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# Happy numbers

(Redirected from Happy Number)
Happy numbers
You are encouraged to solve this task according to the task description, using any language you may know.

A happy number is defined by the following process:
Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals   1   (where it will stay),   or it loops endlessly in a cycle which does not include   1.

Those numbers for which this process end in   1   are       happy   numbers,
while   those numbers   that   do   not   end in   1   are   unhappy   numbers.

Find and print the first   8   happy numbers.

## 11l

Translation of: Python
`F happy(=n)   Set[Int] past   L n != 1      n = sum(String(n).map(с -> Int(с)^2))      I n C past         R 0B      past.add(n)   R 1B print((0.<500).filter(x -> happy(x))[0.<8])`
Output:
```[1, 7, 10, 13, 19, 23, 28, 31]
```

## 8080 Assembly

This is not just a demonstration of 8080 assembly, but also of why it pays to look closely at the problem domain. The following program only does 8-bit unsigned integer math, which not only fits the 8080's instruction set very well, it also means the cycle detection can be done using only an array of 256 flags, and all other state fits in the registers. This makes the program a good deal simpler than it would've been otherwise.

In general, 8-bit math is not good enough for numerical problems, but in this particular case, the problem only asks for the first eight happy numbers, none of which (nor any of the unhappy numbers in between) have a cycle that ever goes above 145, so eight bits is good enough. In fact, for any input under 256, the cycle never goes above 163; this program could be trivially changed to print up to 39 happy numbers.

`flags:	equ	2	; 256-byte page in which to keep track of cyclesputs:	equ	9	; CP/M print stringbdos:	equ	5 	; CP/M entry point	org	100h	lxi	d,0108h	; D=current number to test, E=amount of numbers	;;;	Is D happy?number:	mvi	a,1	; We haven't seen any numbers yet, set flags to 1	lxi	h,256*flagsinit:	mov	m,a	inr	l	jnz	init	mov	a,d	; Get digitsstep:	call	digits	mov	l,a	; L = D1 * D1	mov	h,a	xra	asqr1:	add	h	dcr	l	jnz	sqr1	mov	l,a	mov	h,b	; L += D10 * D10	xra	asqr10:	add	h	dcr 	b	jnz	sqr10	add	l	mov	l,a	mov	h,c	; L += D100 * D100	xra	asqr100:	add	h	dcr	c	jnz	sqr100	add	l	mov	l,a	mvi	h,flags	; Look up corresponding flag	dcr	m 	; Will give 0 the first time and not-0 afterwards	mov	a,l	; If we haven't seen the number before, another step	jz 	step 	dcr	l	; If we _had_ seen it, then is it 1?	jz	happy	; If so, it is happynext:	inr	d	; Afterwards, try next number	jmp	numberhappy:	mov	a,d	; D is happy - get its digits (for output)	lxi	h,string+3	call	digits	; Write digits into string for output	call	sdgt	; Ones digit,	mov	a,b	; Tens digit,	call	sdgt	mov	a,c	; Hundreds digit	call	sdgt	push	d	; Keep counters on stack	mvi	c,puts	; Print string using CP/M call	xchg		call	bdos	pop	d	; Restore counters	dcr	e	; One fewer happy number left	jnz	next	; If we need more, do the next one	ret	;;;	Store A as ASCII digit in [HL] and go to previous digitsdgt:	adi	'0'	dcx	h	mov	m,a	ret	;;;	Get digits of 8-bit number in A.	;;;	Input: A = number	;;;	Output: C=100s digit, B=10s digit, A=1s digitdigits:	lxi	b,-1	; Set B and C to -1 (correct for extra loop cycle)d100:	inr	c	; Calculate hundreds digit	sui	100	; By trial subtraction of 100	jnc	d100	; Until underflow occurs	adi	100	; Loop runs one cycle too many, so add 100 backd10:	inr	b	; Calculate 10s digit in the same way	sui	10	jnc	d10	adi	10	ret		; 1s digit is left in A afterwardsstring:	db	'000',13,10,'\$'`
Output:
```001
007
010
013
019
023
028
031```

## 8th

` : until!  "not while!" eval i; with: wwith: n : sumsqd  \ n -- n    0 swap repeat        0; 10 /mod -rot sqr + swap    again ; : cycle \ n xt -- n    >r    dup [email protected] exec  \ -- tortoise, hare    repeat        swap [email protected] exec        swap [email protected] exec [email protected] exec    2dup = until!    rdrop drop ; : happy?  ' sumsqd cycle 1 = ; : .happy \ n --    1 repeat        dup happy? if  dup . space  swap 1- swap  then 1+    over 0 > while!    2drop cr ; ;with;with `
Output:
```ok> 8 .happy
1 7 10 13 19 23 28 31
```

## ACL2

`(include-book "arithmetic-3/top" :dir :system) (defun sum-of-digit-squares (n)   (if (zp n)       0       (+ (expt (mod n 10) 2)          (sum-of-digit-squares (floor n 10))))) (defun is-happy-r (n seen)   (let ((next (sum-of-digit-squares n)))        (cond ((= next 1) t)              ((member next seen) nil)              (t (is-happy-r next (cons next seen)))))) (defun is-happy (n)   (is-happy-r n nil)) (defun first-happy-nums-r (n i)   (cond ((zp n) nil)         ((is-happy i)          (cons i (first-happy-nums-r (1- n) (1+ i))))         (t (first-happy-nums-r n (1+ i))))) (defun first-happy-nums (n)   (first-happy-nums-r n 1))`

Output:

`(1 7 10 13 19 23 28 31)`

## Action!

`BYTE FUNC SumOfSquares(BYTE x)  BYTE sum,d   sum=0  WHILE x#0  DO    d=x MOD 10    d==*d    sum==+d    x==/10  ODRETURN (sum) BYTE FUNC Contains(BYTE ARRAY a BYTE count,x)  BYTE i   FOR i=0 TO count-1  DO    IF a(i)=x THEN RETURN (1) FI  ODRETURN (0) BYTE FUNC IsHappyNumber(BYTE x)  BYTE ARRAY cache(100)  BYTE count   count=0  WHILE x#1  DO    cache(count)=x    count==+1    x=SumOfSquares(x)    IF Contains(cache,count,x) THEN      RETURN (0)    FI  ODRETURN (1) PROC Main()  BYTE x,count   x=1 count=0  WHILE count<8  DO    IF IsHappyNumber(x) THEN      count==+1      PrintF("%I: %I%E",count,x)    FI    x==+1  ODRETURN`
Output:
```1: 1
2: 7
3: 10
4: 13
5: 19
6: 23
7: 28
8: 31
```

## ActionScript

`function sumOfSquares(n:uint){	var sum:uint = 0;	while(n != 0)	{		sum += (n%10)*(n%10);		n /= 10;	}	return sum;}function isInArray(n:uint, array:Array){	for(var k = 0; k < array.length; k++)		if(n == array[k]) return true;	return false;}function isHappy(n){	var sequence:Array = new Array();	while(n != 1)	{		sequence.push(n);		n = sumOfSquares(n);		if(isInArray(n,sequence))return false;	}	return true;}function printHappy(){	var numbersLeft:uint = 8;	var numberToTest:uint = 1;	while(numbersLeft != 0)	{		if(isHappy(numberToTest))		{			trace(numberToTest);			numbersLeft--;		}		numberToTest++;	}}printHappy();`

Sample output:

```1
7
10
13
19
23
28
31
```

`with Ada.Text_IO;  use Ada.Text_IO;with Ada.Containers.Ordered_Sets; procedure Test_Happy_Digits is   function Is_Happy (N : Positive) return Boolean is      package Sets_Of_Positive is new Ada.Containers.Ordered_Sets (Positive);      use Sets_Of_Positive;      function Next (N : Positive) return Natural is         Sum   : Natural := 0;         Accum : Natural := N;      begin         while Accum > 0 loop            Sum   := Sum + (Accum mod 10) ** 2;            Accum := Accum / 10;         end loop;         return Sum;      end Next;      Current : Positive := N;      Visited : Set;   begin      loop         if Current = 1 then            return True;         elsif Visited.Contains (Current) then            return False;         else            Visited.Insert (Current);            Current := Next (Current);         end if;      end loop;   end Is_Happy;   Found : Natural := 0;begin   for N in Positive'Range loop      if Is_Happy (N) then         Put (Integer'Image (N));         Found := Found + 1;         exit when Found = 8;      end if;   end loop;end Test_Happy_Digits;`

Sample output:

``` 1 7 10 13 19 23 28 31
```

## ALGOL 68

Works with: ALGOL 68 version Standard - no extensions to language used
Works with: ALGOL 68G version Any - tested with release mk15-0.8b.fc9.i386
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8.8d.fc9.i386
`INT base10 = 10, num happy = 8; PROC next = (INT in n)INT: (  INT n := in n;  INT out := 0;  WHILE n NE 0 DO    out +:= ( n MOD base10 ) ** 2;    n := n OVER base10  OD;  out); PROC is happy = (INT in n)BOOL: (  INT n := in n;  FOR i WHILE n NE 1 AND n NE 4 DO n := next(n) OD;  n=1); INT count := 0;FOR i WHILE count NE num happy DO  IF is happy(i) THEN    count +:= 1;    print((i, new line))  FIOD`

Output:

```         +1
+7
+10
+13
+19
+23
+28
+31
```

## ALGOL-M

`begininteger function mod(a,b);integer a,b;mod := a-(a/b)*b; integer function sumdgtsq(n);integer n;sumdgtsq :=    if n = 0 then 0    else mod(n,10)*mod(n,10) + sumdgtsq(n/10); integer function happy(n);integer n;begin    integer i;    integer array seen[0:200];    for i := 0 step 1 until 200 do seen[i] := 0;     while seen[n] = 0 do    begin        seen[n] := 1;        n := sumdgtsq(n);    end;    happy := if n = 1 then 1 else 0;end; integer i, n;i := n := 0;while n < 8 dobegin    if happy(i) = 1 then    begin        write(i);        n := n + 1;    end;    i := i + 1;end;end`
Output:
```     1
7
10
13
19
23
28
31```

## ALGOL W

`begin    % find some happy numbers                                      %    % returns true if n is happy, false otherwise; n must be >= 0  %    logical procedure isHappy( integer value n ) ;        if n < 2 then true        else begin            % seen is used to hold the values of the cycle of the  %            % digit square sums, as noted in the Batch File        %            % version, we do not need a large array. The digit     %            % square sum of 9 999 999 999 is 810...                %            integer array seen( 0 :: 32 );            integer number, trys;            number             := n;            trys               := -1;            while begin                logical terminated;                integer tPos;                terminated     := false;                tPos           := 0;                while not terminated and tPos <= trys do begin                    terminated := seen( tPos ) = number;                    tPos       := tPos + 1                end while_not_terminated_and_tPos_lt_trys ;                number > 1 and not terminated            end do begin                integer sum;                trys           := trys + 1;                seen( trys )   := number;                sum            := 0;                while number > 0 do begin                    integer digit;                    digit      := number rem 10;                    number     := number div 10;                    sum        := sum + ( digit * digit )                end while_number_gt_0 ;                number         := sum            end while_number_gt_1_and_not_terminated ;            number = 1        end isHappy ;    % print the first 8 happy numbers                               %    begin        integer       happyCount, n;        happyCount := 0;        n          := 1;        write( "first 8 happy numbers: " );        while happyCount < 8 do begin            if isHappy( n ) then begin                writeon( i_w := 1, " ", n );                happyCount := happyCount + 1            end if_isHappy_n ;            n := n + 1        end while_happyCount_lt_8    endend.`
Output:
```first 8 happy numbers:  1   7   10   13   19   23   28   31
```

## APL

`     ∇ HappyNumbers arg;⎕IO;∆roof;∆first;bin;iroof[1]   ⍝0: Happy number[2]   ⍝1: http://rosettacode.org/wiki/Happy_numbers[3]    ⎕IO←1                              ⍝ Index origin[4]    ∆roof ∆first←2↑arg,10              ⍝[5][6]    bin←{[7]        ⍺←⍬                            ⍝ Default left arg[8]        ⍵=1:1                          ⍝ Always happy![9][10]       numbers←⍎¨1⊂⍕⍵                 ⍝ Split numbers into parts[11]       next←+/{⍵*2}¨numbers           ⍝ Sum and square of numbers[12][13]       next∊⍺:0                       ⍝ Return 0, if already exists[14]       (⍺,next)∇ next                 ⍝ Check next number (recursive)[15][16]   }¨iroof←⍳∆roof                     ⍝ Does all numbers upto ∆root smiles?[17][18]   ⎕←~∘0¨∆first↑bin/iroof             ⍝ Show ∆first numbers, but not 0     ∇`
```      HappyNumbers 100 8
1  7  10  13  19  23  28  31
```

### Dfn

`  HappyNumbers←{          ⍝ return the first ⍵ Happy Numbers     ⍺←⍬                 ⍝ initial list     ⍵=+/⍺:⍸⍺            ⍝ 1's mark happy numbers     sq←×⍨               ⍝ square function (times selfie)     isHappy←{           ⍝ is ⍵ a happy number?         ⍺←⍬             ⍝ previous sums         ⍵=1:1           ⍝ if we get to 1, it's happy         n←+/sq∘⍎¨⍕⍵     ⍝ sum of the square of the digits         n∊⍺:0           ⍝ if we hit this sum before, it's not happy         (⍺,n)∇ n}       ⍝ recurse until it's happy or not     (⍺,isHappy 1+≢⍺)∇ ⍵ ⍝ recurse until we have ⍵ happy numbers }      HappyNumbers 81 7 10 13 19 23 28 31 `

## AppleScript

### Iteration

`on run    set howManyHappyNumbers to 8    set happyNumberList to {}    set globalCounter to 1     repeat howManyHappyNumbers times        repeat while not isHappy(globalCounter)            set globalCounter to globalCounter + 1        end repeat        set end of happyNumberList to globalCounter        set globalCounter to globalCounter + 1    end repeat    log happyNumberListend run on isHappy(numberToCheck)    set localCycle to {}    repeat while (numberToCheck ≠ 1)        if localCycle contains numberToCheck then            exit repeat        end if        set end of localCycle to numberToCheck        set tempNumber to 0        repeat while (numberToCheck > 0)            set digitOfNumber to numberToCheck mod 10            set tempNumber to tempNumber + (digitOfNumber ^ 2)            set numberToCheck to (numberToCheck - digitOfNumber) / 10        end repeat        set numberToCheck to tempNumber    end repeat    return (numberToCheck = 1)end isHappy`
```Result: (*1, 7, 10, 13, 19, 23, 28, 31*)
```

### Functional composition

Translation of: JavaScript
`---------------------- HAPPY NUMBERS ----------------------- -- isHappy :: Int -> Boolon isHappy(n)     -- endsInOne :: [Int] -> Int -> Bool    script endsInOne         -- sumOfSquaredDigits :: Int -> Int        script sumOfSquaredDigits             -- digitSquared :: Int -> Int -> Int            script digitSquared                on |λ|(a, x)                    (a + (x as integer) ^ 2) as integer                end |λ|            end script             on |λ|(n)                foldl(digitSquared, 0, splitOn("", n as string))            end |λ|        end script         -- [Int] -> Int -> Bool        on |λ|(s, n)            if n = 1 then                true            else                if s contains n then                    false                else                    |λ|(s & n, |λ|(n) of sumOfSquaredDigits)                end if            end if        end |λ|    end script     endsInOne's |λ|({}, n)end isHappy --------------------------- TEST ---------------------------on run     -- seriesLength :: {n:Int, xs:[Int]} -> Bool    script seriesLength        property target : 8         on |λ|(rec)            length of xs of rec = target of seriesLength        end |λ|    end script     -- succTest :: {n:Int, xs:[Int]} -> {n:Int, xs:[Int]}    script succTest        on |λ|(rec)            tell rec to set {xs, n} to {its xs, its n}             script testResult                on |λ|(x)                    if isHappy(x) then                        xs & x                    else                        xs                    end if                end |λ|            end script             {n:n + 1, xs:testResult's |λ|(n)}        end |λ|    end script     xs of |until|(seriesLength, succTest, {n:1, xs:{}})     --> {1, 7, 10, 13, 19, 23, 28, 31}end run  -------------------- GENERIC FUNCTIONS --------------------- -- foldl :: (a -> b -> a) -> a -> [b] -> aon foldl(f, startValue, xs)    tell mReturn(f)        set v to startValue        set lng to length of xs        repeat with i from 1 to lng            set v to |λ|(v, item i of xs, i, xs)        end repeat        return v    end tellend foldl  -- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Scripton mReturn(f)    if class of f is script then        f    else        script            property |λ| : f        end script    end ifend mReturn  -- splitOn :: String -> String -> [String]on splitOn(pat, src)    set {dlm, my text item delimiters} to ¬        {my text item delimiters, pat}    set xs to text items of src    set my text item delimiters to dlm    return xsend splitOn  -- until :: (a -> Bool) -> (a -> a) -> a -> aon |until|(p, f, x)    set mp to mReturn(p)    set v to x     tell mReturn(f)        repeat until mp's |λ|(v)            set v to |λ|(v)        end repeat    end tell    return vend |until|`
Output:
`{1, 7, 10, 13, 19, 23, 28, 31}`

## Applesoft BASIC

` 0 C = 8: DIM S(16):B = 10: PRINT "THE FIRST "C" HAPPY NUMBERS": FOR R = C TO 0 STEP 0:N = H: GOSUB 1: PRINT  MID\$ (" " +  STR\$ (H),1 + (R = C),255 * I);:R = R - I:H = H + 1: NEXT R: END  1 S = 0: GOSUB 3:I = N = 1: IF  NOT Q THEN  RETURN  2  FOR Q = 1 TO 0 STEP 0:S(S) = N:S = S + 1: GOSUB 6:N = T: GOSUB 3: NEXT Q:I = N = 1: RETURN  3 Q = N > 1: IF  NOT Q OR  NOT S THEN  RETURN  4 Q = 0: FOR I = 0 TO S - 1: IF N = S(I) THEN  RETURN  5  NEXT I:Q = 1: RETURN  6 T = 0: FOR I = N TO 0 STEP 0:M =  INT (I / B):T =  INT (T + (I - M * B) ^ 2):I = M: NEXT I: RETURN`
Output:
```THE FIRST 8 HAPPY NUMBERS
1 7 10 13 19 23 28 31
```

## Arturo

Translation of: Nim
`ord0: to :integer `0`happy?: function [x][    n: x    past: new []     while [n <> 1][        s: to :string n        n: 0        loop s 'c [            i: (to :integer c) - ord0            n: n + i * i        ]        if contains? past n -> return false        'past ++ n    ]    return true] loop 0..31 'x [    if happy? x -> print x]`
Output:
```1
7
10
13
19
23
28
31```

## AutoHotkey

`Loop {  If isHappy(A_Index) {    out .= (out="" ? "" : ",") . A_Index    i ++    If (i = 8) {      MsgBox, The first 8 happy numbers are: %out%      ExitApp    }  }} isHappy(num, list="") {  list .= (list="" ? "" : ",") . num  Loop, Parse, num    sum += A_LoopField ** 2  If (sum = 1)    Return true  Else If sum in %list%    Return false  Else Return isHappy(sum, list)}`
```The first 8 happy numbers are: 1,7,10,13,19,23,28,31
```

### Alternative version

`while h < 8    if (Happy(A_Index)) {        Out .= A_Index A_Space        h++    }MsgBox, % Out Happy(n) {    Loop, {        Loop, Parse, n            t += A_LoopField ** 2        if (t = 89)            return, 0        if (t = 1)            return, 1        n := t, t := 0    }}`
`1 7 10 13 19 23 28 31`

## AutoIt

` \$c = 0\$k = 0While \$c < 8	\$k += 1	\$n = \$k	While \$n <> 1		\$s = StringSplit(\$n, "")		\$t = 0		For \$i = 1 To \$s[0]			\$t += \$s[\$i] ^ 2		Next		\$n = \$t		Switch \$n			Case 4,16,37,58,89,145,42,20			ExitLoop		EndSwitch	WEnd	If \$n = 1 Then		ConsoleWrite(\$k & " is Happy" & @CRLF)		\$c += 1	EndIfWEnd `
```Use a set of numbers (4,16,37,58,89,145,42,20) to indicate a loop and exit.
Output:
1 is Happy
7 is Happy
10 is Happy
13 is Happy
19 is Happy
23 is Happy
28 is Happy
31 is Happy
```

### Alternative version

` \$c = 0\$k = 0While \$c < 8	\$a = ObjCreate("System.Collections.ArrayList")	\$k += 1	\$n = \$k	While \$n <> 1		If \$a.Contains(\$n) Then			ExitLoop		EndIf		\$a.add(\$n)		\$s = StringSplit(\$n, "")		\$t = 0		For \$i = 1 To \$s[0]			\$t += \$s[\$i] ^ 2		Next		\$n = \$t	WEnd	If \$n = 1 Then		ConsoleWrite(\$k & " is Happy" & @CRLF)		\$c += 1	EndIf	\$a.ClearWEnd `
```Saves all numbers in a list, duplicate entry indicates a loop.
Output:
1 is Happy
7 is Happy
10 is Happy
13 is Happy
19 is Happy
23 is Happy
28 is Happy
31 is Happy
```

## AWK

`function is_happy(n){  if ( n in happy ) return 1;  if ( n in unhappy ) return 0;  cycle[""] = 0  while( (n!=1) && !(n in cycle) ) {    cycle[n] = n    new_n = 0    while(n>0) {      d = n % 10      new_n += d*d      n = int(n/10)    }    n = new_n  }  if ( n == 1 ) {    for (i_ in cycle) {      happy[cycle[i_]] = 1      delete cycle[i_]    }    return 1  } else {    for (i_ in cycle) {      unhappy[cycle[i_]] = 1      delete cycle[i_]    }    return 0  }} BEGIN {  cnt = 0  happy[""] = 0  unhappy[""] = 0  for(j=1; (cnt < 8); j++) {    if ( is_happy(j) == 1 ) {      cnt++      print j    }  }}`

Result:

```1
7
10
13
19
23
28
31```

### Alternative version

Alternately, for legibility one might write:

`BEGIN {    for (i = 1; i < 50; ++i){        if (isHappy(i)) {            print i;        }    }    exit} function isHappy(n,    seen) {    delete seen;    while (1) {        n = sumSqrDig(n)        if (seen[n]) {            return n == 1        }        seen[n] = 1    }} function sumSqrDig(n,     d, tot) {    while (n) {        d = n % 10        tot += d * d        n = int(n/10)    }    return tot}`

## BASIC256

`n = 1 : cnt = 0print "The first 8 isHappy numbers are:"print while cnt < 8    if isHappy(n) = 1 then        cnt += 1        print cnt; " => "; n    end if    n += 1end while function isHappy(num)    isHappy = 0    cont = 0    while cont < 50 and isHappy <> 1        num\$ = string(num)        cont += 1        isHappy	= 0        for i = 1 to length(num\$)            isHappy += int(mid(num\$,i,1)) ^ 2        next i        num = isHappy    end whileend function`

## Batch File

happy.bat

`@echo offsetlocal enableDelayedExpansion::Define a list with 10 terms as a convenience for defining a loopset "L10=0 1 2 3 4 5 6 7 8 9"shift /1 & goto %1exit /b  :list min count:: This routine prints all happy numbers > min (arg1):: until it finds count (arg2) happy numbers.set /a "n=%~1, cnt=%~2"call :listInternalexit /b  :test min [max]:: This routine sequentially tests numbers between min (arg1) and max (arg2):: to see if they are happy. If max is not specified then it defaults to min.set /a "min=%~1"if "%~2" neq "" (set /a "max=%~2") else set max=%min%::The FOR /L loop does not detect integer overflow, so must protect against::an infinite loop when max=0x7FFFFFFFFset end=%max%if %end% equ 2147483647 set /a end-=1for /l %%N in (%min% 1 %end%) do (  call :testInternal %%N && (echo %%N is happy :^)) || echo %%N is sad :()if %end% neq %max% call :testInternal %max% && (echo %max% is happy :^)) || echo %max% is sad :(exit /b    :listInternal  :: This loop sequentially tests each number >= n. The loop conditionally  :: breaks within the body once cnt happy numbers have been found, or if  :: the max integer value is reached. Performance is improved by using a  :: FOR loop to perform most of the looping, with a GOTO only needed once  :: per 100 iterations.  for %%. in (    %L10% %L10% %L10% %L10% %L10% %L10% %L10% %L10% %L10% %L10%  ) do (    call :testInternal !n! && (      echo !n!      set /a cnt-=1      if !cnt! leq 0 exit /b 0    )    if !n! equ 2147483647 (      >&2 echo ERROR: Maximum integer value reached      exit /b 1    )    set /a n+=1  )  goto :listInternal    :testInternal n  :: This routine loops until the sum of squared digits converges on 1 (happy)  :: or it detects a cycle (sad). It exits with errorlevel 0 for happy and 1 for sad.  :: Performance is improved by using a FOR loop for the looping instead of a GOTO.  :: Numbers less than 1000 never neeed more than 20 iterations, and any number  :: with 4 or more digits shrinks by at least one digit each iteration.  :: Since Windows batch can't handle more than 10 digits, allowance for 27  :: iterations is enough, and 30 is more than adequate.  setlocal  set n=%1  for %%. in (%L10% %L10% %L10%) do (    if !n!==1 exit /b 0    %= Only numbers < 1000 can cycle =%    if !n! lss 1000 (      if defined t.!n! exit /b 1      set t.!n!=1    )    %= Sum the squared digits                                          =%    %= Batch can't handle numbers greater than 10 digits so we can use =%    %= a constrained FOR loop and avoid a slow goto                    =%    set sum=0    for /l %%N in (1 1 10) do (      if !n! gtr 0 set /a "sum+=(n%%10)*(n%%10), n/=10"    )    set /a n=sum  )`

Sample usage and output

```>happy list 1 8
1
7
10
13
19
23
28
31

>happy list 1000000000 10
1000000000
1000000003
1000000009
1000000029
1000000030
1000000033
1000000039
1000000067
1000000076
1000000088

>happy test 30

>happy test 31
31 is happy :)

>happy test 1 10
1 is happy :)
7 is happy :)
10 is happy :)

>happy test "50 + 10 * 5"
100 is happy :)

>happy test 0x7fffffff

>happy test 0x7ffffffd
2147483645 is happy :)

>happy list 0x7ffffff0 10
2147483632
2147483645
ERROR: Maximum integer value reached
```

## BBC BASIC

`      number% = 0      total% = 0      REPEAT        number% += 1        IF FNhappy(number%) THEN          PRINT number% " is a happy number"          total% += 1        ENDIF      UNTIL total% = 8      END       DEF FNhappy(num%)      LOCAL digit&()      DIM digit&(10)      REPEAT        digit&() = 0        \$\$^digit&(0) = STR\$(num%)        digit&() AND= 15        num% = MOD(digit&())^2 + 0.5      UNTIL num% = 1 OR num% = 4      = (num% = 1)`

Output:

```         1 is a happy number
7 is a happy number
10 is a happy number
13 is a happy number
19 is a happy number
23 is a happy number
28 is a happy number
31 is a happy number```

## BCPL

`get "libhdr" let sumdigitsq(n) =     n=0 -> 0, (n rem 10)*(n rem 10)+sumdigitsq(n/10) let happy(n) = valof\$(  let seen = vec 255    for i = 0 to 255 do i!seen := false    \$(  n!seen := true        n := sumdigitsq(n)    \$) repeatuntil n!seen    resultis 1!seen\$) let start() be\$(  let n, i = 0, 0    while n < 8 do    \$(  if happy(i) do        \$(  n := n + 1            writef("%N ",i)        \$)        i := i + 1    \$)    wrch('*N')\$)`
Output:
`1 7 10 13 19 23 28 31`

## Bori

`bool isHappy (int n){   ints cache;    while (n != 1)   {      int sum = 0;       if (cache.contains(n))         return false;       cache.add(n);      while (n != 0)      {         int digit = n % 10;         sum += (digit * digit);         n = (int)(n / 10);      }      n = sum;   }   return true;} void test (){   int num = 1;               ints happynums;    while (happynums.count() < 8)               {      if (isHappy(num))         happynums.add(num);      num++;   }   puts("First 8 happy numbers : " + str.newline + happynums);}`

Output:

```First 8 happy numbers :
[1, 7, 10, 13, 19, 23, 28, 31]```

## BQN

`SumSqDgt ← +´2⋆˜ •Fmt-'0'˙Happy ← ⟨⟩{𝕨((⊑∊˜ )◶⟨∾𝕊(SumSqDgt⊢),1=⊢⟩)𝕩}⊢8↑Happy¨⊸/↕50`
Output:
`⟨ 1 7 10 13 19 23 28 31 ⟩`

## Brat

`include :set happiness = set.new 1sadness = set.new sum_of_squares_of_digits = { num |  num.to_s.dice.reduce 0 { sum, n | sum = sum + n.to_i ^ 2 }} happy? = { n, seen = set.new |  when {true? happiness.include? n } { happiness.merge seen << n; true }    { true? sadness.include? n } { sadness.merge seen; false }    { true? seen.include? n } { sadness.merge seen; false }    { true } { seen << n; happy? sum_of_squares_of_digits(n), seen }} num = 1happies = [] while { happies.length < 8 } {  true? happy?(num)    { happies << num }   num = num + 1} p "First eight happy numbers: #{happies}"p "Happy numbers found: #{happiness.to_array.sort}"p "Sad numbers found: #{sadness.to_array.sort}"`

Output:

```First eight happy numbers: [1, 7, 10, 13, 19, 23, 28, 31]
Happy numbers found: [1, 7, 10, 13, 19, 23, 28, 31, 49, 68, 82, 97, 100, 130]
Sad numbers found: [2, 3, 4, 5, 6, 8, 9, 11, 12, 14, 15, 16, 17, 18, 20, 21, 22, 24, 25, 26, 27, 29, 30, 34, 36, 37, 40, 41, 42, 45, 50, 52, 53, 58, 61, 64, 65, 81, 85, 89, 145]```

## C

Recursively look up if digit square sum is happy.

`#include <stdio.h> #define CACHE 256enum { h_unknown = 0, h_yes, h_no };unsigned char buf[CACHE] = {0, h_yes, 0}; int happy(int n){	int sum = 0, x, nn;	if (n < CACHE) {		if (buf[n]) return 2 - buf[n];		buf[n] = h_no;	} 	for (nn = n; nn; nn /= 10) x = nn % 10, sum += x * x; 	x = happy(sum);	if (n < CACHE) buf[n] = 2 - x;	return x;} int main(){	int i, cnt = 8;	for (i = 1; cnt || !printf("\n"); i++)		if (happy(i)) --cnt, printf("%d ", i); 	printf("The %dth happy number: ", cnt = 1000000);	for (i = 1; cnt; i++)		if (happy(i)) --cnt || printf("%d\n", i); 	return 0;}`
output
```1 7 10 13 19 23 28 31
The 1000000th happy number: 7105849```

Without caching, using cycle detection:

`#include <stdio.h> int dsum(int n){	int sum, x;	for (sum = 0; n; n /= 10) x = n % 10, sum += x * x;	return sum;} int happy(int n){	int nn;	while (n > 999) n = dsum(n); /* 4 digit numbers can't cycle */	nn = dsum(n);	while (nn != n && nn != 1)		n = dsum(n), nn = dsum(dsum(nn));	return n == 1;} int main(){	int i, cnt = 8;	for (i = 1; cnt || !printf("\n"); i++)		if (happy(i)) --cnt, printf("%d ", i); 	printf("The %dth happy number: ", cnt = 1000000);	for (i = 1; cnt; i++)		if (happy(i)) --cnt || printf("%d\n", i); 	return 0;}`
Output is same as above, but much slower.

## C#

`using System;using System.Collections.Generic;using System.Linq;using System.Text; namespace HappyNums{    class Program    {        public static bool ishappy(int n)        {            List<int> cache = new List<int>();            int sum = 0;            while (n != 1)            {                if (cache.Contains(n))                {                    return false;                }                cache.Add(n);                while (n != 0)                {                    int digit = n % 10;                    sum += digit * digit;                    n /= 10;                }                n = sum;                sum = 0;            }           return true;                    }         static void Main(string[] args)        {            int num = 1;            List<int> happynums = new List<int>();             while (happynums.Count < 8)            {                if (ishappy(num))                {                    happynums.Add(num);                }                num++;            }            Console.WriteLine("First 8 happy numbers : " + string.Join(",", happynums));        }    }}`
```First 8 happy numbers : 1,7,10,13,19,23,28,31
```

### Alternate (cacheless)

Instead of caching and checking for being stuck in a loop, one can terminate on the "unhappy" endpoint of 89. One might be temped to try caching the so-far-found happy and unhappy numbers and checking the cache to speed things up. However, I have found that the cache implementation overhead reduces performance compared to this cacheless version.

Reaching 10 million, the <34 second computation time was from Tio.run. It takes under 5 seconds on a somewhat modern CPU. If you edit it to max out at 100 million, it takes about 50 seconds (on the somewhat modern CPU).
`using System;using System.Collections.Generic;class Program{     static int[] sq = { 1, 4, 9, 16, 25, 36, 49, 64, 81 };     static bool isOne(int x)    {        while (true)        {            if (x == 89) return false;            int s = 0, t;            do if ((t = (x % 10) - 1) >= 0) s += sq[t]; while ((x /= 10) > 0);            if (s == 1) return true;            x = s;        }    }     static void Main(string[] args)    {        const int Max = 10_000_000; DateTime st = DateTime.Now;        Console.Write("---Happy Numbers---\nThe first 8:");        int c = 0, i; for (i = 1; c < 8; i++)            if (isOne(i)) Console.Write("{0} {1}", c == 0 ? "" : ",", i, ++c);        for (int m = 10; m <= Max; m *= 10)        {            Console.Write("\nThe {0:n0}th: ", m);            for (; c < m; i++) if (isOne(i)) c++;            Console.Write("{0:n0}", i - 1);        }        Console.WriteLine("\nComputation time {0} seconds.", (DateTime.Now - st).TotalSeconds);    }}`
Output:
```---Happy Numbers---
The first 8: 1, 7, 10, 13, 19, 23, 28, 31
The 10th: 44
The 100th: 694
The 1,000th: 6,899
The 10,000th: 67,169
The 100,000th: 692,961
The 1,000,000th: 7,105,849
The 10,000,000th: 71,313,350
Computation time 33.264518 seconds.```

## C++

Translation of: Python
`#include <map>#include <set> bool happy(int number) {  static std::map<int, bool> cache;   std::set<int> cycle;  while (number != 1 && !cycle.count(number)) {    if (cache.count(number)) {      number = cache[number] ? 1 : 0;      break;    }    cycle.insert(number);    int newnumber = 0;    while (number > 0) {      int digit = number % 10;      newnumber += digit * digit;      number /= 10;    }    number = newnumber;  }  bool happiness = number == 1;  for (std::set<int>::const_iterator it = cycle.begin();       it != cycle.end(); it++)    cache[*it] = happiness;  return happiness;} #include <iostream> int main() {  for (int i = 1; i < 50; i++)    if (happy(i))      std::cout << i << std::endl;  return 0;}`

Output:

```1
7
10
13
19
23
28
31
32
44
49```

Alternative version without caching:

`unsigned int happy_iteration(unsigned int n){  unsigned int result = 0;  while (n > 0)  {    unsigned int lastdig = n % 10;    result += lastdig*lastdig;    n /= 10;  }  return result;} bool is_happy(unsigned int n){  unsigned int n2 = happy_iteration(n);  while (n != n2)  {    n = happy_iteration(n);    n2 = happy_iteration(happy_iteration(n2));  }  return n == 1;} #include <iostream> int main(){  unsigned int current_number = 1;   unsigned int happy_count = 0;  while (happy_count != 8)  {    if (is_happy(current_number))    {      std::cout << current_number << " ";      ++happy_count;    }    ++current_number;  }  std::cout << std::endl;}`

Output:

`1 7 10 13 19 23 28 31 `

Cycle detection in `is_happy()` above is done using Floyd's cycle-finding algorithm.

## Clojure

`(defn happy? [n]  (loop [n n, seen #{}]    (cond       (= n 1)  true      (seen n) false      :else        (recur (->> (str n)                    (map #(Character/digit % 10))                    (map #(* % %))                    (reduce +))               (conj seen n))))) (def happy-numbers (filter happy? (iterate inc 1))) (println (take 8 happy-numbers))`
Output:
`(1 7 10 13 19 23 28 31)`

### Alternate Version (with caching)

`(require '[clojure.set :refer [union]]) (def ^{:private true} cache {:happy (atom #{}) :sad (atom #{})}) (defn break-apart [n]  (->> (str n)       (map str)       (map #(Long/parseLong %)))) (defn next-number [n]  (->> (break-apart n)       (map #(* % %))       (apply +))) (defn happy-or-sad? [prev n]  (cond (or (= n 1) ((deref (:happy cache)) n)) :happy	(or ((deref (:sad cache)) n) (some #(= % n) prev)) :sad	:else :unknown)) (defn happy-algo [n]  (let [get-next (fn [[prev n]] [(conj prev n) (next-number n)])	my-happy-or-sad? (fn [[prev n]] [(happy-or-sad? prev n) (conj prev n)])	unknown? (fn [[res nums]] (= res :unknown))	[res nums] (->> [#{} n]			(iterate get-next)			(map my-happy-or-sad?)			(drop-while unknown?)			first)	_ (swap! (res cache) union nums)]    res)) (def happy-numbers (->> (iterate inc 1)                         (filter #(= :happy (happy-algo %))))) (println (take 8 happy-numbers))`

Same output.

## CLU

`sum_dig_sq = proc (n: int) returns (int)    sum_sq: int := 0    while n > 0 do        sum_sq := sum_sq + (n // 10) ** 2        n := n / 10    end    return (sum_sq)end sum_dig_sq is_happy = proc (n: int) returns (bool)    nn: int := sum_dig_sq(n)    while nn ~= n cand nn ~= 1 do        n := sum_dig_sq(n)        nn := sum_dig_sq(sum_dig_sq(nn))    end    return (nn = 1)end is_happy  happy_numbers = iter (start, num: int) yields (int)    n: int := start    while num > 0 do        if is_happy(n) then            yield (n)            num := num-1        end        n := n+1    endend happy_numbers start_up = proc ()    po: stream := stream\$primary_output()     for i: int in happy_numbers(1, 8) do        stream\$putl(po, int\$unparse(i))    endend start_up `
Output:
```1
7
10
13
19
23
28
31```

## CoffeeScript

`happy = (n) ->  seen = {}  while true    n = sum_digit_squares(n)    return true if n == 1    return false if seen[n]    seen[n] = true sum_digit_squares = (n) ->  sum = 0  for c in n.toString()    d = parseInt(c)    sum += d*d  sum i = 1cnt = 0while cnt < 8  if happy(i)    console.log i    cnt += 1  i += 1`

output

```> coffee happy.coffee
1
7
10
13
19
23
28
31
```

## Common Lisp

`(defun sqr (n)  (* n n)) (defun sum-of-sqr-dgts (n)  (loop for i = n then (floor i 10)        while (plusp i)        sum (sqr (mod i 10)))) (defun happy-p (n &optional cache)  (or (= n 1)       (unless (find n cache)        (happy-p (sum-of-sqr-dgts n)                 (cons n cache))))) (defun happys (&aux (happys 0))  (loop for i from 1        while (< happys 8)        when (happy-p i)        collect i and do (incf happys))) (print (happys)) `
Output:
`(1 7 10 13 19 23 28 31)`

## Cowgol

`include "cowgol.coh"; sub sumDigitSquare(n: uint8): (s: uint8) is    s := 0;    while n != 0 loop        var d := n % 10;        s := s + d * d;        n := n / 10;    end loop;end sub; sub isHappy(n: uint8): (h: uint8) is    var seen: uint8[256];    MemZero(&seen[0], @bytesof seen);     while seen[n] == 0 loop        seen[n] := 1;        n := sumDigitSquare(n);    end loop;     if n == 1 then        h := 1;    else        h := 0;    end if;end sub; var n: uint8 := 1;var seen: uint8 := 0; while seen < 8 loop    if isHappy(n) != 0 then        print_i8(n);        print_nl();        seen := seen + 1;    end if;    n := n + 1;end loop;`
Output:
```1
7
10
13
19
23
28
31```

## Crystal

Translation of: Ruby
`def happy?(n)  past = [] of Int32 | Int64  until n == 1    sum = 0; while n > 0; sum += (n % 10) ** 2; n //= 10 end    return false if past.includes? (n = sum)    past << n  end  trueend i = count = 0until count == 8; (puts i; count += 1) if happy?(i += 1) endputs(99999999999900..99999999999999).each { |i| puts i if happy?(i) }`
Output:
```1
7
10
13
19
23
28
31

99999999999901
99999999999910
99999999999914
99999999999915
99999999999916
99999999999937
99999999999941
99999999999951
99999999999956
99999999999961
99999999999965
99999999999973```

## D

`bool isHappy(int n) pure nothrow {    int[int] past;     while (true) {        int total = 0;        while (n > 0) {            total += (n % 10) ^^ 2;            n /= 10;        }        if (total == 1)            return true;        if (total in past)            return false;        n = total;        past[total] = 0;    }} void main() {    import std.stdio, std.algorithm, std.range;     int.max.iota.filter!isHappy.take(8).writeln;}`
Output:
`[1, 7, 10, 13, 19, 23, 28, 31]`

### Alternative Version

`import std.stdio, std.algorithm, std.range, std.conv, std.string; bool isHappy(int n) pure nothrow {    int[int] seen;     while (true) {        immutable t = n.text.representation.map!q{(a - '0') ^^ 2}.sum;        if (t == 1)            return true;        if (t in seen)            return false;        n = t;        seen[t] = 0;    }} void main() {    int.max.iota.filter!isHappy.take(8).writeln;}`

Same output.

## Dart

`main() {  HashMap<int,bool> happy=new HashMap<int,bool>();  happy[1]=true;   int count=0;  int i=0;   while(count<8) {    if(happy[i]==null) {      int j=i;      Set<int> sequence=new Set<int>();      while(happy[j]==null && !sequence.contains(j)) {        sequence.add(j);        int sum=0;        int val=j;        while(val>0) {          int digit=val%10;          sum+=digit*digit;          val=(val/10).toInt();        }        j=sum;      }      bool sequenceHappy=happy[j];      Iterator<int> it=sequence.iterator();      while(it.hasNext()) {        happy[it.next()]=sequenceHappy;      }    }    if(happy[i]) {      print(i);      count++;    }    i++;  }}`

## dc

`[lcI~rscd*+lc0<H]sH[0rsclHxd4<h]sh[lIp]s_0sI[lI1+dsIlhx2>_z8>s]dssx`

Output:

```1
7
10
13
19
23
28
31```

## DCL

`\$ happy_1 = 1\$ found = 0\$ i = 1\$ loop1:\$  n = i\$  seen_list = ","\$  loop2:\$   if f\$type( happy_'n ) .nes. "" then \$ goto happy\$   if f\$type( unhappy_'n ) .nes. "" then \$ goto unhappy\$   if f\$locate( "," + n + ",", seen_list ) .eq. f\$length( seen_list )\$   then\$    seen_list = seen_list + f\$string( n ) + ","\$   else\$    goto unhappy\$   endif\$   ns = f\$string( n )\$   nl = f\$length( ns )\$   j = 0\$   sumsq = 0\$   loop3:\$    digit = f\$integer( f\$extract( j, 1, ns ))\$    sumsq = sumsq + digit * digit\$    j = j + 1\$    if j .lt. nl then \$ goto loop3\$    n = sumsq\$   goto loop2\$  unhappy:\$  j = 1\$  loop4:\$   x = f\$element( j, ",", seen_list )\$   if x .eqs. "" then \$ goto continue\$   unhappy_'x = 1\$   j = j + 1\$   goto loop4\$  happy:\$  found = found + 1\$  found_'found = i\$  if found .eq. 8 then \$ goto done\$  j = 1\$  loop5:\$   x = f\$element( j, ",", seen_list )\$   if x .eqs. "" then \$ goto continue\$   happy_'x = 1\$   j = j + 1\$   goto loop5\$  continue:\$  i = i + 1\$  goto loop1\$ done:\$ show symbol found*`
Output:
```  FOUND = 8   Hex = 00000008  Octal = 00000000010
FOUND_1 = 1   Hex = 00000001  Octal = 00000000001
FOUND_2 = 7   Hex = 00000007  Octal = 00000000007
FOUND_3 = 10   Hex = 0000000A  Octal = 00000000012
FOUND_4 = 13   Hex = 0000000D  Octal = 00000000015
FOUND_5 = 19   Hex = 00000013  Octal = 00000000023
FOUND_6 = 23   Hex = 00000017  Octal = 00000000027
FOUND_7 = 28   Hex = 0000001C  Octal = 00000000034
FOUND_8 = 31   Hex = 0000001F  Octal = 00000000037```

## Delphi

Library: Boost.Int

Adaptation of #Pascal. The lib Boost.Int can be found here [1]

` program Happy_numbers; {\$APPTYPE CONSOLE} uses  System.SysUtils,  Boost.Int; type  TIntegerDynArray = TArray<Integer>;   TIntHelper = record helper for Integer    function IsHappy: Boolean;    procedure Next;  end; { TIntHelper } function TIntHelper.IsHappy: Boolean;var  cache: TIntegerDynArray;  sum, n: integer;begin  n := self;  repeat    sum := 0;    while n > 0 do    begin      sum := sum + (n mod 10) * (n mod 10);      n := n div 10;    end;    if sum = 1 then      exit(True);     if cache.Has(sum) then      exit(False);    n := sum;    cache.Add(sum);  until false;end; procedure TIntHelper.Next;begin  inc(self);end; var  count, n: integer; begin  n := 1;  count := 0;  while count < 8 do  begin    if n.IsHappy then    begin      count.Next;      write(n, ' ');    end;    n.Next;  end;  writeln;  readln;end.`
Output:
`1 7 10 13 19 23 28 31`

## Draco

`proc nonrec dsumsq(byte n) byte:    byte r, d;    r := 0;    while n~=0 do        d := n % 10;        n := n / 10;        r := r + d * d    od;    rcorp proc nonrec happy(byte n) bool:    [256] bool seen;    byte i;    for i from 0 upto 255 do seen[i] := false od;    while not seen[n] do        seen[n] := true;        n := dsumsq(n)    od;    seen[1]corp proc nonrec main() void:    byte n, seen;    n := 1;    seen := 0;    while seen < 8 do        if happy(n) then            writeln(n:3);            seen := seen + 1        fi;        n := n + 1    odcorp`
Output:
```  1
7
10
13
19
23
28
31```

## DWScript

`function IsHappy(n : Integer) : Boolean;var   cache : array of Integer;   sum : Integer;begin   while True do begin      sum := 0;      while n>0 do begin         sum += Sqr(n mod 10);         n := n div 10;      end;      if sum = 1 then         Exit(True);      if sum in cache then         Exit(False);      n := sum;      cache.Add(sum);   end;end; var n := 8;var i : Integer; while n>0 do begin   Inc(i);   if IsHappy(i) then begin      PrintLn(i);      Dec(n);   end;end;`

Output:

```1
7
10
13
19
23
28
31```

## Dyalect

`func happy(n) {    var m = []    while n > 1 {        m.Add(n)        var x = n        n = 0        while x > 0 {            var d = x % 10            n += d * d            x /= 10        }        if m.IndexOf(n) != -1 {            return false        }    }    return true} var (n, found) = (1, 0)while found < 8 {    if happy(n) {        print("\(n) ", terminator: "")        found += 1    }    n += 1}print()`
Output:
`1 7 10 13 19 23 28 31`

## Déjà Vu

`next-num:	0	while over:		over		* dup % swap 10		+		swap floor / swap 10 swap	drop swap is-happy happies n:	if has happies n:		return happies! n	local :seq set{ n }	n	while /= 1 dup:		next-num		if has seq dup:			drop			set-to happies n false			return false		if has happies dup:			set-to happies n dup happies!			return		set-to seq over true	drop	set-to happies n true	true local :h {}1 0while > 8 over:	if is-happy h dup:		!print( "A happy number: " over )		swap ++ swap	++dropdrop`
Output:
```A happy number: 1
A happy number: 7
A happy number: 10
A happy number: 13
A happy number: 19
A happy number: 23
A happy number: 28
A happy number: 31```

## E

 This example does not show the output mentioned in the task description on this page (or a page linked to from here). Please ensure that it meets all task requirements and remove this message. Note that phrases in task descriptions such as "print and display" and "print and show" for example, indicate that (reasonable length) output be a part of a language's solution.

`def isHappyNumber(var x :int) {  var seen := [].asSet()  while (!seen.contains(x)) {    seen with= x    var sum := 0    while (x > 0) {      sum += (x % 10) ** 2      x //= 10    }    x := sum    if (x == 1) { return true }  }  return false} var count := 0for x ? (isHappyNumber(x)) in (int >= 1) {  println(x)  if ((count += 1) >= 8) { break }}`

## Eiffel

` class	APPLICATION create	make feature {NONE} -- Initialization 	make			-- Run application.		local			l_val: INTEGER		do			from				l_val := 1			until				l_val > 100			loop				if is_happy_number (l_val) then					print (l_val.out)					print ("%N")				end				l_val := l_val + 1			end		end feature -- Happy number 	is_happy_number (a_number: INTEGER): BOOLEAN			-- Is `a_number' a happy number?		require			positive_number: a_number > 0		local			l_number: INTEGER			l_set: ARRAYED_SET [INTEGER]		do			from				l_number := a_number				create l_set.make (10)			until				l_number = 1 or l_set.has (l_number)			loop				l_set.put (l_number)				l_number := square_sum_of_digits (l_number)			end 			Result := (l_number = 1)		end feature{NONE} -- Implementation 	square_sum_of_digits (a_number: INTEGER): INTEGER			-- Sum of the sqares of digits of `a_number'.		require			positive_number: a_number > 0		local			l_number, l_digit: INTEGER		do			from				l_number := a_number			until				l_number = 0			loop				l_digit := l_number \\ 10				Result := Result + l_digit * l_digit				l_number := l_number // 10			end		end end  `

## Elena

Translation of: C#

ELENA 4.x :

`import extensions;import system'collections;import system'routines; isHappy(int n){    auto cache := new List<int>(5);    int sum := 0;    int num := n;    while (num != 1)    {        if (cache.indexOfElement:num != -1)        {            ^ false        };        cache.append(num);        while (num != 0)        {            int digit := num.mod:10;            sum += (digit*digit);            num /= 10        };        num := sum;        sum := 0    };     ^ true} public program(){    auto happynums  := new List<int>(8);    int num := 1;    while (happynums.Length < 8)    {        if (isHappy(num))        {            happynums.append(num)        };         num += 1    };    console.printLine("First 8 happy numbers: ", happynums.asEnumerable())}`
Output:
```First 8 happy numbers: 1,7,10,13,19,23,28,31
```

## Elixir

`defmodule Happy do  def task(num) do    Process.put({:happy, 1}, true)    Stream.iterate(1, &(&1+1))    |> Stream.filter(fn n -> happy?(n) end)    |> Enum.take(num)  end   defp happy?(n) do    sum = square_sum(n, 0)    val = Process.get({:happy, sum})    if val == nil do      Process.put({:happy, sum}, false)      val = happy?(sum)      Process.put({:happy, sum}, val)    end    val  end   defp square_sum(0, sum), do: sum  defp square_sum(n, sum) do    r = rem(n, 10)    square_sum(div(n, 10), sum + r*r)  endend IO.inspect Happy.task(8)`
Output:
```[1, 7, 10, 13, 19, 23, 28, 31]
```

## Erlang

`-module(tasks).-export([main/0]).-import(lists, [map/2, member/2, sort/1, sum/1]). is_happy(X, XS) ->    if	X == 1 ->	    true;	X < 1 ->	    false;	true ->	    case member(X, XS) of		true -> false;		false ->		    is_happy(sum(map(fun(Z) -> Z*Z end, 				     [Y - 48 || Y <- integer_to_list(X)])),			     [X|XS])	    end    end. main(X, XS) ->    if	length(XS) == 8 ->	    io:format("8 Happy Numbers: ~w~n", [sort(XS)]);	true ->	    case is_happy(X, []) of		true -> main(X + 1, [X|XS]);		false -> main(X + 1, XS)	    end    end.main() ->    main(0, []). `
Command:
`erl -run tasks main -run init stop -noshell`
Output:
`8 Happy Numbers: [1,7,10,13,19,23,28,31]`

In a more functional style (assumes integer_to_list/1 will convert to the ASCII value of a number, which then has to be converted to the integer value by subtracting 48):

`-module(tasks). -export([main/0]). main() -> io:format("~w ~n", [happy_list(1, 8, [])]). happy_list(_, N, L) when length(L) =:= N -> lists:reverse(L);happy_list(X, N, L) -> 	Happy = is_happy(X),	if Happy -> happy_list(X + 1, N, [X|L]);	true -> happy_list(X + 1, N, L) end. is_happy(1) -> true;is_happy(4) -> false;is_happy(N) when N > 0 ->	N_As_Digits = [Y - 48 || Y <- integer_to_list(N)],	is_happy(lists:foldl(fun(X, Sum) -> (X * X) + Sum end, 0, N_As_Digits));is_happy(_) -> false.`

Output:

`[1,7,10,13,19,23,28,31]`

## Euphoria

`function is_happy(integer n)    sequence seen    integer k    seen = {}    while n > 1 do        seen &= n        k = 0        while n > 0 do            k += power(remainder(n,10),2)            n = floor(n/10)        end while        n = k        if find(n,seen) then            return 0        end if    end while    return 1end function integer n,countn = 1count = 0while count < 8 do    if is_happy(n) then        ? n        count += 1    end if    n += 1end while`

Output:

```1
7
10
13
19
23
28
31
```

## F#

This requires the F# power pack to be referenced and the 2010 beta of F#

`open System.Collections.Genericopen Microsoft.FSharp.Collections let answer =    let sqr x = x*x                                                 // Classic square definition    let rec AddDigitSquare n =        match n with        | 0 -> 0                                                    // Sum of squares for 0 is 0        | _ -> sqr(n % 10) + (AddDigitSquare (n / 10))              // otherwise add square of bottom digit to recursive call    let dict = new Dictionary<int, bool>()                          // Dictionary to memoize values    let IsHappy n =        if dict.ContainsKey(n) then                                 // If we've already discovered it            dict.[n]                                                // Return previously discovered value        else            let cycle = new HashSet<_>(HashIdentity.Structural)     // Set to keep cycle values in            let rec isHappyLoop n =                if cycle.Contains n then n = 1                      // If there's a loop, return true if it's 1                else                    cycle.Add n |> ignore                           // else add this value to the cycle                    isHappyLoop (AddDigitSquare n)                  // and check the next number in the cycle            let f = isHappyLoop n                                   // Keep track of whether we're happy or not            cycle |> Seq.iter (fun i -> dict.[i] <- f)              // and apply it to all the values in the cycle            f                                                       // Return the boolean     1                                                               // Starting with 1,    |> Seq.unfold (fun i -> Some (i, i + 1))                        // make an infinite sequence of consecutive integers     |> Seq.filter IsHappy                                           // Keep only the happy ones    |> Seq.truncate 8                                               // Stop when we've found 8    |> Seq.iter (Printf.printf "%d\n")				    // Print results `

Output:

```1
7
10
13
19
23
28
31
```

## Factor

`USING: combinators kernel make math sequences ; : squares ( n -- s )    0 [ over 0 > ] [ [ 10 /mod sq ] dip + ] while nip ; : (happy?) ( n1 n2 -- ? )    [ squares ] [ squares squares ] bi* {        { [ dup 1 = ] [ 2drop t ] }        { [ 2dup = ] [ 2drop f ] }        [ (happy?) ]    } cond ; : happy? ( n -- ? )    dup (happy?) ; : happy-numbers ( n -- seq )    [        0 [ over 0 > ] [            dup happy? [ dup , [ 1 - ] dip ] when 1 +        ] while 2drop    ] { } make ;`
Output:
`8 happy-numbers ! { 1 7 10 13 19 23 28 31 }`

## FALSE

`[\$10/\$10*@\-\$*\]m:             {modulo squared and division}[\$m;![\$9>][m;[email protected]@+\]#\$*+]s:     {sum of squares}[\$0[1ø1>][1ø3+ø3ø=|\1-\]#\%]f: {look for duplicates} {check happy number}[  \$1[f;!~2ø1=~&][1+\s;[email protected]]#     {loop over sequence until 1 or duplicate}  1ø1=                         {return value}  \[\$0=~][@%1-]#%              {drop sequence and counter}]h: 0 1"Happy numbers:"[1ø8=~][h;![" "\$.\1+\]?1+]#%%`
Output:
`Happy numbers: 1 7 10 13 19 23 28 31`

## Fantom

`class Main{  static Bool isHappy (Int n)  {    Int[] record := [,]    while (n != 1 && !record.contains(n))    {       record.add (n)      // find sum of squares of digits      newn := 0      while (n > 0)      {         newn += (n.mod(10) * n.mod(10))        n = n.div(10)      }      n = newn    }    return (n == 1)  }   public static Void main ()  {    i := 1    count := 0    while (count < 8)    {      if (isHappy (i))       {        echo (i)        count += 1      }      i += 1     }  }} `

Output:

```1
7
10
13
19
23
28
31
```

## FOCAL

`01.10 S J=0;S N=1;T %201.20 D 3;I (K-2)1.501.30 S N=N+101.40 I (J-8)1.2;Q01.50 T N,!01.60 S J=J+101.70 G 1.3 02.10 S A=K;S R=002.20 S B=FITR(A/10)02.30 S R=R+(A-10*B)^202.40 S A=B02.50 I (-A)2.2 03.10 F X=0,162;S S(X)=-103.20 S K=N03.30 S S(K)=003.40 D 2;S K=R03.50 I (S(K))3.3`
Output:
```=  1
=  7
= 10
= 13
= 19
= 23
= 28
= 31```

## Forth

`: next ( n -- n )  0 swap begin 10 /mod >r  dup * +  r> ?dup 0= until ; : cycle? ( n -- ? )  here dup @ cells +  begin dup here >  while 2dup @ = if 2drop true exit then        1 cells -  repeat  1 over +!  dup @ cells + !  false ; : happy? ( n -- ? )  0 here !  begin next dup cycle? until  1 = ; : happy-numbers ( n -- )  0 swap 0 do    begin 1+ dup happy? until dup .  loop drop ; 8 happy-numbers  \ 1 7 10 13 19 23 28 31`

### Lookup Table

Every sequence either ends in 1, or contains a 4 as part of a cycle. Extending the table through 9 is a (modest) optimization/memoization. This executes '500000 happy-numbers' about 5 times faster than the above solution.

`CREATE HAPPINESS 0 C, 1 C, 0 C, 0 C, 0 C, 0 C, 0 C, 1 C, 0 C, 0 C,: next ( n -- n')   0 swap BEGIN dup WHILE 10 /mod >r  dup * +  r> REPEAT drop ;: happy? ( n -- t|f)   BEGIN dup 10 >= WHILE next REPEAT  chars HAPPINESS + [email protected] 0<> ;: happy-numbers ( n --)  >r 0   BEGIN [email protected] WHILE     BEGIN 1+ dup happy? UNTIL dup . r> 1- >r   REPEAT r> drop drop ;8 happy-numbers`
Output:
`1 7 10 13 19 23 28 31`

Produces the 1 millionth happy number with:

`: happy-number ( n -- n')  \ produce the nth happy number    >r 0  BEGIN [email protected] WHILE     BEGIN 1+ dup happy? UNTIL  r> 1- >r           REPEAT r> drop ;1000000 happy-number .  \ 7105849`

## Fortran

`program happy   implicit none  integer, parameter :: find = 8  integer :: found  integer :: number   found = 0  number = 1  do    if (found == find) then      exit    end if    if (is_happy (number)) then      found = found + 1      write (*, '(i0)') number    end if    number = number + 1  end do contains   function sum_digits_squared (number) result (result)     implicit none    integer, intent (in) :: number    integer :: result    integer :: digit    integer :: rest    integer :: work     result = 0    work = number    do      if (work == 0) then        exit      end if      rest = work / 10      digit = work - 10 * rest      result = result + digit * digit      work = rest    end do   end function sum_digits_squared   function is_happy (number) result (result)     implicit none    integer, intent (in) :: number    logical :: result    integer :: turtoise    integer :: hare     turtoise = number    hare = number    do      turtoise = sum_digits_squared (turtoise)      hare = sum_digits_squared (sum_digits_squared (hare))      if (turtoise == hare) then        exit      end if    end do    result = turtoise == 1   end function is_happy end program happy`

Output:

```1
7
10
13
19
23
28
31```

## FreeBASIC

`' FB 1.05.0 Win64 Function isHappy(n As Integer) As Boolean  If n < 0 Then Return False  ' Declare a dynamic array to store previous sums.  ' If a previous sum is duplicated before a sum of 1 is reached  ' then the number can't be "happy" as the cycle will just repeat  Dim prevSums() As Integer   Dim As Integer digit, ub, sum = 0  Do    While n > 0      digit = n Mod 10      sum += digit * digit      n \= 10    Wend    If sum = 1 Then Return True    ub = UBound(prevSums)    If ub > -1 Then      For i As Integer = 0 To ub         If sum = prevSums(i) Then Return False      Next    End If    ub += 1     Redim Preserve prevSums(0 To ub)    prevSums(ub) = sum    n = sum    sum  = 0  Loop  End Function Dim As Integer n = 1, count = 0 Print "The first 8 happy numbers are : "PrintWhile count < 8  If isHappy(n) Then    count += 1     Print count;" =>"; n  End If  n += 1Wend PrintPrint "Press any key to quit"Sleep`
Output:
``` 1 => 1
2 => 7
3 => 10
4 => 13
5 => 19
6 => 23
7 => 28
8 => 31
```

## Frege

Works with: Frege version 3.21.586-g026e8d7
`module Happy where import Prelude.Math-- ugh, since Frege doesn't have Set, use Map insteadimport Data.Map (member, insertMin, empty emptyMap) digitToInteger :: Char -> IntegerdigitToInteger c = fromInt \$ (ord c) - (ord '0') isHappy :: Integer -> BoolisHappy = p emptyMap  where p _ 1n = true        p s n | n `member` s = false              | otherwise  = p (insertMin n () s) (f n)        f = sum . map (sqr . digitToInteger) . unpacked . show main _ = putStrLn \$ unwords \$ map show \$ take 8 \$ filter isHappy \$ iterate (+ 1n) 1n`
Output:
```1 7 10 13 19 23 28 31
runtime 0.614 wallclock seconds.
```

## Fōrmulæ

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.

Programs in Fōrmulæ are created/edited online in its website, However they run on execution servers. By default remote servers are used, but they are limited in memory and processing power, since they are intended for demonstration and casual use. A local server can be downloaded and installed, it has no limitations (it runs in your own computer). Because of that, example programs can be fully visualized and edited, but some of them will not run if they require a moderate or heavy computation/memory resources, and no local server is being used.

## Go

`package main import "fmt" func happy(n int) bool {	m := make(map[int]bool)	for n > 1 {		m[n] = true		var x int		for x, n = n, 0; x > 0; x /= 10 {			d := x % 10			n += d * d		}		if m[n] {			return false		}	}	return true} func main() {	for found, n := 0, 1; found < 8; n++ {		if happy(n) {			fmt.Print(n, " ")			found++		}	}	fmt.Println()}`
Output:
```1 7 10 13 19 23 28 31
```

## Groovy

`Number.metaClass.isHappy = {    def number = delegate as Long    def cycle = new HashSet<Long>()    while (number != 1 && !cycle.contains(number)) {        cycle << number        number = (number as String).collect { d = (it as Long); d * d }.sum()    }    number == 1} def matches = []for (int i = 0; matches.size() < 8; i++) {    if (i.happy) { matches << i }}println matches`
Output:
`[1, 7, 10, 13, 19, 23, 28, 31]`

## Harbour

`PROCEDURE Main()   LOCAL i := 8, nH := 0    ? hb_StrFormat( "The first %d happy numbers are:", i )   ?    WHILE i > 0      IF IsHappy( ++nH )	?? hb_NtoS( nH ) + " "	--i      ENDIF   END    RETURN STATIC FUNCTION IsHappy( nNumber )   STATIC aUnhappy := {}   LOCAL nDigit, nSum := 0, cNumber := hb_NtoS( nNumber )    FOR EACH nDigit IN cNumber      nSum += Val( nDigit ) ^ 2   NEXT    IF nSum == 1      aUnhappy := {}      RETURN .T.   ELSEIF AScan( aUnhappy, nSum ) > 0     RETURN .F.   ENDIF    AAdd( aUnhappy, nSum )    RETURN IsHappy( nSum )`

Output:

``` The first 8 happy numbers are:
1 7 10 13 19 23 28 31
```

`import Data.Char (digitToInt)import Data.Set (member, insert, empty) isHappy :: Integer -> BoolisHappy = p empty  where    p _ 1 = True    p s n      | n `member` s = False      | otherwise = p (insert n s) (f n)    f = sum . fmap ((^ 2) . toInteger . digitToInt) . show main :: IO ()main = mapM_ print \$ take 8 \$ filter isHappy [1 ..]`
Output:
```1
7
10
13
19
23
28
31```

We can create a cache for small numbers to greatly speed up the process:

`import Data.Array (Array, (!), listArray) happy :: Int -> Boolhappy x  | xx <= 150 = seen ! xx  | otherwise = happy xx  where    xx = dsum x    seen :: Array Int Bool    seen =      listArray (1, 150) \$ True : False : False : False : (happy <\$> [5 .. 150])    dsum n      | n < 10 = n * n      | otherwise =        let (q, r) = n `divMod` 10        in r * r + dsum q main :: IO ()main = print \$ sum \$ take 10000 \$ filter happy [1 ..]`
Output:
`327604323`

## Icon and Unicon

`procedure main(arglist)local nn := arglist[1] | 8    # limiting number of happy numbers to generate, default=8writes("The first ",n," happy numbers are:")every writes(" ", happy(seq()) \ n )write()end procedure happy(i)    #: returns i if i is happylocal n     if  4 ~= (0 <= i) then { # unhappy if negative, 0, or 4        if i = 1 then return i        every (n := 0) +:= !i ^ 2        if happy(n) then return i        }end`

Usage and Output:

```| happynum.exe

The first 8 happy numbers are: 1 7 10 13 19 23 28 31
```

## J

`   8{. (#~1=+/@(*:@(,.&.":))^:(1&~:*.4&~:)^:_ "0) 1+i.1001 7 10 13 19 23 28 31`

This is a repeat while construction

` f ^: cond ^: _   input`

that produces an array of 1's and 4's, which is converted to 1's and 0's forming a binary array having a 1 for a happy number. Finally the happy numbers are extracted by a binary selector.

` (binary array) # 1..100`

So for easier reading the solution could be expressed as:

`   cond=: 1&~: *. 4&~:     NB. not equal to 1 and not equal to 4   sumSqrDigits=: +/@(*:@(,.&.":))    sumSqrDigits 123        NB. test sum of squared digits14   8{. (#~ 1 = sumSqrDigits ^: cond ^:_ "0) 1 + i.1001 7 10 13 19 23 28 31`

## Java

Works with: Java version 1.5+
Translation of: JavaScript
`import java.util.HashSet;public class Happy{   public static boolean happy(long number){       long m = 0;       int digit = 0;       HashSet<Long> cycle = new HashSet<Long>();       while(number != 1 && cycle.add(number)){           m = 0;           while(number > 0){               digit = (int)(number % 10);               m += digit*digit;               number /= 10;           }           number = m;       }       return number == 1;   }    public static void main(String[] args){       for(long num = 1,count = 0;count<8;num++){           if(happy(num)){               System.out.println(num);               count++;           }       }   }}`

Output:

```1
7
10
13
19
23
28
31```

### Java 1.8

Works with: Java version 1.8
Translation of: Java
`  import java.util.Arrays;import java.util.HashSet;import java.util.List; public class HappyNumbers {      public static void main(String[] args) {         for (int current = 1, total = 0; total < 8; current++)            if (isHappy(current)) {                System.out.println(current);                total++;            }    }      public static boolean isHappy(int number) {        HashSet<Integer> cycle = new HashSet<>();        while (number != 1 && cycle.add(number)) {            List<String> numStrList = Arrays.asList(String.valueOf(number).split(""));            number = numStrList.stream().map(i -> Math.pow(Integer.parseInt(i), 2)).mapToInt(i -> i.intValue()).sum();        }        return number == 1;    }}`

Output:

```1
7
10
13
19
23
28
31```

## JavaScript

### ES5

#### Iteration

`function happy(number) {    var m, digit ;    var cycle = [] ;     while(number != 1 && cycle[number] !== true) {        cycle[number] = true ;        m = 0 ;        while (number > 0) {            digit = number % 10 ;            m += digit * digit ;            number = (number  - digit) / 10 ;        }        number = m ;    }    return (number == 1) ;} var cnt = 8 ;var number = 1 ; while(cnt-- > 0) {    while(!happy(number))        number++ ;    document.write(number + " ") ;    number++ ;}`

Output:

`1 7 10 13 19 23 28 31 `

### ES6

#### Functional composition

`(() => {     // isHappy :: Int -> Bool    const isHappy = n => {        const f = n =>            foldl(                (a, x) => a + raise(read(x), 2), // ^2                0,                splitOn('', show(n))            ),            p = (s, n) => n === 1 ? (                true            ) : member(n, s) ? (                false            ) : p(                insert(n, s), f(n)            );        return p(new Set(), n);    };     // GENERIC FUNCTIONS ------------------------------------------------------     // enumFromTo :: Int -> Int -> [Int]    const enumFromTo = (m, n) =>        Array.from({            length: Math.floor(n - m) + 1        }, (_, i) => m + i);     // filter :: (a -> Bool) -> [a] -> [a]    const filter = (f, xs) => xs.filter(f);     // foldl :: (b -> a -> b) -> b -> [a] -> b    const foldl = (f, a, xs) => xs.reduce(f, a);     // insert :: Ord a => a -> Set a -> Set a    const insert = (e, s) => s.add(e);     // member :: Ord a => a -> Set a -> Bool    const member = (e, s) => s.has(e);     // read :: Read a => String -> a    const read = JSON.parse;     // show :: a -> String    const show = x => JSON.stringify(x);     // splitOn :: String -> String -> [String]    const splitOn = (cs, xs) => xs.split(cs);     // raise :: Num -> Int -> Num    const raise = (n, e) => Math.pow(n, e);     // take :: Int -> [a] -> [a]    const take = (n, xs) => xs.slice(0, n);     // TEST -------------------------------------------------------------------    return show(        take(8, filter(isHappy, enumFromTo(1, 50)))    );})()`
Output:
`[1, 7, 10, 13, 19, 23, 28, 31]`

Or, to stop immediately at the 8th member of the series, we can preserve functional composition while using an iteratively implemented until() function:

`(() => {     // isHappy :: Int -> Bool    const isHappy = n => {        const f = n =>            foldl(                (a, x) => a + raise(read(x), 2), // ^2                0,                splitOn('', show(n))            ),            p = (s, n) => n === 1 ? (                true            ) : member(n, s) ? (                false            ) : p(                insert(n, s), f(n)            );        return p(new Set(), n);    };     // GENERIC FUNCTIONS ------------------------------------------------------     // filter :: (a -> Bool) -> [a] -> [a]    const filter = (f, xs) => xs.filter(f);     // foldl :: (b -> a -> b) -> b -> [a] -> b    const foldl = (f, a, xs) => xs.reduce(f, a);     // insert :: Ord a => a -> Set a -> Set a    const insert = (e, s) => s.add(e);     // member :: Ord a => a -> Set a -> Bool    const member = (e, s) => s.has(e);     // read :: Read a => String -> a    const read = JSON.parse;     // show :: a -> String    const show = x => JSON.stringify(x);     // splitOn :: String -> String -> [String]    const splitOn = (cs, xs) => xs.split(cs);     // raise :: Num -> Int -> Num    const raise = (n, e) => Math.pow(n, e);     // until :: (a -> Bool) -> (a -> a) -> a -> a    const until = (p, f, x) => {        let v = x;        while (!p(v)) v = f(v);        return v;    };     // TEST -------------------------------------------------------------------    return show(        until(            m => m.xs.length === 8,            m => {                const n = m.n;                return {                    n: n + 1,                    xs: isHappy(n) ? m.xs.concat(n) : m.xs                };            }, {                n: 1,                xs: []            }        )        .xs    );})();`
Output:
`[1, 7, 10, 13, 19, 23, 28, 31]`

## jq

Works with: jq version 1.4
`def is_happy_number:  def next: tostring | explode | map( (. - 48) | .*.) | add;  def last(g): reduce g as \$i (null; \$i);  # state: either 1 or [i, o]   # where o is an an object with the previously encountered numbers as keys  def loop:   recurse( if      . == 1 then empty    # all done            elif .[0] == 1 then 1        # emit 1            else (.[0]| next) as \$n            | if \$n == 1 then 1              elif .[1]|has(\$n|tostring) then empty              else [\$n, (.[1] + {(\$n|tostring):true}) ]              end            end );  1 == last( [.,{}] | loop );`

Emit a stream of the first n happy numbers:

`# Set n to -1 to continue indefinitely:def happy(n):  def subtask:  # state: [i, found]    if .[1] == n then empty    else .[0] as \$n    | if (\$n | is_happy_number) then \$n, ([ \$n+1, .[1]+1 ] | subtask)       else  (.[0] += 1) | subtask      end    end;    [0,0] | subtask; happy(\$n|tonumber)`
Output:
`\$ jq --arg n 8 -n -f happy.jq17101319232831 `

## Julia

` function happy(x)	happy_ints = ref(Int)	int_try = 1	while length(happy_ints) < x		n = int_try		past = ref(Int)		while n != 1	        	n = sum([y^2 for y in digits(n)])	        	contains(past,n) ? break : push!(past,n)	    	end		n == 1 && push!(happy_ints,int_try)		int_try += 1	end	return happy_intsend`

Output

``` julia> happy(8)
8-element Int32 Array:
1
7
10
13
19
23
28
31```

A recursive version:

`sumhappy(n) = sum(x->x^2, digits(n)) function ishappy(x, mem = [])  x == 1?   true :  x in mem? false :  ishappy(sumhappy(x),[mem ; x])end nexthappy (x) = ishappy(x+1) ? x+1 : nexthappy(x+1) happy(n) = [z = 1 ; [z = nexthappy(z) for i = 1:n-1]] `
Output:
```julia> show(happy(8))
[1,7,10,13,19,23,28,31,32]```

Alternate, Translation of C
Faster with use of cache

Translation of: C
`const CACHE = 256buf = zeros(Int,CACHE)buf[1] = 1#happy(n) returns 1 if happy, 0 if notfunction happy(n)	if n < CACHE		buf[n] > 0 && return 2-buf[n]		buf[n] = 2	end	sum = 0	nn = n	while nn != 0		x = nn%10		sum += x*x		nn = int8(nn/10)	end	x = happy(sum)	n < CACHE && (buf[n] = 2-x)	return xendfunction main()	i = 1; counter = 1000000	while counter > 0		if happy(i) == 1			counter -= 1		end		i += 1	end	return i-1end`

## K

`  hpy: {[email protected]&1={~|/x=1 4}{_+/_sqr 0\$'\$x}//:x}   hpy 1+!1001 7 10 13 19 23 28 31 32 44 49 68 70 79 82 86 91 94 97 100   8#hpy 1+!1001 7 10 13 19 23 28 31`

Another implementation which is easy to follow is given below:

` / happynum.k / sum of squares of digits of an integerdgtsmsqr: {d::(); (0<){d::d,x!10; x%:10}/x; +/d*d}/ Test if an integer is a Happy numberisHappy: {s::(); while[1<x;a:(dgtsmsqr x); :[(a _in s); :0; s::s,a]; x:a];:1} / Returns 1 if Happy/ Generate first x Happy numbers and display the listhnum: {[x]; h::();i:1;while[(#h)<x; :[(isHappy i); h::(h,i)]; i+:1]; `0: ,"List of ", (\$x), " Happy Numbers"; h}  `

The output of a session with this implementation is given below:

Output:
```K Console - Enter \ for help

\l happynum
hnum 8
List of 8 Happy Numbers
1 7 10 13 19 23 28 31
```

## Kotlin

Translation of: C#
`// version 1.0.5-2 fun isHappy(n: Int): Boolean {    val cache = mutableListOf<Int>()    var sum = 0    var nn = n    var digit: Int    while (nn != 1) {        if (nn in cache) return false        cache.add(nn)        while (nn != 0) {            digit = nn % 10            sum += digit * digit            nn /= 10        }        nn = sum        sum = 0    }    return true } fun main(args: Array<String>) {    var num = 1    val happyNums = mutableListOf<Int>()    while (happyNums.size < 8) {        if (isHappy(num)) happyNums.add(num)        num++    }    println("First 8 happy numbers : " + happyNums.joinToString(", "))}`
Output:
```First 8 happy numbers : 1, 7, 10, 13, 19, 23, 28, 31
```

## Lasso

`#!/usr/bin/lasso9 define isHappy(n::integer) => {  local(past = set)  while(#n != 1) => {    #n = with i in string(#n)->values sum math_pow(integer(#i), 2)    #past->contains(#n) ? return false | #past->insert(#n)  }  return true} with x in generateSeries(1, 500)  where isHappy(#x)  take 8select #x`

Output:

`1, 7, 10, 13, 19, 23, 28, 31`

## Liberty BASIC

`    ct = 0    n = 0    DO        n = n + 1        IF HappyN(n, sqrInt\$) = 1 THEN            ct = ct + 1            PRINT ct, n        END IF    LOOP UNTIL ct = 8END FUNCTION HappyN(n, sqrInts\$)    n\$ = Str\$(n)    sqrInts = 0    FOR i = 1 TO Len(n\$)        sqrInts = sqrInts + Val(Mid\$(n\$, i, 1)) ^ 2    NEXT i    IF sqrInts = 1 THEN        HappyN = 1        EXIT FUNCTION    END IF    IF Instr(sqrInts\$, ":";Str\$(sqrInts);":") > 0 THEN        HappyN = 0        EXIT FUNCTION    END IF    sqrInts\$ = sqrInts\$ + Str\$(sqrInts) + ":"    HappyN = HappyN(sqrInts, sqrInts\$)END FUNCTION`

Output:-

```1             1
2             7
3             10
4             13
5             19
6             23
7             28
8             31
```

## Locomotive Basic

`10 mode 1:defint a-z20 for i=1 to 10030 i2=i40 for l=1 to 2050 a\$=str\$(i2)60 i2=070 for j=1 to len(a\$)80 d=val(mid\$(a\$,j,1))90 i2=i2+d*d100 next j110 if i2=1 then print i;"is a happy number":n=n+1:goto 150120 if i2=4 then 150 ' cycle found130 next l140 ' check if we have reached 8 numbers yet150 if n=8 then end160 next i`

## Logo

`to sum_of_square_digits :number  output (apply "sum (map [[d] d*d] ` :number))end  to is_happy? :number [:seen []]  output cond [    [ [:number = 1] "true ]    [ [member? :number :seen] "false ]    [ else (is_happy? (sum_of_square_digits :number) (lput :number :seen))]  ]end to n_happy :count [:start 1] [:result []]  output cond [    [ [:count <= 0] :result ]    [ [is_happy? :start]      (n_happy (:count-1) (:start+1) (lput :start :result)) ]    [ else      (n_happy :count (:start+1) :result) ]  ]end print n_happy 8bye`

Output:

`1 7 10 13 19 23 28 31`

## LOLCODE

Works with: lci 0.10.3
`OBTW   Happy Numbers Rosetta Code task in LOLCODE  Requires 1.3 for BUKKIT availabilityTLDRHAI 1.3CAN HAS STDIO? BTW Simple list implementation.  BTW Used for the list of numbers already seen in IZHAPPY BTW Create a listHOW IZ I MAEKLIST   I HAS A LIST ITZ A BUKKIT  LIST HAS A LENGTH ITZ 0  FOUND YR LISTIF U SAY SO BTW Append an item to listHOW IZ I PUTIN YR LIST AN YR ITEM  LIST HAS A SRS LIST'Z LENGTH ITZ ITEM  LIST'Z LENGTH R SUM OF LIST'Z LENGTH AN 1IF U SAY SO BTW Check for presence of an item in the listHOW IZ I DUZLISTHAS YR HAYSTACK AN YR NEEDLE  IM IN YR BARN UPPIN YR INDEX WILE DIFFRINT INDEX AN HAYSTACK'Z LENGTH    I HAS A ITEM ITZ HAYSTACK'Z SRS INDEX    BOTH SAEM ITEM AN NEEDLE    O RLY?      YA RLY        FOUND YR WIN    OIC  IM OUTTA YR BARN  FOUND YR FAILIF U SAY SO BTW Calculate the next number using the happy formulaHOW IZ I HAPPYSTEP YR NUM  I HAS A NEXT ITZ 0  IM IN YR LOOP     BOTH SAEM NUM AN 0    O RLY?      YA RLY        GTFO    OIC    I HAS A DIGIT ITZ MOD OF NUM AN 10    NUM R QUOSHUNT OF NUM AN 10    I HAS A SQUARE ITZ PRODUKT OF DIGIT AN DIGIT    NEXT R SUM OF NEXT AN SQUARE  IM OUTTA YR LOOP  FOUND YR NEXTIF U SAY SO BTW Check to see if a number is happyHOW IZ I IZHAPPY YR NUM  I HAS A SEENIT ITZ I IZ MAEKLIST MKAY  IM IN YR LOOP     BOTH SAEM NUM AN 1    O RLY?      YA RLY           FOUND YR WIN    OIC    I IZ DUZLISTHAS YR SEENIT AN YR NUM MKAY    O RLY?      YA RLY        FOUND YR FAIL    OIC    I IZ PUTIN YR SEENIT AN YR NUM MKAY    NUM R I IZ HAPPYSTEP YR NUM MKAY  IM OUTTA YR LOOPIF U SAY SO BTW Print out the first 8 happy numbersI HAS A KOUNT ITZ 0IM IN YR LOOP UPPIN YR NUM WILE DIFFRINT KOUNT AN 8  I IZ IZHAPPY YR NUM MKAY  O RLY?    YA RLY      KOUNT R SUM OF KOUNT AN 1      VISIBLE NUM  OICIM OUTTA YR LOOPKTHXBYE`
Output:
```1
7
10
13
19
23
28
31```

## Lua

`function digits(n)  if n > 0 then return n % 10, digits(math.floor(n/10)) endendfunction sumsq(a, ...)  return a and a ^ 2 + sumsq(...) or 0endlocal happy = setmetatable({true, false, false, false}, {      __index = function(self, n)         self[n] = self[sumsq(digits(n))]         return self[n]      end } )i, j = 0, 1repeat   i, j = happy[j] and (print(j) or i+1) or i, j + 1until i == 8`

Output:

```1
7
10
13
19
23
28
31```

## M2000 Interpreter

Translation of: ActionScript

Lambda Function PrintHappy has a closure another lambda function IsHappy which has a closure of another lambda function the sumOfSquares.

` Function FactoryHappy {      sumOfSquares= lambda (n) ->{                  k\$=str\$(abs(n),"")                  Sum=0                  For i=1 to len(k\$)                        sum+=val(mid\$(k\$,i,1))**2                  Next i                  =sum      }      IsHappy=Lambda sumOfSquares (n) ->{            Inventory sequence            While n<>1 {                  Append sequence, n                  n=sumOfSquares(n)                   if exist(sequence, n) then =false : Break            }            =True      }      =Lambda IsHappy ->{                  numleft=8                  numToTest=1                  While numleft {                        if ishappy(numToTest) Then {                              Print numToTest                              numleft--                        }                        numToTest++                  }      }}PrintHappy=factoryHappy()Call PrintHappy() `
Output:
``` 1
7
10
13
19
23
28
31```

`            NORMAL MODE IS INTEGER            BOOLEAN CYCLE            DIMENSION CYCLE(200)            VECTOR VALUES OUTFMT = \$I2*\$             SEEN = 0            I = 0 NEXNUM      THROUGH ZERO, FOR K=0, 1, K.G.200ZERO        CYCLE(K) = 0B            I = I + 1            SUMSQR = ICHKLP       N = SUMSQR            SUMSQR = 0SUMLP       DIG = N-N/10*10            SUMSQR = SUMSQR + DIG*DIG            N = N/10            WHENEVER N.NE.0, TRANSFER TO SUMLP            WHENEVER SUMSQR.E.1, TRANSFER TO HAPPY            WHENEVER CYCLE(SUMSQR), TRANSFER TO NEXNUM            CYCLE(SUMSQR) = 1B            TRANSFER TO CHKLP HAPPY       PRINT FORMAT OUTFMT,I            SEEN = SEEN+1            WHENEVER SEEN.L.8, TRANSFER TO NEXNUM             END OF PROGRAM `
Output:
``` 1
7
10
13
19
23
28
31```

## Maple

To begin, here is a procedure to compute the sum of the squares of the digits of a positive integer. It uses the built-in procedure irem, which computes the integer remainder and, if passed a name as the optional third argument, assigns it the corresponding quotient. (In other words, it performs integer division with remainder. There is also a dual, companion procedure iquo, which returns the integer quotient and assigns the remainder to the (optional) third argument.)

`SumSqDigits := proc( n :: posint )        local s := 0;        local m := n;        while m <> 0 do                s := s + irem( m, 10, 'm' )^2        end do;        send proc:`

(Note that the unevaluation quotes on the third argument to irem are essential here, as that argument must be a name and, if m were passed without quotes, it would evaluate to a number.)

For example,

` > SumSqDigits( 1234567890987654321 );                                  570 `

We can check this by computing it another way (more directly).

` > n := 1234567890987654321:> `+`( op( map( parse, StringTools:-Explode( convert( n, 'string' ) ) )^~2) );                                  570 `

The most straight-forward way to check whether a number is happy or sad seems also to be the fastest (that I could think of).

`Happy? := proc( n )        if n = 1 then                true        elif n = 4 then                false        else                local s := SumSqDigits( n );                while not ( s in { 1, 4 } ) do                        s := SumSqDigits( s )                end do;                evalb( s = 1 )        end ifend proc:`

We can use this to determine the number of happy (H) and sad (S) numbers up to one million as follows.

` > H, S := selectremove( Happy?, [seq]( 1 .. N ) ):> nops( H ), nops( S );                             143071, 856929 `

Finally, to solve the stated problem, here is a completely straight-forward routine to locate the first N happy numbers, returning them in a set.

`FindHappiness := proc( N )        local count := 0;        local T := table();        for local i while count < N do                if Happy?( i ) then                        count := 1 + count;                        T[ count ] := i                end if        end do;        {seq}( T[ i ], i = 1 .. count )end proc:`

With input equal to 8, we get

` > FindHappiness( 8 );                     {1, 7, 10, 13, 19, 23, 28, 31} `

For completeness, here is an implementation of the cycle detection algorithm for recognizing happy numbers. It is much slower, however.

`Happy? := proc( n :: posint )        local a, b;        a, b := n, SumSqDigits( n );        while a <> b do                a := SumSqDigits( a );                b := ([email protected]@2)( b )        end do;        evalb( a = 1 )end proc:`

## Mathematica / Wolfram Language

Custom function HappyQ:

`AddSumSquare[input_]:=Append[input,Total[IntegerDigits[Last[input]]^2]]NestUntilRepeat[a_,f_]:=NestWhile[f,{a},!MemberQ[Most[Last[{##}]],Last[Last[{##}]]]&,All]HappyQ[a_]:=Last[NestUntilRepeat[a,AddSumSquare]]==1`

Examples for a specific number:

`HappyQ[1337]HappyQ[137]`

gives back:

`TrueFalse`

Example finding the first 8:

`m = 8;n = 1;i = 0;happynumbers = {};While[n <= m, i++; If[HappyQ[i],  n++;  AppendTo[happynumbers, i]  ] ]happynumbers`

gives back:

`{1, 7, 10, 13, 19, 23, 28, 31}`

## MATLAB

Recursive version:

`function findHappyNumbers    nHappy = 0;    k = 1;    while nHappy < 8        if isHappyNumber(k, [])            fprintf('%d ', k)            nHappy = nHappy+1;        end        k = k+1;    end    fprintf('\n')end function hap = isHappyNumber(k, prev)    if k == 1        hap = true;    elseif ismember(k, prev)        hap = false;    else        hap = isHappyNumber(sum((sprintf('%d', k)-'0').^2), [prev k]);    endend`
Output:
`1 7 10 13 19 23 28 31 `

## MAXScript

` fn isHappyNumber n =(	local pastNumbers = #()	while n != 1 do	(		n = n as string		local newNumber = 0		for i = 1 to n.count do		(			local digit = n[i] as integer			newNumber += pow digit 2		)		n = newNumber 		if (finditem pastNumbers n) != 0 do return false		append pastNumbers newNumber	)	n == 1)printed = 0for i in (for h in 1 to 500 where isHappyNumber h collect h) do(	if printed == 8 do exit	print i as string	printed += 1 ) `

Output:

` 17101319232831 `

## Mercury

`:- module happy.:- interface.:- import_module io. :- pred main(io::di, io::uo) is det. :- implementation.:- import_module int, list, set_tree234. main(!IO) :-    print_line(get_n_happy_numbers(8, 1), !IO). :- func get_n_happy_numbers(int, int) = list(int). get_n_happy_numbers(NumToFind, N) =    ( if NumToFind > 0 then       ( if is_happy(N, init)       then [N | get_n_happy_numbers(NumToFind - 1, N + 1)]       else get_n_happy_numbers(NumToFind, N + 1)       )    else       []    ). :- pred is_happy(int::in, set_tree234(int)::in) is semidet. is_happy(1, _).is_happy(N, !.Seen) :-   not member(N, !.Seen),   insert(N, !Seen),   is_happy(sum_sqr_digits(N), !.Seen). :- func sum_sqr_digits(int) = int. sum_sqr_digits(N) =   ( if N < 10 then sqr(N) else sqr(N mod 10) + sum_sqr_digits(N div 10) ). :- func sqr(int) = int. sqr(X) = X * X.`
Output:
`[1, 7, 10, 13, 19, 23, 28, 31]`

## MiniScript

This solution uses the observation that any infinite cycle of this algorithm hits the number 89, and so that can be used to know when we've found an unhappy number.

`isHappy = function(x)    while true        if x == 89 then return false        sum = 0        while x > 0            sum = sum + (x % 10)^2            x = floor(x / 10)        end while        if sum == 1 then return true        x = sum    end whileend function found = []i = 1while found.len < 8    if isHappy(i) then found.push i    i = i + 1end whileprint "First 8 happy numbers: " + found`
Output:
`First 8 happy numbers: [1, 7, 10, 13, 19, 23, 28, 31]`

## ML

### mLite

`(*A happy number is defined by the following process. Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers, while those that do not end in 1 are unhappy numbers. Display an example of your output here.*) local	fun get_digits 			(d, s) where (d = 0) = s		| 	(d, s) = get_digits( d div 10, (d mod 10) :: s)		| 	n = get_digits( n div 10, [n mod 10] )	;	fun mem 			(x, []) = false		| 	(x, a :: as) where (x = a) = true		| 	(x, _ :: as) = mem (x, as)in	fun happy			1 = "happy"		|	n =				let 					val this = (fold (+,0) ` map (fn n = n ^ 2) ` get_digits n);					val sads = [2, 4, 16, 37, 58, 89, 145, 42, 20]				in					if (mem (n,sads)) then						"unhappy"					else						happy this				endend; foreach (fn n = (print n; print " is "; println ` happy n)) ` iota 10; `

Output:

```1 is happy
2 is unhappy
3 is unhappy
4 is unhappy
5 is unhappy
6 is unhappy
7 is happy
8 is unhappy
9 is unhappy
10 is happy```

## Modula-2

`MODULE HappyNumbers;FROM InOut IMPORT WriteCard, WriteLn; CONST Amount = 8;VAR seen, num: CARDINAL; PROCEDURE SumDigitSquares(n: CARDINAL): CARDINAL;VAR sum, digit: CARDINAL;BEGIN    sum := 0;    WHILE n>0 DO        digit := n MOD 10;        n := n DIV 10;        sum := sum + digit * digit;    END;    RETURN sum;END SumDigitSquares; PROCEDURE Happy(n: CARDINAL): BOOLEAN;VAR i: CARDINAL;    seen: ARRAY [0..255] OF BOOLEAN;BEGIN    FOR i := 0 TO 255 DO         seen[i] := FALSE;     END;    REPEAT        seen[n] := TRUE;        n := SumDigitSquares(n);    UNTIL seen[n];    RETURN seen[1];END Happy; BEGIN    seen := 0;    num := 0;    WHILE seen < Amount DO        IF Happy(num) THEN            INC(seen);            WriteCard(num,2);            WriteLn();        END;        INC(num);    END;END HappyNumbers.`
Output:
``` 1
7
10
13
19
23
28
31```

## MUMPS

`ISHAPPY(N) ;Determines if a number N is a happy number ;Note that the returned strings do not have a leading digit unless it is a happy number IF (N'=N\1)!(N<0) QUIT "Not a positive integer" NEW SUM,I ;SUM is the sum of the square of each digit ;I is a loop variable ;SEQ is the sequence of previously checked SUMs from the original N ;If it isn't set already, initialize it to an empty string IF \$DATA(SEQ)=0 NEW SEQ SET SEQ="" SET SUM=0 FOR I=1:1:\$LENGTH(N) DO .SET SUM=SUM+(\$EXTRACT(N,I)*\$EXTRACT(N,I)) QUIT:(SUM=1) SUM QUIT:\$FIND(SEQ,SUM)>1 "Part of a sequence not containing 1" SET SEQ=SEQ_","_SUM QUIT \$\$ISHAPPY(SUM)HAPPY(C) ;Finds the first C happy numbers NEW I ;I is a counter for what integer we're looking at WRITE !,"The first "_C_" happy numbers are:" FOR I=1:1 QUIT:C<1  SET Q=+\$\$ISHAPPY(I) WRITE:Q !,I SET:Q C=C-1 KILL I QUIT`
Output:
```USER>D HAPPY^ROSETTA(8)

The first 8 happy numbers are:
1
7
10
13
19
23
28
31
USER>W:+\$\$ISHAPPY^ROSETTA(320) "Happy Number"
Happy Number
USER>W:+\$\$ISHAPPY^ROSETTA(321) "Happy Number"

USER>
```

## NetRexx

Translation of: REXX
`/*NetRexx program to display the 1st 8 (or specified arg) happy numbers*/limit	 = arg[0]                        /*get argument for  LIMIT.        */say limitif limit = null, limit ='' then limit=8  /*if not specified, set LIMIT to 8*/haps	 = 0                             /*count of happy numbers so far.  */ loop n=1 while haps < limit              /*search integers starting at one.*/  q=n                                    /*Q may or may not be "happy".    */  a=0   loop forever                           /*see if  Q  is a happy number.   */    if q==1 then do                      /*if  Q  is unity, then it's happy*/      haps = haps + 1                    /*bump the count of happy numbers.*/      say n                              /*display the number.             */      iterate n                          /*and then keep looking for more. */    end     sum=0                                /*initialize sum to zero.         */     loop j=1 for q.length                /*add the squares of the numerals.*/      sum = sum + q.substr(j,1) ** 2    end     if a[sum] then iterate n             /*if already summed, Q is unhappy.*/    a[sum]=1                             /*mark the sum as being found.    */    q=sum                                /*now, lets try the  Q  sum.      */  endend`
Output
```1
7
10
13
19
23
28
31
```

Sample output when 100 is specified as the program's argument.

```1
7
10
13
19
23
28
31
32
44
49
68
70
79
82
86
91
94
97
100
103
109
129
130
133
139
167
176
188
190
192
193
203
208
219
226
230
236
239
262
263
280
291
293
301
302
310
313
319
320
326
329
331
338
356
362
365
367
368
376
379
383
386
391
392
397
404
409
440
446
464
469
478
487
490
496
536
556
563
565
566
608
617
622
623
632
635
637
638
644
649
653
655
656
665
671
673
680
683
694
```

## Nim

Translation of: Python
`import intsets proc happy(n: int): bool =  var    n = n    past = initIntSet()  while n != 1:    let s = \$n    n = 0    for c in s:      let i = ord(c) - ord('0')      n += i * i    if n in past:      return false    past.incl(n)  return true for x in 0..31:  if happy(x):    echo x`

Output:

```1
7
10
13
19
23
28
31```

## Objeck

`use IO;use Structure; bundle Default {  class HappyNumbers {    function : native : IsHappy(n : Int) ~ Bool {      cache := IntVector->New();        sum := 0;        while(n <> 1) {          if(cache->Has(n)) {            return false;          };           cache->AddBack(n);          while(n <> 0) {            digit := n % 10;            sum += (digit * digit);            n /= 10;          };           n := sum;          sum := 0;        };         return true;                  }       function : Main(args : String[]) ~ Nil {        num := 1;        happynums := IntVector->New();         while(happynums->Size() < 8) {          if(IsHappy(num)) {            happynums->AddBack(num);        };         num += 1;      };       Console->Print("First 8 happy numbers: ");      each(i : happynums) {        Console->Print(happynums->Get(i))->Print(",");      };      Console->PrintLine("");    }  }}`

output:

`First 8 happy numbers: 1,7,10,13,19,23,28,31,`

## OCaml

`open Num let step =	let rec aux s n =	if n =/ Int 0 then s else		let q = quo_num n (Int 10)		and r = mod_num n (Int 10)		in aux (s +/ (r */ r)) q	in aux (Int 0) ;; let happy n =	let rec aux x y =		if x =/ y then x else aux (step x) (step (step y))	in (aux n (step n)) =/ Int 1 ;; let first n =	let rec aux v x n =		if n = 0 then v else			if happy x			then aux (x::v) (x +/ Int 1) (n - 1)			else aux v (x +/ Int 1) n	in aux [ ] (Int 1) n ;; List.iter print_endline (	List.rev_map string_of_num (first 8)) ;;`

Output:

```\$ ocaml nums.cma happy_numbers.ml
1
7
10
13
19
23
28
31```

## Oforth

`: isHappy(n)| cycle |   ListBuffer new ->cycle    while(n 1 <>) [      cycle include(n) ifTrue: [ false return ]      cycle add(n)      0 n asString apply(#[ asDigit sq + ]) ->n      ]   true ; : happyNum(N)| numbers |   ListBuffer new ->numbers   1 while(numbers size N <>) [ dup isHappy ifTrue: [ dup numbers add ] 1+ ]   numbers println ;`

Output:

```>happyNum(8)
[1, 7, 10, 13, 19, 23, 28, 31]
```

## ooRexx

` count = 0say "First 8 happy numbers are:"loop i = 1 while count < 8    if happyNumber(i) then do        count += 1        say i    endend ::routine happyNumber  use strict arg number   -- use to trace previous cycle results  previous = .set~new  loop forever      -- stop when we hit the target      if number = 1 then return .true      -- stop as soon as we start cycling      if previous[number] \== .nil then return .false      previous~put(number)      next = 0      -- loop over all of the digits      loop digit over number~makearray('')          next += digit * digit      end      -- and repeat the cycle      number = next  end `
```First 8 happy numbers are:
1
7
10
13
19
23
28
31
```

## Oz

`functorimport  Systemdefine  fun {IsHappy N}     {IsHappy2 N nil}  end   fun {IsHappy2 N Seen}     if     N == 1          then true     elseif {Member N Seen} then false     else	Next = {Sum {Map {Digits N} Square}}     in	{IsHappy2 Next N|Seen}     end  end   fun {Sum Xs}     {FoldL Xs Number.'+' 0}  end   fun {Digits N}     {Map {Int.toString N} fun {\$ D} D - &0 end}  end   fun {Square N} N*N end   fun lazy {Nat I}     I|{Nat I+1}  end   %% List.filter is eager. But we need a lazy Filter:  fun lazy {LFilter Xs P}     case Xs of X|Xr andthen {P X} then X|{LFilter Xr P}     [] _|Xr then {LFilter Xr P}     [] nil then nil     end  end   HappyNumbers = {LFilter {Nat 1} IsHappy}in  {System.show {List.take HappyNumbers 8}}end`

Output:

`[1 7 10 13 19 23 28 31]`

## PARI/GP

Works with: PARI/GP version 2.4.3 and above
This code uses the select() function, which was added in PARI version 2.4.2. The order of the arguments changed between versions; to use in 2.4.2 change `select(function, vector)` to `select(vector, function)`.

If the number has more than three digits, the sum of the squares of its digits has fewer digits than the number itself. If the number has three digits, the sum of the squares of its digits is at most 3 * 9^2 = 243. A simple solution is to look up numbers up to 243 and calculate the sum of squares only for larger numbers.

`H=[1,7,10,13,19,23,28,31,32,44,49,68,70,79,82,86,91,94,97,100,103,109,129,130,133,139,167,176,188,190,192,193,203,208,219,226,230,236,239];isHappy(n)={  if(n<262,    setsearch(H,n)>0  ,    n=eval(Vec(Str(n)));    isHappy(sum(i=1,#n,n[i]^2))  )};select(isHappy, vector(31,i,i))`

Output:

`%1 = [1, 7, 10, 13, 19, 23, 28, 31]`

## Pascal

`Program HappyNumbers (output); uses  Math; function find(n: integer; cache: array of integer): boolean;  var    i: integer;  begin    find := false;    for i := low(cache) to high(cache) do      if cache[i] = n then        find := true;  end; function is_happy(n: integer): boolean;  var    cache: array of integer;    sum: integer;  begin    setlength(cache, 1);    repeat      sum := 0;      while n > 0 do      begin        sum := sum + (n mod 10)**2;        n := n div 10;      end;      if sum = 1 then      begin        is_happy := true;        break;      end;      if find(sum, cache) then      begin        is_happy := false;        break;      end;      n := sum;      cache[high(cache)]:= sum;      setlength(cache, length(cache)+1);    until false;  end; var  n, count: integer; begin  n := 1;  count := 0;  while count < 8 do  begin    if is_happy(n) then    begin      inc(count);      write(n, ' ');    end;    inc(n);  end;  writeln;end.`

Output:

```:> ./HappyNumbers
1 7 10 13 19 23 28 31
```

### alternative for counting fast

Works with: Free Pascal

The Cache is limited to maximum value of the sum of squared digits and filled up in a blink of an eye.Even for cDigit2=1e9 takes 0.7s.Calculation of sum of squared digits is improved.Saving this SqrdSumCache speeds up tremendous. So i am able to check if the 1'000'000 th happy number is 7105849 as stated in C language.This seems to be true. Extended to 10e18 Tested with Free Pascal 3.0.4

`Program HappyNumbers (output);{\$IFDEF FPC}  {\$MODE DELPHI}  {\$OPTIMIZATION ON,All}{\$ELSE}  {\$APPLICATION CONSOLE}{\$ENDIF}//{\$DEFINE Use1E9}uses  sysutils,//Timing  strutils;//Numb2USA const  base = 10;  HighCache = 20*(sqr(base-1));//sum of sqr digit of Uint64{\$IFDEF Use1E9}  cDigit1  = sqr(base)*sqr(base);//must be power of base  cDigit2  = Base*sqr(cDigit1);// 1e9  cMaxPot  = 18;{\$ELSE}  cDigit1  = base*sqr(base);//must be power of base  cDigit2  = sqr(cDigit1);// 1e6  cMaxPot  = 14;{\$ENDIF} type  tSumSqrDgts    = array[0..cDigit2] of word;  tCache         = array[0..2*HighCache] of word;  tSqrdSumCache  = array[0..2*HighCache] of Uint32; var  SumSqrDgts :tSumSqrDgts;  Cache : tCache;   SqrdSumCache1,  SqrdSumCache2 :tSqrdSumCache;   T1,T0 : TDateTime;  MAX2,Max1 : NativeInt; procedure InitSumSqrDgts;//calc all sum of squared digits 0..cDigits2//using already calculated valuesvar  i,j,n,sq,Base1: NativeInt;begin  For i := 0 to Base-1 do    SumSqrDgts[i] := i*i;  Base1 := Base;  n := Base;  repeat    For i := 1 to base-1 do    Begin      sq := SumSqrDgts[i];      For j := 0 to base1-1 do      Begin        SumSqrDgts[n] := sq+SumSqrDgts[j];        inc(n);      end;    end;    Base1 := Base1*base;  until Base1 >= cDigit2;  SumSqrDgts[n] := 1;end; function SumSqrdDgt(n: Uint64):NativeUint;inline;var  r: Uint64;begin  result := 0;  while n>cDigit2 do  Begin    r := n;    n := n div cDigit2;    r := r-n*cDigit2;    inc(result,SumSqrDgts[r]);  end;  inc(result,SumSqrDgts[n]);end; procedure CalcSqrdSumCache1;var  Count : tSqrdSumCache;  i,sq,result : NativeInt;begin  For i :=High(Count) downto 0 do    Count[i] := 0;  //count the manifold  For i := cDigit1-1 downto 0 do    inc(count[SumSqrDgts[i]]);  For i := High(Count) downto 0 do    if count[i] <> 0 then    Begin      Max1 := i;      BREAK;    end;  For sq := 0 to (20-3)*81 do  Begin    result := 0;    For i := Max1 downto 0 do      inc(result,Count[i]*Cache[sq+i]);    SqrdSumCache1[sq] := result;  end;end; procedure CalcSqrdSumCache2;var  Count : tSqrdSumCache;  i,sq,result : NativeInt;begin  For i :=High(Count) downto 0 do    Count[i] := 0;  For i := cDigit2-1 downto 0 do    inc(count[SumSqrDgts[i]]);  For i := High(Count) downto 0 do    if count[i] <> 0 then    Begin      Max2 := i;      BREAK;    end;  For sq := 0 to (20-6)*81 do  Begin    result := 0;    For i := Max2 downto 0 do      inc(result,Count[i]*Cache[sq+i]);    SqrdSumCache2[sq] := result;  end;end; procedure Inithappy;var  n,s,p : NativeUint;Begin  fillchar(SqrdSumCache1,SizeOf(SqrdSumCache1),#0);  fillchar(SqrdSumCache2,SizeOf(SqrdSumCache2),#0);  InitSumSqrDgts;  fillChar(Cache,SizeOf(Cache),#0);   Cache[1] := 1;  For n := 1 to High(Cache) do  Begin    If Cache[n] = 0 then    Begin      //start a linked list      Cache[n] := n;      p := n;      s := SumSqrdDgt(p);      while Cache[s] = 0 do      Begin        Cache[s] := p;        p := s;        s := SumSqrdDgt(p);      end;      //mark linked list backwards as happy number      IF Cache[s] = 1 then      Begin        repeat          s := Cache[p];          Cache[p] := 1;          p := s;        until s = n;        Cache[n] := 1;      end;    end;  end;  //mark all unhappy numbers with 0  For n := 1 to High(Cache) do    If Cache[n] <> 1 then      Cache[n] := 0;   CalcSqrdSumCache1;   CalcSqrdSumCache2;end; function is_happy(n: NativeUint): boolean;inline;begin  is_happy := Boolean(Cache[SumSqrdDgt(n)])end; function nthHappy(Limit: Uint64):Uint64;var  d,e,sE: NativeUint;begin  result := 0;  d := 0;  e := 0;  sE := SumSqrDgts[e];  //big steps  while Limit >= cDigit2 do  begin    dec(Limit,SqrdSumCache2[SumSqrDgts[d]+sE]);    inc(result,cDigit2);    inc(d);    IF d >=cDigit2 then    Begin      inc(e);      sE := SumSqrdDgt(e);//SumSqrDgts[e];      d :=0;    end;  end;  //small steps  while Limit >= cDigit1 do  Begin    dec(Limit,SqrdSumCache1[SumSqrdDgt(result)]);    inc(result,cDigit1);  end;  //ONE BY ONE  while Limit > 0 do  begin    dec(Limit,Cache[SumSqrdDgt(result)]);    inc(result);  end;  result -= 1;end; var  n, count :Uint64;  Limit: NativeUint;begin  write('cDigit1 = ',Numb2USA(IntToStr(cDigit1)));  writeln('  cDigit2 = ',Numb2USA(IntToStr(cDigit2)));  T0 := now;  Inithappy;  writeln('Init takes ',FormatDateTime(' HH:NN:SS.ZZZ',now-T0));  n := 1;  count := 0;  while count < 10  do  begin    if is_happy(n) then    begin      inc(count);      write(n, ' ');    end;    inc(n);  end;  writeln;   T0 := now;  T1 := T0;  n := 1;  Limit := 10;  repeat    writeln('1E',n:2,' n.th happy number ',Numb2USA(IntToStr(nthHappy(Limit))):26,      FormatDateTime(' HH:NN:SS.ZZZ',now-T1));    T1 := now;    inc(n);    Limit := limit*10;  until n> cMaxPot;  writeln('Total time counting ',FormatDateTime('HH:NN:SS.ZZZ',now-T0));end. `
output
```cDigit1 = 1,000  cDigit2 = 1,000,000
Init takes  00:00:00.004
1 7 10 13 19 23 28 31 32 44
1E 1 n.th happy number                         44 00:00:00.000
1E 2 n.th happy number                        694 00:00:00.000
1E 3 n.th happy number                      6,899 00:00:00.000
1E 4 n.th happy number                     67,169 00:00:00.000
1E 5 n.th happy number                    692,961 00:00:00.000
1E 6 n.th happy number                  7,105,849 00:00:00.000
1E 7 n.th happy number                 71,313,350 00:00:00.000
1E 8 n.th happy number                698,739,425 00:00:00.000
1E 9 n.th happy number              6,788,052,776 00:00:00.000
1E10 n.th happy number             66,305,148,869 00:00:00.000
1E11 n.th happy number            660,861,957,662 00:00:00.001
1E12 n.th happy number          6,745,877,698,967 00:00:00.008
1E13 n.th happy number         70,538,879,028,725 00:00:00.059
1E14 n.th happy number        744,083,563,164,178 00:00:00.612
Total time counting 00:00:00.680

real    0m0,685s

cDigit1 = 10,000  cDigit2 = 1,000,000,000
Init takes  00:00:02.848
1 7 10 13 19 23 28 31 32 44
1E 1 n.th happy number                         44 00:00:00.000
1E 2 n.th happy number                        694 00:00:00.000
1E 3 n.th happy number                      6,899 00:00:00.000
1E 4 n.th happy number                     67,169 00:00:00.000
1E 5 n.th happy number                    692,961 00:00:00.000
1E 6 n.th happy number                  7,105,849 00:00:00.000
1E 7 n.th happy number                 71,313,350 00:00:00.000
1E 8 n.th happy number                698,739,425 00:00:00.001
1E 9 n.th happy number              6,788,052,776 00:00:00.008
1E10 n.th happy number             66,305,148,869 00:00:00.010
1E11 n.th happy number            660,861,957,662 00:00:00.009
1E12 n.th happy number          6,745,877,698,967 00:00:00.008
1E13 n.th happy number         70,538,879,028,725 00:00:00.008
1E14 n.th happy number        744,083,563,164,178 00:00:00.011
1E15 n.th happy number      7,888,334,045,397,315 00:00:00.019
1E16 n.th happy number     82,440,929,809,838,249 00:00:00.079
1E17 n.th happy number    845,099,936,580,193,833 00:00:00.698
1E18 n.th happy number  8,489,964,903,498,345,213 00:00:06.920
Total time counting 00:00:07.771

real    0m10,627s
```

## Perl

Since all recurrences end with 1 or repeat (37,58,89,145,42,20,4,16), we can do this test very quickly without having to make hashes of seen numbers.

`use List::Util qw(sum); sub ishappy {  my \$s = shift;  while (\$s > 6 && \$s != 89) {    \$s = sum(map { \$_*\$_ } split(//,\$s));  }  \$s == 1;} my \$n = 0;print join(" ", map { 1 until ishappy(++\$n); \$n; } 1..8), "\n";`
Output:
`1 7 10 13 19 23 28 31`

Or we can solve using only the rudimentary task knowledge as below. Note the slightly different ways of doing the digit sum and finding the first 8 numbers where ishappy(n) is true -- this shows there's more than one way to do even these small sub-tasks.

Translation of: Raku
`use List::Util qw(sum);sub is_happy {    my (\$n) = @_;    my %seen;    while (1) {        \$n = sum map { \$_ ** 2 } split //, \$n;        return 1 if \$n == 1;        return 0 if \$seen{\$n}++;    }} my \$n;is_happy( ++\$n ) and print "\$n " or redo for 1..8;`
Output:
`1 7 10 13 19 23 28 31`

## Phix

Copy of Euphoria tweaked to give a one-line output

```function is_happy(integer n)
sequence seen = {}
while n>1 do
seen &= n
integer k = 0
while n>0 do
k += power(remainder(n,10),2)
n = floor(n/10)
end while
n = k
if find(n,seen) then
return false
end if
end while
return true
end function

integer n = 1
sequence s = {}
while length(s)<8 do
if is_happy(n) then
s &= n
end if
n += 1
end while
?s
```
Output:
```{1,7,10,13,19,23,28,31}
```

## PHP

Translation of: D
`function isHappy(\$n) {    while (1) {        \$total = 0;        while (\$n > 0) {            \$total += pow((\$n % 10), 2);            \$n /= 10;        }        if (\$total == 1)            return true;        if (array_key_exists(\$total, \$past))            return false;        \$n = \$total;        \$past[\$total] = 0;    }} \$i = \$cnt = 0;while (\$cnt < 8) {    if (isHappy(\$i)) {        echo "\$i ";        \$cnt++;    }    \$i++;}`
`1 7 10 13 19 23 28 31 `

## Picat

`go =>   println(happy_len(8)). happy(N) =>    S = [N],   Happy = 1,   while (Happy == 1, N > 1)     N := sum([to_integer(I)**2 : I in N.to_string()]),     if member(N,S) then       Happy := 0     else        S := S ++ [N]     end   end,   Happy == 1. happy_len(Limit) = S =>    S = [],   N = 1,   while (S.length < Limit)       if happy(N) then         S := S ++ [N]      end,      N := N + 1   end.`
Output:
`[1,7,10,13,19,23,28,31]`

## PicoLisp

`(de happy? (N)   (let Seen NIL      (loop         (T (= N 1) T)         (T (member N Seen))         (setq N            (sum '((C) (** (format C) 2))               (chop (push 'Seen N)) ) ) ) ) ) (let H 0   (do 8      (until (happy? (inc 'H)))      (printsp H) ) )`

Output:

`1 7 10 13 19 23 28 31`

## PILOT

`C :max=8  :n=0  :i=0*testU :*happyT (a=1):#nC (a=1):i=i+1C :n=n+1J (i<max):*testE : *happyC :a=n  :x=nU :*sumsqC :b=s*loopC :x=aU :*sumsqC :a=sC :x=bU :*sumsqC :x=sU :*sumsqC :b=sJ (a<>b):*loopE : *sumsqC :s=0*digitC :y=x/10  :z=x-y*10  :s=s+z*#z  :x=yJ (x):*digitE :`
Output:
```1
7
10
13
19
23
28
31```

## PL/I

`test: proc options (main); /* 19 November 2011 */   declare (i, j, n, m, nh initial (0) ) fixed binary (31); main_loop:   do j = 1 to 100;      n = j;      do i = 1 to 100;         m = 0;         /* Form the sum of squares of the digits. */         do until (n = 0);            m = m + mod(n, 10)**2;            n = n/10;         end;         if m = 1 then            do;               put skip list (j || ' is a happy number');               nh = nh + 1;               if nh = 8 then return;               iterate main_loop;            end;         n = m; /* Replace n with the new number formed from digits. */      end;   end;end test; `

OUTPUT:

```             1 is a happy number
7 is a happy number
10 is a happy number
13 is a happy number
19 is a happy number
23 is a happy number
28 is a happy number
31 is a happy number
```

## PL/M

`100H: /* FIND SUM OF SQUARE OF DIGITS OF NUMBER */DIGIT\$SQUARE: PROCEDURE (N) BYTE;    DECLARE (N, T, D) BYTE;    T = 0;    DO WHILE N > 0;        D = N MOD 10;        T = T + D * D;        N = N / 10;    END;    RETURN T;END DIGIT\$SQUARE; /* CHECK IF NUMBER IS HAPPY */HAPPY: PROCEDURE (N) BYTE;    DECLARE (N, I) BYTE;    DECLARE FLAG (256) BYTE;     DO I=0 TO 255;        FLAG(I) = 0;    END;     DO WHILE NOT FLAG(N);        FLAG(N) = 1;        N = DIGIT\$SQUARE(N);    END;     RETURN N = 1;END HAPPY; /* CP/M BDOS CALL */BDOS: PROCEDURE (FN, ARG);    DECLARE FN BYTE, ARG ADDRESS;    GO TO 5;END BDOS; /* PRINT STRING */PRINT: PROCEDURE (STR);    DECLARE STR ADDRESS;    CALL BDOS(9, STR);END PRINT; /* PRINT NUMBER */PRINT\$NUMBER: PROCEDURE (N);    DECLARE S (6) BYTE INITIAL ('...',13,10,'\$');    DECLARE P ADDRESS;    DECLARE (N, C BASED P) BYTE;    P = .S(3);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; /* FIND FIRST 8 HAPPY NUMBERS */DECLARE SEEN BYTE INITIAL (0);DECLARE N BYTE INITIAL (1); DO WHILE SEEN < 8;    IF HAPPY(N) THEN DO;        CALL PRINT\$NUMBER(N);        SEEN = SEEN + 1;    END;    N = N + 1;END; CALL BDOS(0,0);EOF`
Output:
```1
7
10
13
19
23
28
31```

## Potion

`sqr = (n): n * n. isHappy = (n) :   loop :      if (n == 1): return true.      if (n == 4): return false.      sum = 0      n = n string      n length times (i): sum = sum + sqr(n(i) number integer).      n = sum   .. firstEight = ()i = 0while (firstEight length < 8) :   i++   if (isHappy(i)): firstEight append(i)..firstEight string print`

## PowerShell

`function happy([int] \$n) {    \$a[email protected]()    for(\$i=2;\$a.count -lt \$n;\$i++) {        \$sum=\$i        \$hist[email protected]{}        while( \$hist[\$sum] -eq \$null ) {            if(\$sum -eq 1) {                \$a+=\$i            }            \$hist[\$sum]=\$sum            \$sum2=0            foreach(\$j in \$sum.ToString().ToCharArray()) {                \$k=([int]\$j)-0x30                \$sum2+=\$k*\$k            }            \$sum=\$sum2        }     }    \$a -join ','}`

Output :

`happy(8)7,10,13,19,23,28,31,32`

## Prolog

Works with: SWI-Prolog
`happy_numbers(L, Nb) :-    % creation of the list    length(L, Nb),    % Process of this list    get_happy_number(L, 1).  % the game is overget_happy_number([], _). % querying the newt happy_numberget_happy_number([H | T], N) :-     N1 is N+1,    (is_happy_number(N) ->         H = N,         get_happy_number(T, N1);         get_happy_number([H | T], N1)). % we must memorized the numbers reached is_happy_number(N) :-    is_happy_number(N, [N]). % a number is happy when we get 1is_happy_number(N, _L) :-    get_next_number(N, 1), !. % or when this number is not already reached !is_happy_number(N, L) :-    get_next_number(N, NN),    \+member(NN, L),    is_happy_number(NN, [NN | L]). % Process of the next number from Nget_next_number(N, NewN) :-    get_list_digits(N, LD),    maplist(square, LD, L),    sumlist(L, NewN). get_list_digits(N, LD) :-	number_chars(N, LCD),	maplist(number_chars_, LD, LCD). number_chars_(D, CD) :-	number_chars(D, [CD]). square(N, SN) :-	SN is N * N.`

Output :

` ?- happy_numbers(L, 8).L = [1,7,10,13,19,23,28,31].`

## PureBasic

`#ToFind=8#MaxTests=100#True = 1: #False = 0Declare is_happy(n) If OpenConsole()  Define i=1,Happy  Repeat    If is_happy(i)      Happy+1      PrintN("#"+Str(Happy)+RSet(Str(i),3))    EndIf    i+1  Until Happy>=#ToFind  ;  Print(#CRLF\$+#CRLF\$+"Press ENTER to exit"): Input()  CloseConsole()EndIf Procedure is_happy(n)  Protected i,j=n,dig,sum  Repeat    sum=0    While j      dig=j%10      j/10      sum+dig*dig    Wend    If sum=1: ProcedureReturn #True: EndIf      j=sum    i+1  Until i>#MaxTests  ProcedureReturn #FalseEndProcedure`

Sample output:

```#1  1
#2  7
#3 10
#4 13
#5 19
#6 23
#7 28
#8 31```

## Python

### Procedural

`>>> def happy(n):    past = set()			    while n != 1:        n = sum(int(i)**2 for i in str(n))        if n in past:            return False        past.add(n)    return True >>> [x for x in xrange(500) if happy(x)][:8][1, 7, 10, 13, 19, 23, 28, 31]`

### Composition of pure functions

Drawing 8 terms from a non finite stream, rather than assuming prior knowledge of the finite sample size required:

`'''Happy numbers''' from itertools import islice  # main :: IO ()def main():    '''Test'''    print(        take(8)(            happyNumbers()        )    )  # happyNumbers :: Gen [Int]def happyNumbers():    '''Generator :: non-finite stream of happy numbers.'''    x = 1    while True:        x = until(isHappy)(succ)(x)        yield x        x = succ(x)  # isHappy :: Int -> Booldef isHappy(n):    '''Happy number sequence starting at n reaches 1 ?'''    seen = set()     # p :: Int -> Bool    def p(x):        if 1 == x or x in seen:            return True        else:            seen.add(x)            return False     # f :: Int -> Int    def f(x):        return sum(int(d)**2 for d in str(x))     return 1 == until(p)(f)(n)  # GENERIC ------------------------------------------------- # succ :: Int -> Intdef succ(x):    '''The successor of an integer.'''    return 1 + x  # take :: Int -> [a] -> [a]# take :: Int -> String -> Stringdef take(n):    '''The prefix of xs of length n,       or xs itself if n > length xs.'''    return lambda xs: (        xs[0:n]        if isinstance(xs, list)        else list(islice(xs, n))    )  # until :: (a -> Bool) -> (a -> a) -> a -> adef until(p):    '''The result of repeatedly applying f until p holds.       The initial seed value is x.'''    def go(f, x):        v = x        while not p(v):            v = f(v)        return v    return lambda f: lambda x: go(f, x)  if __name__ == '__main__':    main()`
Output:
`[1, 7, 10, 13, 19, 23, 28, 31]`

## Quackery

`   [ 0 swap    [ 10 /mod 2 **      rot + swap      dup 0 = until ]    drop ]                  is digitsquare ( n --> n )   [ [ digitsquare      dup  1 != while      dup 42 != while      again ]    1 = ]                   is happy       ( n --> b )   [ [] 1                   [ dip       [ 2dup size > ]      swap while      dup happy if        [ tuck join swap ]      1+ again ]     drop nip ]             is happies      ( n --> [ )   8 happies echo`
Output:
`[ 1 7 10 13 19 23 28 31 ]`

## R

`is.happy <- function(n){   stopifnot(is.numeric(n) && length(n)==1)   getdigits <- function(n)   {      as.integer(unlist(strsplit(as.character(n), "")))   }   digits <- getdigits(n)   previous <- c()   repeat   {      sumsq <- sum(digits^2, na.rm=TRUE)      if(sumsq==1L)      {         happy <- TRUE         break            } else if(sumsq %in% previous)      {         happy <- FALSE         attr(happy, "cycle") <- previous         break      } else      {         previous <- c(previous, sumsq)         digits <- getdigits(sumsq)      }   }   happy}`

Example usage

`is.happy(2)`
```[1] FALSE
attr(,"cycle")
[1]   4  16  37  58  89 145  42  20
```
`#Find happy numbers between 1 and 50which(apply(rbind(1:50), 2, is.happy))`
```1  7 10 13 19 23 28 31 32 44 49
```
`#Find the first 8 happy numbershappies <- c()i <- 1Lwhile(length(happies) < 8L){   if(is.happy(i)) happies <- c(happies, i)   i <- i + 1L}happies`
```1  7 10 13 19 23 28 31
```

## Racket

`#lang racket(define (sum-of-squared-digits number (result 0))  (if (zero? number)      result      (sum-of-squared-digits (quotient number 10)                             (+ result (expt (remainder number 10) 2))))) (define (happy-number? number (seen null))  (define next (sum-of-squared-digits number))  (cond ((= 1 next)         #t)        ((memq next seen)         #f)        (else         (happy-number? next (cons number seen))))) (define (get-happys max)  (for/list ((x (in-range max))             #:when (happy-number? x))    x)) (display (take (get-happys 100) 8)) ;displays (1 7 10 13 19 23 28 31)`

## Raku

(formerly Perl 6)

Works with: rakudo version 2015-09-13
`sub happy (Int \$n is copy --> Bool) {  loop {      state %seen;      \$n = [+] \$n.comb.map: { \$_ ** 2 }      return True  if \$n == 1;      return False if %seen{\$n}++;  }} say join ' ', grep(&happy, 1 .. *)[^8];`
Output:
`1 7 10 13 19 23 28 31`

Here's another approach that uses a different set of tricks including lazy lists, gather/take, repeat-until, and the cross metaoperator X.

`my @happy = lazy gather for 1..* -> \$number {    my %stopper = 1 => 1;    my \$n = \$number;    repeat until %stopper{\$n}++ {        \$n = [+] \$n.comb X** 2;    }    take \$number if \$n == 1;} say ~@happy[^8];`

Output is the same as above.

Here is a version using a subset and an anonymous recursion (we cheat a little bit by using the knowledge that 7 is the second happy number):

`subset Happy of Int where sub (\$n) {    \$n == 1 ?? True  !!    \$n < 7  ?? False !!    &?ROUTINE([+] \$n.comb »**» 2);} say (grep Happy, 1 .. *)[^8];`

Again, output is the same as above. It is not clear whether this version returns in finite time for any integer, though.

There's more than one way to do it...

## Relation

` function happy(x)set y = xset lasty = 0set found = " "while y !=  1 and not (found regex "\s".y."\s") set found = found . y . " "set m = 0while y  > 0set digit = y mod 10set m = m + digit * digitset y = (y - digit) / 10end whileset y = format(m,"%1d")end whileset found = found . y . " "if y = 1 set result = 1elseset result = 0end ifend function set c = 0set i = 1while c < 8 and i < 100if happy(i)echo iset c = c + 1end ifset i = i + 1end while `
```1
7
10
13
19
23
28
31
```

## REXX

### unoptimized

`/*REXX program  computes  and  displays  a  specified  amount  of   happy   numbers.    */parse arg limit .                                /*obtain optional argument from the CL.*/if limit=='' | limit==","  then limit=8          /*Not specified?  Then use the default.*/haps=0                                           /*count of the happy numbers  (so far).*/   do n=1  while haps<limit;    @.=0;  q=n        /*search the integers starting at unity*/         do  until q==1                          /*determine if   Q   is a happy number.*/         s=0                                     /*prepare to add squares of digits.    */                 do j=1  for length(q)           /*sum the squares of the decimal digits*/                 s=s + substr(q, j, 1) **2       /*add the square  of  a  decimal digit.*/                 end   /*j*/          if @.s  then iterate n                  /*if already summed,   Q   is unhappy. */         @.s=1;  q=s                             /*mark the sum as found;   try  Q  sum.*/         end   /*until*/  say n                                          /*display the number    (N  is happy). */  haps=haps+1                                    /*bump the  count  of  happy numbers.  */  end          /*n*/                                                 /*stick a fork in it,  we're all done. */`
output   when using the input of:     8
```1
7
10
13
19
23
28
31
```

### optimized, vertical list

This REXX code uses additional memorization (by keeping track of happy and unhappy numbers),
it's about   2 1/2   times faster than the unoptimized version.

This REXX version also accepts a   range   of happy numbers to be shown,   that is,
it can show the 2000th through the 2032nd (inclusive) happy numbers   (as shown below).

`/*REXX program   computes  and  displays   a specified  range  of   happy   numbers.    */parse arg L H .                                  /*obtain optional arguments from the CL*/if L=='' | L==","  then L=8                      /*Not specified?  Then use the default.*/if H=='' | H==","  then do;  H=L; L=1;  end      /*use a range for the displaying of #s.*/            do i=0  to 9;  #.i=i**2;  end /*i*/  /*build a squared decimal digit table. */@.=0;   @.1=1;       !.[email protected].;    !.2=1;    !.4=1   /*sparse array:   @≡happy,  !≡unhappy. */haps=0                                           /*count of the happy numbers  (so far).*/     do n=1  while  haps<H                        /*search integers starting at unity.   */    if !.n  then iterate                         /*if  N  is unhappy, then try another. */    q=n                                          /* [↓]    Q  is the number being tested*/          do  until q==1;    s=0                 /*see if  Q  is a  happy number.       */          ?=q                                    /* [↓]    ?  is destructively parsed.  */               do length(q)                      /*parse all the    decimal digits of ? */               parse var  ?  _  +1 ?             /*obtain a  single decimal digit  of ? */               s=s + #._                         /*add the square of that decimal digit.*/               end   /*length(q)*/               /* [↑]  perform the    DO    W  times. */          if !.s  then do; !.n=1; iterate n; end /*is  S  unhappy?    Then  Q  is also. */          if @.s  then leave                     /*Have we found a  happy  number?      */          q=s                                    /*try the  Q  sum to see if it's happy.*/          end   /*until*/    @.n=1                                        /*mark      N      as a   happy number.*/    haps=haps+1                                  /*bump the counter of the happy numbers*/    if haps<L  then iterate                      /*don't display  if    N    is too low.*/    say  right(n, 30)                            /*display right justified happy number.*/    end        /*n*/                                                 /*stick a fork in it,  we're all done. */`
output   when using the input of:     2000   2032
```                         13141
13142
13148
13158
13177
13182
13184
13185
13188
13203
13212
13214
13218
13221
13228
13230
13233
13241
13247
13248
13258
13266
13274
13281
13282
13284
13285
13299
13300
13302
13303
13305
13307
```

### optimized, horizontal list

This REXX version is identical to the optimized version,   but displays the numbers in a horizontal list.

`/*REXX program   computes  and  displays   a specified  range  of   happy   numbers.    */sw=linesize() - 1                                /*obtain the screen width  (less one). */parse arg limit .                                /*obtain optional argument from the CL.*/if L=='' | L==","  then L=8                      /*Not specified?  Then use the default.*/if H=='' | H==","  then do;  H=L; L=1;  end      /*use a range for the displaying of #s.*/            do i=0  to 9;  #.i=i**2;  end /*i*/  /*build a squared decimal digit table. */@.=0;   @.1=1;        !.[email protected].;    !.2=1;    !.4=1  /*sparse array:   @≡happy,  !≡unhappy. */haps=0                                           /*count of the happy numbers  (so far).*/\$=    do n=1  while  haps<H                        /*search integers starting at  unity.  */    if !.n  then iterate                         /*if  N  is unhappy, then try another. */    q=n                                          /*(below)  Q  is the number tested.    */          do  until q==1;          s=0           /*see if   Q  is a happy number.       */          ?=q                                    /* [↓]    ?  is destructively PARSEd.  */               do length(q)                      /*parse all the    decimal digits of ? */               parse var  ?  _  +1 ?             /*obtain a  single decimal digit  of ? */               s=s + #._                         /*add the square of that decimal digit.*/               end   /*length(q)*/               /* [↑]  perform the   DO    W   times. */           if !.s  then do; !.n=1; iterate n; end /*is  S  unhappy?    Then  Q  is also. */          if @.s  then leave                     /*Have we found a  happy  number?      */           q=s                                   /*try the  Q  sum to see if it's happy.*/           end   /*until*/    @.n=1                                        /*mark     N     as a   happy number.  */    haps=haps+1                                  /*bump the count of the happy numbers. */    if haps<L  then iterate                      /*don't display it,   N   is too low.  */    \$=\$ n                                        /*add   N   to the horizontal list.    */    if length(\$ n)>sw  then do                   /*if the list is too long, then split  */                            say strip(\$)         /*     ··· and display what we've got. */                            \$=n                  /*Set the next line to overflow.       */                            end                  /* [↑]  new line now contains overflow.*/    end     /*n*/if \$\=''  then say  strip(\$)                     /*display any residual happy numbers.  */                                                 /*stick a fork in it,  we're all done. */`

This REXX program makes use of   linesize   REXX program (or BIF) which is used to determine the screen width (or linesize) of the terminal (console).

Some REXXes don't have this BIF, so the   linesize.rex   REXX program is included here   ──►   LINESIZE.REX.

output   when using the input of:     1   15002

(The   linesize   for the terminal being used for this example was   200.)

(Shown at two-thirds size.)

```1 7 10 13 19 23 28 31 32 44 49 68 70 79 82 86 91 94 97 100 103 109 129 130 133 139 167 176 188 190 192 193 203 208 219 226 230 236 239 262 263 280 291 293 301 302 310 313 319 320 326 329 331 338 356
362 365 367 368 376 379 383 386 391 392 397 404 409 440 446 464 469 478 487 490 496 536 556 563 565 566 608 617 622 623 632 635 637 638 644 649 653 655 656 665 671 673 680 683 694 700 709 716 736 739
748 761 763 784 790 793 802 806 818 820 833 836 847 860 863 874 881 888 899 901 904 907 910 912 913 921 923 931 932 937 940 946 964 970 973 989 998 1000 1003 1009 1029 1030 1033 1039 1067 1076 1088
1090 1092 1093 1112 1114 1115 1121 1122 1125 1128 1141 1148 1151 1152 1158 1177 1182 1184 1185 1188 1209 1211 1212 1215 1218 1221 1222 1233 1247 1251 1257 1258 1274 1275 1277 1281 1285 1288 1290 1299
1300 1303 1309 1323 1330 1332 1333 1335 1337 1339 1353 1366 1373 1390 1393 1411 1418 1427 1444 1447 1448 1457 1472 1474 1475 1478 1481 1484 1487 1511 1512 1518 1521 1527 1528 1533 1547 1557 1572 1574
1575 1578 1581 1582 1587 1599 1607 1636 1663 1666 1670 1679 1697 1706 1717 1724 1725 1727 1733 1742 1744 1745 1748 1752 1754 1755 1758 1760 1769 1771 1772 1784 1785 1796 1808 1812 1814 1815 1818 1821
1825 1828 1841 1844 1847 1851 1852 1857 1874 1875 1880 1881 1882 1888 1900 1902 1903 1920 1929 1930 1933 1959 1967 1976 1992 1995 2003 2008 2019 2026 2030 2036 2039 2062 2063 2080 2091 2093 2109 2111
2112 2115 2118 2121 2122 2133 2147 2151 2157 2158 2174 2175 2177 2181 2185 2188 2190 2199 2206 2211 2212 2221 2224 2242 2245 2254 2257 2258 2260 2275 2285 2300 2306 2309 2313 2331 2333 2338 2339 2360
2369 2383 2390 2393 2396 2417 2422 2425 2448 2452 2455 2457 2458 2471 2475 2478 2484 2485 2487 2511 2517 2518 2524 2527 2528 2542 2545 2547 2548 2554 2555 2557 2568 2571 2572 2574 2575 2581 2582 2584
2586 2602 2603 2620 2630 2639 2658 2685 2693 2714 2715 2717 2725 2741 2745 2748 2751 2752 2754 2755 2771 2784 2800 2811 2815 2818 2825 2833 2844 2845 2847 2851 2852 2854 2856 2865 2874 2881 2899 2901
2903 2910 2919 2930 2933 2936 2963 2989 2991 2998 3001 3002 3010 3013 3019 3020 3026 3029 3031 3038 3056 3062 3065 3067 3068 3076 3079 3083 3086 3091 3092 3097 3100 3103 3109 3123 3130 3132 3133 3135
3137 3139 3153 3166 3173 3190 3193 3200 3206 3209 3213 3231 3233 3238 3239 3260 3269 3283 3290 3293 3296 3301 3308 3310 3312 3313 3315 3317 3319 3321 3323 3328 3329 3331 3332 3338 3346 3351 3355 3356
3364 3365 3367 3371 3376 3380 3382 3383 3391 3392 3436 3456 3463 3465 3466 3506 3513 3531 3535 3536 3546 3553 3560 3563 3564 3602 3605 3607 3608 3616 3620 3629 3634 3635 3637 3643 3645 3646 3650 3653
3654 3661 3664 3667 3670 3673 3676 3680 3689 3692 3698 3706 3709 3713 3731 3736 3760 3763 3766 3779 3789 3790 3797 3798 3803 3806 3823 3830 3832 3833 3860 3869 3879 3896 3897 3901 3902 3907 3910 3913
3920 3923 3926 3931 3932 3962 3968 3970 3977 3978 3986 3987 4004 4009 4040 4046 4064 4069 4078 4087 4090 4096 4111 4118 4127 4144 4147 4148 4157 4172 4174 4175 4178 4181 4184 4187 4217 4222 4225 4248
4252 4255 4257 4258 4271 4275 4278 4284 4285 4287 4336 4356 4363 4365 4366 4400 4406 4414 4417 4418 4428 4441 4447 4449 4455 4460 4471 4474 4477 4481 4482 4494 4517 4522 4525 4527 4528 4536 4545 4552
4554 4555 4558 4563 4571 4572 4577 4582 4585 4599 4604 4609 4633 4635 4636 4640 4653 4663 4690 4708 4712 4714 4715 4718 4721 4725 4728 4741 4744 4747 4751 4752 4757 4774 4775 4780 4781 4782 4788 4807
4811 4814 4817 4824 4825 4827 4841 4842 4852 4855 4870 4871 4872 4878 4887 4888 4900 4906 4944 4959 4960 4995 5036 5056 5063 5065 5066 5111 5112 5118 5121 5127 5128 5133 5147 5157 5172 5174 5175 5178
5181 5182 5187 5199 5211 5217 5218 5224 5227 5228 5242 5245 5247 5248 5254 5255 5257 5268 5271 5272 5274 5275 5281 5282 5284 5286 5306 5313 5331 5335 5336 5346 5353 5360 5363 5364 5417 5422 5425 5427
5428 5436 5445 5452 5454 5455 5458 5463 5471 5472 5477 5482 5485 5499 5506 5517 5524 5525 5527 5533 5542 5544 5545 5548 5552 5554 5555 5558 5560 5569 5571 5572 5584 5585 5596 5603 5605 5606 5628 5630
5633 5634 5643 5650 5659 5660 5666 5682 5695 5712 5714 5715 5718 5721 5722 5724 5725 5741 5742 5747 5751 5752 5774 5781 5789 5798 5799 5811 5812 5817 5821 5822 5824 5826 5842 5845 5854 5855 5862 5871
5879 5897 5919 5949 5956 5965 5978 5979 5987 5991 5994 5997 6008 6017 6022 6023 6032 6035 6037 6038 6044 6049 6053 6055 6056 6065 6071 6073 6080 6083 6094 6107 6136 6163 6166 6170 6179 6197 6202 6203
6220 6230 6239 6258 6285 6293 6302 6305 6307 6308 6316 6320 6329 6334 6335 6337 6343 6345 6346 6350 6353 6354 6361 6364 6367 6370 6373 6376 6380 6389 6392 6398 6404 6409 6433 6435 6436 6440 6453 6463
6490 6503 6505 6506 6528 6530 6533 6534 6543 6550 6559 6560 6566 6582 6595 6605 6613 6616 6631 6634 6637 6643 6650 6656 6661 6665 6673 6701 6703 6710 6719 6730 6733 6736 6763 6789 6791 6798 6800 6803
6825 6830 6839 6852 6879 6893 6897 6899 6904 6917 6923 6932 6938 6940 6955 6971 6978 6983 6987 6989 6998 7000 7009 7016 7036 7039 7048 7061 7063 7084 7090 7093 7106 7117 7124 7125 7127 7133 7142 7144
7145 7148 7152 7154 7155 7158 7160 7169 7171 7172 7184 7185 7196 7214 7215 7217 7225 7241 7245 7248 7251 7252 7254 7255 7271 7284 7306 7309 7313 7331 7336 7360 7363 7366 7379 7389 7390 7397 7398 7408
7412 7414 7415 7418 7421 7425 7428 7441 7444 7447 7451 7452 7457 7474 7475 7480 7481 7482 7488 7512 7514 7515 7518 7521 7522 7524 7525 7541 7542 7547 7551 7552 7574 7581 7589 7598 7599 7601 7603 7610
7619 7630 7633 7636 7663 7689 7691 7698 7711 7712 7721 7739 7744 7745 7754 7788 7793 7804 7814 7815 7824 7839 7840 7841 7842 7848 7851 7859 7869 7878 7884 7887 7893 7895 7896 7900 7903 7916 7930 7937
7938 7958 7959 7961 7968 7973 7983 7985 7986 7995 8002 8006 8018 8020 8033 8036 8047 8060 8063 8074 8081 8088 8099 8108 8112 8114 8115 8118 8121 8125 8128 8141 8144 8147 8151 8152 8157 8174 8175 8180
8181 8182 8188 8200 8211 8215 8218 8225 8233 8244 8245 8247 8251 8252 8254 8256 8265 8274 8281 8299 8303 8306 8323 8330 8332 8333 8360 8369 8379 8396 8397 8407 8411 8414 8417 8424 8425 8427 8441 8442
8452 8455 8470 8471 8472 8478 8487 8488 8511 8512 8517 8521 8522 8524 8526 8542 8545 8554 8555 8562 8571 8579 8597 8600 8603 8625 8630 8639 8652 8679 8693 8697 8699 8704 8714 8715 8724 8739 8740 8741
8742 8748 8751 8759 8769 8778 8784 8787 8793 8795 8796 8801 8808 8810 8811 8812 8818 8821 8847 8848 8874 8877 8880 8881 8884 8909 8929 8936 8937 8957 8963 8967 8969 8973 8975 8976 8990 8992 8996 9001
9004 9007 9010 9012 9013 9021 9023 9031 9032 9037 9040 9046 9064 9070 9073 9089 9098 9100 9102 9103 9120 9129 9130 9133 9159 9167 9176 9192 9195 9201 9203 9210 9219 9230 9233 9236 9263 9289 9291 9298
9301 9302 9307 9310 9313 9320 9323 9326 9331 9332 9362 9368 9370 9377 9378 9386 9387 9400 9406 9444 9459 9460 9495 9519 9549 9556 9565 9578 9579 9587 9591 9594 9597 9604 9617 9623 9632 9638 9640 9655
9671 9678 9683 9687 9689 9698 9700 9703 9716 9730 9737 9738 9758 9759 9761 9768 9773 9783 9785 9786 9795 9809 9829 9836 9837 9857 9863 9867 9869 9873 9875 9876 9890 9892 9896 9908 9912 9915 9921 9928
9945 9951 9954 9957 9968 9975 9980 9982 9986 10000 10003 10009 10029 10030 10033 10039 10067 10076 10088 10090 10092 10093 10112 10114 10115 10121 10122 10125 10128 10141 10148 10151 10152 10158
10177 10182 10184 10185 10188 10209 10211 10212 10215 10218 10221 10222 10233 10247 10251 10257 10258 10274 10275 10277 10281 10285 10288 10290 10299 10300 10303 10309 10323 10330 10332 10333 10335
10337
```

## Ring

`n = 1found = 0 While found < 8    If IsHappy(n)           found += 1          see string(found) + " : " + string(n) + nl    ok    n += 1End Func IsHappy n     cache = []    While n != 1        Add(cache,n)        t = 0        strn = string(n)        for e in strn            t += pow(number(e),2)        next        n = t        If find(cache,n) Return False ok    End    Return True `
Output:
```1 : 1
2 : 7
3 : 10
4 : 13
5 : 19
6 : 23
7 : 28
8 : 31
```

## Ruby

Works with: Ruby version 2.1
`require 'set' # Set: Fast array lookup / Simple existence hash @seen_numbers = Set.new@happy_numbers = Set.new def happy?(n)  return true if n == 1 # Base case  return @happy_numbers.include?(n) if @seen_numbers.include?(n) # Use performance cache, and stop unhappy cycles   @seen_numbers << n  digit_squared_sum = n.to_s.each_char.inject(0) { |sum, c| sum + c.to_i**2 } # In Rails: n.to_s.each_char.sum { c.to_i**2 }   if happy?(digit_squared_sum)    @happy_numbers << n    true # Return true  else    false # Return false  endend`

Helper method to produce output:

`def print_happy  happy_numbers = []   1.step do |i|    break if happy_numbers.length >= 8    happy_numbers << i if happy?(i)  end   p happy_numbersend print_happy`
Output:
`[1, 7, 10, 13, 19, 23, 28, 31]`

### Alternative version

`@memo = [0,1]def happy(n)  sum = n.to_s.chars.map{|c| c.to_i**2}.inject(:+)  return @memo[sum] if @memo[sum]==0 or @memo[sum]==1  @memo[sum] = 0                        # for the cycle check  @memo[sum] = happy(sum)               # return 1:Happy number, 0:otherend i = count = 0while count < 8  i += 1  puts i or count+=1 if happy(i)==1end putsfor i in 99999999999900..99999999999999  puts i if happy(i)==1end`
Output:
```1
7
10
13
19
23
28
31

99999999999901
99999999999910
99999999999914
99999999999915
99999999999916
99999999999937
99999999999941
99999999999951
99999999999956
99999999999961
99999999999965
99999999999973
```

### Simpler Alternative

Translation of: Python
`def happy?(n)  past = []			  until n == 1    n = n.digits.sum { |d| d * d }    return false if past.include? n    past << n  end  trueend i = count = 0until count == 8; puts i or count += 1 if happy?(i += 1) endputs(99999999999900..99999999999999).each { |i| puts i if happy?(i) }`
Output:
```1
7
10
13
19
23
28
31

99999999999901
99999999999910
99999999999914
99999999999915
99999999999916
99999999999937
99999999999941
99999999999951
99999999999956
99999999999961
99999999999965
99999999999973```

## Run BASIC

`for i = 1 to 100 if happy(i) = 1 then   cnt = cnt + 1  PRINT cnt;". ";i;" is a happy number "  if cnt = 8 then end end ifnext i FUNCTION happy(num)while count < 50 and happy <> 1  num\$	= str\$(num)  count	= count + 1  happy	= 0  for i = 1 to len(num\$)    happy = happy + val(mid\$(num\$,i,1)) ^ 2  next i  num = happywendend function`
```1. 1 is a happy number
2. 7 is a happy number
3. 10 is a happy number
4. 13 is a happy number
5. 19 is a happy number
6. 23 is a happy number
7. 28 is a happy number
8. 31 is a happy number
```

## Rust

In Rust, using a tortoise/hare cycle detection algorithm (generic for integer types)

`#![feature(core)] fn sumsqd(mut n: i32) -> i32 {    let mut sq = 0;    while n > 0 {        let d = n % 10;        sq += d*d;        n /= 10    }    sq} use std::num::Int;fn cycle<T: Int>(a: T, f: fn(T) -> T) -> T {    let mut t = a;    let mut h = f(a);     while t != h {        t = f(t);        h = f(f(h))    }    t} fn ishappy(n: i32) -> bool {    cycle(n, sumsqd) == 1} fn main() {    let happy = std::iter::count(1, 1)                    .filter(|&n| ishappy(n))                    .take(8)                    .collect::<Vec<i32>>();     println!("{:?}", happy)}`
Output:
```[1, 7, 10, 13, 19, 23, 28, 31]
```

## Salmon

`variable happy_count := 0;outer:iterate(x; [1...+oo])  {    variable seen := <<(* --> false)>>;    variable now := x;    while (true)      {        if (seen[now])          {            if (now == 1)              {                ++happy_count;                print(x, " is happy.\n");                if (happy_count == 8)                    break from outer;;              };            break;          };        seen[now] := true;        variable new := 0;        while (now != 0)          {            new += (now % 10) * (now % 10);            now /::= 10;          };        now := new;      };  };`

This Salmon program produces the following output:

```1 is happy.
7 is happy.
10 is happy.
13 is happy.
19 is happy.
23 is happy.
28 is happy.
31 is happy.```

## Scala

`scala> def isHappy(n: Int) = {     |   new Iterator[Int] {     |   val seen = scala.collection.mutable.Set[Int]()     |   var curr = n     |   def next = {     |     val res = curr     |     curr = res.toString.map(_.asDigit).map(n => n * n).sum     |     seen += res     |     res     |   }     |   def hasNext = !seen.contains(curr)     | }.toList.last == 1     | }isHappy: (n: Int)Boolean scala> Iterator from 1 filter isHappy take 8 foreach println17101319232831 `

## Scheme

`(define (number->list num)  (do ((num num (quotient num 10))       (lst '() (cons (remainder num 10) lst)))    ((zero? num) lst))) (define (happy? num)  (let loop ((num num) (seen '()))    (cond ((= num 1) #t)          ((memv num seen) #f)          (else (loop (apply + (map (lambda (x) (* x x)) (number->list num)))                      (cons num seen)))))) (display "happy numbers:")(let loop ((n 1) (more 8))  (cond ((= more 0) (newline))        ((happy? n) (display " ") (display n) (loop (+ n 1) (- more 1)))        (else (loop (+ n 1) more))))`

The output is:

`happy numbers: 1 7 10 13 19 23 28 31`

## Scratch

Scratch is a free visual programming language. Click the link, then "See inside" to view the code.

This code will allow you to check if a positive interger (<=9999) is a happy number. It will also output a list of the first 8 happy numbers. (1 7 10 13 19 23 28 31)

## Seed7

`\$ include "seed7_05.s7i"; const type: cacheType is hash [integer] boolean;var cacheType: cache is cacheType.value; const func boolean: happy (in var integer: number) is func  result    var boolean: isHappy is FALSE;  local    var bitset: cycle is bitset.value;    var integer: newnumber is 0;    var integer: cycleNum is 0;  begin    while number > 1 and number not in cycle do      if number in cache then        number := ord(cache[number]);      else        incl(cycle, number);        newnumber := 0;        while number > 0 do          newnumber +:= (number rem 10) ** 2;          number := number div 10;        end while;        number := newnumber;      end if;    end while;    isHappy := number = 1;    for cycleNum range cycle do      cache @:= [cycleNum] isHappy;    end for;  end func; const proc: main is func  local    var integer: number is 0;  begin    for number range 1 to 50 do      if happy(number) then        writeln(number);      end if;    end for;  end func;`

Output:

```1
7
10
13
19
23
28
31
32
44
49
```

## SequenceL

`import <Utilities/Math.sl>;import <Utilities/Conversion.sl>; main(argv(2)) := findHappys(stringToInt(head(argv))); findHappys(count) := findHappysHelper(count, 1, []); findHappysHelper(count, n, happys(1)) :=        happys when size(happys) = count    else        findHappysHelper(count, n + 1, happys ++ [n]) when isHappy(n)    else        findHappysHelper(count, n + 1, happys); isHappy(n) := isHappyHelper(n, []); isHappyHelper(n, cache(1)) :=    let        digits[i] := (n / integerPower(10, i - 1)) mod 10                     foreach i within 1 ... ceiling(log(10, n + 1));        newN := sum(integerPower(digits, 2));    in         false when some(n = cache)    else        true when n = 1    else        isHappyHelper(newN, cache ++ [n]);`
Output:
```\$>happy.exe 8
[1,7,10,13,19,23,28,31]
```

## SETL

`proc is_happy(n);  s := [n];  while n > 1 loop    if (n := +/[val(i)**2: i in str(n)]) in s then        return false;    end if;    s with:= n;  end while;  return true;end proc;`
`happy := [];n := 1;until #happy = 8 loop  if is_happy(n) then happy with:= n; end if;  n +:= 1;end loop; print(happy);`

Output:

`[1 7 10 13 19 23 28 31]`

Alternative version:

`print([n : n in [1..100] | is_happy(n)](1..8));`

Output:

`[1 7 10 13 19 23 28 31]`

## Sidef

`func happy(n) is cached {    static seen = Hash()     return true  if n.is_one    return false if seen.exists(n)     seen{n} = 1    happy(n.digits.sum { _*_ })} say happy.first(8)`
Output:
```[1, 7, 10, 13, 19, 23, 28, 31]
```

## Smalltalk

Works with: GNU Smalltalk
Translation of: Python

In addition to the "Python's cache mechanism", the use of a Bag assures that found e.g. the happy 190, we already have in cache also the happy 910 and 109, and so on.

`Object subclass: HappyNumber [  |cache negativeCache|  HappyNumber class >> new [ |me|    me := super new.    ^ me init  ]  init [ cache := Set new. negativeCache := Set new. ]   hasSad: aNum [    ^ (negativeCache includes: (self recycle: aNum))  ]  hasHappy: aNum [    ^ (cache includes: (self recycle: aNum))  ]  addHappy: aNum [    cache add: (self recycle: aNum)  ]  addSad: aNum [    negativeCache add: (self recycle: aNum)  ]   recycle: aNum [ |r n| r := Bag new.    n := aNum.    [ n > 0 ]    whileTrue: [ |d|      d := n rem: 10.      r add: d.      n := n // 10.    ].    ^r  ]   isHappy: aNumber [ |cycle number newnumber|    number := aNumber.    cycle := Set new.    [ (number ~= 1) & ( (cycle includes: number) not ) ]    whileTrue: [      (self hasHappy: number)      ifTrue: [ ^true ]      ifFalse: [        (self hasSad: number) ifTrue: [ ^false ].        cycle add: number.        newnumber := 0.        [ number > 0 ]        whileTrue: [ |digit|          digit := number rem: 10.          newnumber := newnumber + (digit * digit). 	  number := (number - digit) // 10.        ].        number := newnumber.      ]    ].    (number = 1)    ifTrue: [      cycle do: [ :e | self addHappy: e ].      ^true    ]    ifFalse: [       cycle do: [ :e | self addSad: e ].      ^false     ]  ]].`
`|happy|happy := HappyNumber new. 1 to: 31 do: [ :i |  (happy isHappy: i)  ifTrue: [ i displayNl ]].`

Output:

```1
7
10
13
19
23
28
31
```

an alternative version is:

Works with: Smalltalk/X
`|next isHappy happyNumbers| next :=     [:n |         (n printString collect:[:ch | ch digitValue squared] as:Array) sum    ]. isHappy :=     [:n |  | t already |         already := Set new.        t := n.        [ t == 1 or:[ (already includes:t)]] whileFalse:[            already add:t.            t := next value:t.        ].        t == 1    ]. happyNumbers := OrderedCollection new.try := 1.[happyNumbers size < 8] whileTrue:[      (isHappy value:try) ifTrue:[ happyNumbers add:try].      try := try + 1].happyNumbers printCR`

Output: OrderedCollection(1 7 10 13 19 23 28 31)

## Swift

`func isHappyNumber(var n:Int) -> Bool {    var cycle = [Int]()     while n != 1 && !cycle.contains(n) {        cycle.append(n)        var m = 0        while n > 0 {            let d = n % 10            m += d * d            n = (n  - d) / 10        }        n = m    }    return n == 1} var found = 0var count = 0while found != 8 {    if isHappyNumber(count) {        print(count)        found++    }    count++}`
Output:
```1
7
10
13
19
23
28
31```

## Tcl

using code from Sum of squares#Tcl

`proc is_happy n {    set seen [list]    while {\$n > 1 && [lsearch -exact \$seen \$n] == -1} {        lappend seen \$n        set n [sum_of_squares [split \$n ""]]    }    return [expr {\$n == 1}]} set happy [list]set n -1while {[llength \$happy] < 8} {    if {[is_happy \$n]} {lappend happy \$n}    incr n}puts "the first 8 happy numbers are: [list \$happy]"`
`the first 8 happy numbers are: {1 7 10 13 19 23 28 31}`

## TUSCRIPT

`\$\$ MODE TUSCRIPTSECTION check  IF (n!=1) THEN   n = STRINGS (n,":>/:")    LOOP/CLEAR nr=n     square=nr*nr     n=APPEND (n,square)    ENDLOOP   n=SUM(n)   r_table=QUOTES (n)   BUILD R_TABLE/word/EXACT chk=r_table   IF (seq.ma.chk) THEN    status="next"   ELSE    seq=APPEND (seq,n)   ENDIF   RELEASE r_table chk  ELSE    PRINT checkednr," is a happy number"    happynrs=APPEND (happynrs,checkednr)    status="next"  ENDIFENDSECTION happynrs="" LOOP n=1,100sz_happynrs=SIZE(happynrs)IF (sz_happynrs==8) EXITcheckednr=VALUE(n)status=seq="" LOOP  IF (status=="next") EXIT  DO check ENDLOOPENDLOOP`

Output:

```1 is a happy number
7 is a happy number
10 is a happy number
13 is a happy number
19 is a happy number
23 is a happy number
28 is a happy number
31 is a happy number
```

## uBasic/4tH

` ' ************************' MAIN' ************************ PROC _PRINT_HAPPY(20)END ' ************************' END MAIN' ************************ ' ************************' SUBS & FUNCTIONS' ************************ ' --------------------_is_happy PARAM(1)' --------------------LOCAL (5)  [email protected] = 100  [email protected] = [email protected]  [email protected] = 0   DO WHILE [email protected] < [email protected]    [email protected] = 0     DO WHILE [email protected]      [email protected] = [email protected] % 10      [email protected] = [email protected] / 10      [email protected] = [email protected] + ([email protected] * [email protected])    LOOP   UNTIL [email protected] = 1    [email protected] = [email protected]    [email protected] = [email protected] + 1  LOOP RETURN([email protected] < [email protected]) ' --------------------_PRINT_HAPPY PARAM(1)' --------------------LOCAL (2)  [email protected] = 1  [email protected] = 0   DO     IF FUNC (_is_happy([email protected])) THEN       [email protected] = [email protected] + 1       PRINT [email protected]    ENDIF     [email protected] = [email protected] + 1    UNTIL [email protected] + 1 > [email protected]  LOOP RETURN ' ************************' END SUBS & FUNCTIONS' ************************ `

## UNIX Shell

Works with: Bourne Again SHell
`#!/bin/bashfunction sum_of_square_digits{  local -i n="\$1" sum=0  while (( n )); do    local -i d=n%10    let sum+=d*d    let n=n/10  done  echo "\$sum"} function is_happy?{   local -i n="\$1"   local seen=()   while (( n != 1 )); do     if [ -n "\${seen[\$n]}" ]; then        return 1     fi     seen[n]=1     let n="\$(sum_of_square_digits "\$n")"   done   return 0} function first_n_happy{  local -i count="\$1"  local -i n  for (( n=0; count; n+=1 )); do  if is_happy? "\$n"; then    echo "\$n"    let count-=1  fi  done  return 0} first_n_happy 8`
Output:
```1
7
10
13
19
23
28
31```

## Ursala

The happy function is a predicate testing whether a given number is happy, and first(p) defines a function mapping a number n to the first n positive naturals having property p.

`#import std#import nat happy = ==1+ ^== sum:-0+ product*iip+ %np*hiNCNCS+ %nP first "p" = ~&i&& iota; ~&lrtPX/&; [email protected]>lrx ^|\~& ^/[email protected] ^|T\~& "p"&& ~&iNC #cast %nL main = (first happy) 8`

output:

`<1,7,10,13,19,23,28,31>`

## Vala

Library: Gee
`using Gee; /* function to sum the square of the digits */int sum(int input){	// convert input int to string	string input_str = input.to_string();	int total = 0;	// read through each character in string, square them and add to total	for (int x = 0; x < input_str.length; x++){		// char.digit_value converts char to the decimal value the char it represents holds		int digit = input_str[x].digit_value();		total += (digit * digit);	} 	return total;} // end sum /* function to decide if a number is a happy number */bool is_happy(int total){	var past = new HashSet<int>();	while(true){		total = sum(total);		if (total == 1){			return true;} 		if (total in past){			return false;} 		past.add(total);	} // end while loop} // end happy public static void main(){	var happynums = new ArrayList<int>();	int x = 1; 	while (happynums.size < 8){		if (is_happy(x) == true)			happynums.add(x);		x++;	} 	foreach(int num in happynums)		stdout.printf("%d ", num);	stdout.printf("\n");} // end main`

The output is:

```1 7 10 13 19 23 28 31
```

## VBA

` Option Explicit Sub Test_Happy()Dim i&, Cpt&     For i = 1 To 100        If Is_Happy_Number(i) Then            Debug.Print "Is Happy : " & i            Cpt = Cpt + 1            If Cpt = 8 Then Exit For        End If    NextEnd Sub Public Function Is_Happy_Number(ByVal N As Long) As BooleanDim i&, Number\$, Cpt&    Is_Happy_Number = False 'default value    Do        Cpt = Cpt + 1       'Count Loops        Number = CStr(N)    'conversion Long To String to be able to use Len() function        N = 0        For i = 1 To Len(Number)            N = N + CInt(Mid(Number, i, 1)) ^ 2        Next i        'If Not N = 1 after 50 Loop ==> Number Is Not Happy        If Cpt = 50 Then Exit Function    Loop Until N = 1    Is_Happy_Number = TrueEnd Function `
Output:
```Is Happy : 1
Is Happy : 7
Is Happy : 10
Is Happy : 13
Is Happy : 19
Is Happy : 23
Is Happy : 28
Is Happy : 31```

## VBScript

` count = 0firsteigth=""For i = 1 To 100	If IsHappy(CInt(i)) Then		firsteight = firsteight & i & ","		count = count + 1	End If	If count = 8 Then		Exit For	End IfNextWScript.Echo firsteight Function IsHappy(n)	IsHappy = False	m = 0	Do Until m = 60		sum = 0		For j = 1 To Len(n)			sum = sum + (Mid(n,j,1))^2		Next		If sum = 1 Then			IsHappy = True			Exit Do		Else			n = sum			m = m + 1		End If	LoopEnd Function `
Output:
`1,7,10,13,19,23,28,31,`

## Visual Basic .NET

This version uses Linq to carry out the calculations.

`Module HappyNumbers    Sub Main()        Dim n As Integer = 1        Dim found As Integer = 0         Do Until found = 8            If IsHappy(n) Then                found += 1                Console.WriteLine("{0}: {1}", found, n)            End If            n += 1        Loop         Console.ReadLine()    End Sub     Private Function IsHappy(ByVal n As Integer)        Dim cache As New List(Of Long)()         Do Until n = 1            cache.Add(n)            n = Aggregate c In n.ToString() _                Into Total = Sum(Int32.Parse(c) ^ 2)            If cache.Contains(n) Then Return False        Loop         Return True    End FunctionEnd Module`

The output is:

```1: 1
2: 7
3: 10
4: 13
5: 19
6: 23
7: 28
8: 31```

### Cacheless version

Translation of: C#

Curiously, this runs in about two thirds of the time of the cacheless C# version on Tio.run.

`Module Module1     Dim sq As Integer() = {1, 4, 9, 16, 25, 36, 49, 64, 81}     Function isOne(x As Integer) As Boolean        While True            If x = 89 Then Return False            Dim t As Integer, s As Integer = 0            Do                t = (x Mod 10) - 1 : If t >= 0 Then s += sq(t)                x \= 10            Loop While x > 0            If s = 1 Then Return True            x = s        End While        Return False    End Function     Sub Main(ByVal args As String())        Const Max As Integer = 10_000_000        Dim st As DateTime = DateTime.Now        Console.Write("---Happy Numbers---" & vbLf & "The first 8:")        Dim i As Integer = 1, c As Integer = 0        While c < 8            If isOne(i) Then Console.Write("{0} {1}", If(c = 0, "", ","), i, c) : c += 1            i += 1        End While        Dim m As Integer = 10        While m <= Max            Console.Write(vbLf & "The {0:n0}th: ", m)            While c < m                If isOne(i) Then c += 1                i += 1            End While            Console.Write("{0:n0}", i - 1)            m = m * 10        End While        Console.WriteLine(vbLf & "Computation time {0} seconds.", (DateTime.Now - st).TotalSeconds)    End SubEnd Module`
Output:
```---Happy Numbers---
The first 8: 1, 7, 10, 13, 19, 23, 28, 31
The 10th: 44
The 100th: 694
The 1,000th: 6,899
The 10,000th: 67,169
The 100,000th: 692,961
The 1,000,000th: 7,105,849
The 10,000,000th: 71,313,350
Computation time 19.235551 seconds.```

## Vlang

Translation of: go
`fn happy(h int) bool {    mut m := map[int]bool{}    mut n := h    for n > 1 {        m[n] = true        mut x := 0        for x, n = n, 0; x > 0; x /= 10 {            d := x % 10            n += d * d        }        if m[n] {            return false        }    }    return true} fn main() {    for found, n := 0, 1; found < 8; n++ {        if happy(n) {            print("\$n ")            found++        }    }    println('')}`
Output:
```1 7 10 13 19 23 28 31
```

## Wren

Translation of: Go
`var happy = Fn.new { |n|    var m = {}    while (n > 1) {        m[n] = true        var x = n        n = 0        while (x > 0) {            var d = x % 10            n = n + d*d            x = (x/10).floor        }        if (m[n] == true) return false // m[n] will be null if 'n' is not a key    }    return true} var found = 0var n = 1while (found < 8) {    if (happy.call(n)) {        System.write("%(n) ")        found = found + 1    }    n = n + 1}System.print()`
Output:
```1 7 10 13 19 23 28 31
```

## XPL0

The largest possible 32-bit integer is less than 9,999,999,999. The sum of the squares of these ten digits is 10*9^2 = 810. If a cycle consisted of all the values smaller than 810, an array size of 810 would still be sufficiently large to hold them. Actually, tests show that the array only needs to hold 16 numbers.

`int List(810);          \list of numbers in a cycleint Inx;                \index for Listinclude c:\cxpl\codes;  func HadNum(N);         \Return 'true' if number N is in the Listint N;int I;[for I:= 0 to Inx-1 do        if N = List(I) then return true;return false;]; \HadNum  func SqDigits(N);       \Return the sum of the squares of the digits of Nint N;int S, D;[S:= 0;while N do        [N:= N/10;        D:= rem(0);        S:= S + D*D;        ];return S;]; \SqDigits  int N0, N, C;[N0:= 0;                \starting numberC:= 0;                  \initialize happy (starting) number counterrepeat  N:= N0;        Inx:= 0;        \reset List index        loop    [N:= SqDigits(N);                if N = 1 then                   \happy number                        [IntOut(0, N0);  CrLf(0);                        C:= C+1;                        quit;                        ];                if HadNum(N) then quit;         \if unhappy number then quit                List(Inx):= N;                  \if neither, add it to the List                Inx:= Inx+1;                    \ and continue the cycle                ];        N0:= N0+1;                              \next starting numberuntil   C=8;            \done when 8 happy numbers have been found]`

Output:

```1
7
10
13
19
23
28
31
```

## Zig

` const std = @import("std");const stdout = std.io.getStdOut().outStream(); pub fn main() !void {    try stdout.print("The first 8 happy numbers are: ", .{});    var n: u32 = 1;    var c: u4 = 0;    while (c < 8) {        if (isHappy(n)) {            c += 1;            try stdout.print("{} ", .{n});        }        n += 1;    }    try stdout.print("\n", .{});} fn isHappy(n: u32) bool {    var t = n;    var h = sumsq(n);    while (t != h) {        t = sumsq(t);        h = sumsq(sumsq(h));    }    return t == 1;} fn sumsq(n0: u32) u32 {    var s: u32 = 0;    var n = n0;    while (n > 0) : (n /= 10) {        const m = n % 10;        s += m * m;    }    return s;} `
Output:
```The first 8 happy numbers are: 1 7 10 13 19 23 28 31
```

## zkl

Here is a function that generates a continuous stream of happy numbers. Given that there are lots of happy numbers, caching them doesn't seem like a good idea memory wise. Instead, a num of squared digits == 4 is used as a proxy for a cycle (see the Wikipedia article, there are several number that will work).

Translation of: Icon and Unicon
`fcn happyNumbers{  // continously spew happy numbers   foreach N in ([1..]){       n:=N; while(1){	 n=n.split().reduce(fcn(p,n){ p + n*n },0);	 if(n==1) { vm.yield(N); break; }	 if(n==4) break;  // unhappy cycle      }   }}`
`h:=Utils.Generator(happyNumbers);h.walk(8).println();`
Output:
`L(1,7,10,13,19,23,28,31)`

Get the one million-th happy number. Nobody would call this quick.

`Utils.Generator(happyNumbers).drop(0d1_000_000-1).next().println();`
Output:
`7105849`

## ZX Spectrum Basic

Translation of: Run_BASIC
`10 FOR i=1 TO 10020 GO SUB 100030 IF isHappy=1 THEN PRINT i;" is a happy number"40 NEXT i50 STOP 1000 REM Is Happy?1010 LET isHappy=0: LET count=0: LET num=i1020 IF count=50 OR isHappy=1 THEN RETURN 1030 LET n\$=STR\$ (num)1040 LET count=count+11050 LET isHappy=01060 FOR j=1 TO LEN n\$1070 LET isHappy=isHappy+VAL n\$(j)^21080 NEXT j1090 LET num=isHappy1100 GO TO 1020`