# Two identical strings

Two identical strings 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.

Find and display   (here on this page)   positive integers, n, whose base 2 representation is the concatenation of two identical binary strings,
where   n <   1,00010       (one thousand).

For each such integer, show its decimal and binary forms.

## 11l

Translation of: Python
```F bits(=n)
‘Count the amount of bits required to represent n’
V r = 0
L n != 0
n >>= 1
r++
R r

F concat(n)
‘Concatenate the binary representation of n to itself’
R n << bits(n) [|] n

V n = 1
L concat(n) <= 1000
print(‘#.: #.’.format(concat(n), bin(concat(n))))
n++```
Output:
```3: 11
10: 1010
15: 1111
36: 100100
45: 101101
54: 110110
63: 111111
136: 10001000
153: 10011001
170: 10101010
187: 10111011
204: 11001100
221: 11011101
238: 11101110
255: 11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110
```

## 8080 Assembly

```	;;;	Print positive integers whose base-2 representation
;;;	is the concatenation of two identical binary strings,
;;;	for 1 < n < 1000
puts:	equ	9	; CP/M syscall to print a string
org	100h
lxi	b,1	; Counter
loop:	mov	h,b	; HL = counter
mov	l,c
call	concat	; Get current concatenated number
lxi	d,1000	; Reached the end yet?
call	cmp16
rnc		; Stop when >1000
push	b	; Keep the counter
push	h	; And the concatenated number
call 	hldec	; Print decimal value
pop	h	; Restore number
call	hlbin	; Print binary value
lxi	d,nl	; Print newline
mvi	c,puts
call	5
pop	b	; Restore counter
inx	b	; Increment counter
jmp 	loop
;;;	16-bit compare HL to DE
cmp16:	mov	a,h
cmp	d
rnz
mov	a,l
cmp	e
ret
;;;	Concatenate HL with itself
concat:	push	h	; Keep a copy of HL on the stack
mov	d,h	; DE = copy of HL
mov	e,l
ctloop:	mov	a,d	; When DE=0, we are done
ora	e
jz	ctdone
mov	a,d	; Rotate DE left
rar
mov	d,a
mov	a,e
rar
mov	e,a
jmp	ctloop
ctdone:	pop	d	; Retrieve old HL
ret
;;;	Print HL as a decimal value
hldec:	lxi	d,outbuf
push	d	; Output pointer on the stack
lxi	b,-10	; Divisor
decdgt:	lxi	d,-1	; Quotient
div10:	inx	d	; Divide HL by 10 using trial subtraction
jc	div10
mvi	a,'0'+10
add	l	; L contains remainder - 10
pop	h		; Retrieve output pointer
dcx	h	; Store digit
mov	m,a
push	h
xchg		; Continue with quotient
mov	a,h	; If any digits left
ora	l
jnz	decdgt	; Find the next digits
pop	d	; Otherwise, retrieve pointer
mvi	c,puts	; And print result using CP/M
jmp	5
;;;	Print HL as a binary value
hlbin:	lxi	d,outbuf
ora	a	; Zero the carry flag
bindgt:	mov	a,h	; Rotate HL right
rar
mov	h,a
mov	a,l
rar
mov	l,a
mvi	a,0	; A = '0' + carry flag (i.e. lowest bit)
aci	'0'
dcx	d	; Store digit
stax	d
mov	a,h	; Any more digits?
ora	l
jnz	bindgt	; If so, find next digits
mvi	c,puts	; Otherwise, print the result
jmp	5
db	'***********'
outbuf:	db	9,'\$'
nl:	db	13,10,'\$'```
Output:
```3       11
10      1010
15      1111
36      100100
45      101101
54      110110
63      111111
136     10001000
153     10011001
170     10101010
187     10111011
204     11001100
221     11011101
238     11101110
255     11111111
528     1000010000
561     1000110001
594     1001010010
627     1001110011
660     1010010100
693     1010110101
726     1011010110
759     1011110111
792     1100011000
825     1100111001
858     1101011010
891     1101111011
924     1110011100
957     1110111101
990     1111011110```

## 8086 Assembly

```puts:	equ	9
cpu	8086
org	100h
section	.text
main:	mov	di,1		; Counter
.loop:	mov	ax,di
call	concat		; Concatenate current number to itself
cmp	ax,1000
jge	.done		; Stop when >= 1000
mov	si,ax		; Keep a copy of AX
call	pdec		; Print decimal value
mov	ax,si
call	pbin		; Print binary value
mov	bx,nl		; Print newline
call	pstr
inc	di		; Next number
jmp	.loop
.done:	ret
;;;	Concatenate AX to itself
concat:	mov	bx,ax		; Store a copy of AX in BP
mov	cx,ax		; Store a copy of AX in CX
.loop:	shl	ax,1		; Shift AX left
shr	cx,1		; Shift CX right
jnz 	.loop		; Keep going until CX is zero
or	ax,bx		; OR original AX with shifted AX
ret
;;;	Print AX as decimal
pdec:	mov	bp,10		; Divisor
mov	bx,outbuf	; Buffer pointer
.loop:	xor	dx,dx
div	bp
dec	bx		; Store digit
mov	[bx],dl
test	ax,ax		; Any more digits?
jnz	.loop
jmp	pstr		; When done, print the result
;;;	Print AX as binary
pbin:	mov	bx,outbuf	; Buffer pointer
.loop:	shr	ax,1		; Shift AX
mov	dl,'0'		; ASCII 0 or 1
dec	bx
mov	[bx],dl		; Store digit
test	ax,ax
jnz	.loop
pstr:	mov	ah,puts		; When done, print the result
mov	dx,bx
int	21h
ret
section	.data
nl:	db	13,10,'\$'
db	'****************'
outbuf:	db	9,'\$'
```
Output:
```3       11
10      1010
15      1111
36      100100
45      101101
54      110110
63      111111
136     10001000
153     10011001
170     10101010
187     10111011
204     11001100
221     11011101
238     11101110
255     11111111
528     1000010000
561     1000110001
594     1001010010
627     1001110011
660     1010010100
693     1010110101
726     1011010110
759     1011110111
792     1100011000
825     1100111001
858     1101011010
891     1101111011
924     1110011100
957     1110111101
990     1111011110```

## Action!

```INT FUNC Encode(INT x CHAR ARRAY s)
BYTE len,i
CHAR tmp

len=0 tmp=x
WHILE tmp#0
DO
len==+1
s(len)='0+(tmp&1)
tmp==RSH 1
OD
FOR i=1 TO len RSH 1
DO
tmp=s(i) s(i)=s(len-i+1) s(len-i+1)=tmp
OD
FOR i=1 TO len
DO
s(i+len)=s(i)
OD
s(0)=len
RETURN (x LSH len+x)

PROC Main()
CHAR ARRAY s(20)
INT i=[1],res,count=[0]

DO
res=Encode(i,s)
IF res>=1000 THEN
EXIT
FI
Position((count&1)*15+2,count RSH 1+1)
PrintF("%I: %S",res,s)
i==+1 count==+1
OD
RETURN```
Output:
```3: 1           10: 10
15: 11         36: 100
45: 101        54: 110
63: 111        136: 1000
153: 1001      170: 1010
187: 1011      204: 1100
221: 1101      238: 1110
255: 1111      528: 10000
561: 10001     594: 10010
627: 10011     660: 10100
693: 10101     726: 10110
759: 10111     792: 11000
825: 11001     858: 11010
891: 11011     924: 11100
957: 11101     990: 11110
```

```with Ada.Text_Io;        use Ada.Text_Io;

procedure Two_Identical is
package Integer_Io is
use Integer_Io;

Image : String (1 .. 16);
Pos_1, Pos_2 : Natural;
Mid  : Natural;
begin
for N in 1 .. 1000 loop
Put (Image, N, Base => 2);
Pos_1 := Index (Image, "#");
Pos_2 := Index (Image, "#", Pos_1 + 1);
Mid := (Pos_1 + Pos_2) / 2;
if Image (Pos_1 + 1 .. Mid) = Image (Mid + 1 .. Pos_2 - 1) then
Put (N, Width => 3); Put ("  "); Put (Image); New_Line;
end if;
end loop;
end Two_Identical;
```
Output:
```  3             2#11#
10           2#1010#
15           2#1111#
36         2#100100#
45         2#101101#
54         2#110110#
63         2#111111#
136       2#10001000#
153       2#10011001#
170       2#10101010#
187       2#10111011#
204       2#11001100#
221       2#11011101#
238       2#11101110#
255       2#11111111#
528     2#1000010000#
561     2#1000110001#
594     2#1001010010#
627     2#1001110011#
660     2#1010010100#
693     2#1010110101#
726     2#1011010110#
759     2#1011110111#
792     2#1100011000#
825     2#1100111001#
858     2#1101011010#
891     2#1101111011#
924     2#1110011100#
957     2#1110111101#
990     2#1111011110#```

## ALGOL 68

```BEGIN # show the decimal and binary representations of numbers that are of the concatenation of #
# two identical binary strings                                                            #
# returns a binary representation of v #
OP TOBINSTRING = ( INT v )STRING:
IF v = 0 THEN "0"
ELSE
STRING result := "";
INT rest := v;
WHILE rest > 0 DO
IF ODD rest THEN "1" ELSE "0" FI +=: result;
rest OVERAB 2
OD;
result
FI # TOBINSTRING # ;
INT power of 2 := 1;
FOR b WHILE IF b = power of 2 THEN
power of 2 *:= 2
FI;
INT cat value = ( b * power of 2 ) + b;
cat value < 1000
DO
print( ( whole( cat value, -4 ), ": ", TOBINSTRING cat value, newline ) )
OD
END```
Output:
```   3: 11
10: 1010
15: 1111
36: 100100
45: 101101
54: 110110
63: 111111
136: 10001000
153: 10011001
170: 10101010
187: 10111011
204: 11001100
221: 11011101
238: 11101110
255: 11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110
```

## ALGOL-M

```begin
integer function concatself(n);
integer n;
begin
integer shift, copy, test;
test := n;
shift := n;
while test > 0 do
begin
test := test / 2;
shift := shift * 2;
end;
concatself := shift + n;
end;

procedure writebits(n);
integer n;
begin
if n>1 then writebits(n/2);
writeon(if n-n/2*2=0 then "0" else "1");
end;

integer n, m;
n := 1;
m := concatself(n);
while m < 1000 do
begin
write(m, ": ");
writebits(m);
n := n + 1;
m := concatself(n);
end;
end```
Output:
```     3: 11
10: 1010
15: 1111
36: 100100
45: 101101
54: 110110
63: 111111
136: 10001000
153: 10011001
170: 10101010
187: 10111011
204: 11001100
221: 11011101
238: 11101110
255: 11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110```

## ALGOL W

```BEGIN
INTEGER PROCEDURE BITSP ( INTEGER VALUE BT ) ;
BEGIN
INTEGER BITN, BITRSL, BITIDX;
BITN   := BT;
BITRSL := 0;
BITIDX := 1;
WHILE BITN > 0 DO BEGIN
INTEGER BITNX;
BITNX  := BITN DIV 2;
BITRSL := BITRSL + BITIDX*(BITN-BITNX*2);
BITN   := BITNX;
BITIDX := BITIDX*10
END;
BITRSL
END BITSP ;

INTEGER PROCEDURE DPLBIT ( INTEGER VALUE DVAL ) ;
BEGIN
INTEGER DTEMP, DSHFT;
DTEMP := DVAL;
DSHFT := DVAL;
WHILE DTEMP > 0 DO BEGIN
DSHFT := DSHFT  *  2;
DTEMP := DTEMP DIV 2;
END;
DSHFT + DVAL
END DPLBIT ;

BEGIN
INTEGER N;
N := 0;
WHILE BEGIN
N := N + 1;
DPLBIT(N) < 1000
END DO WRITE( S_W := 0, I_W := 3, DPLBIT(N), ": ", I_W := 10, BITSP(DPLBIT(N)) )
END
END.```
Output:
```  3:         11
10:       1010
15:       1111
36:     100100
45:     101101
54:     110110
63:     111111
136:   10001000
153:   10011001
170:   10101010
187:   10111011
204:   11001100
221:   11011101
238:   11101110
255:   11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110
```

## APL

Works with: Dyalog APL
```↑(((⊂2∘⊥),⊂)(,⍨2∘⊥⍣¯1))¨⍳30
```
Output:
```  3  1 1
10  1 0 1 0
15  1 1 1 1
36  1 0 0 1 0 0
45  1 0 1 1 0 1
54  1 1 0 1 1 0
63  1 1 1 1 1 1
136  1 0 0 0 1 0 0 0
153  1 0 0 1 1 0 0 1
170  1 0 1 0 1 0 1 0
187  1 0 1 1 1 0 1 1
204  1 1 0 0 1 1 0 0
221  1 1 0 1 1 1 0 1
238  1 1 1 0 1 1 1 0
255  1 1 1 1 1 1 1 1
528  1 0 0 0 0 1 0 0 0 0
561  1 0 0 0 1 1 0 0 0 1
594  1 0 0 1 0 1 0 0 1 0
627  1 0 0 1 1 1 0 0 1 1
660  1 0 1 0 0 1 0 1 0 0
693  1 0 1 0 1 1 0 1 0 1
726  1 0 1 1 0 1 0 1 1 0
759  1 0 1 1 1 1 0 1 1 1
792  1 1 0 0 0 1 1 0 0 0
825  1 1 0 0 1 1 1 0 0 1
858  1 1 0 1 0 1 1 0 1 0
891  1 1 0 1 1 1 1 0 1 1
924  1 1 1 0 0 1 1 1 0 0
957  1 1 1 0 1 1 1 1 0 1
990  1 1 1 1 0 1 1 1 1 0```

## AppleScript

### Functional

Drawing members of the sequence from a non-finite list, up to a given limit.

```------ CONCATENATION OF TWO IDENTICAL BINARY STRINGS -----

-- binaryTwin :: Int -> (Int, String)
on binaryTwin(n)
-- A tuple of an integer m and a string s, where
-- s is a self-concatenation of the binary
-- represention of n, and m is the integer value of s.

set b to showBinary(n)
set s to b & b
end binaryTwin

--------------------------- TEST -------------------------
on run
script p
on |λ|(pair)
1000 > item 1 of pair
end |λ|
end script

script format
on |λ|(pair)
set {n, s} to pair

(n as string) & " -> " & s
end |λ|
end script

unlines(map(format, ¬
takeWhile(p, ¬
fmap(binaryTwin, enumFrom(1)))))
end run

------------------------- GENERIC ------------------------

-- enumFrom :: Int -> [Int]
on enumFrom(x)
script
property v : missing value
on |λ|()
if missing value is not v then
set v to 1 + v
else
set v to x
end if
return v
end |λ|
end script
end enumFrom

-- fmap <\$> :: (a -> b) -> Gen [a] -> Gen [b]
on fmap(f, gen)
script
property g : mReturn(f)
on |λ|()
set v to gen's |λ|()
if v is missing value then
v
else
g's |λ|(v)
end if
end |λ|
end script
end fmap

-- foldr :: (a -> b -> b) -> b -> [a] -> b
on foldr(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from lng to 1 by -1
set v to |λ|(item i of xs, v, i, xs)
end repeat
return v
end tell
end foldr

-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
-- The list obtained by applying f
-- to each element of xs.
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map

-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
-- 2nd class handler function lifted into 1st class script wrapper.
if script is class of f then
f
else
script
property |λ| : f
end script
end if
end mReturn

-- quotRem :: Int -> Int -> (Int, Int)
on quotRem(m, n)
{m div n, m mod n}
end quotRem

-- readBinary :: String -> Int
-- The integer value of the binary string s
script go
on |λ|(c, en)
set {e, n} to en
set v to ((id of c) - 48)

{2 * e, v * e + n}
end |λ|
end script

item 2 of foldr(go, {1, 0}, s)

-- showBinary :: Int -> String
on showBinary(n)
script binaryChar
on |λ|(n)
character id (48 + n)
end |λ|
end script
showIntAtBase(2, binaryChar, n, "")
end showBinary

-- showIntAtBase :: Int -> (Int -> Char) -> Int -> String -> String
on showIntAtBase(base, toDigit, n, rs)
script go
property f : mReturn(toDigit)
on |λ|(nd_, r)
set {n, d} to nd_
set r_ to f's |λ|(d) & r
if n > 0 then
|λ|(quotRem(n, base), r_)
else
r_
end if
end |λ|
end script
|λ|(quotRem(n, base), rs) of go
end showIntAtBase

-- takeWhile :: (a -> Bool) -> Generator [a] -> [a]
on takeWhile(p, xs)
set ys to {}
set v to |λ|() of xs
tell mReturn(p)
repeat while (|λ|(v))
set end of ys to v
set v to xs's |λ|()
end repeat
end tell
return ys
end takeWhile

-- unlines :: [String] -> String
on unlines(xs)
-- A single string formed by the intercalation
-- of a list of strings with the newline character.
set {dlm, my text item delimiters} to ¬
{my text item delimiters, linefeed}
set s to xs as text
set my text item delimiters to dlm
s
end unlines
```
Output:
```3 -> 11
10 -> 1010
15 -> 1111
36 -> 100100
45 -> 101101
54 -> 110110
63 -> 111111
136 -> 10001000
153 -> 10011001
170 -> 10101010
187 -> 10111011
204 -> 11001100
221 -> 11011101
238 -> 11101110
255 -> 11111111
528 -> 1000010000
561 -> 1000110001
594 -> 1001010010
627 -> 1001110011
660 -> 1010010100
693 -> 1010110101
726 -> 1011010110
759 -> 1011110111
792 -> 1100011000
825 -> 1100111001
858 -> 1101011010
891 -> 1101111011
924 -> 1110011100
957 -> 1110111101
990 -> 1111011110```

### Idiomatic

```on task(maxN)
set startWith0 to false -- Change to true to start with 0 and "00".
set rhv to -(startWith0 as integer) -- Start value of "right hand" string.
script o
property bits : {rhv}
property output : {}
end script

set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to ""
repeat
-- Add 1 to the binary-digit list's LSD and perform any carries.
set carry to 1
repeat with i from (count o's bits) to 1 by -1
set columnSum to (item i of o's bits) + carry
set item i of o's bits to columnSum mod 2
set carry to columnSum div 2
if (carry = 0) then exit repeat
end repeat
if (carry = 1) then set beginning of o's bits to carry
-- Add 1 to the "right-hand" value and work out the corresponding n.
set rhv to rhv + 1
set n to rhv * (2 ^ (count o's bits)) div 1 + rhv
-- Unless n exceeds maxN, append it and its binary form to the output.
if (n > maxN) then exit repeat
set end of o's output to (n as text) & ":  " & o's bits & o's bits
end repeat
set AppleScript's text item delimiters to linefeed
set o's output to o's output as text
set AppleScript's text item delimiters to astid

return o's output

```
Output:
```"3:  11
10:  1010
15:  1111
36:  100100
45:  101101
54:  110110
63:  111111
136:  10001000
153:  10011001
170:  10101010
187:  10111011
204:  11001100
221:  11011101
238:  11101110
255:  11111111
528:  1000010000
561:  1000110001
594:  1001010010
627:  1001110011
660:  1010010100
693:  1010110101
726:  1011010110
759:  1011110111
792:  1100011000
825:  1100111001
858:  1101011010
891:  1101111011
924:  1110011100
957:  1110111101
990:  1111011110"
```

## Arturo

```loop 0..1000 'i [
bin: as.binary i
if even? size bin [
half: (size bin)/2
if equal? slice bin 0 dec half
slice bin half dec size bin ->
print [pad to :string i 4 ":" bin]
]
]
```
Output:
```   3 : 11
10 : 1010
15 : 1111
36 : 100100
45 : 101101
54 : 110110
63 : 111111
136 : 10001000
153 : 10011001
170 : 10101010
187 : 10111011
204 : 11001100
221 : 11011101
238 : 11101110
255 : 11111111
528 : 1000010000
561 : 1000110001
594 : 1001010010
627 : 1001110011
660 : 1010010100
693 : 1010110101
726 : 1011010110
759 : 1011110111
792 : 1100011000
825 : 1100111001
858 : 1101011010
891 : 1101111011
924 : 1110011100
957 : 1110111101
990 : 1111011110```

## AutoHotkey

```n:=0
while (n++<=1000)
{
bin := LTrim(dec2bin(n), "0")
l := Strlen(bin)
if (l/2 <> Floor(l/2))
continue
if (SubStr(bin, 1, l/2) = SubStr(bin, l/2+1))
result .= n "`t" bin "`n"
}
MsgBox % result
return

Dec2Bin(i, s=0, c=0) {
l := StrLen(i := Abs(i + u := i < 0))
Loop, % Abs(s) + !s * l << 2
b := u ^ 1 & i // (1 << c++) . b
Return, b
}
```
Output:
```3	11
10	1010
15	1111
36	100100
45	101101
54	110110
63	111111
136	10001000
153	10011001
170	10101010
187	10111011
204	11001100
221	11011101
238	11101110
255	11111111
528	1000010000
561	1000110001
594	1001010010
627	1001110011
660	1010010100
693	1010110101
726	1011010110
759	1011110111
792	1100011000
825	1100111001
858	1101011010
891	1101111011
924	1110011100
957	1110111101
990	1111011110```

## AWK

```# syntax: GAWK -f TWO_IDENTICAL_STRINGS.AWK
BEGIN {
for (i=1; i<1000; i++) {
b = dec2bin(i)
leng = length(b)
if (leng % 2 == 0) {
if (substr(b,1,leng/2) == substr(b,leng/2+1)) {
printf("%4d %10s\n",i,b)
count++
}
}
}
printf("count: %d\n",count)
exit(0)
}
function dec2bin(n,  str) {
while (n) {
str = ((n%2 == 0) ? "0" : "1") str
n = int(n/2)
}
if (str == "") {
str = "0"
}
return(str)
}
```
Output:
```   3         11
10       1010
15       1111
36     100100
45     101101
54     110110
63     111111
136   10001000
153   10011001
170   10101010
187   10111011
204   11001100
221   11011101
238   11101110
255   11111111
528 1000010000
561 1000110001
594 1001010010
627 1001110011
660 1010010100
693 1010110101
726 1011010110
759 1011110111
792 1100011000
825 1100111001
858 1101011010
891 1101111011
924 1110011100
957 1110111101
990 1111011110
count: 30
```

## BASIC

```10 DEFINT A-Z: DIM B(15)
20 N=0
30 N=N+1
40 C=0: X=N
50 C=C+1
60 X=X\2
70 IF X>0 THEN 50
80 K=N+2^C*N
90 IF K>1000 THEN END
100 PRINT K,
110 FOR I=C*2 TO 1 STEP -1
120 B(I)=K AND 1
130 K=K\2
140 NEXT I
150 FOR I=1 TO C*2
160 PRINT USING "#";B(I);
170 NEXT I
180 PRINT
190 GOTO 30
```
Output:
``` 3            11
10           1010
15           1111
36           100100
45           101101
54           110110
63           111111
136          10001000
153          10011001
170          10101010
187          10111011
204          11001100
221          11011101
238          11101110
255          11111111
528          1000010000
561          1000110001
594          1001010010
627          1001110011
660          1010010100
693          1010110101
726          1011010110
759          1011110111
792          1100011000
825          1100111001
858          1101011010
891          1101111011
924          1110011100
957          1110111101
990          1111011110```

### BASIC256

```n = 1
k = 0
p = 2

while True
if n >= p then p += p
k = n + n * p
if k < 1000 then print k; " = "; tobinary(k) else end
n += 1
end while```

### QBasic

Works with: QBasic version 1.1
Works with: QuickBasic version 4.5
```FUNCTION tobin\$ (d)
s\$ = ""
DO WHILE d <> 0
r = d MOD 2
s\$ = STR\$(r) + s\$
d = d \ 2
LOOP
tobin\$ = s\$
END FUNCTION

k = 0 : n = 1 : p = 2
DO
IF n >= p THEN p = p + p
k = n + n * p
IF k < 1000 THEN
PRINT k; " = "; tobin\$(k)
ELSE
EXIT DO
END IF
n = n + 1
LOOP
```

### PureBasic

```OpenConsole()
n.i = 1
k.i = 0
p.i= 2

While #True
If n >= p
p + p
EndIf
k = n + n * p
If k < 1000
PrintN(Str(k) + " = " + Bin(k))
Else
Break
EndIf
n + 1
Wend

Input()
CloseConsole()```

### True BASIC

```FUNCTION tobin\$(d)
LET s\$ = ""
DO WHILE d <> 0
LET r = REMAINDER(ROUND(d),2)
LET s\$ = STR\$(r) & s\$
LET d = IP(ROUND(d)/2)
LOOP
LET tobin\$ = s\$
END FUNCTION

LET n = 1
LET k = 0
LET p = 2
DO
IF n >= p THEN LET p = p+p
LET k = n+n*p
IF k < 1000 THEN
PRINT k; " = "; tobin\$(k)
ELSE
EXIT DO
END IF
LET n = n+1
LOOP
END
```

## BCPL

```get "libhdr"

let bitlength(n) = n=0 -> 0, 1 + bitlength(n >> 1)
let concat(n,m) = (n << bitlength(n)) | m;

let writebits(n) be
\$(  if n>1 then writebits(n >> 1)
wrch('0' + (n & 1))
\$)

let start() be
\$(  let n = 1 and conc = concat(n,n)
while conc < 1000 do
\$(  writef("%I4: ", conc)
writebits(conc)
wrch('*N')
n := n + 1
conc := concat(n,n)
\$)
\$)```
Output:
```   3: 11
10: 1010
15: 1111
36: 100100
45: 101101
54: 110110
63: 111111
136: 10001000
153: 10011001
170: 10101010
187: 10111011
204: 11001100
221: 11011101
238: 11101110
255: 11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110```

## Befunge

```1>:::>:#v_++:91+v
>^   /  >\vv**::<
^2\*2<>`#@_v
v_v#!:<\+19\0.:<
\$ :   ^ /2<
+v<>2%68*+\^
1:^
\$!,
^_^
```
Output:
```3 11
10 1010
15 1111
36 100100
45 101101
54 110110
63 111111
136 10001000
153 10011001
170 10101010
187 10111011
204 11001100
221 11011101
238 11101110
255 11111111
528 1000010000
561 1000110001
594 1001010010
627 1001110011
660 1010010100
693 1010110101
726 1011010110
759 1011110111
792 1100011000
825 1100111001
858 1101011010
891 1101111011
924 1110011100
957 1110111101
990 1111011110```

## C

```#include <stdio.h>
#include <stdint.h>

uint8_t bit_length(uint32_t n) {
uint8_t r;
for (r=0; n; r++) n >>= 1;
return r;
}

uint32_t concat_bits(uint32_t n) {
return (n << bit_length(n)) | n;
}

char *bits(uint32_t n) {
static char buf[33];
char *ptr = &buf[33];
*--ptr = 0;
do {
*--ptr = '0' + (n & 1);
} while (n >>= 1);
return ptr;
}

int main() {
uint32_t n, r;
for (n=1; (r = concat_bits(n)) < 1000; n++) {
printf("%d: %s\n", r, bits(r));
}
return 0;
}
```
Output:
```3: 11
10: 1010
15: 1111
36: 100100
45: 101101
54: 110110
63: 111111
136: 10001000
153: 10011001
170: 10101010
187: 10111011
204: 11001100
221: 11011101
238: 11101110
255: 11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110```

## C#

Translation of: Visual Basic .NET
```using System; using static System.Console;
class Program { static void Main() { int c = 0, lmt = 1000;
for (int n = 1, p = 2, k; n <= lmt; n++)
if ((k = n + n * (p += n >= p ? p : 0)) > lmt) break;
else Console.Write("{0,3} ({1,-10})  {2}", k,
Convert.ToString(k, 2), ++c % 5 == 0 ? "\n" : "");
Write("\nFound {0} numbers whose base 2 representation is the " +
"concatenation of two identical binary strings.", c); } }
```
Output:

Same as Visual Basic. NET

## C++

```#include <iostream>
#include <string>

// Given the base 2 representation of a number n, transform it into the base 2
// representation of n + 1.
void base2_increment(std::string& s) {
size_t z = s.rfind('0');
if (z != std::string::npos) {
s[z] = '1';
size_t count = s.size() - (z + 1);
s.replace(z + 1, count, count, '0');
} else {
s.assign(s.size() + 1, '0');
s[0] = '1';
}
}

int main() {
std::cout << "Decimal\tBinary\n";
std::string s("1");
for (unsigned int n = 1; ; ++n) {
unsigned int i = n + (n << s.size());
if (i >= 1000)
break;
std::cout << i << '\t' << s << s << '\n';
base2_increment(s);
}
}
```
Output:
```Decimal	Binary
3	11
10	1010
15	1111
36	100100
45	101101
54	110110
63	111111
136	10001000
153	10011001
170	10101010
187	10111011
204	11001100
221	11011101
238	11101110
255	11111111
528	1000010000
561	1000110001
594	1001010010
627	1001110011
660	1010010100
693	1010110101
726	1011010110
759	1011110111
792	1100011000
825	1100111001
858	1101011010
891	1101111011
924	1110011100
957	1110111101
990	1111011110
```

## CLU

```nth_ident = proc (n: int) returns (int)
h: int := n
l: int := n
while l>0 do h, l := h*2, l/2 end
return(h + n)
end nth_ident

bits = proc (n: int) returns (string)
p: string := ""
if n>1 then p := bits(n/2) end
return(p || int\$unparse(n//2))
end bits

start_up = proc ()
po: stream := stream\$primary_output()
n: int := 0
while true do
n := n+1
ident: int := nth_ident(n)
if ident>=1000 then break end
stream\$putright(po, int\$unparse(ident), 3)
stream\$putl(po, ": " || bits(ident))
end
end start_up```
Output:
```  3: 11
10: 1010
15: 1111
36: 100100
45: 101101
54: 110110
63: 111111
136: 10001000
153: 10011001
170: 10101010
187: 10111011
204: 11001100
221: 11011101
238: 11101110
255: 11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110```

## COBOL

```       IDENTIFICATION DIVISION.
PROGRAM-ID.  IDENTICAL-STRINGS.

DATA DIVISION.
WORKING-STORAGE SECTION.
01 VARIABLES.
03 INPUT-NUMBER            PIC 9(4).
03 BINARY-REPRESENTATION   PIC 9(10).
03 BIT                     REDEFINES BINARY-REPRESENTATION,
PIC 9 OCCURS 10 TIMES.
03 FIRST-SET-BIT           PIC 99.
03 CURRENT-BIT             PIC 99.
03 SECOND-BIT              PIC 99.
03 BIT-VALUE               PIC 9(4).
03 OUTPUT-NUMBER           PIC 9(4) VALUE ZERO.
03 REMAINING-BITS          PIC 9(4)V9.
03 FILLER                  REDEFINES REMAINING-BITS.
05 REST                 PIC 9(4).
05 FILLER               PIC 9.
88 BIT-IS-SET        VALUE 5.

01 OUTPUT-FORMAT.
03 DECIMAL-OUTPUT          PIC Z(3)9.
03 BINARY-OUTPUT           PIC Z(9)9.

PROCEDURE DIVISION.
BEGIN.
PERFORM IDENTICAL-STRING
VARYING INPUT-NUMBER FROM 1 BY 1
UNTIL OUTPUT-NUMBER IS GREATER THAN 1000.
STOP RUN.

IDENTICAL-STRING.
MOVE ZERO TO BINARY-REPRESENTATION.
MOVE 10 TO CURRENT-BIT.
MOVE INPUT-NUMBER TO REMAINING-BITS.
PERFORM EXTRACT-BIT UNTIL REMAINING-BITS EQUAL ZERO.
MOVE CURRENT-BIT TO FIRST-SET-BIT.
MOVE 10 TO SECOND-BIT.
PERFORM COPY-BIT UNTIL SECOND-BIT IS EQUAL TO FIRST-SET-BIT.
MOVE ZERO TO OUTPUT-NUMBER.
IF CURRENT-BIT IS EQUAL TO ZERO, MOVE 1 TO CURRENT-BIT.
PERFORM INSERT-BIT
VARYING CURRENT-BIT FROM CURRENT-BIT BY 1
UNTIL CURRENT-BIT IS GREATER THAN 10.
MOVE OUTPUT-NUMBER TO DECIMAL-OUTPUT.
MOVE BINARY-REPRESENTATION TO BINARY-OUTPUT.
IF OUTPUT-NUMBER IS LESS THAN 1000,
DISPLAY DECIMAL-OUTPUT ": " BINARY-OUTPUT.

EXTRACT-BIT.
DIVIDE 2 INTO REMAINING-BITS.
IF BIT-IS-SET, MOVE 1 TO BIT(CURRENT-BIT).
SUBTRACT 1 FROM CURRENT-BIT.
MOVE REST TO REMAINING-BITS.

COPY-BIT.
MOVE BIT(SECOND-BIT) TO BIT(CURRENT-BIT).
SUBTRACT 1 FROM SECOND-BIT.
SUBTRACT 1 FROM CURRENT-BIT.

INSERT-BIT.
COMPUTE BIT-VALUE = 2 ** (10 - CURRENT-BIT)
MULTIPLY BIT(CURRENT-BIT) BY BIT-VALUE.
```
Output:
```   3:         11
10:       1010
15:       1111
36:     100100
45:     101101
54:     110110
63:     111111
136:   10001000
153:   10011001
170:   10101010
187:   10111011
204:   11001100
221:   11011101
238:   11101110
255:   11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110```

## Cowgol

```include "cowgol.coh";

sub bitLength(n: uint32): (l: uint8) is
l := 0;
while n != 0 loop
n := n >> 1;
l := l + 1;
end loop;
end sub;

sub concatBits(n: uint32): (r: uint32) is
r := (n << bitLength(n)) | n;
end sub;

sub printBits(n: uint32) is
var buf: uint8[33];
var ptr := &buf[32];
[ptr] := 0;
loop
ptr := @prev ptr;
[ptr] := '0' + (n as uint8 & 1);
n := n >> 1;
if n == 0 then break; end if;
end loop;
print(ptr);
end sub;

var n: uint32 := 1;
loop
var r := concatBits(n);
if r > 1000 then break; end if;
print_i32(r);
print(": ");
printBits(r);
print_nl();
n := n + 1;
end loop;```
Output:
```3: 11
10: 1010
15: 1111
36: 100100
45: 101101
54: 110110
63: 111111
136: 10001000
153: 10011001
170: 10101010
187: 10111011
204: 11001100
221: 11011101
238: 11101110
255: 11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110```

## Crystal

Translation of: Ruby
```(0..1000).each do |i|
bin = i.to_s(2)
if bin.size.even?
half = bin.size // 2
if bin[0..half-1] == bin[half..]
print "%3d: %10s\n" % [i, bin]
end
end
end
```
Output:
```  3:         11
10:       1010
15:       1111
36:     100100
45:     101101
54:     110110
63:     111111
136:   10001000
153:   10011001
170:   10101010
187:   10111011
204:   11001100
221:   11011101
238:   11101110
255:   11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110```

## Delphi

Works with: Delphi version 6.0

```function IntToBinStr(IValue: Int64) : string;
{Convert integer to binary string, with no leading zero}
var I: integer;
begin
Result:='';
I:=IntPower2(32-1);
while I <> 0 do
begin
if (IValue and I)<>0 then Result:=Result + '1'
else if Length(Result)>0 then Result:=Result + '0';
I:=I shr 1;
end;
if Result='' then Result:='0';
end;

procedure IdenticalBinaryStrings(Memo: TMemo);
var S,S1,S2: string;
var Len,I: integer;
begin
for I:=2 to 1000-1 do
begin
{Get binary String}
S:=IntToBinStr(I);
{Only look at string evenly divisible by 2}
Len:=Length(S);
if (Len and 1)=0 then
begin
{Split string into equal pieces}
S1:=LeftStr(S,Len div 2);
S2:=RightStr(S,Len div 2);
{Each half should be the same}
end;
end;
end;
```
Output:
```3: 11
10: 1010
15: 1111
36: 100100
45: 101101
54: 110110
63: 111111
136: 10001000
153: 10011001
170: 10101010
187: 10111011
204: 11001100
221: 11011101
238: 11101110
255: 11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110

Elapsed Time: 58.641 ms.

```

## Draco

```proc nonrec concat_self(word n) word:
word rslt, k;
k := n;
rslt := n;
while k ~= 0 do
k := k >> 1;
rslt := rslt << 1
od;
rslt | n
corp

proc nonrec main() void:
word n, conc;
n := 1;
while
conc := concat_self(n);
conc < 1000
do
writeln(conc:3, ": ", conc:b:10);
n := n + 1
od
corp```
Output:
```  3:         11
10:       1010
15:       1111
36:     100100
45:     101101
54:     110110
63:     111111
136:   10001000
153:   10011001
170:   10101010
187:   10111011
204:   11001100
221:   11011101
238:   11101110
255:   11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110```

## EasyLang

Translation of: BASIC256
```func\$ tobin k .
if k > 0
.
.
p = 2
repeat
n += 1
if n >= p
p += p
.
k = n + n * p
until k >= 1000
print k & ": " & tobin k
.```

## F#

```// Nigel Galloway. April 5th., 2021
let fN g=let rec fN g=function n when n<2->(char(n+48))::g |n->fN((char(n%2+48))::g)(n/2) in fN [] g|>Array.ofList|>System.String
Seq.unfold(fun(n,g,l)->Some((n<<<l)+n,if n=g-1 then (n+1,g*2,l+1) else (n+1,g,l)))(1,2,1)|>Seq.takeWhile((>)1000)|>Seq.iter(fun g->printfn "%3d %s" g (fN g))
```
Output:
```  3 11
10 1010
15 1111
36 100100
45 101101
54 110110
63 111111
136 10001000
153 10011001
170 10101010
187 10111011
204 11001100
221 11011101
238 11101110
255 11111111
528 1000010000
561 1000110001
594 1001010010
627 1001110011
660 1010010100
693 1010110101
726 1011010110
759 1011110111
792 1100011000
825 1100111001
858 1101011010
891 1101111011
924 1110011100
957 1110111101
990 1111011110
```

## Factor

Works with: Factor version 0.99 2021-02-05
```USING: formatting kernel lists lists.lazy math math.parser
sequences ;

1 lfrom [ >bin dup append bin> ] lmap-lazy [ 1000 < ] lwhile
[ dup "%d %b\n" printf ] leach
```
Output:
```3 11
10 1010
15 1111
36 100100
45 101101
54 110110
63 111111
136 10001000
153 10011001
170 10101010
187 10111011
204 11001100
221 11011101
238 11101110
255 11111111
528 1000010000
561 1000110001
594 1001010010
627 1001110011
660 1010010100
693 1010110101
726 1011010110
759 1011110111
792 1100011000
825 1100111001
858 1101011010
891 1101111011
924 1110011100
957 1110111101
990 1111011110
```

## FALSE

```1[\$\$\$[\$][2/\2*\]#%|\$1000>~][
\$.": "
0\10\[\$1&'0+\2/\$][]#%
[\$][,]#%
1+
]#%%```
Output:
```3: 11
10: 1010
15: 1111
36: 100100
45: 101101
54: 110110
63: 111111
136: 10001000
153: 10011001
170: 10101010
187: 10111011
204: 11001100
221: 11011101
238: 11101110
255: 11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110```

## FOCAL

```01.10 S N=0
01.20 S N=N+1
01.30 D 3
01.40 I (K-1000)1.5;Q
01.50 T %3,K,": "
01.60 D 4
01.70 G 1.2

02.10 S BC=0;S BT=N
02.20 S BC=BC+1
02.30 S BT=FITR(BT/2)
02.40 I (-BT)2.2

03.10 D 2;S I=BC;S BT=N
03.20 S BX=FITR(BT/2)
03.30 S I=I-1
03.40 S B(I)=BT-BX*2
03.50 S BT=BX
03.60 I (-I)3.2,3.2
03.70 F I=0,BC-1;S B(BC+I)=B(I)
03.80 S BC=BC*2;S K=0
03.90 F I=0,BC-1;S K=K*2+B(I)

04.10 F I=0,BC-1;D 4.3
04.20 T !;R
04.30 I (B(I))4.4,4.5,4.4
04.40 T "1"
04.50 T "0"```
Output:
```=   3: 11
=  10: 1010
=  15: 1111
=  36: 100100
=  45: 101101
=  54: 110110
=  63: 111111
= 136: 10001000
= 153: 10011001
= 170: 10101010
= 187: 10111011
= 204: 11001100
= 221: 11011101
= 238: 11101110
= 255: 11111111
= 528: 1000010000
= 561: 1000110001
= 594: 1001010010
= 627: 1001110011
= 660: 1010010100
= 693: 1010110101
= 726: 1011010110
= 759: 1011110111
= 792: 1100011000
= 825: 1100111001
= 858: 1101011010
= 891: 1101111011
= 924: 1110011100
= 957: 1110111101
= 990: 1111011110```

## Forth

```: concat-self
dup dup
begin dup while
1 rshift
swap 1 lshift swap
repeat
drop or
;

: print-bits
0 swap
begin
dup 1 and '0 +
swap 1 rshift
dup 0= until drop
begin dup while emit repeat drop
;

: to1000
1
begin dup concat-self dup 1000 < while
dup . 9 emit print-bits cr
1+
repeat
2drop
;

to1000 bye
```
Output:
```3       11
10      1010
15      1111
36      100100
45      101101
54      110110
63      111111
136     10001000
153     10011001
170     10101010
187     10111011
204     11001100
221     11011101
238     11101110
255     11111111
528     1000010000
561     1000110001
594     1001010010
627     1001110011
660     1010010100
693     1010110101
726     1011010110
759     1011110111
792     1100011000
825     1100111001
858     1101011010
891     1101111011
924     1110011100
957     1110111101
990     1111011110```

## Fortran

```      program IdentStr
implicit none
integer n, concat, bits

n = 1
100     if (concat(n) .lt. 1000) then
write (*,'(I3,2X,I11)') concat(n), bits(concat(n))
n = n + 1
goto 100
end if
stop
end

C     Concatenate binary representation of number with itself
integer function concat(num)
integer num, sl, sr
sl = num
sr = num
100     if (sr .gt. 0) then
sl = sl * 2
sr = sr / 2
goto 100
end if
concat = num + sl
end

C     Calculate binary representation of number
integer function bits(num)
integer num, n, bx
n = num
bits = 0
bx = 1
100     if (n .gt. 0) then
bits = bits + bx * mod(n,2)
bx = bx * 10
n = n / 2
goto 100
end if
end
```
Output:
```  3           11
10         1010
15         1111
36       100100
45       101101
54       110110
63       111111
136     10001000
153     10011001
170     10101010
187     10111011
204     11001100
221     11011101
238     11101110
255     11111111
528   1000010000
561   1000110001
594   1001010010
627   1001110011
660   1010010100
693   1010110101
726   1011010110
759   1011110111
792   1100011000
825   1100111001
858   1101011010
891   1101111011
924   1110011100
957   1110111101
990   1111011110```

## FreeBASIC

```dim as uinteger n=1, k=0
do
k = n + 2*n*2^int(log(n)/log(2))
if k<1000 then print k, bin(k) else end
n=n+1
loop```
Output:
```3             11
10            1010
15            1111
36            100100
45            101101
54            110110
63            111111
136           10001000
153           10011001
170           10101010
187           10111011
204           11001100
221           11011101
238           11101110
255           11111111
528           1000010000
561           1000110001
594           1001010010
627           1001110011
660           1010010100
693           1010110101
726           1011010110
759           1011110111
792           1100011000
825           1100111001
858           1101011010
891           1101111011
924           1110011100
957           1110111101
990           1111011110```

### Alternate

No log() function required.

```dim as uinteger n = 1, k = 0, p = 2
do
if n >= p then p = p + p
k = n + n * p
if k < 1000 then print k, bin(k) else end
n = n + 1
loop```
Output:

Same as log() version.

## Frink

```for n = 1 to 999
if base2[n] =~ %r/^(\d+)\1\$/
println["\$n\t" + base2[n]]```
Output:
```3	11
10	1010
15	1111
36	100100
45	101101
54	110110
63	111111
136	10001000
153	10011001
170	10101010
187	10111011
204	11001100
221	11011101
238	11101110
255	11111111
528	1000010000
561	1000110001
594	1001010010
627	1001110011
660	1010010100
693	1010110101
726	1011010110
759	1011110111
792	1100011000
825	1100111001
858	1101011010
891	1101111011
924	1110011100
957	1110111101
990	1111011110
```

## FutureBasic

```defstr byte

NSUInteger n = 1, k = 0, p = 2

while (1)
if n >= p then p += p
k = n + n * p
if k < 1000 then printf @"%4lu   %@", k, bin(k) else exit while
n++
wend

HandleEvents```
Output:
```   3   00000011
10   00001010
15   00001111
36   00100100
45   00101101
54   00110110
63   00111111
136   10001000
153   10011001
170   10101010
187   10111011
204   11001100
221   11011101
238   11101110
255   11111111
528   00010000
561   00110001
594   01010010
627   01110011
660   10010100
693   10110101
726   11010110
759   11110111
792   00011000
825   00111001
858   01011010
891   01111011
924   10011100
957   10111101
990   11011110
```

## Go

Translation of: Wren
```package main

import (
"fmt"
"strconv"
)

func main() {
i := int64(1)
for {
b2 := strconv.FormatInt(i, 2)
b2 += b2
d, _ := strconv.ParseInt(b2, 2, 64)
if d >= 1000 {
break
}
fmt.Printf("%3d : %s\n", d, b2)
i++
}
fmt.Println("\nFound", i-1, "numbers.")
}
```
Output:
```  3 : 11
10 : 1010
15 : 1111
36 : 100100
45 : 101101
54 : 110110
63 : 111111
136 : 10001000
153 : 10011001
170 : 10101010
187 : 10111011
204 : 11001100
221 : 11011101
238 : 11101110
255 : 11111111
528 : 1000010000
561 : 1000110001
594 : 1001010010
627 : 1001110011
660 : 1010010100
693 : 1010110101
726 : 1011010110
759 : 1011110111
792 : 1100011000
825 : 1100111001
858 : 1101011010
891 : 1101111011
924 : 1110011100
957 : 1110111101
990 : 1111011110

Found 30 numbers.
```

### Data.Bits

```import Control.Monad (join)
import Data.Bits
finiteBitSize,
shift,
(.|.)
)
import Text.Printf (printf)

-- Find the amount of bits required to represent a number
nBits :: Int -> Int
nBits = (-) . finiteBitSize <*> countLeadingZeros

-- Concatenate the bits of a number to itself
concatSelf :: Int -> Int
concatSelf = (.|.) =<< shift <*> nBits

-- Integers whose base-2 representation is the concatenation
-- of two identical binary strings
identStrInts :: [Int]
identStrInts = map concatSelf [1 ..]

main :: IO ()
main =
putStr \$
unlines \$
map (join \$ printf "%d: %b") to1000
where
to1000 = takeWhile (<= 1000) identStrInts
```
Output:
```3: 11
10: 1010
15: 1111
36: 100100
45: 101101
54: 110110
63: 111111
136: 10001000
153: 10011001
170: 10101010
187: 10111011
204: 11001100
221: 11011101
238: 11101110
255: 11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110```

### Data.Char

As an alternative to Data.Bits, we could also express this in terms of Data.Char

```import Control.Monad (join)
import Data.Char (digitToInt, intToDigit)
import Numeric (showIntAtBase)

------ CONCATENATION OF TWO IDENTICAL BINARY STRINGS -----

binaryTwin :: Int -> (Int, String)
binaryTwin = ((,) =<< readBinary) . join (<>) showBinary

--------------------------- TEST -------------------------
main :: IO ()
main =
mapM_ print \$
takeWhile ((1000 >) . fst) \$
binaryTwin <\$> [1 ..]

------------------------- GENERIC ------------------------

showBinary :: Int -> String
showBinary = flip (showIntAtBase 2 intToDigit) []

snd \$
foldr
(\c (e, n) -> (2 * e, digitToInt c * e + n))
(1, 0)
s
```
```(3,"11")
(10,"1010")
(15,"1111")
(36,"100100")
(45,"101101")
(54,"110110")
(63,"111111")
(136,"10001000")
(153,"10011001")
(170,"10101010")
(187,"10111011")
(204,"11001100")
(221,"11011101")
(238,"11101110")
(255,"11111111")
(528,"1000010000")
(561,"1000110001")
(594,"1001010010")
(627,"1001110011")
(660,"1010010100")
(693,"1010110101")
(726,"1011010110")
(759,"1011110111")
(792,"1100011000")
(825,"1100111001")
(858,"1101011010")
(891,"1101111011")
(924,"1110011100")
(957,"1110111101")
(990,"1111011110")```

## J

```(":,': ',":@#:)@(,~&.#:)"0 (>:i.30)
```
Output:
```3: 1 1
10: 1 0 1 0
15: 1 1 1 1
36: 1 0 0 1 0 0
45: 1 0 1 1 0 1
54: 1 1 0 1 1 0
63: 1 1 1 1 1 1
136: 1 0 0 0 1 0 0 0
153: 1 0 0 1 1 0 0 1
170: 1 0 1 0 1 0 1 0
187: 1 0 1 1 1 0 1 1
204: 1 1 0 0 1 1 0 0
221: 1 1 0 1 1 1 0 1
238: 1 1 1 0 1 1 1 0
255: 1 1 1 1 1 1 1 1
528: 1 0 0 0 0 1 0 0 0 0
561: 1 0 0 0 1 1 0 0 0 1
594: 1 0 0 1 0 1 0 0 1 0
627: 1 0 0 1 1 1 0 0 1 1
660: 1 0 1 0 0 1 0 1 0 0
693: 1 0 1 0 1 1 0 1 0 1
726: 1 0 1 1 0 1 0 1 1 0
759: 1 0 1 1 1 1 0 1 1 1
792: 1 1 0 0 0 1 1 0 0 0
825: 1 1 0 0 1 1 1 0 0 1
858: 1 1 0 1 0 1 1 0 1 0
891: 1 1 0 1 1 1 1 0 1 1
924: 1 1 1 0 0 1 1 1 0 0
957: 1 1 1 0 1 1 1 1 0 1
990: 1 1 1 1 0 1 1 1 1 0```

## Java

```public class TwoIdenticalStrings {
public static void main(String[] args) {
System.out.println("Decimal Binary");
for (int i = 0; i < 1_000; i++) {
String binStr = Integer.toBinaryString(i);
if (binStr.length() % 2 == 0) {
int len = binStr.length() / 2;
if (binStr.substring(0, len).equals(binStr.substring(len))) {
System.out.printf("%7d %s%n", i, binStr);
}
}
}
}
}
```
Output:
```Decimal Binary
3 11
10 1010
15 1111
36 100100
45 101101
54 110110
63 111111
136 10001000
153 10011001
170 10101010
187 10111011
204 11001100
221 11011101
238 11101110
255 11111111
528 1000010000
561 1000110001
594 1001010010
627 1001110011
660 1010010100
693 1010110101
726 1011010110
759 1011110111
792 1100011000
825 1100111001
858 1101011010
891 1101111011
924 1110011100
957 1110111101
990 1111011110```

## jq

Works with: jq

Works with gojq, the Go implementation of jq

```def binary_digits:
if . == 0 then 0
else [recurse( if . == 0 then empty else ./2 | floor end ) % 2 | tostring]
| reverse
| .[1:] # remove the leading 0
| join("")
end ;

range(1;1000)
| . as \$i
| binary_digits
| select(length % 2 == 0)
| (length/2) as \$half
| select(.[\$half:] == .[:\$half])
| [\$i, .]```
Output:
```[3,"11"]
[10,"1010"]
[15,"1111"]
[36,"100100"]
[45,"101101"]
[54,"110110"]
[63,"111111"]
[136,"10001000"]
[153,"10011001"]
[170,"10101010"]
[187,"10111011"]
[204,"11001100"]
[221,"11011101"]
[238,"11101110"]
[255,"11111111"]
[528,"1000010000"]
[561,"1000110001"]
[594,"1001010010"]
[627,"1001110011"]
[660,"1010010100"]
[693,"1010110101"]
[726,"1011010110"]
[759,"1011110111"]
[792,"1100011000"]
[825,"1100111001"]
[858,"1101011010"]
[891,"1101111011"]
[924,"1110011100"]
[957,"1110111101"]
[990,"1111011110"]```

## Julia

```function twoidenticalstringsinbase(base, maxnum, verbose=true)
found = Int[]
for i in 1:maxnum
dig = digits(i; base)
k = length(dig)
iseven(k) && dig[begin:begin+k÷2-1] == dig[begin+k÷2:end] && push!(found, i)
end
if verbose
println("\nDecimal  Base \$base")
for n in found
end
end
return found
end

twoidenticalstringsinbase(2, 999)
twoidenticalstringsinbase(16, 999)
```
Output:
```Decimal  Base 2
3        11
10       1010
15       1111
36       100100
45       101101
54       110110
63       111111
136      10001000
153      10011001
170      10101010
187      10111011
204      11001100
221      11011101
238      11101110
255      11111111
528      1000010000
561      1000110001
594      1001010010
627      1001110011
660      1010010100
693      1010110101
726      1011010110
759      1011110111
792      1100011000
825      1100111001
858      1101011010
891      1101111011
924      1110011100
957      1110111101
990      1111011110

Decimal  Base 16
17       11
34       22
51       33
68       44
85       55
102      66
119      77
136      88
153      99
170      aa
187      bb
204      cc
221      dd
238      ee
255      ff
```

### Generator version

```function twoidenticalstringsinbase(base, mx, verbose = true)
gen = filter(x -> x < mx,
reduce(vcat, [[i * (base^d + 1) for i in base^(d-1):base^d-1] for d in 1:ndigits(mx; base) ÷ 2]))
if verbose
println("\nDecimal  Base \$base")
foreach(n-> println(rpad(n, 9), string(n, base = base)), gen)
end
return gen
end

twoidenticalstringsinbase(2, 999)
twoidenticalstringsinbase(16, 999)
```
Output:
`Same as filter version above.`

## Liberty BASIC

```maxNumber = 1000
maxN = Int(Len(DecToBin\$(maxNumber))/ 2)
Print "Value","    Binary"
'Since 1 is obviously not applicable,
'just count to ((2^maxN) - 2); Using "- 2" because
'we know that ((2^5) - 1) = 1023 which is > 1000
For i = 1 To ((2^maxN) - 2)
bin\$ = DecToBin\$(i)
'Let's format the output nicely
Print (((2^Len(bin\$))*i) + i),Space\$((maxN * 2) - Len(bin\$;bin\$));bin\$;bin\$
Next i
End

Function DecToBin\$(decNum)
While decNum
DecToBin\$ = (decNum Mod 2);DecToBin\$
decNum = Int(decNum/ 2)
Wend
If (DecToBin\$ = "") Then DecToBin\$ = "0"
End Function```
Output:
```Value             Binary
3                     11
10                  1010
15                  1111
36                100100
45                101101
54                110110
63                111111
136             10001000
153             10011001
170             10101010
187             10111011
204             11001100
221             11011101
238             11101110
255             11111111
528           1000010000
561           1000110001
594           1001010010
627           1001110011
660           1010010100
693           1010110101
726           1011010110
759           1011110111
792           1100011000
825           1100111001
858           1101011010
891           1101111011
924           1110011100
957           1110111101
990           1111011110```

## MACRO-11

```        .TITLE  IDNSTR
.MCALL  .TTYOUT,.EXIT
IDNSTR::CLR     R4
BR      2\$
1\$:     MOV     R3,R0
JSR     PC,PRDEC
MOV     R3,R0
JSR     PC,PRBIN
2\$:     INC     R4
JSR     PC,IDENT
CMP     R3,#^D1000
BLT     1\$
.EXIT

; LET R3 BE R4'TH IDENTICAL NUMBER
IDENT:  MOV     R4,R3
MOV     R4,R2
1\$:     ASL     R3
ASR     R2
BNE     1\$
BIS     R4,R3
RTS     PC

; PRINT NUMBER IN R0 AS DECIMAL
PRDEC:  MOV     #4\$,R1
1\$:     MOV     #-1,R2
2\$:     INC     R2
SUB     #12,R0
BCC     2\$
MOVB    R0,-(R1)
MOV     R2,R0
BNE     1\$
3\$:     MOVB    (R1)+,R0
.TTYOUT
BNE     3\$
RTS     PC
.ASCII  /...../
4\$:     .BYTE   11,0
.EVEN

; PRINT NUMBER IN R0 AS BINARY
PRBIN:  MOV     #3\$,R1
1\$:     MOV     #60,R2
ASR     R0
MOVB    R2,-(R1)
TST     R0
BNE     1\$
2\$:     MOVB    (R1)+,R0
.TTYOUT
BNE     2\$
RTS     PC
.BLKB   20
3\$:     .BYTE   15,12,0
.END    IDNSTR```
Output:
```3       11
10      1010
15      1111
36      100100
45      101101
54      110110
63      111111
136     10001000
153     10011001
170     10101010
187     10111011
204     11001100
221     11011101
238     11101110
255     11111111
528     1000010000
561     1000110001
594     1001010010
627     1001110011
660     1010010100
693     1010110101
726     1011010110
759     1011110111
792     1100011000
825     1100111001
858     1101011010
891     1101111011
924     1110011100
957     1110111101
990     1111011110```

```            NORMAL MODE IS INTEGER

INTERNAL FUNCTION(BT)
ENTRY TO BITS.
BITN = BT
BITRSL = 0
BITIDX = 1
GETBIT      WHENEVER BITN.G.0
BITNX = BITN/2
BITRSL = BITRSL + BITIDX*(BITN-BITNX*2)
BITN = BITNX
BITIDX = BITIDX*10
TRANSFER TO GETBIT
END OF CONDITIONAL
FUNCTION RETURN BITRSL
END OF FUNCTION

INTERNAL FUNCTION(DVAL)
ENTRY TO DPLBIT.
DTEMP = DVAL
DSHFT = DVAL
DSTEP       WHENEVER DTEMP.G.0
DSHFT = DSHFT * 2
DTEMP = DTEMP / 2
TRANSFER TO DSTEP
END OF CONDITIONAL
FUNCTION RETURN DSHFT + DVAL
END OF FUNCTION

THROUGH NUM, FOR N=1, 1, DPLBIT.(N).GE.1000
NUM         PRINT FORMAT NFMT, DPLBIT.(N), BITS.(DPLBIT.(N))

VECTOR VALUES NFMT = \$I3,2H: ,I10*\$
END OF PROGRAM```
Output:
```  3:         11
10:       1010
15:       1111
36:     100100
45:     101101
54:     110110
63:     111111
136:   10001000
153:   10011001
170:   10101010
187:   10111011
204:   11001100
221:   11011101
238:   11101110
255:   11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110```

## Mathematica/Wolfram Language

```max = 1000;
maxbin = Ceiling[Ceiling[Log2[max]]/2];
s = Table[
id = IntegerDigits[i, 2];
s = Join[id, id];
{FromDigits[s, 2], StringJoin[ToString /@ s]}
,
{i, 2^maxbin - 1}
];
Select[s, First/*LessThan[1000]] // Grid
```
Output:
```3	11
10	1010
15	1111
36	100100
45	101101
54	110110
63	111111
136	10001000
153	10011001
170	10101010
187	10111011
204	11001100
221	11011101
238	11101110
255	11111111
528	1000010000
561	1000110001
594	1001010010
627	1001110011
660	1010010100
693	1010110101
726	1011010110
759	1011110111
792	1100011000
825	1100111001
858	1101011010
891	1101111011
924	1110011100
957	1110111101
990	1111011110```

## Miranda

```main :: [sys_message]
main = [Stdout (lay (map display (takewhile (< 1000) identicals)))]
where display n = shownum n ++ ": " ++ bits n

bits :: num->[char]
bits = bits' []
where bits' acc 0 = acc
bits' acc n = bits' (decode (48 + n mod 2) : acc) (n div 2)

identicals :: [num]
identicals = map identical [1..]

identical :: num->num
identical n = n + n * 2^size n

size :: num->num
size 0 = 0
size n = 1 + size (n div 2)```
Output:
```3: 11
10: 1010
15: 1111
36: 100100
45: 101101
54: 110110
63: 111111
136: 10001000
153: 10011001
170: 10101010
187: 10111011
204: 11001100
221: 11011101
238: 11101110
255: 11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110```

## Modula-2

```MODULE IdenticalStrings;
FROM InOut IMPORT WriteString, WriteCard, WriteLn;

VAR n: CARDINAL;

PROCEDURE identical(n: CARDINAL): CARDINAL;
VAR shiftL, shiftR: CARDINAL;
BEGIN
shiftL := n;
shiftR := n;
WHILE shiftR > 0 DO
shiftL := shiftL * 2;
shiftR := shiftR DIV 2;
END;
RETURN shiftL + n;
END identical;

PROCEDURE WriteBits(n: CARDINAL);
BEGIN
IF n>1 THEN WriteBits(n DIV 2); END;
WriteCard(n MOD 2, 1);
END WriteBits;

BEGIN
n := 1;
WHILE identical(n) < 1000 DO
WriteCard(identical(n), 3);
WriteString(": ");
WriteBits(identical(n));
WriteLn;
INC(n);
END;
END IdenticalStrings.
```
Output:
```  3: 11
10: 1010
15: 1111
36: 100100
45: 101101
54: 110110
63: 111111
136: 10001000
153: 10011001
170: 10101010
187: 10111011
204: 11001100
221: 11011101
238: 11101110
255: 11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110```

## Nim

```import strformat

func isConcat(s: string): bool =
if (s.len and 1) != 0: return false
let half = s.len shr 1
result = s[0..<half] == s[half..^1]

for n in 0..999:
let b = &"{n:b}"
if b.isConcat: echo &"{n:3} {b}"
```
Output:
```  3 11
10 1010
15 1111
36 100100
45 101101
54 110110
63 111111
136 10001000
153 10011001
170 10101010
187 10111011
204 11001100
221 11011101
238 11101110
255 11111111
528 1000010000
561 1000110001
594 1001010010
627 1001110011
660 1010010100
693 1010110101
726 1011010110
759 1011110111
792 1100011000
825 1100111001
858 1101011010
891 1101111011
924 1110011100
957 1110111101
990 1111011110```

## OCaml

```let rec bin_of_int = function
| n when n < 2 -> string_of_int n
| n -> Printf.sprintf "%s%u" (bin_of_int (n lsr 1)) (n land 1)

let rec next n l m () =
if n = l
then next n (l + l) (succ (l + l)) ()
else Seq.Cons (n * m, next (succ n) l m)
in next 1 2 3

let () =
let show n = Printf.printf "%u: %s\n" n (bin_of_int n) in
seq_task |> Seq.take_while ((>) 1000) |> Seq.iter show
```
Output:
```3: 11
10: 1010
15: 1111
36: 100100
45: 101101
54: 110110
63: 111111
136: 10001000
153: 10011001
170: 10101010
187: 10111011
204: 11001100
221: 11011101
238: 11101110
255: 11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110
```

## Pascal

Works with: Turbo Pascal
```program IdenticalStrings;
const
LIMIT = 1000;
var
n: Integer;

function BitLength(n: Integer): Integer;
var count: Integer;
begin
count := 0;
while n > 0 do
begin
n := n shr 1;
count := count + 1;
end;
BitLength := count;
end;

function Concat(n: Integer): Integer;
begin
Concat := n shl BitLength(n) or n;
end;

procedure WriteBits(n: Integer);
var bit: Integer;
begin
bit := 1 shl (BitLength(n)-1);
while bit > 0 do
begin
if (bit and n) <> 0 then Write('1')
else Write('0');
bit := bit shr 1;
end;
end;

begin
n := 1;
while Concat(n) < LIMIT do
begin
Write(Concat(n));
Write(': ');
WriteBits(Concat(n));
WriteLn;
n := n + 1;
end;
end.
```
Output:
```3: 11
10: 1010
15: 1111
36: 100100
45: 101101
54: 110110
63: 111111
136: 10001000
153: 10011001
170: 10101010
187: 10111011
204: 11001100
221: 11011101
238: 11101110
255: 11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110```

## PL/I

```identstr: procedure options(main);
bitLength: procedure(nn) returns(fixed);
declare (n, nn, r) fixed;
r = 0;
do n = nn repeat(n / 2) while(n > 0);
r = r + 1;
end;
return(r);
end bitLength;

concat: procedure(nn, m) returns(fixed);
declare (i, steps, nn, n, m) fixed;
steps = bitLength(m);
n = nn;
do i=1 to steps;
n = n * 2;
end;
return(n + m);
end concat;

printBits: procedure(nn);
declare (nn, n) fixed, bits char(16) varying;
bits = '';
do n = nn repeat(n / 2) while(n > 0);
if mod(n,2) = 0 then
bits = '0' || bits;
else
bits = '1' || bits;
end;
put list(bits);
end printBits;

declare n fixed;
do n=1 repeat(n+1) while(concat(n,n) < 1000);
put skip list(concat(n,n));
call printBits(concat(n,n));
end;
end identstr;```
Output:
```        3 11
10 1010
15 1111
36 100100
45 101101
54 110110
63 111111
136 10001000
153 10011001
170 10101010
187 10111011
204 11001100
221 11011101
238 11101110
255 11111111
528 1000010000
561 1000110001
594 1001010010
627 1001110011
660 1010010100
693 1010110101
726 1011010110
759 1011110111
792 1100011000
825 1100111001
858 1101011010
891 1101111011
924 1110011100
957 1110111101
990 1111011110```

## PL/M

```100H:

/* BINARY CONCATENATION OF A NUMBER WITH ITSELF */
J = I;
K = I;
DO WHILE J > 0;
J = SHR(J,1);
K = SHL(K,1);
END;
RETURN I OR K;
END CONCAT\$SELF;

/* CP/M BDOS CALL */
BDOS: PROCEDURE (FN, ARG);
GO TO 5;
END BDOS;

/* PRINT STRING */
PRINT: PROCEDURE (STR);
CALL BDOS(9, STR);
END PRINT;

/* PRINT NUMBER IN GIVEN BASE (MAX 10) */
PRINT\$BASE: PROCEDURE (BASE, N);
DECLARE S (17) BYTE INITIAL ('................\$');
DECLARE P ADDRESS, C BASED P BYTE;
P = .S(16);
DIGIT:
P = P - 1;
C = N MOD BASE + '0';
N = N / BASE;
IF N > 0 THEN GO TO DIGIT;
CALL PRINT(P);
END PRINT\$BASE;

DO WHILE (C := CONCAT\$SELF(N)) < 1000;
CALL PRINT\$BASE(10, C);
CALL PRINT(.(9,'\$'));
CALL PRINT\$BASE(2, C);
CALL PRINT(.(13,10,'\$'));
N = N + 1;
END;

CALL BDOS(0,0);
EOF```
Output:
```3       11
10      1010
15      1111
36      100100
45      101101
54      110110
63      111111
136     10001000
153     10011001
170     10101010
187     10111011
204     11001100
221     11011101
238     11101110
255     11111111
528     1000010000
561     1000110001
594     1001010010
627     1001110011
660     1010010100
693     1010110101
726     1011010110
759     1011110111
792     1100011000
825     1100111001
858     1101011010
891     1101111011
924     1110011100
957     1110111101
990     1111011110```

## Python

### Python: Procedural

```def bits(n):
"""Count the amount of bits required to represent n"""
r = 0
while n:
n >>= 1
r += 1
return r

def concat(n):
"""Concatenate the binary representation of n to itself"""
return n << bits(n) | n

n = 1
while concat(n) <= 1000:
print("{0}: {0:b}".format(concat(n)))
n += 1
```
Output:
```3: 11
10: 1010
15: 1111
36: 100100
45: 101101
54: 110110
63: 111111
136: 10001000
153: 10011001
170: 10101010
187: 10111011
204: 11001100
221: 11011101
238: 11101110
255: 11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110```

### Python: Functional

A variant composed from pure functions, as an alternative to using mutable variables and a loop.

Values are drawn, up to a limit, from a non-finite list.

```'''Two identical strings'''

from itertools import count, takewhile

# binaryTwin :: Int -> (Int, String)
def binaryTwin(n):
'''A tuple of an integer m and a string s, where
s is a self-concatenation of the binary
represention of n, and m is the integer value of s.
'''
s = bin(n)[2:] * 2
return int(s, 2), s

# ------------------------- TEST -------------------------
def main():
'''Numbers defined by duplicated binary sequences,
up to a limit of decimal 1000.
'''
print(
'\n'.join([
repr(pair) for pair
in takewhile(
lambda x: 1000 > x[0],
map(binaryTwin, count(1))
)
])
)

# MAIN ---
if __name__ == '__main__':
main()
```
Output:
```(3, '11')
(10, '1010')
(15, '1111')
(36, '100100')
(45, '101101')
(54, '110110')
(63, '111111')
(136, '10001000')
(153, '10011001')
(170, '10101010')
(187, '10111011')
(204, '11001100')
(221, '11011101')
(238, '11101110')
(255, '11111111')
(528, '1000010000')
(561, '1000110001')
(594, '1001010010')
(627, '1001110011')
(660, '1010010100')
(693, '1010110101')
(726, '1011010110')
(759, '1011110111')
(792, '1100011000')
(825, '1100111001')
(858, '1101011010')
(891, '1101111011')
(924, '1110011100')
(957, '1110111101')
(990, '1111011110')```

## Perl

```#!/usr/bin/perl

use strict; # https://rosettacode.org/wiki/Two_identical_strings
use warnings;

while( 1 )
{
my \$binary = ( sprintf "%b", ++\$- ) x 2;
(my \$decimal = oct "b\$binary") >= 1000 and last;
printf "%4d  %s\n", \$decimal, \$binary;
}
```
Output:
```   3  11
10  1010
15  1111
36  100100
45  101101
54  110110
63  111111
136  10001000
153  10011001
170  10101010
187  10111011
204  11001100
221  11011101
238  11101110
255  11111111
528  1000010000
561  1000110001
594  1001010010
627  1001110011
660  1010010100
693  1010110101
726  1011010110
759  1011110111
792  1100011000
825  1100111001
858  1101011010
891  1101111011
924  1110011100
957  1110111101
990  1111011110
```

## Phix

```integer n = 1
sequence res = {}
while true do
string binary = sprintf("%b%b",n)
integer decimal = to_number(binary,0,2)
if decimal>1000 then exit end if
res &= {sprintf("%-4d %-10s",{decimal,binary})}
n += 1
end while
printf(1,"Found %d numbers:\n%s\n",{n-1,join_by(res,5,6)})
```
Output:
```Found 30 numbers:
3    11           54   110110       187  10111011     528  1000010000   693  1010110101   858  1101011010
10   1010         63   111111       204  11001100     561  1000110001   726  1011010110   891  1101111011
15   1111         136  10001000     221  11011101     594  1001010010   759  1011110111   924  1110011100
36   100100       153  10011001     238  11101110     627  1001110011   792  1100011000   957  1110111101
45   101101       170  10101010     255  11111111     660  1010010100   825  1100111001   990  1111011110
```

## Picat

```main =>
bp.length(_L,I),
I > 0,
B = to_binary_string(I),
BB = B++B,
( Dec < 1000 -> printf("%4w %10w\n",Dec,BB), fail ; true),
nl.```
Output:
```   3         11
10       1010
15       1111
36     100100
45     101101
54     110110
63     111111
136   10001000
153   10011001
170   10101010
187   10111011
204   11001100
221   11011101
238   11101110
255   11111111
528 1000010000
561 1000110001
594 1001010010
627 1001110011
660 1010010100
693 1010110101
726 1011010110
759 1011110111
792 1100011000
825 1100111001
858 1101011010
891 1101111011
924 1110011100
957 1110111101
990 1111011110```

## Prolog

Works with: SWI Prolog
```main:-
writeln('Decimal\tBinary'),
main(1, 1000).

main(N, Limit):-
format(string(Binary), '~2r', N),
string_length(Binary, Length),
I is N + (N << Length),
I < Limit,
!,
writef('%w\t%w%w\n', [I, Binary, Binary]),
N1 is N + 1,
main(N1, Limit).
main(_, _).
```
Output:
```Decimal	Binary
3	11
10	1010
15	1111
36	100100
45	101101
54	110110
63	111111
136	10001000
153	10011001
170	10101010
187	10111011
204	11001100
221	11011101
238	11101110
255	11111111
528	1000010000
561	1000110001
594	1001010010
627	1001110011
660	1010010100
693	1010110101
726	1011010110
759	1011110111
792	1100011000
825	1100111001
858	1101011010
891	1101111011
924	1110011100
957	1110111101
990	1111011110
```

## Quackery

```  [ 2 base put
number\$ dup size 2 / split =
base release ]               is 2identical ( n --> b )

1000 times
[ i^ 2identical if
[ i^ echo sp
2 base put
i^ echo cr
base release ] ]```
Output:
```3 11
10 1010
15 1111
36 100100
45 101101
54 110110
63 111111
136 10001000
153 10011001
170 10101010
187 10111011
204 11001100
221 11011101
238 11101110
255 11111111
528 1000010000
561 1000110001
594 1001010010
627 1001110011
660 1010010100
693 1010110101
726 1011010110
759 1011110111
792 1100011000
825 1100111001
858 1101011010
891 1101111011
924 1110011100
957 1110111101
990 1111011110
```

## Raku

```my @cat = (1..*).map: { :2([~] .base(2) xx 2) };
say "{+\$_} matching numbers\n{.batch(5)».map({\$_ ~ .base(2).fmt('(%s)')})».fmt('%15s').join: "\n"}\n"
given @cat[^(@cat.first: * > 1000, :k)];
```
Output:
```30 matching numbers
3(11)        10(1010)        15(1111)      36(100100)      45(101101)
54(110110)      63(111111)   136(10001000)   153(10011001)   170(10101010)
187(10111011)   204(11001100)   221(11011101)   238(11101110)   255(11111111)
528(1000010000) 561(1000110001) 594(1001010010) 627(1001110011) 660(1010010100)
693(1010110101) 726(1011010110) 759(1011110111) 792(1100011000) 825(1100111001)
858(1101011010) 891(1101111011) 924(1110011100) 957(1110111101) 990(1111011110)```

## Refal

```\$ENTRY Go {
= <IdentUpTo 1000>;
};

IdentUpTo {
s.M = <IdentUpTo s.M 1>;
s.M s.I, <Ident s.I>: s.Id, <Compare s.Id s.M>: {
'-' = <Prout <Symb s.Id> ': ' <Bin s.Id>>
<IdentUpTo s.M <+ s.I 1>>;
s.X = ;
};
};

Ident {
s.N = <Ident s.N s.N s.N>;
s.N s.R 0 = <+ s.N s.R>;
s.N s.R s.K = <Ident s.N <* s.R 2> <Div s.K 2>>;
};

Bin {
0 = ;
s.N, <Divmod s.N 2>: (s.X) s.Y = <Bin s.X> <Symb s.Y>;
};```
Output:
```3: 11
10: 1010
15: 1111
36: 100100
45: 101101
54: 110110
63: 111111
136: 10001000
153: 10011001
170: 10101010
187: 10111011
204: 11001100
221: 11011101
238: 11101110
255: 11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110```

## REXX

```/*REXX
* program finds/displays decimal numbers
* whose binary version is a doubled literal.
*/
numeric digits 20        /*ensure 'nuff dec. digs for conversion*/
do i=1 to 1000
b= x2b( d2x(i) ) + 0  /*find binary values that can be split.*/
L= length(b)
if L//2  then iterate /*get length of binary;  if odd, skip. */
if left(b, L%2)\==right(b, L%2)  then iterate /*Left half ≡ right half?*/
say right(i, 4)':'   right(b, 12)      /*display number in dec and bin */
end   /*i*/            /*stick a fork in it,  we're all done. */
*/
```
output   (shown at three-quarter size.)
```   3:           11
10:         1010
15:         1111
36:       100100
45:       101101
54:       110110
63:       111111
136:     10001000
153:     10011001
170:     10101010
187:     10111011
204:     11001100
221:     11011101
238:     11101110
255:     11111111
528:   1000010000
561:   1000110001
594:   1001010010
627:   1001110011
660:   1010010100
693:   1010110101
726:   1011010110
759:   1011110111
792:   1100011000
825:   1100111001
858:   1101011010
891:   1101111011
924:   1110011100
957:   1110111101
990:   1111011110
```

### with formatting

```/*REXX program finds/displays decimal numbers whose binary version is a doubled literal.*/
numeric digits 100                               /*ensure hangling of larger integers.  */
parse arg hi cols .                              /*obtain optional argument from the CL.*/
if   hi=='' |   hi==","  then   hi= 1000         /* "      "         "   "   "     "    */
if cols=='' | cols==","  then cols=    4         /* "      "         "   "   "     "    */
w= 20                                            /*width of a number in any column.     */
title= ' decimal integers whose binary version is a doubled binary literal, N  < '   ,
commas(hi)
if cols>0  then say ' index │'center(title,   1 + cols*(w+1)     )
if cols>0  then say '───────┼'center(""   ,   1 + cols*(w+1), '─')
#= 0;                   idx= 1                   /*initialize # of integers and index.  */
\$=                                               /*a list of  nice  primes  (so far).   */
do j=1  for hi-1;  b= x2b( d2x(j) ) + 0     /*find binary values that can be split.*/
L= length(b);      h= L % 2                 /*obtain length of the binary value.   */
if L//2                      then iterate   /*Can binary version be split? No, skip*/
if left(b, h)\==right(b, h)  then iterate   /*Left half match right half?   "    " */
#= # + 1                                    /*bump the number of integers found.   */
if cols<=0                   then iterate   /*Build the list  (to be shown later)? */
c= commas(j) || '(' || b")"                 /*maybe add commas, add binary version.*/
\$= \$  right(c, max(w, length(c) ) )         /*add a nice prime ──► list, allow big#*/
if #//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.*/
if cols>0  then say '───────┴'center(""   ,   1 + cols*(w+1), '─')
say
say 'Found '       commas(#)         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 ?
```
output   when using the default inputs:
``` index │    decimal integers whose binary version is a doubled binary literal, N  <  1,000
───────┼─────────────────────────────────────────────────────────────────────────────────────
1   │                3(11)             10(1010)             15(1111)           36(100100)
5   │           45(101101)           54(110110)           63(111111)        136(10001000)
9   │        153(10011001)        170(10101010)        187(10111011)        204(11001100)
13   │        221(11011101)        238(11101110)        255(11111111)      528(1000010000)
17   │      561(1000110001)      594(1001010010)      627(1001110011)      660(1010010100)
21   │      693(1010110101)      726(1011010110)      759(1011110111)      792(1100011000)
25   │      825(1100111001)      858(1101011010)      891(1101111011)      924(1110011100)
29   │      957(1110111101)      990(1111011110)
───────┴─────────────────────────────────────────────────────────────────────────────────────

Found  30  decimal integers whose binary version is a doubled binary literal, N  <  1,000
```

## Ring

```load "stdlib.ring"

decList = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
baseList = ["0","1","2","3","4","5","6","7","8","9","A","B","C","D","E","F"]

see "working..." + nl
see "Numbers whose base 2 representation is the juxtaposition of two identical strings:" + nl

row = 0
limit1 = 1000

for n = 1 to limit1
bin = decimaltobase(n,2)
ln = len(bin)
if ln & 1 = 0
if left(bin,ln/2) = right(bin,ln/2)
row++
see sfl(n, 3) + " (" + sfrs(bin, 10) + ")  "
if row % 5 = 0 see nl ok
ok
ok
next

? nl + "Found " + row + " numbers whose base 2 representation is the juxtaposition of two identical strings"
? "done..."

func decimaltobase(nr,base)
binList = []
binary = 0
remainder = 1
while(nr != 0)
remainder = nr % base
ind = find(decList,remainder)
rem = baseList[ind]
nr = floor(nr/base)
end
binlist = reverse(binList)
binList = list2str(binList)
binList = substr(binList,nl,"")
return binList

# a very plain string formatter, intended to even up columnar outputs
def sfrs x, y
l = len(x)
x += "            "
if l > y y = l ok
return substr(x, 1, y)

# a very plain string formatter, intended to even up columnar outputs
def sfl x, y
s = string(x) l = len(s)
if l > y y = l ok
return substr("          ", 11 - y + l) + s```
Output:
```working...
Numbers whose base 2 representation is the juxtaposition of two identical strings:
3 (11        )   10 (1010      )   15 (1111      )   36 (100100    )   45 (101101    )
54 (110110    )   63 (111111    )  136 (10001000  )  153 (10011001  )  170 (10101010  )
187 (10111011  )  204 (11001100  )  221 (11011101  )  238 (11101110  )  255 (11111111  )
528 (1000010000)  561 (1000110001)  594 (1001010010)  627 (1001110011)  660 (1010010100)
693 (1010110101)  726 (1011010110)  759 (1011110111)  792 (1100011000)  825 (1100111001)
858 (1101011010)  891 (1101111011)  924 (1110011100)  957 (1110111101)  990 (1111011110)

Found 30 numbers whose base 2 representation is the juxtaposition of two identical strings
done...```

## RPL

### Slow version

Binary versions of numbers from 1 to 999 are converted into strings to check the half-half identity.

Works with: Halcyon Calc version 4.2.8
```≪ R→B →STR 3 OVER SIZE 1 - SUB
IF DUP SIZE 2 MOD THEN DROP 0
ELSE
1 OVER SIZE 2 / SUB
LAST SWAP DROP 1 + OVER SIZE SUB ==
END
≫ ‘TWIN?' STO

≪ BIN { } 1 999 FOR n
IF n TWIN?
THEN n →STR "=" +
n R→B →STR 2 OVER SIZE 1 - SUB + + END NEXT
```
Output:
```1: { "3= 11" "10= 1010" "15= 1111" "36= 100100" "45= 101101" "54= 110110" "63= 111111" "136= 10001000" "153= 10011001" "170= 10101010" "187= 10111011" "204= 11001100" "221= 11011101" "238= 11101110" "255= 11111111" "528= 1000010000" "561= 1000110001" "594= 1001010010" "627= 1001110011" "660= 1010010100" "693= 1010110101" "726= 1011010110" "759= 1011110111" "792= 1100011000" "825= 1100111001" "858= 1101011010" "891= 1101111011" "924= 1110011100" "957= 1110111101" "990= 1111011110" }
```

Runs on an HP-28S in 130 seconds.

### Optimized version

Translation of: FreeBASIC

40% of the code is dedicated to the display of the results.

Works with: Halcyon Calc version 4.2.8
RPL code Comment
``` ≪ BIN { }
2 1
DO
IF DUP2 ≤ THEN OVER ROT + SWAP END
DUP2 * OVER +
4 ROLL OVER DUP →STR "=" +
SWAP R→B →STR 2 OVER SIZE 1 - SUB + + 4 ROLLD
SWAP 1 +
UNTIL SWAP 1000 ≥ END DROP2
```
```TASK (  -- { "results" } )
n = 1, p = 2
do
if n >= p then p = p + p
k = n + n * p
print "k= ";
print "bin(k)"
n = n + 1
loop until k ≥ 1000

```

Runs on an HP-28S in 6 seconds.

## Ruby

Translation of: C
```for i in 0 .. 1000
bin = i.to_s(2)
if bin.length % 2 == 0 then
half = bin.length / 2
if bin[0..half-1] == bin[half..] then
print "%3d: %10s\n" % [i, bin]
end
end
end
```
Output:
```  3:         11
10:       1010
15:       1111
36:     100100
45:     101101
54:     110110
63:     111111
136:   10001000
153:   10011001
170:   10101010
187:   10111011
204:   11001100
221:   11011101
238:   11101110
255:   11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110```

## SETL

```program identical_strings;
loop init i := 0; doing n := ident(i +:= 1); while n<1000 do
end loop;

proc ident(n);
ns := n;
loop init t := n; doing t div:= 2; until t = 0 do
ns *:= 2;
end loop;
return ns + n;
end proc;

proc binary(n);
return {[0,""]}(n) ? binary(n div 2) + str (n mod 2);
end proc;
end program;```
Output:
```   3             11
10           1010
15           1111
36         100100
45         101101
54         110110
63         111111
136       10001000
153       10011001
170       10101010
187       10111011
204       11001100
221       11011101
238       11101110
255       11111111
528     1000010000
561     1000110001
594     1001010010
627     1001110011
660     1010010100
693     1010110101
726     1011010110
759     1011110111
792     1100011000
825     1100111001
858     1101011010
891     1101111011
924     1110011100
957     1110111101
990     1111011110```

## Snobol

```        define("bits(n)")                   :(bits_end)
bits    bits = gt(n,0) remdr(n,2) bits      :f(return)
n = n / 2                           :(bits)
bits_end

define("concat(n)m")                :(concat_end)
concat  concat = n
m = n
c_loop  m = gt(m,0) m / 2                   :f(c_done)
concat = concat * 2                 :(c_loop)
c_done  concat = concat + n                 :(return)
concat_end

n = 0
loop    n = n + 1
m = concat(n)
output = lt(m,1000) m ": " bits(m)  :s(loop)
end
```
Output:
```3: 11
10: 1010
15: 1111
36: 100100
45: 101101
54: 110110
63: 111111
136: 10001000
153: 10011001
170: 10101010
187: 10111011
204: 11001100
221: 11011101
238: 11101110
255: 11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110```

## Swift

```print("Decimal\tBinary")
var n = 1
while (true) {
let binary = String(n, radix: 2)
let i = n + (n << binary.count)
if i >= 1000 {
break
}
print("\(i)\t\(binary)\(binary)")
n += 1
}
```
Output:
```Decimal	Binary
3	11
10	1010
15	1111
36	100100
45	101101
54	110110
63	111111
136	10001000
153	10011001
170	10101010
187	10111011
204	11001100
221	11011101
238	11101110
255	11111111
528	1000010000
561	1000110001
594	1001010010
627	1001110011
660	1010010100
693	1010110101
726	1011010110
759	1011110111
792	1100011000
825	1100111001
858	1101011010
891	1101111011
924	1110011100
957	1110111101
990	1111011110
```

## Visual Basic .NET

Translation of: FreeBASIC

Based on the Alternate version.

```Imports System.Console
Module Module1
Sub Main()
Dim p, c, k, lmt as integer : p = 2 : lmt = 1000
For n As Integer = 1 to lmt
p += If(n >= p, p, 0) : k = n + n * p
If k > lmt Then Exit For Else c += 1
Write("{0,3} ({1,-10})  {2}", k, Convert.ToString( k, 2),
If(c Mod 5 = 0, vbLf, ""))
Next : WriteLine(vbLf + "Found {0} numbers whose base 2 representation is the concatenation of two identical binary strings.", c)
End Sub
End Module
```
Output:
```  3 (11        )   10 (1010      )   15 (1111      )   36 (100100    )   45 (101101    )
54 (110110    )   63 (111111    )  136 (10001000  )  153 (10011001  )  170 (10101010  )
187 (10111011  )  204 (11001100  )  221 (11011101  )  238 (11101110  )  255 (11111111  )
528 (1000010000)  561 (1000110001)  594 (1001010010)  627 (1001110011)  660 (1010010100)
693 (1010110101)  726 (1011010110)  759 (1011110111)  792 (1100011000)  825 (1100111001)
858 (1101011010)  891 (1101111011)  924 (1110011100)  957 (1110111101)  990 (1111011110)

Found 30 numbers whose base 2 representation is the concatenation of two identical binary strings.```

## V (Vlang)

Translation of: Go
```import strconv

fn main() {
mut i := i64(1)
for {
mut b2 := '\${i:b}'
b2 += b2
d := strconv.parse_int(b2,2,16) ?
if d >= 1000 {
break
}
println("\$d : \$b2")
i++
}
println("\nFound \${i-1} numbers.")
}```
Output:
```3 : 11
10 : 1010
15 : 1111
36 : 100100
45 : 101101
54 : 110110
63 : 111111
136 : 10001000
153 : 10011001
170 : 10101010
187 : 10111011
204 : 11001100
221 : 11011101
238 : 11101110
255 : 11111111
528 : 1000010000
561 : 1000110001
594 : 1001010010
627 : 1001110011
660 : 1010010100
693 : 1010110101
726 : 1011010110
759 : 1011110111
792 : 1100011000
825 : 1100111001
858 : 1101011010
891 : 1101111011
924 : 1110011100
957 : 1110111101
990 : 1111011110

Found 30 numbers.
```

## Wren

Library: Wren-fmt
```import "./fmt" for Conv, Fmt

var i = 1
while(true) {
var b2 = Conv.itoa(i, 2)
b2 = b2 + b2
var d = Conv.atoi(b2, 2)
if (d >= 1000) break
Fmt.print("\$3d : \$s", d, b2)
i = i + 1
}
System.print("\nFound %(i-1) numbers.")
```
Output:
```  3 : 11
10 : 1010
15 : 1111
36 : 100100
45 : 101101
54 : 110110
63 : 111111
136 : 10001000
153 : 10011001
170 : 10101010
187 : 10111011
204 : 11001100
221 : 11011101
238 : 11101110
255 : 11111111
528 : 1000010000
561 : 1000110001
594 : 1001010010
627 : 1001110011
660 : 1010010100
693 : 1010110101
726 : 1011010110
759 : 1011110111
792 : 1100011000
825 : 1100111001
858 : 1101011010
891 : 1101111011
924 : 1110011100
957 : 1110111101
990 : 1111011110

Found 30 numbers.
```

## XPL0

```proc BinOut(N);         \Output N in binary
int  N, Rem;
[Rem:= N&1;
N:= N>>1;
if N then BinOut(N);
ChOut(0, Rem+^0);
];

int H, N, M;
[for H:= 1 to 31 do
[N:= H;  M:= H;
while M do
[N:= N<<1;  M:= M>>1];
N:= N+H;
if N < 1000 then
[IntOut(0, N);
Text(0, ": ");
BinOut(N);
CrLf(0);
];
];
]```
Output:
```3: 11
10: 1010
15: 1111
36: 100100
45: 101101
54: 110110
63: 111111
136: 10001000
153: 10011001
170: 10101010
187: 10111011
204: 11001100
221: 11011101
238: 11101110
255: 11111111
528: 1000010000
561: 1000110001
594: 1001010010
627: 1001110011
660: 1010010100
693: 1010110101
726: 1011010110
759: 1011110111
792: 1100011000
825: 1100111001
858: 1101011010
891: 1101111011
924: 1110011100
957: 1110111101
990: 1111011110
```

## Yabasic

```// Rosetta Code problem: http://rosettacode.org/wiki/Two_identical_strings
// by Galileo, 04/2022

for n = 1 to 1000
n\$ = bin\$(n)
if not mod(len(n\$), 2) then
k = len(n\$) / 2
if left\$(n\$, k) = right\$(n\$, k) print n, " = ", n\$
endif
next```
Output:
```3 = 11
10 = 1010
15 = 1111
36 = 100100
45 = 101101
54 = 110110
63 = 111111
136 = 10001000
153 = 10011001
170 = 10101010
187 = 10111011
204 = 11001100
221 = 11011101
238 = 11101110
255 = 11111111
528 = 1000010000
561 = 1000110001
594 = 1001010010
627 = 1001110011
660 = 1010010100
693 = 1010110101
726 = 1011010110
759 = 1011110111
792 = 1100011000
825 = 1100111001
858 = 1101011010
891 = 1101111011
924 = 1110011100
957 = 1110111101
990 = 1111011110
---Program done, press RETURN---```