Numbers in base 10 that are palindromic in bases 2, 4, and 16

From Rosetta Code
Numbers in base 10 that are palindromic in bases 2, 4, and 16 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
Find numbers in base 10 that are palindromic in bases 2, 4, and 16, where n < 25,000



11l[edit]

Translation of: Python
F reverse(=n, base)
   V r = 0
   L n > 0
      r = r * base + n % base
      n I/= base
   R r

F palindrome(n, base)
   R n == reverse(n, base)

V cnt = 0
L(i) 25000
   I all((2, 4, 16).map(base -> palindrome(@i, base)))
      cnt++
      print(‘#5’.format(i), end' " \n"[cnt % 12 == 0])

print()
Output:
    0     1     3     5    15    17    51    85   255   257   273   771
  819  1285  1365  3855  4095  4097  4369 12291 13107 20485 21845 

Action![edit]

BYTE FUNC IsPalindrome(INT x BYTE base)
  CHAR ARRAY digits="0123456789abcdef",s(16)
  BYTE d,i,len

  len=0
  DO
    d=x MOD base
    len==+1
    s(len)=digits(d+1)
    x==/base
  UNTIL x=0
  OD
  s(0)=len

  FOR i=1 TO len/2
  DO
    IF s(i)#s(len-i+1) THEN
      RETURN (0)
    FI
  OD
RETURN (1)

PROC Main()
  INT i

  FOR i=0 TO 24999
  DO
    IF IsPalindrome(i,16)=1 AND IsPalindrome(i,4)=1 AND IsPalindrome(i,2)=1 THEN
      PrintI(i) Put(32)
    FI
  OD
RETURN
Output:

Screenshot from Atari 8-bit computer

0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845

ALGOL 68[edit]

BEGIN # show numbers in decimal that are palindromic in bases 2, 4 and 16 #
    INT max number = 25 000;    # maximum number to consider #
    INT min base   = 2;         # smallest base needed #
    INT max digits = BEGIN      # number of digits max number has in the smallest base #
                         INT d := 1;
                         INT v := max number;
                         WHILE v >= min base DO
                             v OVERAB min base;
                             d PLUSAB 1
                         OD;
                         d
                     END;
    # returns the digits of n in the specified base #
    PRIO DIGITS = 9;
    OP   DIGITS = ( INT n, INT base )[]INT:
         IF   INT v := ABS n;
              v < base
         THEN v # single dogit #
         ELSE   # multiple digits #
            [ 1 : max digits ]INT result;
            INT d pos := UPB result + 1;
            INT v     := ABS n;
            WHILE v > 0 DO
                result[ d pos -:= 1 ] := v MOD base;
                v OVERAB base
            OD;
            result[ d pos : UPB result ]
         FI # DIGITS # ;
    # returns TRUE if the digits in d form a palindrome, FALSE otherwise #
    OP   PALINDROMIC = ( []INT d )BOOL:
         BEGIN
             INT  left := LWB d, right := UPB d;
             BOOL is palindromic := TRUE;
             WHILE left < right AND is palindromic DO
                 is palindromic := d[ left ] = d[ right ];
                 left          +:= 1;
                 right         -:= 1
             OD;
             is palindromic
         END;
    # print the numbers in decimal that are palendromic in bases 2, 4 and 16 #
    FOR n FROM 0 TO max number DO
        IF PALINDROMIC ( n DIGITS 16 ) THEN
            IF PALINDROMIC ( n DIGITS 4 ) THEN
                IF PALINDROMIC ( n DIGITS 2 ) THEN
                    print( ( " ", whole( n, 0 ) ) )
                FI
            FI
        FI
    OD
END
Output:
 0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845

APL[edit]

Works with: Dyalog APL
((/⍨)(2 4 16.((⊢≡⌽)(¯1)))¨)0,⍳24999
Output:
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845

AWK[edit]

# syntax: GAWK -f NUMBERS_IN_BASE_10_THAT_ARE_PALINDROMIC_IN_BASES_2_4_AND_16.AWK
# converted from C
BEGIN {
    start = 0
    stop = 24999
    for (i=start; i<stop; i++) {
      if (palindrome(i,2) && palindrome(i,4) && palindrome(i,16)) {
         printf("%5d%1s",i,++count%10?"":"\n")
      }
    }
    printf("\nBase 10 numbers that are palindromes in bases 2, 4, and 16: %d-%d: %d\n",start,stop,count)
    exit(0)
}
function palindrome(n,base) {
    return n == reverse(n,base)
}
function reverse(n,base,  r) {
    for (r=0; n; n=int(n/base)) {
      r = int(r*base) + n%base
    }
    return(r)
}
Output:
    0     1     3     5    15    17    51    85   255   257
  273   771   819  1285  1365  3855  4095  4097  4369 12291
13107 20485 21845
Base 10 numbers that are palindromes in bases 2, 4, and 16: 0-24999: 23

BASIC[edit]

10 DEFINT A-Z: DEFDBL R
20 FOR I=1 TO 25000
30 B=2: GOSUB 100: IF R<>I GOTO 70
40 B=4: GOSUB 100: IF R<>I GOTO 70
50 B=16: GOSUB 100: IF R<>I GOTO 70
60 PRINT I,
70 NEXT
80 END
100 R=0: N=I
110 IF N=0 THEN RETURN
120 R=R*B+N MOD B
130 N=N\B
140 GOTO 110
Output:
 0             1             3             5             15
 17            51            85            255           257
 273           771           819           1285          1365
 3855          4095          4097          4369          12291
 13107         20485         21845

BCPL[edit]

get "libhdr"
manifest $( MAXIMUM = 25000 $)

let reverse(n, base) = valof
$(  let r = 0
    while n > 0
    $(  r := r*base + n rem base
        n := n / base
    $)
    resultis r
$)

let palindrome(n, base) = n = reverse(n, base)

let start() be 
    for i = 0 to MAXIMUM
        if palindrome(i,2) & palindrome(i,4) & palindrome(i,16)
            do writef("%N*N", i)
Output:
0
1
3
5
15
17
51
85
255
257
273
771
819
1285
1365
3855
4095
4097
4369
12291
13107
20485
21845

C[edit]

#include <stdio.h>
#define MAXIMUM 25000

int reverse(int n, int base) {
    int r;
    for (r = 0; n; n /= base)
        r = r*base + n%base;
    return r;
}

int palindrome(int n, int base) {
    return n == reverse(n, base);
}

int main() {
    int i, c = 0;
    
    for (i = 0; i < MAXIMUM; i++) {
        if (palindrome(i, 2) &&
            palindrome(i, 4) &&
            palindrome(i, 16)) {
            printf("%5d%c", i, ++c % 12 ? ' ' : '\n');
        }
    }
    printf("\n");
    return 0;
}
Output:
    0     1     3     5    15    17    51    85   255   257   273   771
  819  1285  1365  3855  4095  4097  4369 12291 13107 20485 21845

COBOL[edit]

       IDENTIFICATION DIVISION.
       PROGRAM-ID. PALINDROMIC-BASE-2-4-16.
       
       DATA DIVISION.
       WORKING-STORAGE SECTION.
       01 VARIABLES.
          02 CUR-NUM           PIC 9(5).
          02 REV-BASE          PIC 99.
          02 REV-REST          PIC 9(5).
          02 REV-NEXT          PIC 9(5).
          02 REV-DGT           PIC 99.
          02 REVERSED          PIC 9(5).
       
       01 OUTPUT-FORMAT.
          02 OUT-NUM           PIC Z(4)9.
       
       PROCEDURE DIVISION.
       BEGIN.
           PERFORM 2-4-16-PALINDROME
               VARYING CUR-NUM FROM ZERO BY 1
               UNTIL CUR-NUM IS NOT LESS THAN 25000.
           STOP RUN.
       
       2-4-16-PALINDROME.
           MOVE 16 TO REV-BASE, PERFORM REVERSE THRU REV-LOOP
           IF CUR-NUM IS EQUAL TO REVERSED
               MOVE 4 TO REV-BASE, PERFORM REVERSE THRU REV-LOOP
               IF CUR-NUM IS EQUAL TO REVERSED
                   MOVE 2 TO REV-BASE, PERFORM REVERSE THRU REV-LOOP
                   IF CUR-NUM IS EQUAL TO REVERSED
                       MOVE CUR-NUM TO OUT-NUM
                       DISPLAY OUT-NUM.
       
       REVERSE.
           MOVE ZERO TO REVERSED.
           MOVE CUR-NUM TO REV-REST.
       REV-LOOP.
           IF REV-REST IS GREATER THAN ZERO
               DIVIDE REV-BASE INTO REV-REST GIVING REV-NEXT
               COMPUTE REV-DGT = REV-REST - REV-NEXT * REV-BASE
               MULTIPLY REV-BASE BY REVERSED
               ADD REV-DGT TO REVERSED
               MOVE REV-NEXT TO REV-REST
               GO TO REV-LOOP.
Output:
    0
    1
    3
    5
   15
   17
   51
   85
  255
  257
  273
  771
  819
 1285
 1365
 3855
 4095
 4097
 4369
12291
13107
20485
21845

Cowgol[edit]

include "cowgol.coh";
const MAXIMUM := 25000;

sub reverse(n: uint16, base: uint16): (r: uint16) is
    r := 0;
    while n != 0 loop
        r := r * base + n % base;
        n := n / base;
    end loop;
end sub;

var i: uint16 := 0;
var c: uint8 := 0;
while i < MAXIMUM loop
    if reverse(i,2) == i
    and reverse(i,4) == i
    and reverse(i,16) == i 
    then
        c := c + 1;
        print_i16(i);
        if c == 15 then
            print_nl();
            c := 0;
        else
            print_char(' ');
        end if;
    end if;
    i := i + 1;
end loop;
print_nl();
Output:
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365
3855 4095 4097 4369 12291 13107 20485 21845

F#[edit]

// Palindromic numbers in bases 2,4, and 16. Nigel Galloway: June 25th., 2021
let fG n g=let rec fG n g=[yield n%g; if n>=g then yield! fG(n/g) g] in let n=fG n g in n=List.rev n
Seq.initInfinite id|>Seq.takeWhile((>)25000)|>Seq.filter(fun g->fG g 16 && fG g 4 && fG g 2)|>Seq.iter(printf "%d "); printfn ""
Output:
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845

Factor[edit]

Works with: Factor version 0.99 2021-06-02
USING: io kernel math.parser prettyprint sequences ;

25,000 <iota> [
    { 2 4 16 } [ >base ] with map [ dup reverse = ] all?
] filter [ pprint bl ] each nl
Output:
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845 

FreeBASIC[edit]

function ispal( byval n as integer, b as integer ) as boolean
    'determines if n is palindromic in base b
    dim as string ns
    while n
        ns += chr(48+n mod b)       'temporarily represent as a string
        n\=b
    wend
    for i as integer = 1 to len(ns)\2
        if mid(ns,i,1)<>mid(ns,len(ns)-i+1,1) then return false
    next i
    return true
end function

for i as integer = 0 to 25000
    if ispal(i,16) andalso ispal(i,4) andalso ispal(i,2) then print i;"  ";
next i
Output:
0   1   3   5   15   17   51   85   255   257   273   771   819   1285   1365   3855   4095   4097   4369   12291   13107   20485   21845

Go[edit]

Library: Go-rcu
package main

import (
    "fmt"
    "rcu"
    "strconv"
)

func reverse(s string) string {
    chars := []rune(s)
    for i, j := 0, len(chars)-1; i < j; i, j = i+1, j-1 {
        chars[i], chars[j] = chars[j], chars[i]
    }
    return string(chars)
}

func main() {
    fmt.Println("Numbers under 25,000 in base 10 which are palindromic in bases 2, 4 and 16:")
    var numbers []int
    for i := int64(0); i < 25000; i++ {
        b2 := strconv.FormatInt(i, 2)
        if b2 == reverse(b2) {
            b4 := strconv.FormatInt(i, 4)
            if b4 == reverse(b4) {
                b16 := strconv.FormatInt(i, 16)
                if b16 == reverse(b16) {
                    numbers = append(numbers, int(i))
                }
            }
        }
    }
    for i, n := range numbers {
        fmt.Printf("%6s ", rcu.Commatize(n))
        if (i+1)%10 == 0 {
            fmt.Println()
        }
    }
    fmt.Println("\n\nFound", len(numbers), "such numbers.")
}
Output:
Numbers under 25,000 in base 10 which are palindromic in bases 2, 4 and 16:
     0      1      3      5     15     17     51     85    255    257 
   273    771    819  1,285  1,365  3,855  4,095  4,097  4,369 12,291 
13,107 20,485 21,845 

Found 23 such numbers.

J[edit]

   palinbase=: (-: |.)@(#.inv)"0
   I. (2&palinbase * 4&palinbase * 16&palinbase) i.25e3
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845

Julia[edit]

palinbases(n, bases = [2, 4, 16]) = all(b -> (d = digits(n, base = b); d == reverse(d)), bases)

foreach(p -> print(rpad(p[2], 7), p[1] % 11 == 0 ? "\n" : ""), enumerate(filter(palinbases, 1:25000)))
Output:
1      3      5      15     17     51     85     255    257    273    771    
819    1285   1365   3855   4095   4097   4369   12291  13107  20485  21845

Mathematica/Wolfram Language[edit]

ClearAll[PalindromeBaseQ, Palindrom2416Q]
PalindromeBaseQ[n_Integer, b_Integer] := PalindromeQ[IntegerDigits[n, b]]
Palindrom2416Q[n_Integer] := PalindromeBaseQ[n, 2] \[And] PalindromeBaseQ[n, 4] \[And] PalindromeBaseQ[n, 16]
Select[Range[0, 24999], Palindrom2416Q]
Length[%]
Output:
{0, 1, 3, 5, 15, 17, 51, 85, 255, 257, 273, 771, 819, 1285, 1365, 3855, 4095, 4097, 4369, 12291, 13107, 20485, 21845}
23

Nim[edit]

import strutils, sugar

type Digit = 0..15

func toBase(n: Natural; b: Positive): seq[Digit] =
  if n == 0: return @[Digit 0]
  var n = n
  while n != 0:
    result.add n mod b
    n = n div b

func isPalindromic(s: seq[Digit]): bool =
  for i in 1..(s.len div 2):
    if s[i-1] != s[^i]: return false
  result = true

let list = collect(newSeq):
             for n in 0..<25_000:
               if n.toBase(2).isPalindromic and
                  n.toBase(4).isPalindromic and
                  n.toBase(16).isPalindromic: n

echo "Found ", list.len, " numbers which are palindromic in bases 2, 4 and 16:"
echo list.join(" ")
Output:
Found 23 numbers which are palindromic in bases 2, 4 and 16:
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845

Perl[edit]

Library: ntheory
use strict;
use warnings;
use ntheory 'todigitstring';

sub pb { my $s = todigitstring(shift,shift); return $s eq join '', reverse split '', $s }

pb($_,2) and pb($_,4) and pb($_,16) and print "$_ " for 1..25000;
Output:
1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845

Phix[edit]

with javascript_semantics
function palindrome(string s) return s=reverse(s) end function
function p2416(integer n)
    return palindrome(sprintf("%a",{{2,n}}))
       and palindrome(sprintf("%a",{{4,n}}))
       and palindrome(sprintf("%a",{{16,n}}))
end function
sequence res = apply(filter(tagset(25000,0),p2416),sprint)
printf(1,"%d found: %s\n",{length(res),join(res)})
Output:
23 found: 0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845

PL/M[edit]

100H:
/* CP/M CALLS */
BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS;
EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT;
PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9,S); END PRINT;

DECLARE MAXIMUM LITERALLY '25$000';

/* PRINT A NUMBER */
PRINT$NUMBER: PROCEDURE (N);
    DECLARE S (7) BYTE INITIAL ('..... $');
    DECLARE (N, P) ADDRESS, C BASED P BYTE;
    P = .S(5);
DIGIT:
    P = P - 1;
    C = N MOD 10 + '0';
    N = N / 10;
    IF N > 0 THEN GO TO DIGIT;
    CALL PRINT(P);
END PRINT$NUMBER;

/* REVERSE NUMBER GIVEN BASE */
REVERSE: PROCEDURE (N, B) ADDRESS;
    DECLARE (N, R) ADDRESS, B BYTE;
    R = 0;
    DO WHILE N > 0;
        R = R*B + N MOD B;
        N = N/B;
    END;
    RETURN R;
END REVERSE;

/* CHECK IF NUMBER IS PALINDROME */
PALIN: PROCEDURE (N, B) BYTE;
    DECLARE N ADDRESS, B BYTE;
    RETURN N = REVERSE(N, B);
END PALIN;

DECLARE I ADDRESS, C BYTE;
C = 0;
DO I = 0 TO MAXIMUM;
    IF PALIN(I,2) AND PALIN(I,4) AND PALIN(I,16) THEN DO;
        CALL PRINT$NUMBER(I);
        C = C + 1;
        IF C = 15 THEN DO;
            CALL PRINT(.(13,10,'$'));
            C = 0;
        END;
    END;
END;
CALL EXIT;
EOF
Output:
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365
3855 4095 4097 4369 12291 13107 20485 21845

Python[edit]

def reverse(n, base):
    r = 0
    while n > 0:
        r = r*base + n%base
        n = n//base
    return r
    
def palindrome(n, base):
    return n == reverse(n, base)
    
cnt = 0
for i in range(25000):
    if all(palindrome(i, base) for base in (2,4,16)):
        cnt += 1
        print("{:5}".format(i), end=" \n"[cnt % 12 == 0])

print()
Output:
    0     1     3     5    15    17    51    85   255   257   273   771
  819  1285  1365  3855  4095  4097  4369 12291 13107 20485 21845

Raku[edit]

put "{+$_} such numbers:\n", .batch(10)».fmt('%5d').join("\n") given
(^25000).grep: -> $n { all (2,4,16).map: { $n.base($_) eq $n.base($_).flip } }
Output:
23 such numbers:
    0     1     3     5    15    17    51    85   255   257
  273   771   819  1285  1365  3855  4095  4097  4369 12291
13107 20485 21845

REXX[edit]

Programming note:   the conversions of a decimal number to another base (radix) was ordered such that the fastest
base conversion was performed before the other conversions.

The use of REXX's BIFs to convert decimal numbers to binary and hexadecimal were used   (instead of the   base  
function)   because they are much faster).

This REXX version takes advantage that no   even   integers need be tested   (except for the single exception:   zero),
this makes the execution twice as fast.

/*REXX pgm finds non─neg integers that are palindromes in base 2, 4, and 16, where N<25k*/
numeric digits 100                               /*ensure enough dec. digs for large #'s*/
parse arg n cols .                               /*obtain optional argument from the CL.*/
if    n=='' |    n==","  then   n = 25000        /*Not specified?  Then use the default.*/
if cols=='' | cols==","  then cols=    10        /* "      "         "   "   "     "    */
w= 10                                            /*width of a number in any column.     */
title= ' non-negative integers that are palindromes in base 2, 4, and 16,  where  N  < ' ,
       commas(n)
say ' index │'center(title, 1 + cols*(w+1)     ) /*display the title for the output.    */
say '───────┼'center(""   , 1 + cols*(w+1), '─') /*   "     a   sep   "   "     "       */
$= right(0, w+1)                                 /*list of numbers found  (so far).     */
found= 1                                         /*# of finds (so far), the only even #.*/
idx= 1                                           /*set the IDX  (index) to unity.       */
       do j=1  by 2  to n-1                      /*find int palindromes in bases 2,4,16.*/
          h= d2x(j)                              /*convert dec. # to hexadecimal.       */
       if h\==reverse(h)          then iterate   /*Hex    number not palindromic?  Skip.*/    /* ◄■■■■■■■■ a filter. */
          b= x2b( d2x(j) ) + 0                   /*convert dec. # to hex, then to binary*/
       if b\==reverse(b)          then iterate   /*Binary number not palindromic?  Skip.*/    /* ◄■■■■■■■■ a filter. */
          q= base(j, 4)                          /*convert a decimal integer to base 4. */
       if q\==reverse(q)          then iterate   /*Base 4 number not palindromic?  Skip.*/    /* ◄■■■■■■■■ a filter. */
       found= found + 1                          /*bump number of found such numbers.   */
       $= $  right( commas(j), w)                /*add the found 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), '─')     /*display the foot sep for output. */
say
say 'Found '          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 ?
/*──────────────────────────────────────────────────────────────────────────────────────*/
base:   procedure; parse arg #,t,,y;  @= 0123456789abcdefghijklmnopqrstuvwxyz /*up to 36*/
        @@= substr(@, 2);    do while #>=t;   y= substr(@, #//t + 1, 1)y;         #= # % t
                             end;                       return substr(@, #+1, 1)y
output   when using the default inputs:
 index │             non-negative integers that are palindromes in base 2, 4, and 16,  where  N  <  25,000
───────┼───────────────────────────────────────────────────────────────────────────────────────────────────────────────
   1   │          0          1          3          5         15         17         51         85        255        257
  11   │        273        771        819      1,285      1,365      3,855      4,095      4,097      4,369     12,291
  21   │     13,107     20,485     21,845
───────┴───────────────────────────────────────────────────────────────────────────────────────────────────────────────

Found  23  non-negative integers that are palindromes in base 2, 4, and 16,  where  N  <  25,000

Ring[edit]

load "stdlib.ring"
see "working..." + nl
see "Numbers in base 10 that are palindromic in bases 2, 4, and 16:" + nl

row = 0
limit = 25000

for n = 1 to limit
    base2 = decimaltobase(n,2)
    base4 = decimaltobase(n,4)
    base16 = hex(n)
    bool = ispalindrome(base2) and ispalindrome(base4) and ispalindrome(base16)
    if bool = 1
       see "" + n + " "
       row = row + 1
       if row%5 = 0
          see nl
       ok
    ok
next

see nl + "Found " + row + " numbers" + nl
see "done..." + nl

func decimaltobase(nr,base)
     decList = 0:15
     baseList = ["0","1","2","3","4","5","6","7","8","9","A","B","C","D","E","F"]

     binList = [] 
     binary = 0
     remainder = 1
     while(nr != 0)
          remainder = nr % base
          ind = find(decList,remainder)
          rem = baseList[ind]
          add(binList,rem)
          nr = floor(nr/base) 
     end
     binlist = reverse(binList)
     binList = list2str(binList)
     binList = substr(binList,nl,"")  
     return binList
Output:
working...
Numbers in base 10 that are palindromic in bases 2, 4, and 16:
1 3 5 15 17 
51 85 255 257 273 
771 819 1285 1365 3855 
4095 4097 4369 12291 13107 
20485 21845 
Found 22 numbers
done...

Seed7[edit]

$ include "seed7_05.s7i";

const func boolean: palindrome (in string: input) is
  return input = reverse(input);

const proc: main is func
  local
    var integer: n is 1;
  begin
    write("0 ");
    for n range 1 to 24999 step 2 do
      if palindrome(n radix 2) and palindrome(n radix 4) and palindrome(n radix 16) then
        write(n <& " ");
      end if;
    end for;
  end func;
Output:
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845

Sidef[edit]

say gather {
    for (var k = 0; k < 25_000; k = k.next_palindrome(16)) {
        take(k) if [2, 4].all{|b| k.is_palindrome(b) }
    }
}
Output:
[0, 1, 3, 5, 15, 17, 51, 85, 255, 257, 273, 771, 819, 1285, 1365, 3855, 4095, 4097, 4369, 12291, 13107, 20485, 21845]

Wren[edit]

Library: Wren-fmt
Library: Wren-seq
import "/fmt" for Conv, Fmt
import "/seq" for Lst

System.print("Numbers under 25,000 in base 10 which are palindromic in bases 2, 4 and 16:")
var numbers = []
for (i in 0..24999) {
    var b2 = Conv.itoa(i, 2)
    if (b2 == b2[-1..0]) {
        var b4 = Conv.itoa(i, 4)
        if (b4 == b4[-1..0]) {
            var b16 = Conv.itoa(i, 16)
            if (b16 == b16[-1..0]) numbers.add(i)
        }
    }
}
for (chunk in Lst.chunks(numbers, 8)) Fmt.print("$,6d", chunk)
System.print("\nFound %(numbers.count) such numbers.")
Output:
Numbers under 25,000 in base 10 which are palindromic in bases 2, 4 and 16:
     0      1      3      5     15     17     51     85
   255    257    273    771    819  1,285  1,365  3,855
 4,095  4,097  4,369 12,291 13,107 20,485 21,845

Found 23 such numbers.

XPL0[edit]

func Reverse(N, Base);  \Reverse order of digits in N for given Base
int  N, Base, M;
[M:= 0;
repeat  N:= N/Base;
        M:= M*Base + rem(0);
until   N=0;
return M;
];

int Count, N;
[Count:= 0;
for N:= 1 to 25000-1 do
    if N = Reverse(N, 2) &
       N = Reverse(N, 4) &
       N = Reverse(N, 16) then
        [IntOut(0, N);
        Count:= Count+1;
        if rem(Count/10) = 0 then CrLf(0) else ChOut(0, 9\tab\);
        ];
CrLf(0);
IntOut(0, Count);
Text(0, " such numbers found.
");
]
Output:
1       3       5       15      17      51      85      255     257     273
771     819     1285    1365    3855    4095    4097    4369    12291   13107
20485   21845   
22 such numbers found.