# 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
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 cycles
puts:	equ	9	; CP/M print string
bdos:	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*flags
init:	mov	m,a
inr	l
jnz	init
mov	a,d	; Get digits
step:	call	digits
mov	l,a	; L = D1 * D1
mov	h,a
xra	a
dcr	l
jnz	sqr1
mov	l,a
mov	h,b	; L += D10 * D10
xra	a
dcr 	b
jnz	sqr10
mov	l,a
mov	h,c	; L += D100 * D100
xra	a
dcr	c
jnz	sqr100
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 happy
next:	inr	d	; Afterwards, try next number
jmp	number
happy:	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 digit
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 digit
digits:	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 back
d10:	inr	b	; Calculate 10s digit in the same way
sui	10
jnc	d10
ret		; 1s digit is left in A afterwards
string:	db	'000',13,10,'\$'```
Output:
```001
007
010
013
019
023
028
031```

## 8th

```: until!  "not while!" eval i;

with: w
with: n

: sumsqd  \ n -- n
0 swap repeat
0; 10 /mod -rot sqr + swap
again ;

: cycle \ n xt -- n
>r
dup r@ exec  \ -- tortoise, hare
repeat
swap r@ exec
swap r@ exec r@ 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
```

## ABC

```HOW TO RETURN square.digit.sum n:
PUT 0 IN sum
WHILE n>0:
PUT n mod 10 IN digit
PUT sum + digit ** 2 IN sum
PUT floor (n/10) IN n
RETURN sum

HOW TO REPORT happy n:
PUT {} IN seen
WHILE n not.in seen:
INSERT n IN seen
PUT square.digit.sum n IN n
REPORT n=1

HOW TO RETURN next.happy n:
PUT n+1 IN n
WHILE NOT happy n: PUT n+1 IN n
RETURN n

PUT 0 IN n
FOR i IN {1..8}:
PUT next.happy n IN n
WRITE n/```
Output:
```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
OD
RETURN (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
OD
RETURN (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
OD
RETURN (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
OD
RETURN```
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;

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))
FI
OD```

Output:

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

## ALGOL-M

```begin
integer 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 do
begin
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: numbers whose digit-square sums become 1    %
% when repeatedly applied                                              %
% returns true if n is happy, false otherwise                            %
logical procedure isHappy ( integer value n ) ;
begin
% in base ten, numbers either reach 1 or loop around a sequence      %
% containing 4 (see the Wikipedia article)                           %
integer v, dSum, d;
v := abs n;
if v > 1 then begin
while begin
dSum := 0;
while v not = 0 do begin
d    := v rem 10;
v    := v div 10;
dSum := dSum + ( d * d )
end while_v_ne_0 ;
v := dSum;
v not = 1 and v not = 4
end do begin end
end if_v_ne_0 ;
v = 1
end isHappy ;
begin % find the first 8 happy numbers                                   %
integer n, hCount;
hCount := 0;
n      := 1;
while hCount < 8 do begin
if isHappy( n ) then begin
writeon( i_w := 1, s_w := 0, " ", n );
hCount := hCount + 1
end if_isHappy__n ;
n := n + 1
end while_hCount_lt_10
end
end.```
Output:
``` 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 8
1 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 happyNumberList
end 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 -> Bool
on 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] -> a
on 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 tell
end foldl

-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end 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 xs
end splitOn

-- until :: (a -> Bool) -> (a -> a) -> a -> a
on |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 v
end |until|
```
Output:
```{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 = 0
While \$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
EndIf
WEnd
```
```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 = 0
While \$c < 8
\$a = ObjCreate("System.Collections.ArrayList")
\$k += 1
\$n = \$k
While \$n <> 1
If \$a.Contains(\$n) Then
ExitLoop
EndIf
\$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.Clear
WEnd
```
```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)
}
}
```

## BASIC

### 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
```

### BASIC256

```n = 1 : cnt = 0
print "The first 8 isHappy numbers are:"
print

while cnt < 8
if isHappy(n) = 1 then
cnt += 1
print cnt; " => "; n
end if
n += 1
end 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 while
end function
```

### 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```

### Commodore BASIC

The array sizes here are tuned to the minimum values required to find the first 8 happy numbers in numerical order. The H and U arrays are used for memoization, so the subscripts H(n) and U(n) must exist for the highest n encountered. The array N must have room to hold the longest chain examined in the course of determining whether a single number is happy, which thanks to the memoization is only ten elements long.

```100 C=8:DIM H(145),U(145),N(9)
110 PRINT CHR\$(147):PRINT "THE FIRST"C"HAPPY NUMBERS:":PRINT
120 H(1)=1:N=1
130 FOR C=C TO 0 STEP 0
140 : GOSUB 200
150 : IF H THEN PRINT N,:C=C-1
160 : N=N+1
170 NEXT C
180 PRINT
190 END
200 K=0:N(K)=N
210 IF H(N(K)) THEN H=1:FOR J=0 TO K:U(N(J))=0:H(N(J))=1:NEXT J:RETURN
220 IF U(N(K)) THEN H=0:RETURN
230 U(N(K))=1
240 N\$=MID\$(STR\$(N(K)),2)
250 L=LEN(N\$)
260 K=K+1:N(K)=0
270 FOR I=1 TO L
280 : D = VAL(MID\$(N\$,I,1))
290 : N(K) = N(K) + D * D
300 NEXT I
310 GOTO 210
```
Output:
```THE FIRST 8 HAPPY NUMBERS:

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 : "
Print
While count < 8
If isHappy(n) Then
count += 1
Print count;" =>"; n
End If
n += 1
Wend
Print
Print "Press any key to quit"
Sleep
```
Output:
``` 1 => 1
2 => 7
3 => 10
4 => 13
5 => 19
6 => 23
7 => 28
8 => 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 = 8
END

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-z
20 for i=1 to 100
30 i2=i
40 for l=1 to 20
50 a\$=str\$(i2)
60 i2=0
70 for j=1 to len(a\$)
80 d=val(mid\$(a\$,j,1))
90 i2=i2+d*d
100 next j
110 if i2=1 then print i;"is a happy number":n=n+1:goto 150
120 if i2=4 then 150 ' cycle found
130 next l
140 ' check if we have reached 8 numbers yet
150 if n=8 then end
160 next i
```

### PureBasic

```#ToFind=8
#MaxTests=100
#True = 1: #False = 0
Declare 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 #False
EndProcedure
```

Sample output:

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

### 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 if
next 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 = happy
wend
end 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
```

### uBasic/4tH

```' ************************
' MAIN
' ************************

PROC _PRINT_HAPPY(20)
END

' ************************
' END MAIN
' ************************

' ************************
' SUBS & FUNCTIONS
' ************************

' --------------------
_is_happy PARAM(1)
' --------------------
LOCAL (5)
f@ = 100
c@ = a@
b@ = 0

DO WHILE b@ < f@
e@ = 0

DO WHILE c@
d@ = c@ % 10
c@ = c@ / 10
e@ = e@ + (d@ * d@)
LOOP

UNTIL e@ = 1
c@ = e@
b@ = b@ + 1
LOOP

RETURN(b@ < f@)

' --------------------
_PRINT_HAPPY PARAM(1)
' --------------------
LOCAL (2)
b@ = 1
c@ = 0

DO

IF FUNC (_is_happy(b@)) THEN
c@ = c@ + 1
PRINT b@
ENDIF

b@ = b@ + 1
UNTIL c@ + 1 > a@
LOOP

RETURN

' ************************
' END SUBS & FUNCTIONS
' ************************
```

### 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
Next
End Sub

Public Function Is_Happy_Number(ByVal N As Long) As Boolean
Dim 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 = True
End 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 = 0
firsteigth=""
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 If
Next
WScript.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
Loop
End 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

End Sub

Private Function IsHappy(ByVal n As Integer)
Dim cache As New List(Of Long)()

Do Until n = 1
n = Aggregate c In n.ToString() _
Into Total = Sum(Int32.Parse(c) ^ 2)
If cache.Contains(n) Then Return False
Loop

Return True
End Function
End 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 Sub
End 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.```

### ZX Spectrum Basic

Translation of: Run_BASIC
```10 FOR i=1 TO 100
20 GO SUB 1000
30 IF isHappy=1 THEN PRINT i;" is a happy number"
40 NEXT i
50 STOP
1000 REM Is Happy?
1010 LET isHappy=0: LET count=0: LET num=i
1020 IF count=50 OR isHappy=1 THEN RETURN
1030 LET n\$=STR\$ (num)
1040 LET count=count+1
1050 LET isHappy=0
1060 FOR j=1 TO LEN n\$
1070 LET isHappy=isHappy+VAL n\$(j)^2
1080 NEXT j
1090 LET num=isHappy
1100 GO TO 1020
```

## Batch File

happy.bat

```@echo off
setlocal enableDelayedExpansion
::Define a list with 10 terms as a convenience for defining a loop
set "L10=0 1 2 3 4 5 6 7 8 9"
shift /1 & goto %1
exit /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 :listInternal
exit /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=0x7FFFFFFFF
set end=%max%
if %end% equ 2147483647 set /a end-=1
for /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
```

## 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;

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))
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 1

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? seen.include? n } { sadness.merge seen; false }
{ true } { seen << n; happy? sum_of_squares_of_digits(n), seen }
}

num = 1
happies = []

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}"

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 256
enum { 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;
}
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))
{
}
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 +)))

(cond (or (= n 1) ((deref (:happy cache)) n)) :happy
:else :unknown))

(defn happy-algo [n]
(let [get-next (fn [[prev n]] [(conj prev n) (next-number n)])
unknown? (fn [[res nums]] (= res :unknown))
[res nums] (->> [#{} n]
(iterate get-next)
(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
end
end 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))
end
end start_up```
Output:
```1
7
10
13
19
23
28
31```

## COBOL

```       IDENTIFICATION DIVISION.
PROGRAM-ID. HAPPY.

DATA DIVISION.
WORKING-STORAGE SECTION.
01 VARIABLES.
03 CANDIDATE        PIC 9(4).
03 SQSUM-IN         PIC 9(4).
03 FILLER           REDEFINES SQSUM-IN.
05 DIGITS        PIC 9 OCCURS 4 TIMES.
03 SQUARE           PIC 9(4).
03 SUM-OF-SQUARES   PIC 9(4).
03 N                PIC 9.
03 TORTOISE         PIC 9(4).
03 HARE             PIC 9(4).
88 HAPPY         VALUE 1.
03 SEEN             PIC 9 VALUE ZERO.
03 OUT-FMT          PIC ZZZ9.

PROCEDURE DIVISION.
BEGIN.
PERFORM DISPLAY-IF-HAPPY VARYING CANDIDATE FROM 1 BY 1
UNTIL SEEN IS EQUAL TO 8.
STOP RUN.

DISPLAY-IF-HAPPY.
PERFORM CHECK-HAPPY.
IF HAPPY,
MOVE CANDIDATE TO OUT-FMT,
DISPLAY OUT-FMT,

CHECK-HAPPY.
MOVE CANDIDATE TO TORTOISE, SQSUM-IN.
PERFORM CALC-SUM-OF-SQUARES.
MOVE SUM-OF-SQUARES TO HARE.
PERFORM CHECK-HAPPY-STEP UNTIL TORTOISE IS EQUAL TO HARE.

CHECK-HAPPY-STEP.
MOVE TORTOISE TO SQSUM-IN.
PERFORM CALC-SUM-OF-SQUARES.
MOVE SUM-OF-SQUARES TO TORTOISE.
MOVE HARE TO SQSUM-IN.
PERFORM CALC-SUM-OF-SQUARES.
MOVE SUM-OF-SQUARES TO SQSUM-IN.
PERFORM CALC-SUM-OF-SQUARES.
MOVE SUM-OF-SQUARES TO HARE.

CALC-SUM-OF-SQUARES.
MOVE ZERO TO SUM-OF-SQUARES.
PERFORM ADD-DIGIT-SQUARE VARYING N FROM 1 BY 1
UNTIL N IS GREATER THAN 4.

MULTIPLY DIGITS(N) BY DIGITS(N) GIVING SQUARE.
```
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 = 1
cnt = 0
while 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
true
end

i = count = 0
until count == 8; (puts i; count += 1) if happy?(i += 1) end
puts
(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)) {
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;
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;
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;
r
corp

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
od
corp```
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;
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 {
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 0
while > 8 over:
if is-happy h dup:
!print( "A happy number: " over )
swap ++ swap
++
drop
drop```
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 := 0
for x ? (isHappyNumber(x)) in (int >= 1) {
println(x)
if ((count += 1) >= 8) { break }
}```

## EasyLang

```func dsum n .
while n > 0
d = n mod 10
s += d * d
n = n div 10
.
return s
.
func happy n .
while n > 999
n = dsum n
.
len seen[] 999
repeat
n = dsum n
until seen[n] = 1
seen[n] = 1
.
return if n = 1
.
while cnt < 8
n += 1
if happy n = 1
cnt += 1
write n & " "
.
.
```
Output:
```1 7 10 13 19 23 28 31
```

## 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 6.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
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)
end
end

```
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 1
end function

integer n,count
n = 1
count = 0
while count < 8 do
if is_happy(n) then
? n
count += 1
end if
n += 1
end 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.Generic
open Microsoft.FSharp.Collections

let sqr x = x*x                                                 // Classic square definition
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;!@@+\]#\$*+]s:     {sum of squares}
[\$0[1ø1>][1ø3+ø3ø=|\1-\]#\%]f: {look for duplicates}

{check happy number}
[
\$1[f;!~2ø1=~&][1+\s;!@]#     {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))
{
// 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 %2
01.20 D 3;I (K-2)1.5
01.30 S N=N+1
01.40 I (J-8)1.2;Q
01.50 T N,!
01.60 S J=J+1
01.70 G 1.3

02.10 S A=K;S R=0
02.20 S B=FITR(A/10)
02.30 S R=R+(A-10*B)^2
02.40 S A=B
02.50 I (-A)2.2

03.10 F X=0,162;S S(X)=-1
03.20 S K=N
03.30 S S(K)=0
03.40 D 2;S K=R
03.50 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 + C@ 0<> ;
: happy-numbers ( n --)  >r 0
BEGIN r@ 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 r@ 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```

## Frege

Works with: Frege version 3.21.586-g026e8d7
```module Happy where

import Prelude.Math
-- ugh, since Frege doesn't have Set, use Map instead
import Data.Map (member, insertMin, empty emptyMap)

digitToInteger :: Char -> Integer
digitToInteger c = fromInt \$ (ord c) - (ord '0')

isHappy :: Integer -> Bool
isHappy = 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.
```

## FutureBasic

```include "NSLog.incl"

local fn IsHappy( num as NSUInteger ) as NSUInteger
NSUInteger i, happy = 0, count = 0

while ( count < 50 ) and ( happy != 1 )
CFStringRef numStr = str( num )
count++ : happy = 0
for i = 1 to len( numStr )
happy = happy + fn StringIntegerValue( mid( numStr, i, 1 ) ) ^ 2
next
num = happy
wend
end fn = num

void local fn HappyNumbers
NSUInteger i, count = 0

for i = 1 to 100
if ( fn IsHappy(i) == 1 )
count++
NSLog( @"%2lu. %2lu is a happy number", count, i )
if count == 8 then exit fn
end if
next
end fn

fn HappyNumbers

HandleEvents```
Output:
``` 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
```

## 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.

In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation.

Solution.

The following function returns whether a given number is happy or not:

Retrieving the first 8 happy numbers

## 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

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 -> Bool
isHappy = 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 -> Bool
happy 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 n
n := arglist[1] | 8    # limiting number of happy numbers to generate, default=8
writes("The first ",n," happy numbers are:")
every writes(" ", happy(seq()) \ n )
write()
end

procedure happy(i)    #: returns i if i is happy
local 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.100
1 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 digits
14
8{. (#~ 1 = sumSqrDigits ^: cond ^:_ "0) 1 + i.100
1 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>();
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);

// 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);

// 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;

happy(\$n|tonumber)```
Output:
```\$ jq --arg n 8 -n -f happy.jq
1
7
10
13
19
23
28
31
```

## Julia

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

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 = Int[])
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) = accumulate((a, b) -> nexthappy(a), 1:n)
```
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 = 256
buf = zeros(Int, CACHE)
buf[begin] = 1

function happy(n)
if n < CACHE
buf[n] > 0 && return 2-buf[n]
buf[n] = 2
end
sqsum = 0
nn = n
while nn != 0
nn, x = divrem(nn, 10)
sqsum += x * x
end
x = happy(sqsum)
n < CACHE && (buf[n] = 2 - x)
return x
end

function main()
i, counter = 1, 1000000
while counter > 0
if happy(i) != 0
counter -= 1
end
i += 1
end
return i - 1
end
```

## K

```  hpy: {x@&1={~|/x=1 4}{_+/_sqr 0\$'\$x}//:x}

hpy 1+!100
1 7 10 13 19 23 28 31 32 44 49 68 70 79 82 86 91 94 97 100

8#hpy 1+!100
1 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 integer
dgtsmsqr: {d::(); (0<){d::d,x!10; x%:10}/x; +/d*d}
/ Test if an integer is a Happy number
isHappy: {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 list
hnum: {[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
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) {
num++
}
println("First 8 happy numbers : " + happyNums.joinToString(", "))
}
```
Output:
```First 8 happy numbers : 1, 7, 10, 13, 19, 23, 28, 31
```

## Lambdatalk

```{def happy
{def happy.sum
{lambda {:n}
{if {= {W.length :n} 1}
then {pow {W.first :n} 2}
else {+ {pow {W.first :n} 2}
{happy.sum {W.rest :n}}}}}}
{def happy.is
{lambda {:x :a}
{if {= :x 1}
then true
else {if {> {A.in? :x :a} -1}
then false
else {happy.is {happy.sum :x}
{def happy.rec
{lambda {:n :a :i}
{if {= {A.length :a} :n}
then :a
else {happy.rec :n
{if {happy.is :i {A.new}}
else :a}
{+ :i 1}}}}}
{lambda {:n}
{happy.rec :n {A.new} 0}}}
-> happy

{happy 8}
-> [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 8
select #x
```

Output:

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

## 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 8
bye```

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 availability
TLDR
HAI 1.3
CAN HAS STDIO?

BTW Simple list implementation.
BTW Used for the list of numbers already seen in IZHAPPY

BTW Create a list
HOW IZ I MAEKLIST
I HAS A LIST ITZ A BUKKIT
LIST HAS A LENGTH ITZ 0
FOUND YR LIST
IF U SAY SO

BTW Append an item to list
HOW 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 1
IF U SAY SO

BTW Check for presence of an item in the list
HOW 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 FAIL
IF U SAY SO

BTW Calculate the next number using the happy formula
HOW 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 NEXT
IF U SAY SO

BTW Check to see if a number is happy
HOW 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 LOOP
IF U SAY SO

BTW Print out the first 8 happy numbers
I HAS A KOUNT ITZ 0
IM 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
OIC
IM OUTTA YR LOOP
KTHXBYE```

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)) end
end
function sumsq(a, ...)
return a and a ^ 2 + sumsq(...) or 0
end
local happy = setmetatable({true, false, false, false}, {
__index = function(self, n)
self[n] = self[sumsq(digits(n))]
return self[n]
end } )
i, j = 0, 1
repeat
i, j = happy[j] and (print(j) or i+1) or i, j + 1
until 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```

## MACRO-11

```        .TITLE  HAPPY
.MCALL  .TTYOUT,.EXIT
HAPPY:: MOV     #^D8,R5         ; 8 HAPPY NUMBERS
CLR     R4
1\$:     INC     R4
MOV     R4,R0
JSR     PC,CHECK
BNE     1\$
MOV     R4,R0
JSR     PC,PR0
SOB     R5,1\$
.EXIT

; CHECK IF R0 IS HAPPY: ZERO FLAG SET IF TRUE
CHECK:  MOV     #200,R1
MOV     #3\$,R2
1\$:     CLR     (R2)+
SOB     R1,1\$
2\$:     INCB    3\$(R0)
JSR     PC,SUMSQ
TST     3\$(R0)
BEQ     2\$
DEC     R0
RTS     PC
3\$:     .BLKW   200

; LET R0 = SUM OF SQUARES OF DIGITS OF R0
SUMSQ:  CLR     R2
1\$:     MOV     #-1,R1
2\$:     INC     R1
SUB     #12,R0
BCC     2\$
MOVB    3\$(R0),R0
MOV     R1,R0
BNE     1\$
MOV     R2,R0
RTS     PC
3\$:     .BYTE   ^D 0,^D 1,^D 4,^D 9,^D16
.BYTE   ^D25,^D36,^D49,^D64,^D81

; PRINT NUMBER IN R0 AS DECIMAL.
PR0:    MOV     #4\$,R1
1\$:     MOV     #-1,R2
2\$:     INC     R2
SUB     #12,R0
BCC     2\$
MOVB    R0,-(R1)
MOV     R2,R0
BNE     1\$
3\$:     MOVB    (R1)+,R0
.TTYOUT
BNE     3\$
RTS     PC
.ASCII  /...../
4\$:     .BYTE   15,12,0
.END HAPPY```
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.200
ZERO        CYCLE(K) = 0B
I = I + 1
SUMSQR = I
CHKLP       N = SUMSQR
SUMSQR = 0
SUMLP       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;
s
end 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 if
end 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 := (SumSqDigits@@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]
```

Examples for a specific number:

```HappyQ[1337]
HappyQ[137]
```

gives back:

```True
False
```

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]);
end
end
```
Output:
`1 7 10 13 19 23 28 31 `

## Maxima

```/* Function that decomposes te number into a list */
decompose(N) := block(
digits: [],
while N > 0 do
(remainder: mod(N, 10),
digits: cons(remainder, digits),
N: floor(N/10)),
digits
)\$

/* Function that given a number returns the sum of their digits */
sum_squares_digits(n):=block(
decompose(n),
map(lambda([x],x^2),%%),
apply("+",%%))\$

/* Predicate function based on the task iterated digits squaring */
happyp(n):=if n=1 then true else if n=89 then false else block(iter:n,while not member(iter,[1,89]) do iter:sum_squares_digits(iter),iter,if iter=1 then true)\$

/* Test case */
/* First eight happy numbers */
block(
happy:[],i:1,
while length(happy)<8 do (if happyp(i) then happy:endcons(i,happy),i:i+1),
happy);
```
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 = 0
for 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:

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

## 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 while
end function

found = []
i = 1
while found.len < 8
if isHappy(i) then found.push i
i = i + 1
end while
print "First 8 happy numbers: " + found
```
Output:
`First 8 happy numbers: [1, 7, 10, 13, 19, 23, 28, 31]`

## Miranda

```main :: [sys_message]
main = [Stdout (lay (map show (take 8 happynumbers)))]

happynumbers :: [num]
happynumbers = filter ishappy [1..]

ishappy :: num->bool
ishappy n = 1 \$in loop (iterate sumdigitsquares n)

sumdigitsquares :: num->num
sumdigitsquares 0 = 0
sumdigitsquares n = (n mod 10)^2 + sumdigitsquares (n div 10)

loop :: [*]->[*]
loop = loop' []
where loop' mem (a:as) = mem,              if a \$in mem
= loop' (a:mem) as, otherwise

in :: *->[*]->bool
in val []     = False
in val (a:as) = True,       if a=val
= val \$in as, otherwise```
Output:
```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
"unhappy"
else
happy this
end
end
;

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 limit
if 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.      */
end
end```
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;
};

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)) {
};

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 ]
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]
```

## Ol

```(define (number->list num)
(let loop ((num num) (lst #null))
(if (zero? num)
lst
(loop (quotient num 10) (cons (remainder num 10) lst)))))

(define (** x) (* x x))

(define (happy? num)
(let loop ((num num) (seen #null))
(cond
((= num 1) #true)
((memv num seen) #false)
(else
(loop (apply + (map ** (number->list num)))
(cons num seen))))))

(display "happy numbers: ")
(let loop ((n 1) (count 0))
(unless (= count 8)
(if (happy? n)
then
(display n) (display " ")
(loop (+ n 1) (+ count 1))
else
(loop (+ n 1) count))))
(print)```
```happy numbers: 1 7 10 13 19 23 28 31
```

## ooRexx

```count = 0
say "First 8 happy numbers are:"
loop i = 1 while count < 8
if happyNumber(i) then do
count += 1
say i
end
end

::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

```functor
import
System
define
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
var
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
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
```

## PascalABC.NET

```uses School;

function IsHappy(n: integer): boolean;
begin
Result := False;
var cache := new HashSet<integer>;
while n <> 1 do
begin
if n in cache then
exit;
n := n.Digits.Sum(d -> d*d);
end;
Result := True;
end;

begin
var n := 1;
var happyNums := new List<integer>;
while happyNums.Count < 8 do
begin
if IsHappy(n) then
n += 1;
end;
happyNums.Print
end.```
Output:
```1 7 10 13 19 23 28 31
```

## 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
*test
U :*happy
T (a=1):#n
C (a=1):i=i+1
C :n=n+1
J (i<max):*test
E :

*happy
C :a=n
:x=n
U :*sumsq
C :b=s
*loop
C :x=a
U :*sumsq
C :a=s
C :x=b
U :*sumsq
C :x=s
U :*sumsq
C :b=s
J (a<>b):*loop
E :

*sumsq
C :s=0
*digit
C :y=x/10
:z=x-y*10
:s=s+z*#z
:x=y
J (x):*digit
E :```
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);
GO TO 5;
END BDOS;

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

/* PRINT NUMBER */
PRINT\$NUMBER: PROCEDURE (N);
DECLARE S (6) BYTE INITIAL ('...',13,10,'\$');
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 = 0
while (firstEight length < 8) :
i++
if (isHappy(i)): firstEight append(i).
.
firstEight string print```

## PowerShell

```function happy([int] \$n) {
\$a=@()
for(\$i=2;\$a.count -lt \$n;\$i++) {
\$sum=\$i
\$hist=@{}
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 over
get_happy_number([], _).

% querying the newt happy_number
get_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 1
is_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 N
get_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].```

## 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
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 -> Bool
def 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:
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 -> Int
def succ(x):
'''The successor of an integer.'''
return 1 + x

# take :: Int -> [a] -> [a]
# take :: Int -> String -> String
def 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 -> a
def 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 50
which(apply(rbind(1:50), 2, is.happy))```
```1  7 10 13 19 23 28 31 32 44 49
```
```#Find the first 8 happy numbers
happies <- c()
i <- 1L
while(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...

## Refal

```\$ENTRY Go {
= <ShowFirst 8 Happy 1>;
};

ShowFirst {
0   s.F s.I = ;
s.N s.F s.I, <Mu s.F s.I>: T =
<Prout s.I>
<ShowFirst <- s.N 1> s.F <+ s.I 1>>;
s.N s.F s.I =
<ShowFirst s.N s.F <+ s.I 1>>;
};

Happy {
1   e.X = T;
s.N e.X s.N e.Y = F;
s.N e.X = <Happy <SqDigSum s.N> s.N e.X>;
};

SqDigSum {
0 = 0;
s.N, <Symb s.N>: s.Ds e.Rs,
<Numb s.Ds>: s.D,
<Numb e.Rs>: s.R =
<+ <* s.D s.D> <SqDigSum s.R>>;
};```
Output:
```1
7
10
13
19
23
28
31```

## Relation

```function happy(x)
set y = x
set lasty = 0
set found = " "
while y !=  1 and not (found regex "\s".y."\s")
set found = found . y . " "
set m = 0
while y  > 0
set digit = y mod 10
set m = m + digit * digit
set y = (y - digit) / 10
end while
set y = format(m,"%1d")
end while
set found = found . y . " "
if y = 1
set result = 1
else
set result = 0
end if
end function

set c = 0
set i = 1
while c < 8 and i < 100
if happy(i)
echo i
set c = c + 1
end if
set i = i + 1
end while```
```1
7
10
13
19
23
28
31
```

## REXX

```/*REXX program computes and displays a specified range of happy numbers.         */
Call time 'R'
linesize=80
Parse Arg low high                        /* obtain range of happy numbers       */
If low='?' Then Call help
If low='' Then low=10
If high='' Then
Parse Value 1 low With low high
Do i=0 To 9                               /*build a squared decimal digit table. */
square.i=i*i
End
happy.=0                                  /* happy.m=1 - m is a happy number     */
unhappy.=0                                /* unhappy.n=1 - n is an unhappy number*/
hapn=0                                    /* count of the happy numbers          */
ol=''
Do n=1 While hapn<high                    /* test integers starting with 1       */
If unhappy.n Then                       /* if n  is unhappy,                   */
Iterate                               /* then try next number                */
work=n
suml=''                                 /* list of computed sums               */
Do Forever
sum=0
Do length(work)                       /* compute sum of squared digits       */
Parse Var work digit +1 work
sum=sum+square.digit
End
Select
When unhappy.sum |,                 /* sum is known to be unhappy          */
wordpos(sum,suml)>0 Then Do    /* or was already encountered          */
--     If wordpos(sum,suml)>0 Then say 'Loop' n':' suml sum
--     If n<7 Then say n':' suml sum
unhappy.n=1                       /* n is unhappy                        */
Call set suml                     /* amd so are all sums so far          */
Iterate n
End
When sum=1 Then Do                  /* we reached sum=1                    */
hapn+=1                           /* increment number of happy numbers   */
happy.n=1                         /* n is happy                          */
If hapn>=low Then                 /* if it is in specified range         */
Call out n                      /* output it                           */
If hapn=high Then                 /* end of range reached                */
Leave n                         /* we are done                         */
Iterate n                         /* otherwise proceed                   */
End
Otherwise Do                        /* otherwise                           */
suml=suml sum                     /* add sum to list of sums             */
work=sum                          /* proceed with the new sum            */
End
End
End
End
If ol>'' Then                             /* more output data                    */
Say strip(ol)                           /* write to console                    */
-- Say time('E')
Exit

set:                                      /* all intermediate sums are unhappy   */
Parse Arg list
Do While list<>''
Parse Var list s list
unhappy.s=1
End
Return

out:                                      /* output management                   */
Parse Arg hn                            /* the happy number                    */
If length(ol hn)>linesize Then Do       /* if it does not fit                  */
Say strip(ol)                         /* output the line                     */
ol=hn                                 /* and start a new line                */
End
Else                                    /* otherwise                           */
ol=ol hn                              /* append is to the output line        */
Return

help:
Say 'rexx hno n compute and show the first n happy numbers'
Say 'rexx hno low high      show happy numbers from index low to high'
Exit```
Output:
```K:\_B\HN>rexx hno ?
rexx hno n compute and show the first n happy numbers
rexx hno low high      show happy numbers from index low to high

K:\_B\HN>rexx hno 8
1 7 10 13 19 23 28 31

K:\_B\HN>rexx hno 1000 1003
6899 6904 6917 6923
```

## Ring

```n = 1
found = 0

While found < 8
If IsHappy(n)
found += 1
see string(found) + " : " + string(n) + nl
ok
n += 1
End

Func IsHappy n
cache = []
While n != 1
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
```

## RPL

Works with: Halcyon Calc version 4.2.7
```≪ { } SWAP DO
SWAP OVER + 0 ROT
DO
MANT RND DUP IP SQ ROT + SWAP FP
UNTIL DUP NOT END
DROP
UNTIL DUP2 POS END
SWAP DROP 1 ==
≫
'HAPY?' STO

≪ { } 0 DO
1 + IF DUP HAPY? THEN SWAP OVER + SWAP END
UNTIL OVER SIZE 8 == END
≫ EVAL
```
Output:
```1: { 1 7 10 13 19 23 28 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.digits.sum{|n| n*n}

if happy?(digit_squared_sum)
@happy_numbers << n
true # Return true
else
false # Return false
end
end```

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_numbers
end

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

### Alternative version

```@memo = [0,1]
def happy(n)
sum = n.digits.sum{|n| n*n}
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:other
end

i = count = 0
while count < 8
i += 1
puts i or count+=1 if happy(i)==1
end

puts
for i in 99999999999900..99999999999999
puts i if happy(i)==1
end```
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
true
end

i = count = 0
until count == 8; puts i or count += 1 if happy?(i += 1) end
puts
(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```

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

## 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>;

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. ]

^ (negativeCache includes: (self recycle: aNum))
]
hasHappy: aNum [
^ (cache includes: (self recycle: aNum))
]
]
]

recycle: aNum [ |r n| r := Bag new.
n := aNum.
[ n > 0 ]
whileTrue: [ |d|
d := n rem: 10.
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 ].
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: [
^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 |
t := n.
[ t == 1 or:[ (already includes:t)]] whileFalse:[
t := next value:t.
].
t == 1
].

happyNumbers := OrderedCollection new.
try := 1.
[happyNumbers size < 8] whileTrue:[
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 = 0
var count = 0
while 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 -1
while {[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 TUSCRIPT
SECTION 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"
ENDIF
ENDSECTION

happynrs=""

LOOP n=1,100
sz_happynrs=SIZE(happynrs)
IF (sz_happynrs==8) EXIT
checkednr=VALUE(n)
status=seq=""
LOOP
IF (status=="next") EXIT
DO check
ENDLOOP
ENDLOOP```

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
```

## Uiua

Works with: Uiua version 0.10.0-dev.1
```HC ← /+ⁿ2≡⋕°⋕                 # Happiness calc = sum of squares of digits
IH ← |2 memo⟨IH ⊙⊂.|=1⟩∊,, HC # Apply HC until seen value recurs
Happy ← ⟨0◌|∘⟩IH : [1] .      # Pre-load `seen` with 1. Return start number or 0

# Brute force approach isn't too bad with memoisation even for high bounds.
↙8⊚>0≡Happy⇡10000

# But iterative approach is still much faster
NH ← |1 ⟨NH|∘⟩≠0Happy.+1 # Find next Happy number
⇌[⍥(NH.) 7 1]```

## UNIX Shell

Works with: Bourne Again SHell
Works with: Korn Shell
Works with: Z Shell
```function sum_of_square_digits {
typeset -i n=\$1 sum=0 d
while (( n )); do
(( d=n%10, sum+=d*d, n=n/10 ))
done
printf '%d\n' "\$sum"
}

function is_happy {
typeset -i n=\$1
typeset -a seen=()
while (( n != 1 )); do
if [[ -n \${seen[\$n]} ]]; then
return 1
fi
seen[\$n]=1
(( n=\$(sum_of_square_digits "\$n") ))
done
return 0
}

function first_n_happy {
typeset -i count=\$1 n
for (( n=1; count; n+=1 )); do
if is_happy "\$n"; then
printf '%d\n' "\$n"
(( count -= 1 ))
fi
done
return 0
}

first_n_happy 8```

## 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/&; leql@lrPrX->lrx ^|\~& ^/successor@l ^|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);
}

} // 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;}

} // 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)
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
```

## V (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 = 0
var n = 1
while (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 cycle
int Inx;                \index for List
include c:\cxpl\codes;

func HadNum(N);         \Return 'true' if number N is in the List
int N;
int I;
[for I:= 0 to Inx-1 do
if N = List(I) then return true;
return false;

func SqDigits(N);       \Return the sum of the squares of the digits of N
int 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 number
C:= 0;                  \initialize happy (starting) number counter
repeat  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 number
until   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`