Distinct power numbers

Distinct power 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.

Task

Let given all integer combinations of $a^{b}$ for 2 <= a <= 5 and 2 <= b <= 5:

Place them in numerical order, with any repeats removed.

We get the following sequence of 15 distinct terms:
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125

REXX

<lang rexx>/*REXX pgm finds and displays distinct power integers: a^b, where a and b are 2≤both≤5*/ parse arg n lo hi cols . /*obtain optional argument from the CL.*/ if n== | n=="," then n= 15 /*Not specified? Then use the default.*/ if lo== | lo=="," then lo= 2 /* " " " " " " */ if hi== | hi=="," then hi= 5 /* " " " " " " */ if cols== | cols=="," then cols= 10 /* " " " " " " */ w= 10 /*width of a number in any column. */ title= ' distinct power integers, a^b, where a and b are: ' lo "≤ both ≤" hi say ' index │'center(title, 1 + cols*(w+1) ) say '───────┼'center("" , 1 + cols*(w+1), '─') @.= .; mx= 0 /*the default value for the @. array.*/

        do   a=lo  to  hi                       /*traipse through  A  values (LO──►HI).*/
          do b=lo  to  hi                       /*   "       "     B     "     "    "  */
          x= a**b;  if @.x\==.  then iterate    /*Has it been found before?  Then skip.*/
          @.x= x;      mx= max(mx, x)           /*assign the power product; fix the max*/
          end   /*b*/
        end     /*a*/

found= 0; idx= 1 /*initialize # distinct power integers.*/ $= /*a list of distinct power integers. */

    do j=0  for mx+1  until found==n            /*search number range or until N found.*/
    if @.j==. then iterate                      /*this number "found" in 1st  DO  loop?*/
    found= found + 1;           c= commas(j)    /*maybe add commas to the number.      */
    $= $ right(c, max(w, length(c) ) )          /*add a distinct power number ──► list.*/
    if found//cols\==0           then iterate   /*have we populated a line of output?  */
    say center(idx, 7)'│'  substr($, 2);   $=   /*display what we have so far  (cols). */
    idx= idx + cols                             /*bump the  index  count for the output*/
    end   /*j*/

if $\== then say center(idx, 7)"│" substr($, 2) /*possible display residual output.*/ say '───────┴'center("" , 1 + cols*(w+1), '─') say say 'Found ' commas(found) title exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ?</lang>

output   when using the default inputs:

 index │                       distinct power integers, a^b, where  a  and  b  are:  2 ≤ both ≤ 5
───────┼───────────────────────────────────────────────────────────────────────────────────────────────────────────────
   1   │          4          8          9         16         25         27         32         64         81        125
  11   │        243        256        625      1,024      3,125
───────┴───────────────────────────────────────────────────────────────────────────────────────────────────────────────

Found  15  distinct power integers, a^b, where  a  and  b  are:  2 ≤ both ≤ 5

Ring

<lang ring> load "stdlib.ring"

see "working..." + nl see "Distinct powers are:" + nl row = 0 distPow = []

for n = 2 to 5

   for m = 2 to 5
       sum = pow(n,m)
       add(distPow,sum)
   next

next

distPow = sort(distPow)

for n = len(distPow) to 2 step -1

   if distPow[n] = distPow[n-1]
      del(distPow,n-1)
   ok

next

for n = 1 to len(distPow)

   row++
   see "" + distPow[n] + " "
   if row%5 = 0
      see nl
   ok

next

see "Found " + row + " numbers" + nl see "done..." + nl </lang>

Output:

working...
Distinct powers are:
4 8 9 16 25 
27 32 64 81 125 
243 256 625 1024 3125 
Found 15 numbers
done...

Wren

Library: Wren-seq

Library: Wren-fmt

<lang ecmascript>import "/seq" for Lst import "/fmt" for Fmt

var pows = [] for (a in 2..5) {

   var pow = a
   for (b in 2..5) {
       pow = pow * a
       pows.add(pow)
   }

} pows = Lst.distinct(pows).sort() System.print("Ordered distinct values of a ^ b for a in [2..5] and b in [2..5]:") for (chunk in Lst.chunks(pows, 5)) Fmt.print("$,5d", chunk) System.print("\nFound %(pows.count) such numbers.")</lang>

Output:

Ordered distinct values of a ^ b for a in [2..5] and b in [2..5]:
    4     8     9    16    25
   27    32    64    81   125
  243   256   625 1,024 3,125

Found 15 such numbers.