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Numbers which are not the sum of distinct squares

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


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 == 1
     5 == 4 + 1
    25 == 16 + 9
    77 == 36 + 25 + 16
   103 == 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 + 9
   130 == 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    

A finite number can not be generated by any combination of distinct squares:

   2, 3, 6, 7, etc.


Task

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

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


See also


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

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 n2 summables, and we are going to mark every one of those entries plus n2 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*n2 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]

Try it online!

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]

Translation of: C#
Library: Wren-math
Library: Wren-fmt

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.