004: 1675958167 = 1159^3 +  492^3 = 1096^3 +  711^3
2005: 1676926719 = 1095^3 +  714^3 = 1188^3 +   63^3
2006: 1677646971 =  990^3 +  891^3 = 1188^3 +   99^3

JavaScript

<lang JavaScript>var cubes=[], sum={} for (var n=1,e=1200; n<e; n+=1) cubes[n]=n*n*n for (var i=1; i<e-1; i+=1) { var a = cubes[i] for (var j=i; j<e; j+=1) { var b = cubes[j] var s = a+b, abs = sum[s] if (!abs) sum[s] = abs = [] abs.push([i,j]) } }

var i=0 for (var s in sum) { var abs = sum[s] if (abs.length < 2) continue i+=1 if (abs.length == 2 && i > 25 && i < 2000) continue if (i > 2006) break document.write(i, ': ', s) for (var ab of abs) document.write( ' = ', ab[0], '3+', ab[1], '3') document.write('
') }</lang>

Output:
1: 1729 = 13+123 = 93+103
2: 4104 = 23+163 = 93+153
3: 13832 = 23+243 = 183+203
4: 20683 = 103+273 = 193+243
5: 32832 = 43+323 = 183+303
6: 39312 = 23+343 = 153+333
7: 40033 = 93+343 = 163+333
8: 46683 = 33+363 = 273+303
9: 64232 = 173+393 = 263+363
10: 65728 = 123+403 = 313+333
11: 110656 = 43+483 = 363+403
12: 110808 = 63+483 = 273+453
13: 134379 = 123+513 = 383+433
14: 149389 = 83+533 = 293+503
15: 165464 = 203+543 = 383+483
16: 171288 = 173+553 = 243+543
17: 195841 = 93+583 = 223+573
18: 216027 = 33+603 = 223+593
19: 216125 = 53+603 = 453+503
20: 262656 = 83+643 = 363+603
21: 314496 = 43+683 = 303+663
22: 320264 = 183+683 = 323+663
23: 327763 = 303+673 = 513+583
24: 373464 = 63+723 = 543+603
25: 402597 = 423+693 = 563+613
455: 87539319 = 1673+4363 = 2283+4233 = 2553+4143
535: 119824488 = 113+4933 = 903+4923 = 3463+4283
588: 143604279 = 1113+5223 = 3593+4603 = 4083+4233
655: 175959000 = 703+5603 = 1983+5523 = 3153+5253
888: 327763000 = 3003+6703 = 3393+6613 = 5103+5803
1299: 700314552 = 3343+8723 = 4563+8463 = 5103+8283
1398: 804360375 = 153+9303 = 1983+9273 = 2953+9203
1515: 958595904 = 223+9863 = 1803+9843 = 6923+8563
1660: 1148834232 = 2223+10443 = 7183+9203 = 8163+8463
1837: 1407672000 = 1403+11203 = 3963+11043 = 6303+10503
2000: 1671816384 = 4283+11683 = 9403+9443
2001: 1672470592 = 293+11873 = 6323+11243
2002: 1673170856 = 4583+11643 = 8283+10343
2003: 1675045225 = 5223+11533 = 7443+10813
2004: 1675958167 = 4923+11593 = 7113+10963
2005: 1676926719 = 633+11883 = 7143+10953
2006: 1677646971 = 993+11883 = 8913+9903

jq

Works with: jq version 1.4

<lang jq># Output: an array of the form [i^3 + j^3, [i, j]] sorted by the sum.

  1. Only cubes of 1 to ($in-1) are considered; the listing is therefore truncated
  2. as it might not capture taxicab numbers greater than $in ^ 3.

def sum_of_two_cubes:

 def cubed: .*.*.;
 . as $in
 | (cubed + 1) as $limit
 | [range(1;$in) as $i | range($i;$in) as $j
 | [ ($i|cubed) + ($j|cubed), [$i, $j] ] ] | sort
 | map( select( .[0] < $limit ) );
  1. Output a stream of triples [t, d1, d2], in order of t,
  2. where t is a taxicab number, and d1 and d2 are distinct
  3. decompositions [i,j] with i^3 + j^3 == t.
  4. The stream includes each taxicab number once only.

def taxicabs0:

 sum_of_two_cubes as $sums
 | range(1;$sums|length) as $i
 | if $sums[$i][0] == $sums[$i-1][0]
     and ($i==1 or $sums[$i][0] != $sums[$i-2][0])
   then [$sums[$i][0], $sums[$i-1][1], $sums[$i][1]]
   else empty
   end;
  1. Output a stream of $n taxicab triples: [t, d1, d2] as described above,
  2. without repeating t.

def taxicabs:

 # If your jq includes until/2 then the following definition
 # can be omitted:
 def until(cond; next):
   def _until: if cond then . else (next|_until) end;  _until;
 . as $n
 | [10, ($n / 10 | floor)] | max as $increment
 | [20, ($n / 2 | floor)] | max
 | [ ., [taxicabs0] ]
 | until( .[1] | length >= $m; (.[0] + $increment) | [., [taxicabs0]] )
 | .[1][0:$n] ;</lang>

The task <lang jq>2006 | taxicabs as $t | (range(0;25), range(1999;2006)) as $i | "\($i+1): \($t[$i][0]) ~ \($t[$i][1]) and \($t[$i][2])"</lang>

Output:

<lang sh>$ jq -n -r -f Taxicab_numbers.jq 1: 1729 ~ [1,12] and [9,10] 2: 4104 ~ [2,16] and [9,15] 3: 13832 ~ [2,24] and [18,20] 4: 20683 ~ [10,27] and [19,24] 5: 32832 ~ [4,32] and [18,30] 6: 39312 ~ [2,34] and [15,33] 7: 40033 ~ [9,34] and [16,33] 8: 46683 ~ [3,36] and [27,30] 9: 64232 ~ [17,39] and [26,36] 10: 65728 ~ [12,40] and [31,33] 11: 110656 ~ [4,48] and [36,40] 12: 110808 ~ [6,48] and [27,45] 13: 134379 ~ [12,51] and [38,43] 14: 149389 ~ [8,53] and [29,50] 15: 165464 ~ [20,54] and [38,48] 16: 171288 ~ [17,55] and [24,54] 17: 195841 ~ [9,58] and [22,57] 18: 216027 ~ [3,60] and [22,59] 19: 216125 ~ [5,60] and [45,50] 20: 262656 ~ [8,64] and [36,60] 21: 314496 ~ [4,68] and [30,66] 22: 320264 ~ [18,68] and [32,66] 23: 327763 ~ [30,67] and [51,58] 24: 373464 ~ [6,72] and [54,60] 25: 402597 ~ [42,69] and [56,61] 2000: 1671816384 ~ [428,1168] and [940,944] 2001: 1672470592 ~ [29,1187] and [632,1124] 2002: 1673170856 ~ [458,1164] and [828,1034] 2003: 1675045225 ~ [522,1153] and [744,1081] 2004: 1675958167 ~ [492,1159] and [711,1096] 2005: 1676926719 ~ [63,1188] and [714,1095] 2006: 1677646971 ~ [99,1188] and [891,990]</lang>

PARI/GP

<lang parigp>taxicab(n)=my(t); for(k=sqrtnint((n-1)\2,3)+1, sqrtnint(n,3), if(ispower(n-k^3, 3), if(t, return(1), t=1))); 0; cubes(n)=my(t); for(k=sqrtnint((n-1)\2,3)+1, sqrtnint(n,3), if(ispower(n-k^3, 3, &t), print(n" = \t"k"^3\t+ "t"^3"))) select(taxicab, [1..402597]) apply(cubes, %);</lang>

Output:
%1 = [1729, 4104, 13832, 20683, 32832, 39312, 40033, 46683, 64232, 65728, 110656, 110808, 134379, 149389, 165464, 171288, 195841, 216027, 216125, 262656, 314496, 320264, 327763, 373464, 402597]
1729 =          10^3    + 9^3
1729 =          12^3    + 1^3
4104 =          15^3    + 9^3
4104 =          16^3    + 2^3
13832 =         20^3    + 18^3
13832 =         24^3    + 2^3
20683 =         24^3    + 19^3
20683 =         27^3    + 10^3
32832 =         30^3    + 18^3
32832 =         32^3    + 4^3
39312 =         33^3    + 15^3
39312 =         34^3    + 2^3
40033 =         33^3    + 16^3
40033 =         34^3    + 9^3
46683 =         30^3    + 27^3
46683 =         36^3    + 3^3
64232 =         36^3    + 26^3
64232 =         39^3    + 17^3
65728 =         33^3    + 31^3
65728 =         40^3    + 12^3
110656 =        40^3    + 36^3
110656 =        48^3    + 4^3
110808 =        45^3    + 27^3
110808 =        48^3    + 6^3
134379 =        43^3    + 38^3
134379 =        51^3    + 12^3
149389 =        50^3    + 29^3
149389 =        53^3    + 8^3
165464 =        48^3    + 38^3
165464 =        54^3    + 20^3
171288 =        54^3    + 24^3
171288 =        55^3    + 17^3
195841 =        57^3    + 22^3
195841 =        58^3    + 9^3
216027 =        59^3    + 22^3
216027 =        60^3    + 3^3
216125 =        50^3    + 45^3
216125 =        60^3    + 5^3
262656 =        60^3    + 36^3
262656 =        64^3    + 8^3
314496 =        66^3    + 30^3
314496 =        68^3    + 4^3
320264 =        66^3    + 32^3
320264 =        68^3    + 18^3
327763 =        58^3    + 51^3
327763 =        67^3    + 30^3
373464 =        60^3    + 54^3
373464 =        72^3    + 6^3
402597 =        61^3    + 56^3
402597 =        69^3    + 42^3

Pascal

Works with: Free Pascal

searchSameSum takes most of the time.c Version is ~9 times faster aka 43 ms. <lang pascal>program taxiCabNo; uses

 sysutils;

type

 tPot3    = Uint32;
 tPot3Sol = record
              p3Sum : tPot3;
              i1,j1,
              i2,j2 : word;
            end;
tpPot3Sol = ^tPot3Sol;
tPotarr   = array[0..0] of tPot3Sol;
tpPotarr   = ^tPotarr;

var //1290^3 = 2'146'689'000 < 2^31-1

 pot3 : array[0..1290] of tPot3;//
 AllSol : array[0..3000] of tpot3Sol;
 AllSolHigh : NativeInt;
 MaxInsDist : NativeInt;

procedure SolOut(const s:tpot3Sol;no: NativeInt); begin

 with s do
   writeln(no:5,p3Sum:12,' = ',j1:5,'^3 +',i1:5,'^3 =',j2:5,'^3 +',i2:5,'^3');

end;

procedure InsertAllSol; var

 tmp: tpot3Sol;
 p :tpPotarr;
 p3Sum: tPot3;
 i: NativeInt;

Begin

 i := AllSolHigh;
 IF i > 0 then
 Begin
   p := @AllSol[0];
   tmp := p^[i];
   p3Sum := p^[i].p3Sum;
   //search the right place for insertion
   repeat
     dec(i);
     IF (p^[i].p3Sum <= p3Sum) then
       BREAK;
   until  (i<=0);
   IF p^[i].p3Sum = p3Sum then
     EXIT;
   //free the right place by moving one place up
   inc(i);
   IF i<AllSolHigh then
   Begin
     move(p^[i],p^[i+1],SizeOf(AllSol[0])*(AllSolHigh-i));
     AllSol[i]:= tmp;
     IF MaxInsDist < AllSolHigh-i then
       MaxInsDist := AllSolHigh-i;
   end;
 end;
 inc(AllSolHigh);

end;

function searchSameSum(var sol:tpot3Sol):boolean; var

 Sum,
 SumHi: tPot3;
 hi,lo: NativeInt;

Begin

 Sum := sol.p3Sum;
 lo:= Sol.i1;
 hi:= Sol.j1;
 repeat
   dec(hi);
   SumHi := Sum-Pot3[hi];
   while (lo<hi) AND (SumHi>Pot3[lo]) do
     inc(lo);
   IF SumHi = Pot3[lo] then
     BREAK;
 until lo>=hi;
 IF lo< hi then
 Begin
   sol.i2:= lo;
   sol.j2:= hi;
   searchSameSum := true;
 end
 else
   searchSameSum := false;

end;

procedure Search; var

 i,j: LongInt;

Begin

 MaxInsDist := 0;
 AllSolHigh := 0;
 For j := 2 to High(pot3)-2 do
 Begin
   For i := 1 to j-1 do
   Begin
     with AllSol[AllSolHigh] do
     Begin
       p3Sum:= pot3[i]+pot3[j];
       i1:= i;
       j1:= j;
     end;
     IF searchSameSum(AllSol[AllSolHigh]) then
     BEGIN
       InsertAllSol;
       IF AllSolHigh>High(AllSol) then
         EXIT;
     end;
   end;
 end;

end;

var

 i: LongInt;

Begin

 For i := Low(pot3) to High(pot3) do
   pot3[i] := i*i*i;
 AllSolHigh := 0;
 Search;
 For i :=    0 to   24 do SolOut(AllSol[i],i+1);
 For i := 1999 to 2005 do SolOut(AllSol[i],i+1);
 writeln('count of solutions         ',AllSolHigh);
 writeln('maximal insertion distance ', MaxInsDist);

end.</lang>

    1        1729 =    12^3 +    1^3 =   10^3 +    9^3
    2        4104 =    16^3 +    2^3 =   15^3 +    9^3
.....
   24      373464 =    72^3 +    6^3 =   60^3 +   54^3
   25      402597 =    69^3 +   42^3 =   61^3 +   56^3
 2000  1671816384 =  1168^3 +  428^3 =  944^3 +  940^3
 2001  1672470592 =  1187^3 +   29^3 = 1124^3 +  632^3
..
 2005  1676926719 =  1188^3 +   63^3 = 1095^3 +  714^3
 2006  1677646971 =  1188^3 +   99^3 =  990^3 +  891^3
count of solutions         2297
maximal insertion distance 50

real  0m0.383s

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

   2        4104 =    16^3 +    2^3 =   15^3 +    9^3

.....

  24      373464 =    72^3 +    6^3 =   60^3 +   54^3
  25      402597 =    69^3 +   42^3 =   61^3 +   56^3
2000  1671816384 =  1168^3 +  428^3 =  944^3 +  940^3
2001  1672470592 =  1187^3 +   29^3 = 1124^3 +  632^3

..

2005  1676926719 =  1188^3 +   63^3 = 1095^3 +  714^3
2006  1677646971 =  1188^3 +   99^3 =  990^3 +  891^3

count of solutions 2297 maximal insertion distance 50

real 0m0.383s

Perl

Uses segmentation so memory use is constrained as high values are searched for. Also has parameter to look for Ta(3) and Ta(4) numbers (which is when segmentation is really needed). By default shows the first 25 numbers; with one argument shows that many; with two arguments shows results in the range.

<lang perl>my($beg, $end) = (@ARGV==0) ? (1,25) : (@ARGV==1) ? (1,shift) : (shift,shift);

my $lim = 1e14; # Ought to be dynamic as should segment size my @basis = map { $_*$_*$_ } (1 .. int($lim ** (1.0/3.0) + 1)); my $paira = 2; # We're looking for Ta(2) and larger

my ($segsize, $low, $high, $i) = (500_000_000, 0, 0, 0);

while ($i < $end) {

 $low = $high+1;
 die "lim too low" if $low > $lim;
 $high = $low + $segsize - 1;
 $high = $lim if $high > $lim;
 foreach my $p (_find_pairs_segment(\@basis, $paira, $low, $high,
                sub { sprintf("%4d^3 + %4d^3", $_[0], $_[1]) })    ) {
   $i++;
   next if $i < $beg;
   last if $i > $end;
   my $n = shift @$p;
   printf "%4d: %10d  = %s\n", $i, $n, join("  = ", @$p);
 }

}

sub _find_pairs_segment {

 my($p, $len, $start, $end, $formatsub) = @_;
 my $plen = $#$p;
 my %allpairs;
 foreach my $i (0 .. $plen) {
   my $pi = $p->[$i];
   next if ($pi+$p->[$plen]) < $start;
   last if (2*$pi) > $end;
   foreach my $j ($i .. $plen) {
     my $sum = $pi + $p->[$j];
     next if $sum < $start;
     last if $sum > $end;
     push @{ $allpairs{$sum} }, $i, $j;
   }
   # If we wanted to save more memory, we could filter and delete every entry
   # where $n < 2 * $p->[$i+1].  This can cut memory use in half, but is slow.
 }
 my @retlist;
 foreach my $list (grep { scalar @$_ >= $len*2 } values %allpairs) {
   my $n = $p->[$list->[0]] + $p->[$list->[1]];
   my @pairlist;
   while (@$list) {
     push @pairlist, $formatsub->(1 + shift @$list, 1 + shift @$list);
   }
   push @retlist, [$n, @pairlist];
 }
 @retlist = sort { $a->[0] <=> $b->[0] } @retlist;
 return @retlist;

}</lang>

Output:
   1:       1729  =    1^3 +   12^3  =    9^3 +   10^3
   2:       4104  =    2^3 +   16^3  =    9^3 +   15^3
   3:      13832  =    2^3 +   24^3  =   18^3 +   20^3
   4:      20683  =   10^3 +   27^3  =   19^3 +   24^3
   5:      32832  =    4^3 +   32^3  =   18^3 +   30^3
   6:      39312  =    2^3 +   34^3  =   15^3 +   33^3
   7:      40033  =    9^3 +   34^3  =   16^3 +   33^3
   8:      46683  =    3^3 +   36^3  =   27^3 +   30^3
   9:      64232  =   17^3 +   39^3  =   26^3 +   36^3
  10:      65728  =   12^3 +   40^3  =   31^3 +   33^3
  11:     110656  =    4^3 +   48^3  =   36^3 +   40^3
  12:     110808  =    6^3 +   48^3  =   27^3 +   45^3
  13:     134379  =   12^3 +   51^3  =   38^3 +   43^3
  14:     149389  =    8^3 +   53^3  =   29^3 +   50^3
  15:     165464  =   20^3 +   54^3  =   38^3 +   48^3
  16:     171288  =   17^3 +   55^3  =   24^3 +   54^3
  17:     195841  =    9^3 +   58^3  =   22^3 +   57^3
  18:     216027  =    3^3 +   60^3  =   22^3 +   59^3
  19:     216125  =    5^3 +   60^3  =   45^3 +   50^3
  20:     262656  =    8^3 +   64^3  =   36^3 +   60^3
  21:     314496  =    4^3 +   68^3  =   30^3 +   66^3
  22:     320264  =   18^3 +   68^3  =   32^3 +   66^3
  23:     327763  =   30^3 +   67^3  =   51^3 +   58^3
  24:     373464  =    6^3 +   72^3  =   54^3 +   60^3
  25:     402597  =   42^3 +   69^3  =   56^3 +   61^3

With arguments 2000 2006:

2000: 1671816384  =  428^3 + 1168^3  =  940^3 +  944^3
2001: 1672470592  =   29^3 + 1187^3  =  632^3 + 1124^3
2002: 1673170856  =  458^3 + 1164^3  =  828^3 + 1034^3
2003: 1675045225  =  522^3 + 1153^3  =  744^3 + 1081^3
2004: 1675958167  =  492^3 + 1159^3  =  711^3 + 1096^3
2005: 1676926719  =   63^3 + 1188^3  =  714^3 + 1095^3
2006: 1677646971  =   99^3 + 1188^3  =  891^3 +  990^3

Perl 6

This uses a pretty simple search algorithm that doesn't necessarily return the Taxicab numbers in order. Assuming we want all the Taxicab numbers within some range S to N, we'll search until we find N values. When we find the Nth value, we continue to search up to the cube root of the largest Taxicab number found up to that point. That ensures we will find all of them inside the desired range without needing to search arbitrarily or use magic numbers. Defaults to returning the Taxicab numbers from 1 to 25. Pass in a different start and end value if you want some other range. <lang perl6>sub MAIN ($start = 1, $end = 25) {

   my %taxi;
   my $taxis = 0;
   my $terminate = 0;
   for 1 .. * -> $c1 {
       display( %taxi, $start, $end ) and exit if 0 < $terminate < $c1;
       my $c = $c1 ** 3;
       for 1 .. $c1 -> $c2 {
           my $this = $c2 ** 3 + $c;
           %taxi{$this}.push: [$c2, $c1];
           $taxis++ if %taxi{$this}.elems == 2;
   	    $terminate = %taxi.grep( { $_.value.elems > 1 } ).sort( +*.key )[*-1].key**(1/3)
               if $taxis == $end and !$terminate;
       }
   }

}

sub display (%this_stuff, $start, $end) {

   my $i = $start; 
   printf "%4d %10d  =>\t%s\n", $i++, $_.key, 
       ($_.value.map({ sprintf "%4d³ + %-s", $_[0], "$_[1]\³" })).join: ",\t"
       for %this_stuff.grep( { $_.value.elems > 1 } ).sort( +*.key )[$start-1..$end-1];
   1;

}</lang>

Output:

With no passed parameters (default)

   1       1729  =>	   9³ + 10³,	   1³ + 12³
   2       4104  =>	   9³ + 15³,	   2³ + 16³
   3      13832  =>	  18³ + 20³,	   2³ + 24³
   4      20683  =>	  19³ + 24³,	  10³ + 27³
   5      32832  =>	  18³ + 30³,	   4³ + 32³
   6      39312  =>	  15³ + 33³,	   2³ + 34³
   7      40033  =>	  16³ + 33³,	   9³ + 34³
   8      46683  =>	  27³ + 30³,	   3³ + 36³
   9      64232  =>	  26³ + 36³,	  17³ + 39³
  10      65728  =>	  31³ + 33³,	  12³ + 40³
  11     110656  =>	  36³ + 40³,	   4³ + 48³
  12     110808  =>	  27³ + 45³,	   6³ + 48³
  13     134379  =>	  38³ + 43³,	  12³ + 51³
  14     149389  =>	  29³ + 50³,	   8³ + 53³
  15     165464  =>	  38³ + 48³,	  20³ + 54³
  16     171288  =>	  24³ + 54³,	  17³ + 55³
  17     195841  =>	  22³ + 57³,	   9³ + 58³
  18     216027  =>	  22³ + 59³,	   3³ + 60³
  19     216125  =>	  45³ + 50³,	   5³ + 60³
  20     262656  =>	  36³ + 60³,	   8³ + 64³
  21     314496  =>	  30³ + 66³,	   4³ + 68³
  22     320264  =>	  32³ + 66³,	  18³ + 68³
  23     327763  =>	  51³ + 58³,	  30³ + 67³
  24     373464  =>	  54³ + 60³,	   6³ + 72³
  25     402597  =>	  56³ + 61³,	  42³ + 69³

With passed parameters 2000 2006:

2000 1671816384  =>	 940³ + 944³,	 428³ + 1168³
2001 1672470592  =>	 632³ + 1124³,	  29³ + 1187³
2002 1673170856  =>	 828³ + 1034³,	 458³ + 1164³
2003 1675045225  =>	 744³ + 1081³,	 522³ + 1153³
2004 1675958167  =>	 711³ + 1096³,	 492³ + 1159³
2005 1676926719  =>	 714³ + 1095³,	  63³ + 1188³
2006 1677646971  =>	 891³ + 990³,	  99³ + 1188³

Python

(Magic number 1201 found by trial and error) <lang python>from collections import defaultdict from itertools import product from pprint import pprint as pp

cube2n = {x**3:x for x in range(1, 1201)} sum2cubes = defaultdict(set) for c1, c2 in product(cube2n, cube2n): if c1 >= c2: sum2cubes[c1 + c2].add((cube2n[c1], cube2n[c2]))

taxied = sorted((k, v) for k,v in sum2cubes.items() if len(v) >= 2)

  1. pp(len(taxied)) # 2068

for t in enumerate(taxied[:25], 1):

   pp(t)

print('...') for t in enumerate(taxied[2000-1:2000+6], 2000):

   pp(t)</lang>
Output:
(1, (1729, {(12, 1), (10, 9)}))
(2, (4104, {(16, 2), (15, 9)}))
(3, (13832, {(20, 18), (24, 2)}))
(4, (20683, {(27, 10), (24, 19)}))
(5, (32832, {(30, 18), (32, 4)}))
(6, (39312, {(33, 15), (34, 2)}))
(7, (40033, {(33, 16), (34, 9)}))
(8, (46683, {(30, 27), (36, 3)}))
(9, (64232, {(36, 26), (39, 17)}))
(10, (65728, {(33, 31), (40, 12)}))
(11, (110656, {(48, 4), (40, 36)}))
(12, (110808, {(48, 6), (45, 27)}))
(13, (134379, {(51, 12), (43, 38)}))
(14, (149389, {(50, 29), (53, 8)}))
(15, (165464, {(54, 20), (48, 38)}))
(16, (171288, {(54, 24), (55, 17)}))
(17, (195841, {(57, 22), (58, 9)}))
(18, (216027, {(60, 3), (59, 22)}))
(19, (216125, {(60, 5), (50, 45)}))
(20, (262656, {(64, 8), (60, 36)}))
(21, (314496, {(66, 30), (68, 4)}))
(22, (320264, {(66, 32), (68, 18)}))
(23, (327763, {(58, 51), (67, 30)}))
(24, (373464, {(72, 6), (60, 54)}))
(25, (402597, {(69, 42), (61, 56)}))
...
(2000, (1671816384, {(1168, 428), (944, 940)}))
(2001, (1672470592, {(1187, 29), (1124, 632)}))
(2002, (1673170856, {(1164, 458), (1034, 828)}))
(2003, (1675045225, {(1153, 522), (1081, 744)}))
(2004, (1675958167, {(1159, 492), (1096, 711)}))
(2005, (1676926719, {(1188, 63), (1095, 714)}))
(2006, (1677646971, {(990, 891), (1188, 99)}))

Although, for this task it's simply faster to look up the cubes in the sum when we need to print them, because we can now store and sort only the sums: <lang python>cubes, crev = [x**3 for x in range(1,1200)], {}

  1. for cube root lookup

for x,x3 in enumerate(cubes): crev[x3] = x + 1

sums = sorted(x+y for x in cubes for y in cubes if y < x)

idx = 0 for i in range(1, len(sums)-1):

   if sums[i-1] != sums[i] and sums[i] == sums[i+1]:
       idx += 1
       if idx > 25 and idx < 2000 or idx > 2006: continue
       n,p = sums[i],[]
       for x in cubes:
           if n-x < x: break
           if n-x in crev:
               p.append((crev[x], crev[n-x]))
       print "%4d: %10d"%(idx,n),
       for x in p: print " = %4d^3 + %4d^3"%x,
       print</lang>
Output:

Output trimmed to reduce clutter.

   1:       1729  =    1^3 +   12^3  =    9^3 +   10^3
   2:       4104  =    2^3 +   16^3  =    9^3 +   15^3
   3:      13832  =    2^3 +   24^3  =   18^3 +   20^3
   4:      20683  =   10^3 +   27^3  =   19^3 +   24^3
   5:      32832  =    4^3 +   32^3  =   18^3 +   30^3
...
2004: 1675958167  =  492^3 + 1159^3  =  711^3 + 1096^3
2005: 1676926719  =   63^3 + 1188^3  =  714^3 + 1095^3
2006: 1677646971  =   99^3 + 1188^3  =  891^3 +  990^3

Using heapq module

A priority queue that holds cube sums. When consecutive sums come out with the same value, they are taxis. <lang python>from heapq import heappush, heappop

def cubesum():

   h,n = [],1
   while True:
       while not h or h[0][0] > n**3: # could also pre-calculate cubes
           heappush(h, (n**3 + 1, n, 1))
           n += 1
       (s, x, y) = heappop(h)
       yield((s, x, y))
       y += 1
       if y < x:    # should be y <= x?
           heappush(h, (x**3 + y**3, x, y))

def taxis():

   out = [(0,0,0)]
   for s in cubesum():
       if s[0] == out[-1][0]:
           out.append(s)
       else:
           if len(out) > 1: yield(out)
           out = [s]

n = 0 for x in taxis():

   n += 1
   if n >= 2006: break
   if n <= 25 or n >= 2000:
       print(n, x)</lang>
Output:
(1, [(1729, 10, 9), (1729, 12, 1)])
(2, [(4104, 15, 9), (4104, 16, 2)])
(3, [(13832, 20, 18), (13832, 24, 2)])
(4, [(20683, 24, 19), (20683, 27, 10)])
(5, [(32832, 30, 18), (32832, 32, 4)])
(6, [(39312, 33, 15), (39312, 34, 2)])
(7, [(40033, 33, 16), (40033, 34, 9)])
(8, [(46683, 30, 27), (46683, 36, 3)])
(9, [(64232, 36, 26), (64232, 39, 17)])
(10, [(65728, 33, 31), (65728, 40, 12)])
(11, [(110656, 40, 36), (110656, 48, 4)])
(12, [(110808, 45, 27), (110808, 48, 6)])
(13, [(134379, 43, 38), (134379, 51, 12)])
(14, [(149389, 50, 29), (149389, 53, 8)])
(15, [(165464, 48, 38), (165464, 54, 20)])
(16, [(171288, 54, 24), (171288, 55, 17)])
(17, [(195841, 57, 22), (195841, 58, 9)])
(18, [(216027, 59, 22), (216027, 60, 3)])
(19, [(216125, 50, 45), (216125, 60, 5)])
(20, [(262656, 60, 36), (262656, 64, 8)])
(21, [(314496, 66, 30), (314496, 68, 4)])
(22, [(320264, 66, 32), (320264, 68, 18)])
(23, [(327763, 58, 51), (327763, 67, 30)])
(24, [(373464, 60, 54), (373464, 72, 6)])
(25, [(402597, 61, 56), (402597, 69, 42)])
(2000, [(1671816384, 944, 940), (1671816384, 1168, 428)])
(2001, [(1672470592, 1124, 632), (1672470592, 1187, 29)])
(2002, [(1673170856, 1034, 828), (1673170856, 1164, 458)])
(2003, [(1675045225, 1081, 744), (1675045225, 1153, 522)])
(2004, [(1675958167, 1096, 711), (1675958167, 1159, 492)])
(2005, [(1676926719, 1095, 714), (1676926719, 1188, 63)])

Racket

This is the straighforward implementation, so it finds only the first 25 values in a sensible amount of time. <lang Racket>#lang racket

(define (cube x) (* x x x))

floor of cubic root

(define (cubic-root x)

 (let ([aprox (inexact->exact (round (expt x (/ 1 3))))])
   (if (> (cube aprox) x)
       (- aprox 1)
       aprox)))

(let loop ([p 1] [n 1])

 (let ()
   (define pairs
     (for*/list ([j (in-range 1 (add1 (cubic-root (quotient n 2))))]
                 [k (in-value (cubic-root (- n (cube j))))]
                 #:when (= n (+ (cube j) (cube k))))
       (cons j k)))
   (if (>= (length pairs) 2)
     (begin
       (printf "~a: ~a" p n)
       (for ([pair (in-list pairs)])
         (printf " = ~a^3 + ~a^3" (car pair) (cdr pair)))
         (newline)
       (when (< p 25)
         (loop (add1 p) (add1 n))))
     (loop p (add1 n)))))</lang>
Output:
1: 1729 = 1^3 + 12^3 = 9^3 + 10^3
2: 4104 = 2^3 + 16^3 = 9^3 + 15^3
3: 13832 = 2^3 + 24^3 = 18^3 + 20^3
4: 20683 = 10^3 + 27^3 = 19^3 + 24^3
5: 32832 = 4^3 + 32^3 = 18^3 + 30^3
6: 39312 = 2^3 + 34^3 = 15^3 + 33^3
7: 40033 = 9^3 + 34^3 = 16^3 + 33^3
8: 46683 = 3^3 + 36^3 = 27^3 + 30^3
9: 64232 = 17^3 + 39^3 = 26^3 + 36^3
10: 65728 = 12^3 + 40^3 = 31^3 + 33^3
11: 110656 = 4^3 + 48^3 = 36^3 + 40^3
12: 110808 = 6^3 + 48^3 = 27^3 + 45^3
13: 134379 = 12^3 + 51^3 = 38^3 + 43^3
14: 149389 = 8^3 + 53^3 = 29^3 + 50^3
15: 165464 = 20^3 + 54^3 = 38^3 + 48^3
16: 171288 = 17^3 + 55^3 = 24^3 + 54^3
17: 195841 = 9^3 + 58^3 = 22^3 + 57^3
18: 216027 = 3^3 + 60^3 = 22^3 + 59^3
19: 216125 = 5^3 + 60^3 = 45^3 + 50^3
20: 262656 = 8^3 + 64^3 = 36^3 + 60^3
21: 314496 = 4^3 + 68^3 = 30^3 + 66^3
22: 320264 = 18^3 + 68^3 = 32^3 + 66^3
23: 327763 = 30^3 + 67^3 = 51^3 + 58^3
24: 373464 = 6^3 + 72^3 = 54^3 + 60^3
25: 402597 = 42^3 + 69^3 = 56^3 + 61^3

REXX

<lang rexx>/*REXX program displays the specified first (lowest) taxicab numbers (for three ranges).*/ parse arg L.1 H.1 L.2 H.2 L.3 H.3 . /*obtain optional arguments from the CL*/

if L.1== | L.1=="," then L.1= 1 /*L1 is the low part of 1st range. */ if H.1== | H.1=="," then H.1= 25 /*H1 " " high " " " " */

if L.2== | L.2=="," then L.2= 454 /*L2 " " low " " 2nd " */ if H.2== | H.2=="," then H.2= 456 /*H2 " " high " " " " */

if L.3== | L.3=="," then L.3=2000 /*L3 " " low " " 3rd " */ if H.3== | H.3=="," then H.3=2006 /*H3 " " high " " " " */

mx=max(L.1, H.1, L.2, H.2, L.3, H.3) /*find how many taxicab numbers needed.*/ mx=mx+mx%10; ww=length(mx)*3 /*cushion; compensate for the triples.*/ w=ww%2; numeric digits max(9,ww) /*prepare to use some larger numbers. */ @.=.; #=0; @@.=0; @and=" ──and── "; $.= /*set some REXX vars and handy literals*/

                                                /* [↓]  generate extra taxicab numbers.*/
 do j=1  until #>=mx;       C=j**3              /*taxicab numbers may not be in order. */
 !.j=C                                          /*use memoization for cube calculation.*/
       do k=1  for j-1;     s=C+!.k             /*define a whole bunch of cube sums.   */
       if @.s==.  then do                       /*Cube not defined?   Then process it. */
                       @.s=j;  b.s=k            /*define  @.S  and  B.S≡sum  of 2 cubes*/
                       iterate                  /*      ··· and then go and do another.*/
                       end                      /* [↑]  define one cube sum at a time. */
       has=@@.s                                 /*has this number been defined before? */
       if \has  then $.s=right(s,ww)   '───►'  r(@.s,"   +")r(b.s)  @and  r(j,'   +')r(k)
                else $.s=$.s  @and  r(j,'   +')r(k)   /*◄─ build a display string. [↑] */
       @@.s=1                                   /*mark taxicab number as a sum of cubes*/
       if has   then iterate                    /*S  is a triple (or sometimes better).*/
       #=#+1;   #.#=s                           /*bump taxicab counter; define taxicab#*/
       end   /*k*/                              /* [↑]  build the cubes one─at─a─time. */
 end         /*j*/                              /* [↑]  complete with overage numbers. */

call Bsort # /*use a bubble sort on taxicab numbers.*/

            do range=1  to 3;  call tell L.range, H.range;  say;  end /*range*/

exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ r: return right(arg(1),w)'^3'arg(2) /*right─justify a number, append "^3" */ tell: do t=arg(1) to arg(2); _=#.t; say right(t,9)':' $._; end; return /*──────────────────────────────────────────────────────────────────────────────────────*/ Bsort: procedure expose @.; parse arg #; m=#-1 /*bubble sort (in ascending order). */

             do  until ok;  ok=1;       do j=1  for m;  k=j+1;  if @.j<=@.k  then iterate
                                        _=@.j;      @.j=@.k;      @.k=_;      ok=0
                                        end   /*j*/
             end   /*until ··· */
      return</lang>

output   using the default inputs:

 
        1:         1729 ───►     10^3   +     9^3   ──and──      12^3   +     1^3
        2:         4104 ───►     15^3   +     9^3   ──and──      16^3   +     2^3
        3:        13832 ───►     20^3   +    18^3   ──and──      24^3   +     2^3
        4:        20683 ───►     24^3   +    19^3   ──and──      27^3   +    10^3
        5:        32832 ───►     30^3   +    18^3   ──and──      32^3   +     4^3
        6:        39312 ───►     33^3   +    15^3   ──and──      34^3   +     2^3
        7:        40033 ───►     33^3   +    16^3   ──and──      34^3   +     9^3
        8:        46683 ───►     30^3   +    27^3   ──and──      36^3   +     3^3
        9:        64232 ───►     36^3   +    26^3   ──and──      39^3   +    17^3
       10:        65728 ───►     33^3   +    31^3   ──and──      40^3   +    12^3
       11:       110656 ───►     40^3   +    36^3   ──and──      48^3   +     4^3
       12:       110808 ───►     45^3   +    27^3   ──and──      48^3   +     6^3
       13:       134379 ───►     43^3   +    38^3   ──and──      51^3   +    12^3
       14:       149389 ───►     50^3   +    29^3   ──and──      53^3   +     8^3
       15:       165464 ───►     48^3   +    38^3   ──and──      54^3   +    20^3
       16:       171288 ───►     54^3   +    24^3   ──and──      55^3   +    17^3
       17:       195841 ───►     57^3   +    22^3   ──and──      58^3   +     9^3
       18:       216027 ───►     59^3   +    22^3   ──and──      60^3   +     3^3
       19:       216125 ───►     50^3   +    45^3   ──and──      60^3   +     5^3
       20:       262656 ───►     60^3   +    36^3   ──and──      64^3   +     8^3
       21:       314496 ───►     66^3   +    30^3   ──and──      68^3   +     4^3
       22:       320264 ───►     66^3   +    32^3   ──and──      68^3   +    18^3
       23:       327763 ───►     58^3   +    51^3   ──and──      67^3   +    30^3
       24:       373464 ───►     60^3   +    54^3   ──and──      72^3   +     6^3
       25:       402597 ───►     61^3   +    56^3   ──and──      69^3   +    42^3

      454:     87483968 ───►    363^3   +   341^3   ──and──     440^3   +   132^3
      455:     87539319 ───►    414^3   +   255^3   ──and──     423^3   +   228^3   ──and──     436^3   +   167^3
      456:     87579037 ───►    370^3   +   333^3   ──and──     444^3   +    37^3

     2000:   1671816384 ───►    944^3   +   940^3   ──and──    1168^3   +   428^3
     2001:   1672470592 ───►   1124^3   +   632^3   ──and──    1187^3   +    29^3
     2002:   1673170856 ───►   1034^3   +   828^3   ──and──    1164^3   +   458^3
     2003:   1675045225 ───►   1081^3   +   744^3   ──and──    1153^3   +   522^3
     2004:   1675958167 ───►   1096^3   +   711^3   ──and──    1159^3   +   492^3
     2005:   1676926719 ───►   1095^3   +   714^3   ──and──    1188^3   +    63^3
     2006:   1677646971 ───►    990^3   +   891^3   ──and──    1188^3   +    99^3

Ruby

<lang ruby>def taxicab_number(nmax=1200)

 [*1..nmax].repeated_combination(2).group_by{|x,y| x**3 + y**3}.select{|k,v| v.size>1}.sort

end

t = [0] + taxicab_number

[*1..25, *2000...2007].each do |i|

 puts "%4d: %10d" % [i, t[i][0]] + t[i][1].map{|a| " = %4d**3 + %4d**3" % a}.join

end</lang>

Output:
   1:       1729 =    1**3 +   12**3 =    9**3 +   10**3
   2:       4104 =    2**3 +   16**3 =    9**3 +   15**3
   3:      13832 =    2**3 +   24**3 =   18**3 +   20**3
   4:      20683 =   10**3 +   27**3 =   19**3 +   24**3
   5:      32832 =    4**3 +   32**3 =   18**3 +   30**3
   6:      39312 =    2**3 +   34**3 =   15**3 +   33**3
   7:      40033 =    9**3 +   34**3 =   16**3 +   33**3
   8:      46683 =    3**3 +   36**3 =   27**3 +   30**3
   9:      64232 =   17**3 +   39**3 =   26**3 +   36**3
  10:      65728 =   12**3 +   40**3 =   31**3 +   33**3
  11:     110656 =    4**3 +   48**3 =   36**3 +   40**3
  12:     110808 =    6**3 +   48**3 =   27**3 +   45**3
  13:     134379 =   12**3 +   51**3 =   38**3 +   43**3
  14:     149389 =    8**3 +   53**3 =   29**3 +   50**3
  15:     165464 =   20**3 +   54**3 =   38**3 +   48**3
  16:     171288 =   17**3 +   55**3 =   24**3 +   54**3
  17:     195841 =    9**3 +   58**3 =   22**3 +   57**3
  18:     216027 =    3**3 +   60**3 =   22**3 +   59**3
  19:     216125 =    5**3 +   60**3 =   45**3 +   50**3
  20:     262656 =    8**3 +   64**3 =   36**3 +   60**3
  21:     314496 =    4**3 +   68**3 =   30**3 +   66**3
  22:     320264 =   18**3 +   68**3 =   32**3 +   66**3
  23:     327763 =   30**3 +   67**3 =   51**3 +   58**3
  24:     373464 =    6**3 +   72**3 =   54**3 +   60**3
  25:     402597 =   42**3 +   69**3 =   56**3 +   61**3
2000: 1671816384 =  428**3 + 1168**3 =  940**3 +  944**3
2001: 1672470592 =   29**3 + 1187**3 =  632**3 + 1124**3
2002: 1673170856 =  458**3 + 1164**3 =  828**3 + 1034**3
2003: 1675045225 =  522**3 + 1153**3 =  744**3 + 1081**3
2004: 1675958167 =  492**3 + 1159**3 =  711**3 + 1096**3
2005: 1676926719 =   63**3 + 1188**3 =  714**3 + 1095**3
2006: 1677646971 =   99**3 + 1188**3 =  891**3 +  990**3

Scala

<lang scala>import scala.math.pow

implicit class Pairs[A, B]( p:List[(A, B)]) {

 def collectPairs: Map[A, List[B]] = p.groupBy(_._1).mapValues(_.map(_._2)).filterNot(_._2.size<2)

}

// Make a sorted List of Taxi Cab Numbers. Limit it to the cube of 1200 because we know it's high enough. val taxiNums = {

 (1 to 1200).toList            // Start with a sequential list of integers
   .combinations(2).toList     // Find all two number combinations
   .map {
     case a :: b :: nil => ((pow(a, 3) + pow(b, 3)).toInt, (a, b))
     case _ => 0 ->(0, 0)
   }                           // Turn the list into the sum of two cubes and
                               //      remember what we started with, eg. 28->(1,3)
   .collectPairs               // Only keep taxi cab numbers with a duplicate
   .toList.sortBy(_._1)        // Sort the results

}

def output() : Unit = {

 println( "%20s".format( "Taxi Cab Numbers" ) )
 println( "%20s%15s%15s".format( "-"*20, "-"*15, "-"*15 ) )
 taxiNums.take(25) foreach {
   case (p, a::b::Nil) => println( "%20d\t(%d\u00b3 + %d\u00b3)\t\t(%d\u00b3 + %d\u00b3)".format(p,a._1,a._2,b._1,b._2) )
 }
 taxiNums.slice(1999,2007) foreach {
   case (p, a::b::Nil) => println( "%20d\t(%d\u00b3 + %d\u00b3)\t(%d\u00b3 + %d\u00b3)".format(p,a._1,a._2,b._1,b._2) )
 }

} </lang>

Output:
    Taxi Cab Numbers
--------------------------------------------------
                1729	(1³ + 12³)	(9³ + 10³)
                4104	(2³ + 16³)	(9³ + 15³)
               13832	(2³ + 24³)	(18³ + 20³)
               20683	(10³ + 27³)	(19³ + 24³)
               32832	(4³ + 32³)	(18³ + 30³)
               39312	(2³ + 34³)	(15³ + 33³)
               40033	(9³ + 34³)	(16³ + 33³)
               46683	(3³ + 36³)	(27³ + 30³)
               64232	(17³ + 39³)	(26³ + 36³)
               65728	(12³ + 40³)	(31³ + 33³)
              110656	(4³ + 48³)	(36³ + 40³)
              110808	(6³ + 48³)	(27³ + 45³)
              134379	(12³ + 51³)	(38³ + 43³)
              149389	(8³ + 53³)	(29³ + 50³)
              165464	(20³ + 54³)	(38³ + 48³)
              171288	(17³ + 55³)	(24³ + 54³)
              195841	(9³ + 58³)	(22³ + 57³)
              216027	(3³ + 60³)	(22³ + 59³)
              216125	(5³ + 60³)	(45³ + 50³)
              262656	(8³ + 64³)	(36³ + 60³)
              314496	(4³ + 68³)	(30³ + 66³)
              320264	(18³ + 68³)	(32³ + 66³)
              327763	(30³ + 67³)	(51³ + 58³)
              373464	(6³ + 72³)	(54³ + 60³)
              402597	(42³ + 69³)	(56³ + 61³)
          1671816384	(428³ + 1168³)	(940³ + 944³)
          1672470592	(29³ + 1187³)	(632³ + 1124³)
          1673170856	(458³ + 1164³)	(828³ + 1034³)
          1675045225	(522³ + 1153³)	(744³ + 1081³)
          1675958167	(492³ + 1159³)	(711³ + 1096³)
          1676926719	(63³ + 1188³)	(714³ + 1095³)
          1677646971	(99³ + 1188³)	(891³ + 990³)

Sidef

Translation of: Perl 6

<lang ruby>var (start=1, end=25) = ARGV.map{.to_i}...;

func display (h, start, end) {

   var i = start
   h.grep { |_,v| v.len > 1 }.keys.sort_by{.to_i}.ft(start-1, end-1).each {|n|
       printf("%4d %10d  =>\t%s\n", i++, n,
           h{n}.map{ "%4d³ + %-s" % (.first, "#{.last}³") }.join(",\t"))
   }

}

var taxi = Hash() var taxis = 0 var terminate = 0

Inf.times { |c1|

   if (terminate>0 && terminate>c1) {
       display(taxi, start, end)
       break
   }
   var c = c1**3
   c1.times { |c2|
       var this = (c2**3 + c)
       taxi{this} := [] << [c2, c1]
       ++taxis if (taxi{this}.len == 2);
       if (taxis == end && terminate.is_zero) {
           terminate = taxi.grep{|_,v| v.len > 1 }.keys.map{.to_i}.max.root(3)
       }
   }

}</lang>

Output:
   1       1729  =>	   9³ + 10³,	   1³ + 12³
   2       4104  =>	   9³ + 15³,	   2³ + 16³
   3      13832  =>	  18³ + 20³,	   2³ + 24³
   4      20683  =>	  19³ + 24³,	  10³ + 27³
   5      32832  =>	  18³ + 30³,	   4³ + 32³
   6      39312  =>	  15³ + 33³,	   2³ + 34³
   7      40033  =>	  16³ + 33³,	   9³ + 34³
   8      46683  =>	  27³ + 30³,	   3³ + 36³
   9      64232  =>	  26³ + 36³,	  17³ + 39³
  10      65728  =>	  31³ + 33³,	  12³ + 40³
  11     110656  =>	  36³ + 40³,	   4³ + 48³
  12     110808  =>	  27³ + 45³,	   6³ + 48³
  13     134379  =>	  38³ + 43³,	  12³ + 51³
  14     149389  =>	  29³ + 50³,	   8³ + 53³
  15     165464  =>	  38³ + 48³,	  20³ + 54³
  16     171288  =>	  24³ + 54³,	  17³ + 55³
  17     195841  =>	  22³ + 57³,	   9³ + 58³
  18     216027  =>	  22³ + 59³,	   3³ + 60³
  19     216125  =>	  45³ + 50³,	   5³ + 60³
  20     262656  =>	  36³ + 60³,	   8³ + 64³
  21     314496  =>	  30³ + 66³,	   4³ + 68³
  22     320264  =>	  32³ + 66³,	  18³ + 68³
  23     327763  =>	  51³ + 58³,	  30³ + 67³
  24     373464  =>	  54³ + 60³,	   6³ + 72³
  25     402597  =>	  56³ + 61³,	  42³ + 69³

With passed parameters 2000 and 2006:

2000 1671816384  =>	 940³ + 944³,	 428³ + 1168³
2001 1672470592  =>	 632³ + 1124³,	  29³ + 1187³
2002 1673170856  =>	 828³ + 1034³,	 458³ + 1164³
2003 1675045225  =>	 744³ + 1081³,	 522³ + 1153³
2004 1675958167  =>	 711³ + 1096³,	 492³ + 1159³
2005 1676926719  =>	 714³ + 1095³,	  63³ + 1188³
2006 1677646971  =>	 891³ + 990³,	  99³ + 1188³

Tcl

Works with: Tcl version 8.6
Translation of: Python

<lang tcl>package require Tcl 8.6

proc heappush {heapName item} {

   upvar 1 $heapName heap
   set idx [lsearch -bisect -index 0 -integer $heap [lindex $item 0]]
   set heap [linsert $heap [expr {$idx + 1}] $item]

} coroutine cubesum apply {{} {

   yield
   set h {}
   set n 1
   while true {

while {![llength $h] || [lindex $h 0 0] > $n**3} { heappush h [list [expr {$n**3 + 1}] $n 1] incr n } set h [lassign $h item] yield $item lassign $item s x y if {[incr y] < $x} { heappush h [list [expr {$x**3 + $y**3}] $x $y] }

   }

}} coroutine taxis apply {{} {

   yield
   set out Template:0 0 0
   while true {

set s [cubesum] if {[lindex $s 0] == [lindex $out end 0]} { lappend out $s } else { if {[llength $out] > 1} {yield $out} set out [list $s] }

   }

}}

  1. Put a cache in front for convenience

variable taxis {} proc taxi {n} {

   variable taxis
   while {$n > [llength $taxis]} {lappend taxis [taxis]}
   return [lindex $taxis [expr {$n-1}]]

}

set 3 "\u00b3" for {set n 1} {$n <= 25} {incr n} {

   puts ${n}:[join [lmap t [taxi $n] {format " %d = %d$3 + %d$3" {*}$t}] ","]

} for {set n 2000} {$n <= 2006} {incr n} {

   puts ${n}:[join [lmap t [taxi $n] {format " %d = %d$3 + %d$3" {*}$t}] ","]

}</lang>

Output:
1: 1729 = 10³ + 9³, 1729 = 12³ + 1³
2: 4104 = 15³ + 9³, 4104 = 16³ + 2³
3: 13832 = 20³ + 18³, 13832 = 24³ + 2³
4: 20683 = 24³ + 19³, 20683 = 27³ + 10³
5: 32832 = 30³ + 18³, 32832 = 32³ + 4³
6: 39312 = 33³ + 15³, 39312 = 34³ + 2³
7: 40033 = 33³ + 16³, 40033 = 34³ + 9³
8: 46683 = 30³ + 27³, 46683 = 36³ + 3³
9: 64232 = 36³ + 26³, 64232 = 39³ + 17³
10: 65728 = 33³ + 31³, 65728 = 40³ + 12³
11: 110656 = 40³ + 36³, 110656 = 48³ + 4³
12: 110808 = 45³ + 27³, 110808 = 48³ + 6³
13: 134379 = 43³ + 38³, 134379 = 51³ + 12³
14: 149389 = 50³ + 29³, 149389 = 53³ + 8³
15: 165464 = 48³ + 38³, 165464 = 54³ + 20³
16: 171288 = 54³ + 24³, 171288 = 55³ + 17³
17: 195841 = 57³ + 22³, 195841 = 58³ + 9³
18: 216027 = 59³ + 22³, 216027 = 60³ + 3³
19: 216125 = 50³ + 45³, 216125 = 60³ + 5³
20: 262656 = 60³ + 36³, 262656 = 64³ + 8³
21: 314496 = 66³ + 30³, 314496 = 68³ + 4³
22: 320264 = 66³ + 32³, 320264 = 68³ + 18³
23: 327763 = 58³ + 51³, 327763 = 67³ + 30³
24: 373464 = 60³ + 54³, 373464 = 72³ + 6³
25: 402597 = 61³ + 56³, 402597 = 69³ + 42³
2000: 1671816384 = 944³ + 940³, 1671816384 = 1168³ + 428³
2001: 1672470592 = 1124³ + 632³, 1672470592 = 1187³ + 29³
2002: 1673170856 = 1034³ + 828³, 1673170856 = 1164³ + 458³
2003: 1675045225 = 1081³ + 744³, 1675045225 = 1153³ + 522³
2004: 1675958167 = 1096³ + 711³, 1675958167 = 1159³ + 492³
2005: 1676926719 = 1095³ + 714³, 1676926719 = 1188³ + 63³
2006: 1677646971 = 990³ + 891³, 1677646971 = 1188³ + 99³

zkl

Translation of: D

An array of bytes is used to hold n, where array[n³+m³]==n. <lang zkl>fcn taxiCabNumbers{

  const HeapSZ=0d5_000_000;
  iCubes:=[1..120].apply("pow",3);
  sum2cubes:=Data(HeapSZ).fill(0);  // BFheap of 1 byte zeros
  taxiNums:=List();
  foreach i,i3 in ([1..].zip(iCubes)){
     foreach j,j3 in ([i+1..].zip(iCubes[i,*])){
        ij3:=i3+j3; 

if(z:=sum2cubes[ij3]){ taxiNums.append(T(ij3, z,(ij3-z.pow(3)).toFloat().pow(1.0/3).round().toInt(), i,j)); } else sum2cubes[ij3]=i;

     }
  }
  taxiNums.sort(fcn([(a,_)],[(b,_)]){ a<b })

}</lang> <lang zkl>fcn print(n,taxiNums){

  [n..].zip(taxiNums).pump(Console.println,fcn([(n,t)]){ 
     "%4d: %10,d = %2d\u00b3 + %2d\u00b3 =  %2d\u00b3 + %2d\u00b3".fmt(n,t.xplode())
  })

} taxiNums:=taxiCabNumbers(); // 63 pairs taxiNums[0,25]:print(1,_);</lang>

Output:
   1:      1,729 =  1³ + 12³ =   9³ + 10³
   2:      4,104 =  2³ + 16³ =   9³ + 15³
   3:     13,832 =  2³ + 24³ =  18³ + 20³
   4:     20,683 = 10³ + 27³ =  19³ + 24³
   5:     32,832 =  4³ + 32³ =  18³ + 30³
   6:     39,312 =  2³ + 34³ =  15³ + 33³
   7:     40,033 =  9³ + 34³ =  16³ + 33³
   8:     46,683 =  3³ + 36³ =  27³ + 30³
   9:     64,232 = 17³ + 39³ =  26³ + 36³
  10:     65,728 = 12³ + 40³ =  31³ + 33³
  11:    110,656 =  4³ + 48³ =  36³ + 40³
  12:    110,808 =  6³ + 48³ =  27³ + 45³
  13:    134,379 = 12³ + 51³ =  38³ + 43³
  14:    149,389 =  8³ + 53³ =  29³ + 50³
  15:    165,464 = 20³ + 54³ =  38³ + 48³
  16:    171,288 = 17³ + 55³ =  24³ + 54³
  17:    195,841 =  9³ + 58³ =  22³ + 57³
  18:    216,027 =  3³ + 60³ =  22³ + 59³
  19:    216,125 =  5³ + 60³ =  45³ + 50³
  20:    262,656 =  8³ + 64³ =  36³ + 60³
  21:    314,496 =  4³ + 68³ =  30³ + 66³
  22:    320,264 = 18³ + 68³ =  32³ + 66³
  23:    327,763 = 30³ + 67³ =  51³ + 58³
  24:    373,464 =  6³ + 72³ =  54³ + 60³
  25:    402,597 = 42³ + 69³ =  56³ + 61³
Translation of: Python

Using a binary heap: <lang zkl>fcn cubeSum{

  heap,n:=Heap(fcn([(a,_)],[(b,_)]){ a<=b }), 1;  // heap cnt maxes out @ 244
  while(1){
     while(heap.empty or heap.top[0]>n.pow(3)){ # could also pre-calculate cubes

heap.push(T(n.pow(3) + 1, n,1)); n+=1;

     }
     s,x,y:= sxy:=heap.pop();
     vm.yield(sxy);
     y+=1;
     if(y<x)    # should be y <= x?

heap.push(T(x.pow(3) + y.pow(3), x,y));

  }

} fcn taxis{

  out:=List(T(0,0,0));
  foreach s in (Utils.Generator(cubeSum)){
     if(s[0]==out[-1][0]) out.append(s);
     else{

if(out.len()>1) vm.yield(out); out.clear(s)

     }
  }

} n:=0; foreach x in (Utils.Generator(taxis)){

  n += 1;
  if(n >= 2006) break;
  if(n <= 25 or n >= 2000) println(n,": ",x);

}</lang> And a quickie heap implementation: <lang zkl>class Heap{ // binary heap

  fcn init(lteqFcn='<=){
     var [const, private] heap=List().pad(64,Void); // a power of 2
     var cnt=0, cmp=lteqFcn;
  }
  fcn push(v){

// Resize the heap if it is too small to hold another item

     if (cnt==heap.len()) heap.pad(cnt*2,Void);
     index:=cnt; cnt+=1; while(index){	 // Find out where to put the element

parent:=(index - 1)/2; if(cmp(heap[parent],v)) break; heap[index] = heap[parent]; index = parent;

     }
     heap[index] = v;
  }
  fcn pop{  // Remove the biggest element and return it
     if(not cnt) return(Void);
     v,temp:=heap[0], heap[cnt-=1];
     // Reorder the elements
     index:=0; while(1){   // Find the child to swap with

swap:=index*2 + 1; if (swap>=cnt) break; // If there are no children, the heap is reordered other:=swap + 1; if(other<cnt and cmp(heap[other],heap[swap])) swap = other; if(cmp(temp,heap[swap])) break; // If the bigger child is less than or equal to its parent, the heap is reordered

heap[index]=heap[swap]; index = swap;

     }
     heap[index] = temp;
     v
  }
  var [proxy] top=fcn  { if(cnt==0) Void else heap[0] };
  var [proxy] empty=fcn{ (not cnt) };

}</lang>

Output:
1: L(L(1729,10,9),L(1729,12,1))
...
23: L(L(327763,67,30),L(327763,58,51))
24: L(L(373464,60,54),L(373464,72,6))
25: L(L(402597,61,56),L(402597,69,42))
2000: L(L(1671816384,944,940),L(1671816384,1168,428))
2001: L(L(1672470592,1124,632),L(1672470592,1187,29))
2002: L(L(1673170856,1034,828),L(1673170856,1164,458))
2003: L(L(1675045225,1153,522),L(1675045225,1081,744))
2004: L(L(1675958167,1096,711),L(1675958167,1159,492))
2005: L(L(1676926719,1188,63),L(1676926719,1095,714))