Numbers in base 10 that are palindromic in bases 2, 4, and 16: Difference between revisions

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

=={{header|Cowgol}}==
<lang cowgol>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();</lang>
{{out}}
<pre>0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365
3855 4095 4097 4369 12291 13107 20485 21845</pre>


=={{header|Factor}}==
=={{header|Factor}}==

Revision as of 22:30, 24 June 2021

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



ALGOL 68

<lang algol68>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 primes 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</lang>

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

BASIC

<lang basic>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</lang>

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

<lang bcpl>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)</lang>
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

<lang c>#include <stdio.h>

  1. 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;

}</lang>

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

<lang cowgol>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();</lang>

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

Works with: Factor version 0.99 2021-06-02

<lang factor>USING: io kernel math.parser prettyprint sequences ;

25,000 <iota> [ 

   { 2 4 16 } [ >base ] with map [ dup reverse = ] all?

] filter [ pprint bl ] each nl</lang>

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

Library: Go-rcu

<lang go>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.")

}</lang>

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.

Phix

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

<lang plm>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</lang>

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

<lang python>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()</lang>

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

<lang perl6>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 } }</lang>

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

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. <lang rexx>/*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</lang>
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

<lang ring> 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

</lang>

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...

Wren

Library: Wren-fmt
Library: Wren-seq

<lang ecmascript>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.")</lang>

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

<lang XPL0>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. "); ]</lang>

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.
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