You are encouraged to solve this task according to the task description, using any language you may know.

One form of   Pythagorean quadruples   is   (for positive integers   a,   b,   c,   and   d):

a2   +   b2   +   c2     =     d2

An example:

22   +   32   +   62     =     72
which is:
4    +   9    +   36     =     49

For positive integers up   2,200   (inclusive),   for all values of   a,   b,   c,   and   d,
find   (and show here)   those values of   d   that   can't   be represented.

Show the values of   d   on one line of output   (optionally with a title).

Reference

## 11l

Translation of: Python
```F quad(top = 2200)
V r = [0B] * top
V ab = [0B] * (top * 2)^2
L(a) 1 .< top
L(b) a .< top
ab[a * a + b * b] = 1B
V s = 3
L(c) 1 .< top
(V s1, s, V s2) = (s, s + 2, s + 2)
L(d) c + 1 .< top
I ab[s1]
r[d] = 1B
s1 += s2
s2 += 2
R enumerate(r).filter((i, val) -> !val & i).map((i, val) -> i)

Output:
```[1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048]
```

## ALGOL 68

As with the optimised REXX solution, we find the values of d for which there are no a^2 + b^2 = d^2 - c^2.

```BEGIN
# find values of d where d^2 =/= a^2 + b^2 + c^2 for any integers a, b, c #
# where d in [1..2200], a, b, c =/= 0                                     #
# max number to check #
INT max number = 2200;
INT max square = max number * max number;
# table of numbers that can be the sum of two squares #
[ 1 : max square ]BOOL sum of two squares; FOR n TO max square DO sum of two squares[ n ] := FALSE OD;
FOR a TO max number DO
INT a2 = a * a;
FOR b FROM a TO max number WHILE INT sum2 = ( b * b ) + a2;
sum2 <= max square DO
sum of two squares[ sum2 ] := TRUE
OD
OD;
# now find d such that d^2 - c^2 is in sum of two squares #
[ 1 : max number ]BOOL solution; FOR n TO max number DO solution[ n ] := FALSE OD;
FOR d TO max number DO
INT d2 = d * d;
FOR c TO d - 1 WHILE NOT solution[ d ] DO
INT diff2 = d2 - ( c * c );
IF sum of two squares[ diff2 ] THEN
solution[ d ] := TRUE
FI
OD
OD;
# print the numbers whose squares are not the sum of three squares #
FOR d TO max number DO
IF NOT solution[ d ] THEN
print( ( " ", whole( d, 0 ) ) )
FI
OD;
print( ( newline ) )
END```
Output:
``` 1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
```

## Amazing Hopper

Translation of: C

The process runs in 2.9 secs. on an Intel Core 2 Duo at 2.53 or 2.66 GHz. It's SLOW, but believe me, when it comes to Hopper, it's quite a feat! :D

```#include <flow.h>

DEF-MAIN(argv, argc)
SET(N, 2200)
DIM( MUL(MUL(N,N),2) ) AS-ZEROS( temp )
DIM( N ) AS-ZEROS( found )

MSET( a,T1,T2 )
TIC(T1)
SEQ-SPC(1,N,N,a), LET( a := MUL(a,a) )

SET(i,1), SET(r,0)
PERF-UP(i,N,1)
LET( r := ADD( [i] GET( a ), [i:end] CGET(a) ) )
SET-RANGE( r ), SET(temp, 1), CLR-RANGE
NEXT

SET(c,1), SET(s,3), MSET(s1,s2,d)
PERF-UP(c, N, 1)
LET( s1 := s )
s += 2
LET( s2 := s )
PERF-UP(d, N, 1)
COND ( [s1] GET(temp) )
[d] {1} PUT(found)
CEND
s1 += s2
s2 += 2
NEXT
NEXT
TOC(T1, T2), PRNL("Time = ", T2 )
CART( IS-ZERO?( found ) ) MOVE-TO( r )
PRNL( r )

MCLEAR(temp, found, a, r)
END
```
Output:
```Time = 2.88824
1,2,4,5,8,10,16,20,32,40,64,80,128,160,256,320,512,640,1024,1280,2048
```

## AppleScript

```-- double :: Num -> Num
on double(x)
x + x
end double

-- powersOfTwo :: Generator [Int]
on powersOfTwo()
iterate(double, 1)
end powersOfTwo

on run
-- Two infinite lists, from each of which we can draw an arbitrary number of initial terms

set xs to powersOfTwo() -- {1, 2, 4, 8, 16, 32 ...

set ys to fmapGen(timesFive, powersOfTwo()) -- {5, 10, 20, 40, 80, 160 ...

-- Another infinite list, derived from the first two (sorted in rising value)

set zs to mergeInOrder(xs, ys) -- {1, 2, 4, 5, 8, 10 ...

-- Taking terms from the derived list while their value is below 2200 ...

takeWhileGen(le2200, zs)

--> {1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048}
end run

-- le2200 :: Num -> Bool
on le2200(x)
x ≤ 2200
end le2200

-- timesFive :: Num -> Num
on timesFive(x)
5 * x
end timesFive

-- mergeInOrder :: Generator [Int] -> Generator [Int] -> Generator [Int]
on mergeInOrder(ga, gb)
script
property a : uncons(ga)
property b : uncons(gb)
on |λ|()
if (Nothing of a or Nothing of b) then
missing value
else
set ta to Just of a
set tb to Just of b
if |1| of ta < |1| of tb then
set a to uncons(|2| of ta)
return |1| of ta
else
set b to uncons(|2| of tb)
return |1| of tb
end if
end if
end |λ|
end script
end mergeInOrder

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

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

-- iterate :: (a -> a) -> a -> Gen [a]
on iterate(f, x)
script
property v : missing value
property g : mReturn(f)'s |λ|
on |λ|()
if missing value is v then
set v to x
else
set v to g(v)
end if
return v
end |λ|
end script
end iterate

-- Just :: a -> Maybe a
on Just(x)
{type:"Maybe", Nothing:false, Just:x}
end Just

-- length :: [a] -> Int
on |length|(xs)
set c to class of xs
if list is c or string is c then
length of xs
else
(2 ^ 29 - 1) -- (maxInt - simple proxy for non-finite)
end if
end |length|

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

-- Nothing :: Maybe a
on Nothing()
{type:"Maybe", Nothing:true}
end Nothing

-- take :: Int -> [a] -> [a]
-- take :: Int -> String -> String
on take(n, xs)
set c to class of xs
if list is c then
if 0 < n then
items 1 thru min(n, length of xs) of xs
else
{}
end if
else if string is c then
if 0 < n then
text 1 thru min(n, length of xs) of xs
else
""
end if
else if script is c then
set ys to {}
repeat with i from 1 to n
set v to xs's |λ|()
if missing value is v then
return ys
else
set end of ys to v
end if
end repeat
return ys
else
missing value
end if
end take

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

-- Tuple (,) :: a -> b -> (a, b)
on Tuple(a, b)
{type:"Tuple", |1|:a, |2|:b, length:2}
end Tuple

-- uncons :: [a] -> Maybe (a, [a])
on uncons(xs)
set lng to |length|(xs)
if 0 = lng then
Nothing()
else
if (2 ^ 29 - 1) as integer > lng then
if class of xs is string then
set cs to text items of xs
Just(Tuple(item 1 of cs, rest of cs))
else
Just(Tuple(item 1 of xs, rest of xs))
end if
else
Just(Tuple(item 1 of take(1, xs), xs))
end if
end if
end uncons
```
Output:
`{1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048}`

## AWK

```# syntax: GAWK -f PYTHAGOREAN_QUADRUPLES.AWK
# converted from Go
BEGIN {
n = 2200
s = 3
for (a=1; a<=n; a++) {
a2 = a * a
for (b=a; b<=n; b++) {
ab[a2 + b * b] = 1
}
}
for (c=1; c<=n; c++) {
s1 = s
s += 2
s2 = s
for (d=c+1; d<=n; d++) {
if (ab[s1]) {
r[d] = 1
}
s1 += s2
s2 += 2
}
}
for (d=1; d<=n; d++) {
if (!r[d]) {
printf("%d ",d)
}
}
printf("\n")
exit(0)
}
```
Output:
```1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
```

## C

### Version 1

Five seconds on my Intel Linux box.

```#include <stdio.h>
#include <math.h>
#include <string.h>

#define N 2200

int main(int argc, char **argv){
int a,b,c,d;
int r[N+1];
memset(r,0,sizeof(r));	// zero solution array
for(a=1; a<=N; a++){
for(b=a; b<=N; b++){
int aabb;
if(a&1 && b&1) continue;  // for positive odd a and b, no solution.
aabb=a*a + b*b;
for(c=b; c<=N; c++){
int aabbcc=aabb + c*c;
d=(int)sqrt((float)aabbcc);
if(aabbcc == d*d && d<=N) r[d]=1;	// solution
}
}
}
for(a=1; a<=N; a++)
if(!r[a]) printf("%d ",a);	// print non solution
printf("\n");
}
```
Output:
```\$ clang -O3 foo.c -lm
\$ ./a.out
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
```

### Version 2 (much faster)

Translation of second version of FreeBASIC entry. Runs in about 0.15 seconds.

```#include <stdlib.h>
#include <stdio.h>
#include <string.h>

#define N 2200
#define N2 2200 * 2200 * 2

int main(int argc, char **argv) {
int a, b, c, d, a2, s = 3, s1, s2;
int r[N + 1];
memset(r, 0, sizeof(r));
int *ab = calloc(N2 + 1, sizeof(int));  // allocate on heap, zero filled

for (a = 1; a <= N; a++) {
a2 = a * a;
for (b = a; b <= N; b++) ab[a2 + b * b] = 1;
}

for (c = 1; c <= N; c++) {
s1 = s;
s += 2;
s2 = s;
for (d = c + 1; d <= N; d++) {
if (ab[s1]) r[d] = 1;
s1 += s2;
s2 += 2;
}
}

for (d = 1; d <= N; d++) {
if (!r[d]) printf("%d ", d);
}
printf("\n");
free(ab);
return 0;
}
```
Output:
```Same as first version.
```

## C#

Translation of: Java
```using System;

class Program {
const int MAX = 2200;
const int MAX2 = MAX * MAX * 2;

static void Main(string[] args) {
bool[] found = new bool[MAX + 1]; // all false by default
bool[] a2b2 = new bool[MAX2 + 1]; // ditto
int s = 3;

for(int a = 1; a <= MAX; a++) {
int a2 = a * a;
for (int b=a; b<=MAX; b++) {
a2b2[a2 + b * b] = true;
}
}

for (int c = 1; c <= MAX; c++) {
int s1 = s;
s += 2;
int s2 = s;
for (int d = c + 1; d <= MAX; d++) {
if (a2b2[s1]) found[d] = true;
s1 += s2;
s2 += 2;
}
}

Console.WriteLine("The values of d <= {0} which can't be represented:", MAX);
for (int d = 1; d < MAX; d++) {
if (!found[d]) Console.Write("{0}  ", d);
}
Console.WriteLine();
}
}
}
```
Output:
```The values of d <= 2200 which can't be represented:
1  2  4  5  8  10  16  20  32  40  64  80  128  160  256  320  512  640  1024  1280  2048```

## C++

Translation of: D
```#include <iostream>
#include <vector>

constexpr int N = 2200;
constexpr int N2 = 2 * N * N;

int main() {
using namespace std;

vector<bool> found(N + 1);
vector<bool> aabb(N2 + 1);

int s = 3;

for (int a = 1; a < N; ++a) {
int aa = a * a;
for (int b = 1; b < N; ++b) {
aabb[aa + b * b] = true;
}
}

for (int c = 1; c <= N; ++c) {
int s1 = s;
s += 2;
int s2 = s;
for (int d = c + 1; d <= N; ++d) {
if (aabb[s1]) {
found[d] = true;
}
s1 += s2;
s2 += 2;
}
}

cout << "The values of d <= " << N << " which can't be represented:" << endl;
for (int d = 1; d <= N; ++d) {
if (!found[d]) {
cout << d << " ";
}
}
cout << endl;

return 0;
}
```
Output:
```The values of d <= 2200 which can't be represented:
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048```

## Crystal

Translation of: Ruby
```n = 2200
l_add, l = Hash(Int32, Bool).new(false), Hash(Int32, Bool).new(false)
(1..n).each do |x|
x2 = x * x
(x..n).each { |y| l_add[x2 + y * y] = true }
end

s = 3
(1..n).each do |x|
s1 = s
s += 2
s2 = s
((x+1)..n).each do |y|
s1 += s2
s2 += 2
end
end

puts (1..n).reject{ |x| l[x] }.join(" ")
```
Output:
```1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
```

Translation of faster Ruby version using Enumerators.

```squares  = (0..).each.map { |n| 2_u64**n }
squares5 = (0..).each.map { |n| 2_u64**n * 5 }

n = squares.next.as(Int)
m = squares5.next.as(Int)

if n < m
value = n
n = squares.next.as(Int)
else
value = m
m = squares5.next.as(Int)
end
value
end

puts pyth_quad.take_while { |n| n <= 1000000000 }.join(" ")
```
Output:
```1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048 2560 4096 5120 8192 10240 16384 20480 32768 40960 65536 81920 131072 163840 262144 327680 524288 655360 1048576 1310720 2097152 2621440 4194304 5242880 8388608 10485760 16777216 20971520 33554432 41943040 67108864 83886080 134217728 167772160 268435456 335544320 536870912 671088640
```

## D

Translation of: C
```import std.bitmanip : BitArray;
import std.stdio;

enum N = 2_200;
enum N2 = 2*N*N;

void main() {
BitArray found;
found.length = N+1;

BitArray aabb;
aabb.length = N2+1;

uint s=3;

for (uint a=1; a<=N; ++a) {
uint aa = a*a;
for (uint b=1; b<N; ++b) {
aabb[aa + b*b] = true;
}
}

for (uint c=1; c<=N; ++c) {
uint s1 = s;
s += 2;
uint s2 = s;
for (uint d=c+1; d<=N; ++d) {
if (aabb[s1]) {
found[d] = true;
}
s1 += s2;
s2 += 2;
}
}

writeln("The values of d <= ", N, " which can't be represented:");
for (uint d=1; d<=N; ++d) {
if (!found[d]) {
write(d, ' ');
}
}
writeln;
}
```
Output:
```The values of d <= 2200 which can't be represented:
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048```

## EasyLang

Translation of: Nim
```n = 2200
n2 = n * n * 2
len r[] n
len ab[] n2
for a = 1 to n
a2 = a * a
for b = a to n
ab[a2 + b * b] = 1
.
.
s = 3
for c = 1 to n
s1 = s
s += 2
s2 = s
for d = c + 1 to n
if ab[s1] = 1
r[d] = 1
.
s1 += s2
s2 += 2
.
.
for i to n
if r[i] = 0
write i & " "
.
.```
Output:
```1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
```

## EDSAC order code

A solution from first principles would probably take a long time on EDSAC, so we use the theoretical results quoted in MathOverflow. From these it follows easily that if d is a power of 2, or 5 times a power of 2, then d^2 is not the sum of three non-zero squares. The converse does not follow, but if d is a counterexample then d^2 exceeds 5*(10^10), and therefore d exceeds the limit in the task description. The EDSAC output thus consists of two interleaved arrays, as in the AppleScript solution.

```[Pythagorean quadruples - Rosetta Code
EDSAC program, Initial Orders 2]

[Arrange the storage]
T46K P56F       [N parameter: modified library s/r P7 to print integer]
T47K P106F      [M parameter: main routine]

Here, header leaves teleprinter in figures mode.]
PFGKIFAFRDLFUFOFE@A6FG@E8FEZPF
*NUMBERS!WHOSE!SQUARES!ARE!NOT!THE!SUM!
OF!THREE!NONZERO!SQUARES@&MAXIMUM#!V!
..PK            [after header, blank tape and PK (WWG, 1951, page 91)]

[------------------------------------------------------------------------------]
[Main routine]
[Constants]
[0]   P1100F          [limit, right-justified, e.g. P1100F for 2200]
[1]   !F              [teleprinter space]
[2]   @F              [carriage return]
[3]   &F              [line feed]
[4]   K4096F          [teleprinter null]
[5]   PD              [17-bit constant 1]
[6]   P2D             [17-bit constant 5]
[Variables]
[7]   PF              [2^m, where m = 0, 1, 2, ...]
[8]   PF              [5*2^n, where n = 0, 1, 2, ...]
[Enter here, with acc = 0]
[9]   A4@ T1F         [print leading zeros as nulls]
A@ TF           [pass limit to print subroutine in 0F]
[13]   A13@ GN         [call print subroutine; leaves acc clear]
O2@ O3@         [print new line]
[Initialize variables]
A5@ T7@         [2^m := 1]
A6@ T8@         [5*2^n := 5]
[Loop back to here after printing number]
[Print 2^m or 5*2^n, whichever is smaller]
[21]   A7@ S8@         [compare values]
E28@            [jump if 5*2^n is smaller]
A8@             [else restore 2^m in acc]
LD U7@          [double value in memory]
[28]   T4F             [clear acc]
A8@             [acc := 5*2^n]
LD U8@          [double value in memory]
[32]   RD              [common code: undo doubling in acc]
TF              [pass number to print subroutine in 0F]
A@ SF           [test for number > limit]
O1@             [print space before number]
T4F             [clear acc]
[39]   A39@ GN         [call print subroutine; leaves acc clear]
E21@            [loop back for next number]
[Here when done]
[42]   O2@ O3@         [print new line]
O4@             [print null to flush teleprinter buffer]
ZF              [halt the machine]

[------------------------------------------------------------]
[Subroutine to print 17-bit non-negative integer
Parameters: 0F = integer to be printed (not preserved)
1F = character for leading zero
(preserved; typically null, space or zero)
Workspace: 4F, 5F
GKA3FT34@A1FT35@S37@T36@T4DAFT4FH38@V4FRDA4D
R1024FH30@E23@O35@A2FT36@T5FV4DYFL8FT4DA5F
L1024FUFA36@G16@OFTFT35@A36@G17@ZFPFPFP4FZ219D

E25K TM GK      [M parameter again]
E9Z             [define entry point]
PF              [acc = 0 on entry]```
Output:
```NUMBERS WHOSE SQUARES ARE NOT THE SUM OF THREE NONZERO SQUARES
MAXIMUM = 2200
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048

```

## FreeBASIC

Alternate parametrization, second version both A and B even.Time just less then 0.7 second on a AMD Athlon II X4 645 3.34GHz win7 64bit. Program uses one core. When the limit is set to 576 (abs. minimum for 2200), the time is about 0.85 sec.

```' version 12-08-2017
' compile with: fbc -s console

#Define max 2200

Dim As UInteger l, m, n, l2, l2m2
Dim As UInteger limit = max * 4 \ 15
Dim As UInteger max2 = limit * limit * 2
ReDim As Ubyte list_1(max2), list_2(max2 +1)

' prime sieve, list_2(l) contains a 0 if l = prime
For l = 4 To max2 Step 2
list_1(l) = 1
Next
For l = 3 To max2 Step 2
If list_1(l) = 0 Then
For m = l * l To max2 Step l * 2
list_1(m) = 1
Next
End If
Next

' we do not need a and b (a and b are even, l = a \ 2, m = b \ 2)
' we only need to find d
For l = 1 To limit
l2 = l * l
For m = l To limit
l2m2 = l2 + m * m
list_2(l2m2 +1) = 1
' if l2m2 is a prime, no other factors exits
If list_1(l2m2) = 0 Then Continue For
' find possible factors of l2m2
' if l2m2 is odd, we need only to check the odd divisors
For n = 2 + (l2m2 And 1) To Fix(Sqr(l2m2 -1)) Step 1 + (l2m2 And 1)
If l2m2 Mod n = 0 Then
' set list_2(x) to 1 if solution is found
list_2(l2m2 \ n + n) = 1
End If
Next
Next
Next

For l = 1 To max
If list_2(l) = 0 Then Print l; " ";
Next
Print

' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End```
Output:
`1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048`

### Brute force

Based on the second REXX version: A^2 + B^2 = D^2 - C^2. Faster then the first version, about 0.2 second

```' version 14-08-2017
' compile with: fbc -s console

#Define n 2200

Dim As UInteger s = 3, s1, s2, x, x2, y
ReDim As Ubyte l(n), l_add(n * n * 2)

For x = 1 To n
x2 = x * x
For y = x To n
l_add(x2 + y * y) = 1
Next
Next

For x = 1 To n
s1 = s
s += 2
s2 = s
For y = x +1 To n
If l_add(s1) = 1 Then l(y) = 1
s1 += s2
s2 += 2
Next
Next

For x = 1 To n
If l(x) = 0 Then Print Str(x); " ";
Next
Print

' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End```
Output:
`1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048`

## FutureBasic

```_limit = 2200

long x, x2, y, s = 3, s1, s2, counter
long l( _limit )
long ladd( _limit  * _limit * 2 )

for x = 1 to _limit
x2 = x * x
for y = x to _limit
ladd( x2 + y * y ) = 1
next
next

for x = 1 to _limit
s1 = s
s = s + 2
s2 = s
for y = x + 1 to _limit
if ladd(s1) == 1 then l(y) = 1
s1 = s1 + s2
s2 = s2 +  2
next
next

counter = 1
for x = 1 to _limit
if ( l(x) == 0 )
if ( counter mod 7 == 0 )
printf @"%6ld", x : counter == 1 : continue
else
printf @"%6ld\b", x
counter++
end if
end if
next
print

HandleEvents```
Output:
```     1     2     4     5     8    10    16
20    32    40    64    80   128   160
256   320   512   640  1024  1280  2048
```

## Go

Translation of: FreeBASIC
```package main

import "fmt"

const (
N = 2200
N2 = N * N * 2
)

func main() {
s  := 3
var s1, s2 int
var r  [N + 1]bool
var ab [N2 + 1]bool

for a := 1; a <= N; a++ {
a2 := a * a
for b := a; b <= N; b++ {
ab[a2 + b * b] = true
}
}

for c := 1; c <= N; c++ {
s1 = s
s += 2
s2 = s
for d := c + 1; d <= N; d++ {
if ab[s1] {
r[d] = true
}
s1 += s2
s2 += 2
}
}

for d := 1; d <= N; d++ {
if !r[d] {
fmt.Printf("%d ", d)
}
}
fmt.Println()
}
```
Output:
```1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
```

```powersOfTwo :: [Int]
powersOfTwo = iterate (2 *) 1

unrepresentable :: [Int]
unrepresentable = merge powersOfTwo ((5 *) <\$> powersOfTwo)

merge :: [Int] -> [Int] -> [Int]
merge xxs@(x:xs) yys@(y:ys)
| x < y = x : merge xs yys
| otherwise = y : merge xxs ys

main :: IO ()
main = do
putStrLn "The values of d <= 2200 which can't be represented."
print \$ takeWhile (<= 2200) unrepresentable
```
Output:
```The values of d <= 2200 which can't be represented.
[1,2,4,5,8,10,16,20,32,40,64,80,128,160,256,320,512,640,1024,1280,2048]```

## J

Approach: generate the set of all triple sums of squares, then select the legs for which there aren't any squared "d"s. The solution is straightforward interactive play.

```   Filter =: (#~`)(`:6)

B =: *: A =: i. >: i. 2200

S1 =: , B +/ B             NB. S1 is a raveled table of the sums of squares
S1 =: <:&({:B)Filter S1    NB. remove sums of squares exceeding bound
S1 =: ~. S1                NB. remove duplicate entries

S2 =: , B +/ S1
S2 =: <:&({:B)Filter S2
S2 =: ~. S2

RESULT =: (B -.@:e. S2) # A
RESULT
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
```

## Java

Replace translation with Java coded implementation.

Compute sequence directly.

```import java.util.ArrayList;
import java.util.List;

public static void main(String[] args) {
long d = 2200;
System.out.printf("Values of d < %d where a, b, and c are non-zero and a^2 + b^2 + c^2 = d^2 has no solutions:%n%s%n", d, getPythagoreanQuadruples(d));
}

//  See:  https://oeis.org/A094958
private static List<Long> getPythagoreanQuadruples(long max) {
List<Long> list = new ArrayList<>();
long n = -1;
long m = -1;
while ( true ) {
long nTest = (long) Math.pow(2, n+1);
long mTest = (long) (5L * Math.pow(2, m+1));
long test = 0;
if ( nTest > mTest ) {
test = mTest;
m++;
}
else {
test = nTest;
n++;
}
if ( test < max ) {
}
else {
break;
}
}
return list;
}

}
```
Output:
```Values of d < 2200 where a, b, and c are non-zero and a^2 + b^2 + c^2 = d^2 has no solutions:
[1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048]
```

## JavaScript

```(() => {
'use strict';

// main :: IO ()
const main = () => {
const xs = takeWhileGen(
x => 2200 >= x,
mergeInOrder(
powersOfTwo(),
fmapGen(x => 5 * x, powersOfTwo())
)
);

return (
console.log(JSON.stringify(xs)),
xs
);
}

// powersOfTwo :: Gen [Int]
const powersOfTwo = () =>
iterate(x => 2 * x, 1);

// mergeInOrder :: Gen [Int] -> Gen [Int] -> Gen [Int]
const mergeInOrder = (ga, gb) => {
function* go(ma, mb) {
let
a = ma,
b = mb;
while (!a.Nothing && !b.Nothing) {
let
ta = a.Just,
tb = b.Just;
if (fst(ta) < fst(tb)) {
yield(fst(ta));
a = uncons(snd(ta))
} else {
yield(fst(tb));
b = uncons(snd(tb))
}
}
}
return go(uncons(ga), uncons(gb))
};

// GENERIC FUNCTIONS ----------------------------

// fmapGen <\$> :: (a -> b) -> Gen [a] -> Gen [b]
function* fmapGen(f, gen) {
const g = gen;
let v = take(1, g);
while (0 < v.length) {
yield(f(v))
v = take(1, g)
}
}

// fst :: (a, b) -> a
const fst = tpl => tpl[0];

// iterate :: (a -> a) -> a -> Generator [a]
function* iterate(f, x) {
let v = x;
while (true) {
yield(v);
v = f(v);
}
}

// Just :: a -> Maybe a
const Just = x => ({
type: 'Maybe',
Nothing: false,
Just: x
});

// Returns Infinity over objects without finite length
// this enables zip and zipWith to choose the shorter
// argument when one is non-finite, like cycle, repeat etc

// length :: [a] -> Int
const length = xs => xs.length || Infinity;

// Nothing :: Maybe a
const Nothing = () => ({
type: 'Maybe',
Nothing: true,
});

// snd :: (a, b) -> b
const snd = tpl => tpl[1];

// take :: Int -> [a] -> [a]
// take :: Int -> String -> String
const take = (n, xs) =>
xs.constructor.constructor.name !== 'GeneratorFunction' ? (
xs.slice(0, n)
) : [].concat.apply([], Array.from({
length: n
}, () => {
const x = xs.next();
return x.done ? [] : [x.value];
}));

// takeWhileGen :: (a -> Bool) -> Generator [a] -> [a]
const takeWhileGen = (p, xs) => {
const ys = [];
let
nxt = xs.next(),
v = nxt.value;
while (!nxt.done && p(v)) {
ys.push(v);
nxt = xs.next();
v = nxt.value
}
return ys;
};

// Tuple (,) :: a -> b -> (a, b)
const Tuple = (a, b) => ({
type: 'Tuple',
'0': a,
'1': b,
length: 2
});

// uncons :: [a] -> Maybe (a, [a])
const uncons = xs => {
const lng = length(xs);
return (0 < lng) ? (
lng < Infinity ? (
Just(Tuple(xs[0], xs.slice(1))) // Finite list
) : (() => {
const nxt = take(1, xs);
return 0 < nxt.length ? (
Just(Tuple(nxt[0], xs))
) : Nothing();
})() // Lazy generator
) : Nothing();
};

// MAIN ---
return main();
})();
```
Output:
`[1,2,4,5,8,10,16,20,32,40,64,80,128,160,256,320,512,640,1024,1280,2048]`

## jq

The following is a direct solution but with some obvious optimizations. Its main value may be to illustrate how looping with breaks can be accomplished in jq without `foreach`. Notice also how `first/1` is used in `is_pythagorean_quad/0` to avoid unnecessary computation.

```# Emit a proof that the input is a pythagorean quad, or else false
. as \$d
| (.*.) as \$d2
| first(
label \$continue_a | range(1; \$d) | . as \$a | (.*.) as \$a2
|   if 3*\$a2 > \$d2 then break \$continue_a else . end
| label \$continue_b | range(\$a; \$d) | . as \$b | (.*.) as \$b2
|   if \$a2  + 2 * \$b2  > \$d2 then break \$continue_b else . end
| ((\$d2-(\$a2+\$b2)) | sqrt) as \$c
| if (\$c | floor) == \$c then [\$a, \$b, \$c] else empty end )
// false;

[range(1; 2201) | select( is_pythagorean_quad | not )] | join(" ")```

Invocation and Output

```jq -r -n -f program.jq
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048```

## Julia

Translation of: C
```function quadruples(N::Int=2200)
r  = falses(N)
ab = falses(2N ^ 2)

for a in 1:N, b in a:N
ab[a ^ 2 + b ^ 2] = true
end

s = 3
for c in 1:N
s1, s, s2 = s, s + 2, s + 2
for d in c+1:N
if ab[s1] r[d] = true end
s1 += s2
s2 += 2
end
end

return findall(!, r)
end

```
Output:
`Pythagorean quadruples up to 2200: 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048`

## Kotlin

### Version 1

This uses a similar approach to the REXX optimized version. It also takes advantage of a hint in the C entry that there is no solution if both a and b are odd (confirmed by Wikipedia article). Runs in about 7 seconds on my modest laptop which is more than 4 times faster than the brute force version would have been:

```// version 1.1.3

const val MAX = 2200
const val MAX2 = MAX * MAX - 1

fun main(args: Array<String>) {
val found = BooleanArray(MAX + 1)       // all false by default
val p2 = IntArray(MAX + 1) { it * it }  // pre-compute squares

// compute all possible positive values of d * d - c * c and map them back to d
val dc = mutableMapOf<Int, MutableList<Int>>()
for (d in 1..MAX) {
for (c in 1 until d) {
val diff = p2[d] - p2[c]
val v = dc[diff]
if (v == null)
dc.put(diff, mutableListOf(d))
else if (d !in v)
}
}

for (a in 1..MAX) {
for (b in 1..a) {
if ((a and 1) != 0 && (b and 1) != 0) continue
val sum = p2[a] + p2[b]
if (sum > MAX2) continue
val v = dc[sum]
if (v != null) v.forEach { found[it] = true }
}
}
println("The values of d <= \$MAX which can't be represented:")
for (i in 1..MAX) if (!found[i]) print("\$i ")
println()
}
```
Output:
```The values of d <= 2200 which can't be represented:
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
```

### Version 2 (much faster)

This is a translation of the second FreeBASIC version and runs in about the same time (0.2 seconds).

One thing I've noticed about the resulting sequence is that it appears to be an interleaving of the two series 2 ^ n and 5 * (2 ^ n) for n >= 0 though whether it's possible to prove this mathematically I don't know.

```// version 1.1.3

const val MAX = 2200
const val MAX2 = MAX * MAX * 2

fun main(args: Array<String>) {
val found = BooleanArray(MAX + 1)   // all false by default
val a2b2  = BooleanArray(MAX2 + 1)  // ditto
var s = 3

for (a in 1..MAX) {
val a2 = a * a
for (b in a..MAX) a2b2[a2 + b * b] = true
}

for (c in 1..MAX) {
var s1 = s
s += 2
var s2 = s
for (d in (c + 1)..MAX) {
if (a2b2[s1]) found[d] = true
s1 += s2
s2 += 2
}
}

println("The values of d <= \$MAX which can't be represented:")
for (d in 1..MAX) if (!found[d]) print("\$d ")
println()
}
```
Output:
```Same as Version 1.
```

## Lua

```-- initialize
local N = 2200
local ar = {}
for i=1,N do
ar[i] = false
end

-- process
for a=1,N do
for b=a,N do
if (a % 2 ~= 1) or (b % 2 ~= 1) then
local aabb = a * a + b * b
for c=b,N do
local aabbcc = aabb + c * c
local d = math.floor(math.sqrt(aabbcc))
if (aabbcc == d * d) and (d <= N) then
ar[d] = true
end
end
end
end
-- print('done with a='..a)
end

-- print
for i=1,N do
if not ar[i] then
io.write(i.." ")
end
end
print()
```
Output:
`1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048`

## Mathematica/Wolfram Language

```max = 2200;
maxsq = max^2;
d = Range[max]^2;
Dynamic[{a, b, Length[d]}]
Do[
Do[
c = Range[1, Floor[(maxsq - a^2 - b^2)^(1/2)]];
dposs = a^2 + b^2 + c^2;
d = Complement[d, dposs]
,
{b, Floor[(maxsq - a^2)^(1/2)]}
]
,
{a, Floor[maxsq^(1/2)]}
]
Sqrt[d]
```
Output:
`{1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048}`

## Modula-2

Translation of: C
```MODULE PythagoreanQuadruples;
FROM FormatString IMPORT FormatString;
FROM RealMath IMPORT sqrt;

PROCEDURE WriteInteger(i : INTEGER);
VAR buffer : ARRAY[0..16] OF CHAR;
BEGIN
FormatString("%i", buffer, i);
WriteString(buffer)
END WriteInteger;

(* Main *)
CONST N = 2200;
VAR
r : ARRAY[0..N] OF BOOLEAN;
a,b,c,d : INTEGER;
aabb,aabbcc : INTEGER;
BEGIN
(* Initialize *)
FOR a:=0 TO HIGH(r) DO
r[a] := FALSE
END;

(* Process *)
FOR a:=1 TO N DO
FOR b:=a TO N DO
IF (a MOD 2 = 1) AND (b MOD 2 = 1) THEN
(* For positive odd a and b, no solution *)
CONTINUE
END;
aabb := a*a + b*b;
FOR c:=b TO N DO
aabbcc := aabb + c*c;
d := INT(sqrt(FLOAT(aabbcc)));
IF (aabbcc = d*d) AND (d <= N) THEN
(* solution *)
r[d] := TRUE
END
END
END
END;

FOR a:=1 TO N DO
IF NOT r[a] THEN
(* pritn non-solution *)
WriteInteger(a);
WriteString(" ")
END
END;
WriteLn;

```
Output:
`1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048 `

## Nim

Translation of: FreeBasic

### Version 1

```import math

const N = 2_200

template isOdd(n: int): bool = (n and 1) != 0

var r = newSeq[bool](N + 1)

for a in 1..N:
for b in a..N:
if a.isOdd and b.isOdd: continue
let aabb = a * a + b * b
for c in b..N:
let aabbcc = aabb + c * c
d = sqrt(aabbcc.float).int
if aabbcc == d * d and d <= N: r[d] = true

for i in 1..N:
if not r[I]: stdout.write i, " "
echo()
```
Output:
`1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048`

### Version 2

```const N = 2_200
const N2 = N * N * 2

var r = newSeq[bool](N + 1)

var ab = newSeq[bool](N2 + 1)
for a in 1..N:
let a2 = a * a
for b in a..N:
ab[a2 + b * b] = true

var s = 3
for c in 1..N:
var s1 = s
s += 2
var s2 = s
for d in (c+1)..N:
if ab[s1]: r[d] = true
s1 += s2
s2 += 2

for d in 1..N:
if not r[d]: stdout.write d, " "
```
Output:
`1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048 `

## Pascal

Works with: Free Pascal

compiled with fpc 3.2.0 ( 2019.01.10 ) -O4 -Xs

### version 1

Brute froce, but not as brute as Ring.Did it ever run?
Stopping search if limit is reached

```program pythQuad;
//find phythagorean Quadrupel up to a,b,c,d <= 2200
//a^2 + b^2 +c^2 = d^2
//find all values of d which are not possible
//brute force
//split in two procedure to reduce register pressure for CPU32

const
MaxFactor =2200;
limit = MaxFactor*MaxFactor;
type
tIdx = NativeUint;
tSum = NativeUint;

var
check : array[0..MaxFactor] of boolean;
checkCnt : LongWord;

procedure Find2(s:tSum;idx:tSum);
//second sum (a*a+b*b) +c*c =?= d*d
var
s1 : tSum;
d : tSum;
begin
d := trunc(sqrt(s+idx*idx));// calculate first sqrt
For idx := idx to MaxFactor do
Begin
s1 := s+idx*idx;
If s1 <= limit then
Begin
while s1 > d*d do //adjust sqrt
inc(d);
inc(checkCnt);
IF s1=d*d then
check[d] := true;
end
else
Break;
end;
end;

procedure Find1;
//first sum a*a+b*b
var
a,b : tIdx;
s : tSum;
begin
For a := 1 to MaxFactor do
For b := a to MaxFactor do
Begin
s := a*a+b*b;
if s < limit then
Find1(s,b)
else
break;
end;
end;

var
i : NativeUint;
begin
Find1;

For i := 1 to MaxFactor do
If Not(Check[i]) then
write(i,' ');
writeln;
writeln(CheckCnt,' checks were done');
end.
```
Output:
```1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
929605937 checks were done
real    0m2.323s -> 9 cpu-cycles per check on Ryzen 5 1600 3,7 Ghz ( Turbo )
```

### version 2

Using a variant of REXX optimized optimized
As I now see the same as Algol68
a^2 + b^2 is like moving/jumping a rake with tines at a^2 from tine(1) to tine(MaxFactor) and mark their positions
Quite fast.

```program pythQuad_2;
//find phythagorean Quadrupel up to a,b,c,d <= 2200
//a^2 + b^2 +c^2 = d^2
//a^2 + b^2 = d^2-c^2
{\$IFDEF FPC}
{\$R+,O+} //debug purposes, not slower
{\$OPTIMIZATION ON,ALL}
{\$CODEALIGN proc=16}
{\$ELSE}
{\$APPTYPE CONSOLE}
{\$ENDIF}
uses
sysutils;
const
MaxFactor = 2200;//22000;//40960;
limit = MaxFactor*MaxFactor;
type
tIdx = NativeUint;
tSum = NativeUint;
var
// global variables are initiated with 0 at startUp
sumA2B2 :array[0..limit] of byte;
check :  array[0..MaxFactor] of byte;

procedure BuildSumA2B2;
var
a,b,a2,Uplmt: tIdx;
begin
//Uplimt = a*a+b*b < Maxfactor | max(a,b) = Uplmt
Uplmt := Trunc(MaxFactor*sqrt(0.5));
For a := 1 to Uplmt  do
Begin
a2:= a*a;
For b := a downto 1 do
sumA2B2[b*b+a2] := 1
end;
end;

procedure CheckDifD2C2;
var
d,d2,c : tIdx;
begin
For d := 1 to MaxFactor do
Begin
//c < d => (d*d-c*c) > 0
d2 := d*d;
For c := d-1 downto 1 do
Begin
// d*d-c*c == (d+c)*(d-c) nonsense
if sumA2B2[d2-c*c] <> 0 then
Begin
Check[d] := 1;
//first for d found is enough
BREAK;
end;
end;
end;
end;

var
i : NativeUint;
begin
BuildSumA2B2;
CheckDifD2C2;
//FindHoles
For i := 1 to MaxFactor do
If Check[i] = 0  then
write(i,' ');
writeln;
end.
```
Output:
```1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
real  0m0,002s ( Ryzen 2200G 3.7 Ghz )
MaxFactor = 22000
.. 2048 2560 4096 5120 8192 10240 16384 20480
real  0m0,746s
MaxFactor = 40960
.. 2048 2560 4096 5120 8192 10240 16384 20480 32768 40960
real  0m3,222s
```

## Perl

Translation of: Raku
```my \$N = 2200;
push @sq, \$_**2 for 0 .. \$N;
my @not = (0) x \$N;
@not[0] = 1;

for my \$d (1 .. \$N) {
my \$last = 0;
for my \$a (reverse ceiling(\$d/3) .. \$d) {
for my \$b (1 .. ceiling(\$a/2)) {
my \$ab = \$sq[\$a] + \$sq[\$b];
last if \$ab > \$sq[\$d];
my \$x = sqrt(\$sq[\$d] - \$ab);
if (\$x == int \$x) {
\$not[\$d] = 1;
\$last = 1;
last
}
}
last if \$last;
}
}

sub ceiling { int \$_[0] + 1 - 1e-15 }

for (0 .. \$#not) {
\$result .= "\$_ " unless \$not[\$_]
}
print "\$result\n"
```
Output:
`1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048`

## Phix

```with javascript_semantics
constant N = 2200,
N2 = N*N*2
sequence found = repeat(false,N),
squares = repeat(false,N2)
-- first mark all numbers that can be the sum of two squares
for a=1 to N do
integer a2 = a*a
for b=a to N do
squares[a2+b*b] = true
end for
end for
-- now find all d such that d^2 - c^2 is in squares
for d=1 to N do
integer d2 = d*d
for c=1 to d-1 do
if squares[d2-c*c] then
found[d] = true
exit
end if
end for
end for

sequence res = {}
for i=1 to N do
end for
pp(res)
```
Output:
```{1,2,4,5,8,10,16,20,32,40,64,80,128,160,256,320,512,640,1024,1280,2048}
```

## PicoLisp

Translation of: C
```(de quadruples (N)
(let (AB NIL  S 3  R)
(for A N
(for (B A (>= N B) (inc B))
(idx
'AB
(+ (* A A) (* B B))
T ) ) )
(for C N
(let (S1 S  S2)
(inc 'S 2)
(setq S2 S)
(for (D (+ C 1) (>= N D) (inc D))
(and (idx 'AB S1) (idx 'R D T))
(inc 'S1 S2)
(inc 'S2 2) ) ) )
(make
(for A N
(or (idx 'R A) (link A)) ) ) ) )

Output:
`(1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048)`

## PureBasic

Translation of: FreeBASIC
```OpenConsole()
limite.i = 2200
s.i = 3
Dim l.i(limite)
Dim ladd.i(limite * limite * 2)

For x.i = 1 To limite
x2.i = x * x
For y = x To limite
ladd(x2 + y * y) = 1
Next y
Next x

For x.i = 1 To limite
s1.i = s
s.i + 2
s2.i = s
For y = x +1 To limite
l(y) = 1
EndIf
s1 + s2
s2 + 2
Next y
Next x

For x.i = 1 To limite
If l(x) = 0
Print(Str(x) + " ")
EndIf
Next x
Input()
CloseConsole()```
Output:
`1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048`

## Python

### Search

Translation of: Julia
```def quad(top=2200):
r = [False] * top
ab = [False] * (top * 2)**2
for a in range(1, top):
for b in range(a, top):
ab[a * a + b * b] = True
s = 3
for c in range(1, top):
s1, s, s2 = s, s + 2, s + 2
for d in range(c + 1, top):
if ab[s1]:
r[d] = True
s1 += s2
s2 += 2
return [i for i, val in enumerate(r) if not val and i]

if __name__ == '__main__':
n = 2200
print(f"Those values of d in 1..{n} that can't be represented: {quad(n)}")
```
Output:
`Those values of d in 1..2200 that can't be represented: [1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048]`

### Composition of simpler generators

Or, as an alternative to search – a generative solution (defining the generator we need as a composition of simpler generators):

Translation of: JavaScript
Translation of: AppleScript
```'''Pythagorean Quadruples'''

from itertools import islice, takewhile

# unrepresentables :: () -> [Int]
def unrepresentables():
'''A non-finite stream of powers of two which can
not be represented as a Pythagorean quadruple.
'''
return merge(
powersOfTwo()
)(
5 * x for x in powersOfTwo()
)

# powersOfTwo :: Gen [Int]
def powersOfTwo():
'''A non-finite stream of successive powers of two.
'''
def double(x):
return 2 * x

return iterate(double)(1)

# ------------------------- TEST -------------------------
# main :: IO ()
def main():
'''For positive integers up to 2,200 (inclusive)
'''
def p(x):
return 2200 >= x

print(
list(
takewhile(p, unrepresentables())
)
)

# ----------------------- GENERIC ------------------------

# iterate :: (a -> a) -> a -> Gen [a]
def iterate(f):
'''An infinite list of repeated
applications of f to x.
'''
def go(x):
v = x
while True:
yield v
v = f(v)
return go

# merge :: Gen [Int] -> Gen [Int] -> Gen [Int]
def merge(ga):
'''An ordered stream of values drawn from two
other ordered streams.
'''
def go(gb):
def f(ma, mb):
a, b = ma, mb
while a and b:
ta, tb = a, b
if ta[0] < tb[0]:
yield ta[0]
a = uncons(ta[1])
else:
yield tb[0]
b = uncons(tb[1])
return f(uncons(ga), uncons(gb))
return go

# 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.
'''
def go(xs):
return (
xs[0:n]
if isinstance(xs, (list, tuple))
else list(islice(xs, n))
)
return go

# uncons :: [a] -> Maybe (a, [a])
def uncons(xs):
'''The deconstruction of a non-empty list
(or generator stream) into two parts:
a head value, and the remaining values.
'''
if isinstance(xs, list):
return (xs[0], xs[1:]) if xs else None
else:
nxt = take(1)(xs)
return (nxt[0], xs) if nxt else None

# MAIN ---
if __name__ == '__main__':
main()
```
Output:
`[1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048]`

## R

The best solution to this problem - using lots of for loops - is practically language agnostic. The R way of doing this is far less efficient, taking about 10 minutes on my machine, but it's the obvious way to do this in R.

```squares <- d <- seq_len(2200)^2
aAndb <- outer(squares, squares, '+')
aAndb <- aAndb[upper.tri(aAndb, diag = TRUE)]
sapply(squares, function(c) d <<- setdiff(d, aAndb + c))
print(sqrt(d))```
Output:
`[1]    1    2    4    5    8   10   16   20   32   40   64   80  128  160  256  320  512  640 1024 1280 2048`

## Racket

Translation of: Python
```#lang racket

(require data/bit-vector)

(define 1..top (in-range 1 top+1))
(define r (make-bit-vector top+1))
(define ab (make-bit-vector (add1 (sqr (* top 2)))))
(for* ((a 1..top) (b (in-range a top+1))) (bit-vector-set! ab (+ (sqr a) (sqr b)) #t))

(for/fold ((s 3))
((c 1..top))
(for/fold ((s1 s) (s2 (+ s 2)))
(when (bit-vector-ref ab s1)
(bit-vector-set! r d #t))
(values (+ s1 s2) (+ s2 2)))
(+ 2 s))

(for/list ((i (in-naturals 1)) (v (in-bit-vector r 1)) #:unless v) i))

(define (report n)
(printf "Those values of d in 1..~a that can't be represented: ~a~%" n (quadruples n)))

(report 2200)
```
Output:
`Those values of d in 1..2200 that can't be represented: (1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048)`

## Raku

(formerly Perl 6)

Works with: Rakudo version 2018.09
```my \N = 2200;
my @sq = (0 .. N)»²;
my @not = False xx N;
@not[0] = True;

(1 .. N).race.map: -> \$d {
my \$last = 0;
for \$d ... (\$d/3).ceiling -> \$a {
for 1 .. (\$a/2).ceiling -> \$b {
last if (my \$ab = @sq[\$a] + @sq[\$b]) > @sq[\$d];
if (@sq[\$d] - \$ab).sqrt.narrow ~~ Int {
@not[\$d] = True;
\$last = 1;
last
}
}
last if \$last;
}
}

say @not.grep( *.not, :k );
```
Output:
`(1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048)`

## REXX

### brute force

This version is a brute force algorithm, with some optimization (to save compute time)
which pre-computes some of the squares of the positive integers used in the search.

```/*REXX pgm computes/shows (integers),  D  that aren't possible for: a² + b² + c²  =  d² */
parse arg hi .                                   /*obtain optional argument from the CL.*/
if hi=='' | hi==","  then hi=2200;  high= 3 * hi /*Not specified?  Then use the default.*/
@.=.                                             /*array of integers to be squared.     */
!.=.                                             /*  "    "     "    squared.           */
do j=1  for high                          /*precompute possible squares (to max).*/
_= j*j;   !._= j;   if j<=hi  then @.j= _ /*define a square; D  value; squared # */
end   /*j*/
d.=.                                             /*array of possible solutions  (D)     */
do       a=1  for hi-2;  aodd= a//2       /*go hunting for solutions to equation.*/
do    b=a   to hi-1;
if aodd  then  if b//2  then iterate   /*Are  A  and  B  both odd?  Then skip.*/
ab = @.a + @.b                         /*calculate sum of  2  (A,B)   squares.*/
do c=b   to hi;     abc= ab  + @.c  /*    "      "   "  3  (A,B,C)    "    */
if !.abc==.  then iterate           /*Not a square? Then skip it*/
s=!.abc;    d.s=                    /*define this D solution as being found*/
end   /*c*/
end      /*b*/
end         /*a*/
say
say 'Not possible positive integers for   d ≤' hi "  using equation:  a² + b² + c²  =  d²"
say
\$=                                               /* [↓]  find all the  "not possibles". */
do p=1  for hi;   if d.p==.  then \$=\$ p   /*Not possible? Then add it to the list*/
end   /*p*/                               /* [↓]  display list of not-possibles. */
say substr(\$, 2)                                 /*stick a fork in it,  we're all done. */
```
output   when using the default input:
```Not possible positive integers for   d ≤ 2200   using equation:  a² + b² + c²  =  d²

1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
```

### optimized

This REXX version is an optimized version, it solves the formula:

a2   +   b2         =         d2   -   c2

This REXX version is around   60   times faster then the previous version.

Programming note:   testing for   a   and   b   both being   odd   (lines 15 and 16   that each contain a   do   loop)   as
being a case that won't produce any solutions actually slows up the calculations and makes the program execute slower.

```/*REXX pgm computes/shows (integers),  D  that aren't possible for: a² + b² + c²  =  d² */
parse arg hi .                                   /*obtain optional argument from the CL.*/
if hi=='' | hi==","  then hi=2200                /*Not specified?  Then use the default.*/
high= hi * 3                                     /*D  can be three times the  HI  (max).*/
@.= .                                            /*array of integers  (≤ hi)    squared.*/
do s=1  for high;  _= s*s;  r._= s;  @.s=_ /*precompute squares and square roots. */
end  /*s*/
!.=                                              /*array of differences between squares.*/
do    c=1   for high;       cc = @.c       /*precompute possible differences.     */
do d=c+1  to high;       dif= @.d - cc  /*process  D  squared; calc differences*/
!.dif= !.dif cc                         /*add    CC    to the    !.DIF   list. */
end   /*d*/
end      /*c*/
d.=.                                             /*array of the possible solutions (D). */
do     a=1  for hi-2                       /*go hunting for solutions to equation.*/
do  b=a   to hi-1;        ab= @.a + @.b /*calculate sum of two  (A,B)  squares.*/
if !.ab==''  then iterate               /*Not a difference?   Then ignore it.  */
do n=1  for words(!.ab)              /*handle all ints that satisfy equation*/
abc= ab  +  word(!.ab, n)            /*add the  C²  integer  to  A²  +  B²  */
_= r.abc                             /*retrieve the square root  of  C²     */
d._=                                 /*mark the  D  integer as being found. */
end   /*n*/
end      /*b*/
end         /*a*/
say
say 'Not possible positive integers for   d ≤' hi "  using equation:  a² + b² + c²  =  d²"
say
\$=                                               /* [↓]  find all the  "not possibles". */
do p=1  for hi;   if d.p==.  then \$= \$ p  /*Not possible? Then add it to the list*/
end   /*p*/                               /* [↓]  display list of not-possibles. */
say substr(\$, 2)                                 /*stick a fork in it,  we're all done. */
```
output   is the same as the 1st REXX version.

## Ring

```# Project : Pythagorean quadruples

limit = 2200
pq = list(limit)
for n = 1 to limit
for m = 1 to limit
for p = 1 to limit
for x = 1 to limit
if pow(x,2) = pow(n,2) + pow(m,2) + pow(p,2)
pq[x] = 1
ok
next
next
next
next
pqstr = ""
for d = 1 to limit
if pq[d] = 0
pqstr = pqstr + d + " "
ok
next
see pqstr + nl```
Output:
```1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
```

## Ruby

Translation of: VBA
```n = 2200
1.step(n) do |x|
x2 = x*x
x.step(n) {|y| l_add[x2 + y*y] = true}
end

s = 3
1.step(n) do |x|
s1 = s
s += 2
s2 = s
(x+1).step(n) do |y|
s1 += s2
s2 += 2
end
end

puts (1..n).reject{|x| l[x]}.join(" ")
```
Output:
```1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
```

Considering the observations in the Rust and Sidef sections and toying with Enumerators :

```squares  = Enumerator.new{|y| (0..).each{|n| y << 2**n} }
squares5 = Enumerator.new{|y| (0..).each{|n| y << 2**n*5} }

n = squares.next
m = squares5.next
loop do
if n < m
y << n
n = squares.next
else
y << m
m = squares5.next
end
end
end
# this takes less than a millisecond
puts pyth_quad.take_while{|n| n <= 1000000000}.join(" ")
```
Output:
```1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048 2560 4096 5120 8192 10240 16384 20480 32768 40960 65536 81920 131072 163840 262144 327680 524288 655360 1048576 1310720 2097152 2621440 4194304 5242880 8388608 10485760 16777216 20971520 33554432 41943040 67108864 83886080 134217728 167772160 268435456 335544320 536870912 671088640
```

## Rust

 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.

This is equivalent to https://oeis.org/A094958 which simply contains positive integers of the form 2^n or 5*2^n. Multiple implementations are provided.

```use std::collections::BinaryHeap;

fn a094958_iter() -> Vec<u16> {
(0..12)
.map(|n| vec![1 << n, 5 * (1 << n)])
.flatten()
.filter(|x| x < &2200)
.collect::<BinaryHeap<u16>>()
.into_sorted_vec()
}

fn a094958_filter() -> Vec<u16> {
(1..2200) // ported from Sidef
.filter(|n| ((n & (n - 1) == 0) || (n % 5 == 0 && ((n / 5) & (n / 5 - 1) == 0))))
.collect()
}

fn a094958_loop() -> Vec<u16> {
let mut v = vec![];
for n in 0..12 {
v.push(1 << n);
if 5 * (1 << n) < 2200 {
v.push(5 * (1 << n));
}
}
v.sort();
return v;
}

fn main() {
println!("{:?}", a094958_iter());
println!("{:?}", a094958_loop());
println!("{:?}", a094958_filter());
}

#[cfg(test)]
mod tests {
use super::*;
static HAPPY: &str = "[1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048]";
#[test]
fn test_a094958_iter() {
assert!(format!("{:?}", a094958_iter()) == HAPPY);
}
#[test]
fn test_a094958_loop() {
assert!(format!("{:?}", a094958_loop()) == HAPPY);
}
#[test]
fn test_a094958_filter() {
assert!(format!("{:?}", a094958_filter()) == HAPPY);
}
}
```

## Scala

Output:

Best seen running in your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or Scastie (remote JVM).

```object PythagoreanQuadruple extends App {
val MAX = 2200
val MAX2: Int = MAX * MAX * 2
val found = Array.ofDim[Boolean](MAX + 1)
val a2b2 = Array.ofDim[Boolean](MAX2 + 1)
var s = 3
for (a <- 1 to MAX) {
val a2 = a * a

for (b <- a to MAX) a2b2(a2 + b * b) = true
}

for (c <- 1 to MAX) {
var s1 = s
s += 2
var s2 = s
for (d <- (c + 1) to MAX) {
if (a2b2(s1)) found(d) = true
s1 += s2
s2 += 2
}
}

println(f"The values of d <= \${MAX}%d which can't be represented:")
val notRepresented = (1 to MAX).filterNot(d =>  found(d) )
println(notRepresented.mkString(" "))

}
```

## Sidef

```# Finds all solutions (a,b) such that: a^2 + b^2 = n^2
func sum_of_two_squares(n) is cached {

n == 0 && return [[0, 0]]

var prod1 = 1
var prod2 = 1

var prime_powers = []

for p,e in (n.factor_exp) {
if (p % 4 == 3) {                  # p = 3 (mod 4)
e.is_even || return []         # power must be even
prod2 *= p**(e >> 1)
}
elsif (p == 2) {                   # p = 2
if (e.is_even) {               # power is even
prod2 *= p**(e >> 1)
}
else {                         # power is odd
prod1 *= p
prod2 *= p**((e - 1) >> 1)
prime_powers.append([p, 1])
}
}
else {                             # p = 1 (mod 4)
prod1 *= p**e
prime_powers.append([p, e])
}
}

prod1 == 1 && return [[prod2, 0]]
prod1 == 2 && return [[prod2, prod2]]

# All the solutions to the congruence: x^2 = -1 (mod prod1)
var square_roots = gather {
gather {
for p,e in (prime_powers) {
var pp = p**e
var r = sqrtmod(-1, pp)
take([[r, pp], [pp - r, pp]])
}
}.cartesian { |*a|
take(Math.chinese(a...))
}
}

var solutions = []

for r in (square_roots) {

var s = r
var q = prod1

while (s*s > prod1) {
(s, q) = (q % s, s)
}

solutions.append([prod2 * s, prod2 * (q % s)])
}

for p,e in (prime_powers) {
for (var i = e%2; i < e; i += 2) {

var sq = p**((e - i) >> 1)
var pp = p**(e - i)

solutions += (
__FUNC__(prod1 / pp).map { |pair|
pair.map {|r| sq * prod2 * r }
}
)
}
}

solutions.map     {|pair| pair.sort } \
.uniq_by {|pair| pair[0]   } \
.sort_by {|pair| pair[0]   }
}

# Finds all solutions (a,b,c) such that: a^2 + b^2 + c^2 = n^2
func sum_of_three_squares(n) {
gather {
for k in (1 .. n//3) {
var t = sum_of_two_squares(n**2 - k**2) || next
take(t.map { [k, _...] }...)
}
}
}

say gather {
for n in (1..2200) {
sum_of_three_squares(n) || take(n)
}
}
```
Output:
```[1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048]
```

Numbers d that cannot be expressed as a^2 + b^2 + c^2 = d^2, are numbers of the form 2^n or 5*2^n:

```say gather {
for n in (1..2200) {
if ((n & (n-1) == 0) || (n%%5 && ((n/5) & (n/5 - 1) == 0))) {
take(n)
}
}
}
```
Output:
```[1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048]
```

## Swift

Translation of: C
```func missingD(upTo n: Int) -> [Int] {
var a2 = 0, s = 3, s1 = 0, s2 = 0
var res = [Int](repeating: 0, count: n + 1)
var ab = [Int](repeating: 0, count: n * n * 2 + 1)

for a in 1...n {
a2 = a * a

for b in a...n {
ab[a2 + b * b] = 1
}
}

for c in 1..<n {
s1 = s
s += 2
s2 = s

for d in c+1...n {
if ab[s1] != 0 {
res[d] = 1
}

s1 += s2
s2 += 2
}
}

return (1...n).filter({ res[\$0] == 0 })
}

print(missingD(upTo: 2200))
```
Output:
`[1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048]`

## VBA

Translation of: FreeBasic
```Const n = 2200
Public Sub pq()
Dim s As Long, s1 As Long, s2 As Long, x As Long, x2 As Long, y As Long: s = 3
Dim l(n) As Boolean, l_add(9680000) As Boolean '9680000=n * n * 2
For x = 1 To n
x2 = x * x
For y = x To n
l_add(x2 + y * y) = True
Next y
Next x
For x = 1 To n
s1 = s
s = s + 2
s2 = s
For y = x + 1 To n
If l_add(s1) Then l(y) = True
s1 = s1 + s2
s2 = s2 + 2
Next
Next
For x = 1 To n
If Not l(x) Then Debug.Print x;
Next
Debug.Print
End Sub```
Output:
` 1  2  4  5  8  10  16  20  32  40  64  80  128  160  256  320  512  640  1024  1280  2048 `

## Wren

Translation of: FreeBASIC
```var N = 2200
var N2 = N * N * 2
var s = 3
var s1 = 0
var s2 = 0
var r = List.filled(N + 1, false)
var ab = List.filled(N2 + 1, false)

for (a in 1..N) {
var a2 = a * a
for (b in a..N) ab[a2 + b*b] = true
}

for (c in 1..N) {
s1 = s
s = s + 2
s2 = s
var d = c + 1
while (d <= N) {
if (ab[s1]) r[d] = true
s1 = s1 + s2
s2 = s2 + 2
d = d + 1
}
}

for (d in 1..N) {
if (!r[d]) System.write("%(d) ")
}
System.print()
```
Output:
```1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
```

## XPL0

Translation of: C

Twenty-seven seconds on my (cheap) Raspberry Pi 4.

```def  N = 2200;
int  A, B, C, D, AABB, AABBCC;
char R(N+1);
[FillMem(R, 0, N+1);                    \zero solution array
for A:= 1 to N do
[for B:= A to N do
[if (A&1 and B&1) = 0 then      \for positive odd A and B, no solution
[AABB:= A*A + B*B;
for C:= B to N do
[AABBCC:= AABB + C*C;
D:= sqrt(AABBCC);
if AABBCC = D*D and D <= N then R(D):= 1;       \solution
];
];
];
];
for A:= 1 to N do
if R(A) = 0 then
[IntOut(0, A);  ChOut(0, ^ )];  \print non-solutions
CrLf(0);
]```
Output:
```1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
```

## Yabasic

Translation of: FreeBASIC
```limite = 2200
s = 3
dim l(limite)
dim ladd(limite * limite * 2)

for x = 1 to limite
x2 = x * x
for y = x to limite
ladd(x2 + y * y) = 1
next y
next x

for x = 1 to limite
s1 = s
s = s + 2
s2 = s
for y = x +1 to limite
if ladd(s1) = 1  l(y) = 1
s1 = s1 + s2
s2 = s2 + 2
next y
next x

for x = 1 to limite
if l(x) = 0  print str\$(x), " ";
next x
print
end```
Output:
`1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048`

## zkl

Translation of: ALGOL 68
```# find values of d where d^2 =/= a^2 + b^2 + c^2 for any integers a, b, c #
# where d in [1..2200], a, b, c =/= 0                                     #
# max number to check #
const max_number = 2200;
const max_square = max_number * max_number;
# table of numbers that can be the sum of two squares #
sum_of_two_squares:=Data(max_square+1,Int).fill(0);  # 4 meg byte array
foreach a in ([1..max_number]){
a2 := a * a;
foreach b in ([a..max_number]){
sum2 := ( b * b ) + a2;
if(sum2 <= max_square) sum_of_two_squares[ sum2 ] = True;  # True-->1
}
}
# now find d such that d^2 - c^2 is in sum of two squares #
solution:=Data(max_number+1,Int).fill(0);	# another byte array
foreach d in ([1..max_number]){
d2 := d * d;
foreach c in ([1..d-1]){
diff2 := d2 - ( c * c );
if(sum_of_two_squares[ diff2 ]){ solution[ d ] = True; break; }
}
}
# print the numbers whose squares are not the sum of three squares #
foreach d in ([1..max_number]){
if(not solution[ d ]) print(d, " ");
}
println();```
Output:
```1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
```