Numbers in base 10 that are palindromic in bases 2, 4, and 16
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
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!
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
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;
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 #
# as noted by the REXX sample, even numbers ( other than 0 ) aren't #
# applicable as even numbers end in 0 in base 2 so can't be palendromic #
print( ( " 0" ) ); # clearly, 0 is palendromic in all bases #
FOR n BY 2 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;
print( ( newline ) )
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
ALGOL W
begin % find numbers palendromic in bases 2, 4, and 16 %
% returns true if n is palendromic in the specified base, false otherwide %
logical procedure palendromic( integer value n, base ) ;
begin
integer array digit( 1 :: 32 );
integer dPos, v, lPos, rPos;
logical isPalendromic;
dPos := 0;
v := n;
while v > 0 do begin
dPos := dPos + 1;
digit( dPos ) := v rem base;
v := v div base
end while_v_gt_0 ;
isPalendromic := true;
lPos := 1;
rPos := dPos;
while rPos > lPos and isPalendromic do begin
isPalendromic := digit( lPos ) = digit( rPos );
lPos := lPos + 1;
rPos := rPos - 1
end while_rPos_gt_lPos_and_isPalendromic ;
isPalendromic
end palendromic ;
% as noted by the REXX sample, all even numbers end in 0 in base 2 %
% so 0 is the only possible even number, note 0 is palendromic in all bases %
write( " 0" );
for n := 1 step 2 until 24999 do begin
if palendromic( n, 16 ) then begin
if palendromic( n, 4 ) then begin
if palendromic( n, 2 ) then begin
writeon( i_w := 1, s_w := 0, " ", n )
end if_palendromic__n_2
end if_palendromic__n_4
end if_palendromic__n_16
end for_n
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
(⊢(/⍨)(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
Arturo
multiPalindromic?: function [n][
if (digits.base:2 n) <> reverse digits.base:2 n -> return false
if (digits.base:4 n) <> reverse digits.base:4 n -> return false
if (digits.base:16 n) <> reverse digits.base:16 n -> return false
return true
]
mpUpTo25K: select 0..25000 => multiPalindromic?
loop split.every: 12 mpUpTo25K 'x ->
print map x 's -> pad to :string s 5
- 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
# 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
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
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
#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
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
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
Delphi
function GetRadixString(L: Integer; Radix: Byte): string;
{Converts integer a string of any radix}
const RadixChars: array[0..35] Of char =
('0', '1', '2', '3', '4', '5', '6', '7',
'8', '9', 'A', 'B', 'C', 'D', 'E', 'F',
'G','H', 'I', 'J', 'K', 'L', 'M', 'N',
'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V',
'W', 'X', 'Y', 'Z');
var I: integer;
var S: string;
var Sign: string[1];
begin
Result:='';
If (L < 0) then
begin
Sign:='-';
L:=Abs(L);
end
else Sign:='';
S:='';
repeat
begin
I:=L mod Radix;
S:=RadixChars[I] + S;
L:=L div Radix;
end
until L = 0;
Result:=Sign + S;
end;
function IsPalindrome(N, Base: integer): boolean;
{Test if number is the same forward or backward}
{For a specific Radix}
var S1,S2: string;
begin
S1:=GetRadixString(N,Base);
S2:=ReverseString(S1);
Result:=S1=S2;
end;
function IsPalindrome2416(N: integer): boolean;
{Is N palindromic for bases 2, 4 and 16}
begin
Result:=IsPalindrome(N,2) and
IsPalindrome(N,4) and
IsPalindrome(N,16);
end;
procedure ShowPalindrome2416(Memo: TMemo);
{Show all numbers Palindromic for bases 2, 4 and 16}
var S: string;
var I,Cnt: integer;
begin
S:='';
Cnt:=0;
for I:=0 to 25000-1 do
if IsPalindrome2416(I) then
begin
Inc(Cnt);
S:=S+Format('%8D',[I]);
If (Cnt mod 5)=0 then S:=S+#$0D#$0A;
end;
Memo.Lines.Add('Count='+IntToStr(Cnt));
Memo.Lines.Add(S);
end;
- Output:
Count=23
0 1 3 5 15
17 51 85 255 257
273 771 819 1285 1365
3855 4095 4097 4369 12291
13107 20485 21845
EasyLang
func rev n base .
while n > 0
r = r * base + n mod base
n = n div base
.
return r
.
for i = 0 to 25000 - 1
if rev i 2 = i and rev i 4 = i and rev i 16 = i
write 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
Euler
begin
new palendromic; new n; label forN;
palendromic
<- ` formal n; formal base;
begin
new v; new lPos; new rPos; new isPalendromic;
new digit;
label vGT0; label rGTl;
digit <- list 64;
rPos <- 0;
v <- n;
vGT0: if v > 0 then begin
rPos <- rPos + 1;
digit[ rPos ] <- v mod base;
v <- v % base;
goto vGT0
end else 0;
isPalendromic <- true;
lPos <- 1;
rGTl: if rPos > lPos and isPalendromic then begin
isPalendromic <- digit[ lPos ] = digit[ rPos ];
lPos <- lPos + 1;
rPos <- rPos - 1;
goto rGTl
end else 0;
isPalendromic
end
'
;
out 0;
n <- -1;
forN: if [ n <- n + 2 ] < 25000 then begin
if not palendromic( n, 16 ) then 0
else if not palendromic( n, 4 ) then 0
else if palendromic( n, 2 ) then out n
else 0
;
goto forN
end else 0
end $
- Output:
NUMBER 0
NUMBER 1
NUMBER 3
NUMBER 5
NUMBER 15
NUMBER 17
NUMBER 51
NUMBER 85
NUMBER 255
NUMBER 257
NUMBER 273
NUMBER 771
NUMBER 819
NUMBER 1285
NUMBER 1365
NUMBER 3855
NUMBER 4095
NUMBER 4097
NUMBER 4369
NUMBER 12291
NUMBER 13107
NUMBER 20485
NUMBER 21845
F#
// 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
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
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
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
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
jq
Also works with gojq and fq, the Go implementations
With minor tweaks, also works with jaq, the Rust implementation
This entry, which uses a stream-oriented approach to illustrate an economical use of memory, uses `tobase` as found in the Wikipedia article on jq; it works for bases up to 36 inclusive.
Use gojq or fq for unbounded-precision integer arithmetic.
def lpad($len): tostring | ($len - length) as $l | (" " * $l)[:$l] + .;
# nwise/2 assumes that null can be taken as the eos marker
def nwise(stream; $n):
foreach (stream, null) as $x ([];
if length == $n then [$x] else . + [$x] end;
if (.[-1] == null) and length>1 then .[:-1]
elif length == $n then .
else empty
end);
def tobase($b):
def digit: "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[.:.+1];
def mod: . % $b;
def div: ((. - mod) / $b);
def digits: recurse( select(. > 0) | div) | mod ;
# For jq it would be wise to protect against `infinite` as input, but using `isinfinite` confuses gojq
select( (tostring|test("^[0-9]+$")) and 2 <= $b and $b <= 36)
| if . == 0 then "0"
else [digits | digit] | reverse[1:] | add
end;
# boolean
def palindrome: explode as $in | ($in|reverse) == $in;
# boolean
def palindrome($b):
tobase($b) | palindrome;
def task($n):
"Numbers under \($n) in base 10 which are palindromic in bases 2, 4 and 16:",
(nwise(range(0;$n) | select(palindrome(2) and palindrome(4) and palindrome(16)); 5)
| map( lpad(6) ) | join(" "));
task(25000)- Output:
Numbers under 25000 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 1285 1365
3855 4095 4097 4369 12291
13107 20485 21845
Julia
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
Lua
do -- find numbers palendromic in bases 2, 4, and 16
local function palendromic( n, base )
local digits, v = "", n
while v > 0 do
local dPos = ( v % base ) + 1
digits = digits..string.sub( "0123456789abcdef", dPos, dPos )
v = math.floor( v / base )
end
return digits == string.reverse( digits )
end
-- as noted by the REXX sample, all even numbers end in 0 in base 2
-- so 0 is the only possible even number, note 0 is palendromic in all bases
io.write( " 0" )
for n = 1, 24999, 2 do
if palendromic( n, 16 ) then
if palendromic( n, 4 ) then
if palendromic( n, 2 ) then
io.write( " ", n )
end
end
end
end
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
Mathematica /Wolfram Language
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
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
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
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
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
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
Quackery
[ temp put
0
[ over 0 > while
temp share tuck *
dip /mod +
again ]
temp release
nip ] is rev ( n n --> n )
[ dip dup rev = ] is pal ( n n --> b )
[]
25000 times
[ i^ 16 pal while
i^ 4 pal while
i^ 2 pal while
i^ join ]
echo- 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
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
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
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...
RPL
Brute force
« ""
OVER SIZE 1 FOR j
OVER j DUP SUB +
NEXT SWAP DROP
» 'REVSTR' STO
« → base
« ""
WHILE OVER REPEAT
SWAP base MOD LASTARG / IP
"0123456789ABCDEF" ROT 1 + DUP SUB ROT +
END SWAP DROP
» » 'D→B' STO
« CASE
HEX DUP R→B →STR 3 OVER SIZE SUB DUP REVSTR ≠ THEN DROP 0 END
DUP 4 D→B DUP REVSTR ≠ THEN DROP 0 END
BIN DUP R→B →STR 3 OVER SIZE SUB DUP REVSTR ==
END
» 'PAL2416' STO
« { 0 }
1 25000 FOR n
IF n PAL2416 THEN n + END
2 STEP
» 'TASK' STO
Runs in 42 minutes on a HP-48SX.
Much faster approach
The task generates palindromes in base 16, which must then be verified as palindromes in the other two bases.
« BIN 1 SF
R→B →STR 3 OVER SIZE 1 - SUB
0 1 FOR b
IF DUP SIZE b 1 + MOD THEN "0" SWAP + END
""
OVER SIZE b - 1 FOR j
OVER j DUP b + SUB +
-1 b - STEP
IF OVER ≠ THEN 1 CF 1 'b' STO END
NEXT DROP
1 FS?
» 'PAL24?' STO
« HEX R→B →STR → h
« "#"
h SIZE 1 - 3 FOR j
h j DUP SUB +
-1 STEP
"h" + STR→ B→R
» » ‘REVHEX’ STO
« { }
0 15 FOR b
IF b PAL24? THEN b + END NEXT
1 2 FOR x
-1 15 FOR m
16 x 1 - ^ 16 x ^ 1 - FOR b
b
IF m 0 ≥ THEN 16 * m + END
16 x ^ *
b REVHEX +
IF DUP PAL24? THEN +
ELSE IF 25000 ≥ THEN KILL END @ not idiomatic but useful to exit 3 nested loops
END
NEXT NEXT NEXT
SORT
» 'TASK' STO
TASK SORT
Runs in 2 minutes 16 on a HP-48SX: 18 times faster than brute force!
- Output:
1: { 0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845 }
Ruby
res = (0..25000).select do |n|
[2, 4, 16].all? do |base|
b = n.to_s(base)
b == b.reverse
end
end
puts res.join(" ")
- Output:
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845
Seed7
$ 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
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]
Uiua
▽⊸≡(/↧≡(≍⇌.⬚0↯⊟∞)[4 2 1]¤⋯)⇡25000- 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
import "./fmt" for Conv, Fmt
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)
}
}
}
Fmt.tprint("$,6d", numbers, 8)
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
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.