# Numbers which are not the sum of distinct squares

*is a*

**Numbers which are not the sum of distinct squares****draft**programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Integer squares are the set of integers multiplied by themselves: 1 x 1 = **1**, 2 × 2 = **4**, 3 × 3 = **9**, etc. ( 1, 4, 9, 16 ... )

Most positive integers can be generated as the sum of 1 or more distinct integer squares.

1== 15== 4 + 125== 16 + 977== 36 + 25 + 16103== 49 + 25 + 16 + 9 + 4

Many can be generated in multiple ways:

90== 36 + 25 + 16 + 9 + 4 == 64 + 16 + 9 + 1 == 49 + 25 + 16 == 64 + 25 + 1 == 81 + 9130== 64 + 36 + 16 + 9 + 4 + 1 == 49 + 36 + 25 + 16 + 4 == 100 + 16 + 9 + 4 + 1 == 81 + 36 + 9 + 4 == 64 + 49 + 16 + 1 == 100 + 25 + 4 + 1 == 81 + 49 == 121 + 9

The number of positive integers that **cannot** be generated by any combination of distinct squares is in fact finite:

2, 3, 6, 7, etc.

- Task

Find and show here, on this page, **every** positive integer than cannot be generated as the sum of distinct squares.

Do not use magic numbers or pre-determined limits. Justify your answer mathematically.

- See also

## Contents

## C#[edit]

Following in the example set by the **Free Pascal** entry for this task, this C# code is re-purposed from Practical_numbers#C#.

It seems that finding as many (or more) contiguous numbers-that-*are*-the-sum-of-distinct-squares as the highest found gap demonstrates that there is no higher gap, since there is enough overlap among the permutations of the sums of possible squares (once the numbers are large enough).

using System;

using System.Collections.Generic;

using System.Linq;

class Program {

// recursively permutates the list of squares to seek a matching sum

static bool soms(int n, IEnumerable<int> f) {

if (n <= 0) return false;

if (f.Contains(n)) return true;

switch(n.CompareTo(f.Sum())) {

case 1: return false;

case 0: return true;

case -1:

var rf = f.Reverse().Skip(1).ToList();

return soms(n - f.Last(), rf) || soms(n, rf);

}

return false;

}

static void Main() {

var sw = System.Diagnostics.Stopwatch.StartNew();

int c = 0, r, i, g; var s = new List<int>(); var a = new List<int>();

var sf = "stopped checking after finding {0} sequential non-gaps after the final gap of {1}";

for (i = 1, g = 1; g >= (i >> 1); i++) {

if ((r = (int)Math.Sqrt(i)) * r == i) s.Add(i);

if (!soms(i, s)) a.Add(g = i);

}

sw.Stop();

Console.WriteLine("Numbers which are not the sum of distinct squares:");

Console.WriteLine(string.Join(", ", a));

Console.WriteLine(sf, i - g, g);

Console.Write("found {0} total in {1} ms",

a.Count, sw.Elapsed.TotalMilliseconds);

}

}

- Output @Tio.run:

Numbers which are not the sum of distinct squares: 2, 3, 6, 7, 8, 11, 12, 15, 18, 19, 22, 23, 24, 27, 28, 31, 32, 33, 43, 44, 47, 48, 60, 67, 72, 76, 92, 96, 108, 112, 128 stopped checking after finding 130 sequential non-gaps after the final gap of 128 found 31 total in 24.7904 ms

### Alternate Version[edit]

A little quicker, seeks between squares.

using System;

using System.Collections.Generic;

using System.Linq;

class Program {

static List<int> y = new List<int>();

// checks permutations of squares in a binary fashion

static void soms(ref List<int> f, int d) { f.Add(f.Last() + d);

int l = 1 << f.Count, max = f.Last(), min = max - d;

var x = new List<int>();

for (int i = 1; i < l; i++) {

int j = i, k = 0, r = 0; while (j > 0) {

if ((j & 1) == 1 && (r += f[k]) >= max) break;

j >>= 1; k++; } if (r > min && r < max) x.Add(r); }

for ( ; ++min < max; ) if (!x.Contains(min)) y.Add(min); }

static void Main() {

var sw = System.Diagnostics.Stopwatch.StartNew();

var s = new List<int>{ 1 };

var sf = "stopped checking after finding {0} sequential non-gaps after the final gap of {1}";

for (int d = 1; d <= 29; ) soms(ref s, d += 2);

sw.Stop();

Console.WriteLine("Numbers which are not the sum of distinct squares:");

Console.WriteLine(string.Join (", ", y));

Console.WriteLine("found {0} total in {1} ms",

y.Count, sw.Elapsed.TotalMilliseconds);

Console.Write(sf, s.Last()-y.Last(),y.Last());

}

}

- Output @Tio.run:

Numbers which are not the sum of distinct squares: 2, 3, 6, 7, 8, 11, 12, 15, 18, 19, 22, 23, 24, 27, 28, 31, 32, 33, 43, 44, 47, 48, 60, 67, 72, 76, 92, 96, 108, 112, 128 found 31 total in 9.9693 ms stopped checking after finding 128 sequential non-gaps after the final gap of 128

## jq[edit]

**Works with gojq, the Go implementation of jq**

The following program is directly based on the "Spoiler" observation at Raku. In effect, therefore, it computes the "magic number". It is nevertheless quite fast: on a 3GHz machine, u+s ~ 32ms using either the C or Go implementation of jq.

Since the implementation is based on generic concepts (such as runs), the helper functions are potentially independently useful.

#def squares: range(1; infinite) | . * .;

# sums of distinct squares (sods)

# Output: a stream of [$n, $sums] where $sums is the array of sods based on $n

def sods:

# input: [$n, $sums]

def next:

. as [$n, $sums]

| (($n+1)*($n+1)) as $next

| reduce $sums[] as $s ($sums + [$next];

if index($s + $next) then . else . + [$s + $next] end)

| [$n + 1, unique];

[1, [1]]

| recurse(next);

# Input: an array

# Output: the length of the run beginning at $n

def length_of_run($n):

label $out

| (range($n+1; length),null) as $i

| if $i == null then (length-$n)

elif .[$i] == .[$i-1] + 1 then empty

else $i-$n, break $out

end;

# Input: an array

def length_of_longest_run:

. as $in

| first(

foreach (range(0; length),null) as $i (0;

if $i == null then .

else ($in|length_of_run($i)) as $n

| if $n > . then $n else . end

end;

if $i == null or (($in|length) - $i) < . then . else empty end) );

def sq: .*.;

# Output: [$ix, $length]

def relevant_sods:

first(

sods

| . as [$n, $s]

| ($s|length_of_longest_run) as $length

| if $length > ($n+1|sq) then . else empty end );

def not_sum_of_distinct_squares:

relevant_sods

| . as [$ix, $sods]

| [range(2; $ix+2|sq)] - $sods;

not_sum_of_distinct_squares

- Output:

[2,3,6,7,8,11,12,15,18,19,22,23,24,27,28,31,32,33,43,44,47,48,60,67,72,76,92,96,108,112,128]

## Julia[edit]

A true proof of the sketch below would require formal mathematical induction.

#=

Here we show all the 128 < numbers < 400 can be expressed as a sum of distinct squares. Now

11 * 11 < 128 < 12 * 12. It is also true that we need no square less than 144 (12 * 12) to

reduce via subtraction of squares all the numbers above 400 to a number > 128 and < 400 by

subtracting discrete squares of numbers over 12, since the interval between such squares can

be well below 128: for example, |14^2 - 15^2| is 29. So, we can always find a serial subtraction

of discrete integer squares from any number > 400 that targets the interval between 129 and

400. Once we get to that interval, we already have shown in the program below that we can

use the remaining squares under 400 to complete the remaining sum.

=#

using Combinatorics

squares = [n * n for n in 1:20]

possibles = [n for n in 1:500 if all(combo -> sum(combo) != n, combinations(squares))]

println(possibles)

- Output:

[2, 3, 6, 7, 8, 11, 12, 15, 18, 19, 22, 23, 24, 27, 28, 31, 32, 33, 43, 44, 47, 48, 60, 67, 72, 76, 92, 96, 108, 112, 128]

## Pascal[edit]

### Free Pascal[edit]

Modified Practical_numbers#Pascal.

Searching for a block of numbers that are all a possible sum of square numbers.

There is a symmetry of hasSum whether

2,3,6,..108,112,128,

are not reachably nor

SumOfSquare-2, SumOfSquare-3,SumOfSquare-6,...SumOfSquare-108,SumOfSquare-112,SumOfSquare-128

program practicalnumbers;

{$IFDEF Windows} {$APPTYPE CONSOLE} {$ENDIF}

var

HasSum: array of byte;

function FindLongestContiniuosBlock(startIdx,MaxIdx:NativeInt):NativeInt;

var

hs0 : pByte;

l : NativeInt;

begin

l := 0;

hs0 := @HasSum[0];

for startIdx := startIdx to MaxIdx do

Begin

IF hs0[startIdx]=0 then

BREAK;

inc(l);

end;

FindLongestContiniuosBlock := l;

end;

function SumAllSquares(Limit: Uint32):NativeInt;

//mark sum and than shift by next summand == add

var

hs0, hs1: pByte;

idx, j, maxlimit, delta,MaxContiniuos,MaxConOffset: NativeInt;

begin

MaxContiniuos := 0;

MaxConOffset := 0;

maxlimit := 0;

hs0 := @HasSum[0];

hs0[0] := 1; //has sum of 0*0

idx := 1;

writeln('number offset longest sum of');

writeln(' block squares');

while idx <= Limit do

begin

delta := idx*idx;

//delta is within the continiuos range than break

If (MaxContiniuos-MaxConOffset) > delta then

BREAK;

//mark oldsum+ delta with oldsum

hs1 := @hs0[delta];

for j := maxlimit downto 0 do

hs1[j] := hs1[j] or hs0[j];

maxlimit := maxlimit + delta;

j := MaxConOffset;

repeat

delta := FindLongestContiniuosBlock(j,maxlimit);

IF delta>MaxContiniuos then

begin

MaxContiniuos:= delta;

MaxConOffset := j;

end;

inc(j,delta+1);

until j > (maxlimit-delta);

writeln(idx:4,MaxConOffset:7,MaxContiniuos:8,maxlimit:8);

inc(idx);

end;

SumAllSquares:= idx;

end;

var

limit,

sumsquare,

n: NativeInt;

begin

Limit := 25;

sumsquare := 0;

for n := 1 to Limit do

sumsquare := sumsquare+n*n;

writeln('sum of square [1..',limit,'] = ',sumsquare) ;

writeln;

setlength(HasSum,sumsquare+1);

n := SumAllSquares(Limit);

writeln(n);

for Limit := 1 to n*n do

if HasSum[Limit]=0 then

write(Limit,',');

setlength(HasSum,0);

{$IFNDEF UNIX} readln; {$ENDIF}

end.

- Output:

sum of square [1..25] = 5525 number offset longest sum of block squares 1 0 2 1 -> 0,1 2 0 2 5 3 0 2 14 4 0 2 30 5 0 2 55 6 34 9 91 ->34..42 7 49 11 140 ->49..59 8 77 15 204 ->77..91 9 129 28 285 10 129 128 385 11 129 249 506 12 129 393 650 13 2,3,6,7,8,11,12,15,18,19,22,23,24,27,28,31,32,33,43,44,47,48,60,67,72,76,92,96,108,112,128,

## Phix[edit]

As per Raku (but possibly using slightly different logic, and this is using a simple flag array), if we find there will be a block of n^{2} summables,
and we are going to mark every one of those entries plus n^{2} as summable, those regions will marry up or overlap and it is guaranteed to become at least twice that length in the next step, and all subsequent steps since 2*n^{2} is also going to be longer than (n+1)^{2} for all n>1, hence it will (eventually) mark everything beyond that point as summable. You can run this online here.

Strictly speaking the termination test should probably be `if r and sq>r then`

, not that shaving off two pointless iterations makes any difference at all.

with javascript_semantics sequence summable = {true} -- (1 can be expressed as 1*1) integer n = 2 while true do integer sq = n*n summable &= repeat(false,sq) -- (process backwards to avoid adding sq more than once) for i=length(summable)-sq to 1 by -1 do if summable[i] then summable[i+sq] = true end if end for summable[sq] = true integer r = match(repeat(true,(n+1)*(n+1)),summable) if r then summable = summable[1..r-1] exit end if n += 1 end while constant nwansods = "numbers which are not the sum of distinct squares" printf(1,"%s\n",{join(shorten(apply(find_all(false,summable),sprint),nwansods,5))})

- Output:

2 3 6 7 8 ... 92 96 108 112 128 (31 numbers which are not the sum of distinct squares)

## Raku[edit]

**Spoiler:** *(highlight to read)*

Once the longest run of consecutive generated sums is longer the the next square, every number after can be generated by adding the next square to every number in the run. Find the *new* longest run, add the *next* square, etc.

my @squares = ^∞ .map: *²; # Infinite series of squares

for 1..∞ -> $sq { # for every combination of all squares

my @sums = @squares[^$sq].combinations».sum.unique.sort;

my @run;

for @sums {

@run.push($_) and next unless @run.elems;

if $_ == @run.tail + 1 { @run.push: $_ } else { last if @run.elems > @squares[$sq]; @run = () }

}

put grep * ∉ @sums, 1..@run.tail and last if @run.elems > @squares[$sq];

}

- Output:

2 3 6 7 8 11 12 15 18 19 22 23 24 27 28 31 32 33 43 44 47 48 60 67 72 76 92 96 108 112 128

## Wren[edit]

Well I found a proof by induction here that there are only a finite number of numbers satisfying this task but I don't see how we can prove it programatically without using a specialist language such as Agda or Coq.

### Brute force[edit]

This uses a brute force approach to generate the relevant numbers, similar to Julia, except using the same figures as the above proof. Still slow in Wren, around 20 seconds.

var squares = (1..18).map { |i| i * i }.toList

var combs = []

var results = []

// generate combinations of the numbers 0 to n-1 taken m at a time

var combGen = Fn.new { |n, m|

var s = List.filled(m, 0)

var last = m - 1

var rc // recursive closure

rc = Fn.new { |i, next|

var j = next

while (j < n) {

s[i] = j

if (i == last) {

combs.add(s.toList)

} else {

rc.call(i+1, j+1)

}

j = j + 1

}

}

rc.call(0, 0)

}

for (n in 1..324) {

var all = true

for (m in 1..18) {

combGen.call(18, m)

for (comb in combs) {

var tot = (0...m).reduce(0) { |acc, i| acc + squares[comb[i]] }

if (tot == n) {

all = false

break

}

}

if (!all) break

combs.clear()

}

if (all) results.add(n)

}

System.print("Numbers which are not the sum of distinct squares:")

System.print(results)

- Output:

Numbers which are not the sum of distinct squares: [2, 3, 6, 7, 8, 11, 12, 15, 18, 19, 22, 23, 24, 27, 28, 31, 32, 33, 43, 44, 47, 48, 60, 67, 72, 76, 92, 96, 108, 112, 128]

### Quicker[edit]

Hugely quicker in fact - only 24 ms, the same as C# itself.

import "./math" for Nums

import "./fmt" for Fmt

// recursively permutates the list of squares to seek a matching sum

var soms

soms = Fn.new { |n, f|

if (n <= 0) return false

if (f.contains(n)) return true

var sum = Nums.sum(f)

if (n > sum) return false

if (n == sum) return true

var rf = f[-1..0].skip(1).toList

return soms.call(n - f[-1], rf) || soms.call(n, rf)

}

var start = System.clock

var c = 0

var s = []

var a = []

var sf = "\nStopped checking after finding $d sequential non-gaps after the final gap of $d"

var i = 1

var g = 1

var r

while (g >= (i >> 1)) {

if ((r = i.sqrt.floor) * r == i) s.add(i)

if (!soms.call(i, s)) a.add(g = i)

i = i + 1

}

System.print("Numbers which are not the sum of distinct squares:")

System.print(a.join(", "))

Fmt.print(sf, i - g, g)

Fmt.print("Found $d total in $d ms.", a.count, ((System.clock - start)*1000).round)

- Output:

Numbers which are not the sum of distinct squares: 2, 3, 6, 7, 8, 11, 12, 15, 18, 19, 22, 23, 24, 27, 28, 31, 32, 33, 43, 44, 47, 48, 60, 67, 72, 76, 92, 96, 108, 112, 128 Stopped checking after finding 130 sequential non-gaps after the final gap of 128 Found 31 total in 24 ms.