# Is this number Loeschian?

A positive integer k is a Loeschian number if

• k can be expressed as i*i + j*j + i*j for i, j integers.

For example, the first positive Loeschian numbers are: 1 (i=1, j=0); 3 (i=j=1); 4 (i=2, j=0); 7 (i=2, j=1); 9 (i=-3, j=3); ... Note that i, j for a given k are not unique. For example, 9 can also be generated with i=3, j=0.

Other equivalent characterizations of these numbers are:

• k can be expressed as i*i + j*j + i*j for i, j non-negative integers. (For each pair of integers i, j there's a pair of non-negative integers that gives the same k)

• There is a set of k contiguous hexagons that forms a tesselation on a hexagonal grid (see illustrations for k = 4 and for k = 7). (Because of this property, these numbers find application in mobile cellular communication networks.)

• See more characterizations in the OEIS page of the sequence.

## The challenge

Given a positive integer, output a truthy result if it is a Loeschian number, or a falsy result otherwise.

The program or function should handle (say in less than a minute) inputs up to 1000, or up to data type limitations.

Code golf. Shortest wins.

## Test cases

The following numbers should output a truthy result:

1, 4, 7, 12, 13, 108, 109, 192, 516, 999


The following numbers should output a falsy result:

2, 5, 10, 42, 101, 102, 128, 150, 501, 1000

• Related (as noted by @PeterTaylor) – Luis Mendo Aug 4 '16 at 15:35
• note for the brute force algorithm: if you iterate to √k you reduce algorithm complexity from O(n²) to O(n), at expense of some bytes c; – Rod Aug 4 '16 at 17:19
• i, j non-negative integers or 9 (i=-3, j=3) - which one is it? – Titus Jan 6 '17 at 12:32
• @Titus Oh now I see. For each pair of integers i, j there's a non-negative pair that gives the same k – Luis Mendo Jan 6 '17 at 13:13

# Jelly, 11 9 bytes

ÆF‘%3,2ḄȦ


### Background

In Elementary results on the binary quadratic form a² + ab + b², the author proves the following theorem about Löschian numbers.

Theorem 16. The necessary and sufficient condition of any non-negative integer to be in the form a² + ab + b² is that, in its prime factorization, all primes other than 3 that are not in the form (6k + 1) have even exponents.

As noted on the relevant OEIS page, since all integers are are congruent to 0, 1 or 2 modulo 3, the number 3 is the only prime that is congruent to 0, and all numbers of the form (6k + 1) are congruent to 1, the theorem can be stated alternatively as follows.

A non-negative integer n is a Löschian number if and only if all prime factors of n that are congruent to 2 modulo 3 have even exponents.

### How it works

ÆF‘%3,2ḄȦ  Main link. Argument: n (integer)

ÆF         Yield the prime factorization of n, as prime-exponent pairs.
‘        Increment all primes and exponents, turning primes of the form 3k - 2
into multiples of 3 and odd exponents into multiples of 2.
%3,2    Reduce all incremented primes/exponents modulo 3/2.
n is Löschian if and only if this does not result in a [0, 0] pair.
Due to Jelly's form of vectorization, this yields [3, 2] if n = 1.
Ḅ   Unbinary; convert each pair from base 2 to integer.
Note that [x, y] = [0, 0] if and only if 2x + y = 0.
Ȧ  All; return 1 if the result contains no zeroes, 0 otherwise.


## Retina, 66634543 36 bytes

^()(\1(?<1>.\1))+(\1(.(?(4).\4)))*$ Despite the title saying Retina, this is just a plain .NET regex which accepts unary representations of Loeschian numbers. Inputs 999 and 1000 take well under a second. Try it online! (The first line enables a linefeed-separated test suite, and the next two take care of the conversion to unary for convenience.) ### Explanation The solution is based on the classification that the input can be written as i*i + j*(i + j) for positive i and non-negative j (since we don't have to handle input 0), and that n*n is just the sum of the first n odd integers. Golfing this was an interesting exercise in forward references. A "forward reference" is when you put a backreference inside the group it refers to. Of course that doesn't work when the group is used the first time, since there is nothing to be backreferenced yet, but if you put this in a loop, then the backreference gets the previous iteration's capture each time. This in turn, let's you build up a larger capture with each iteration. This can be used to craft very compact patterns for things like triangular numbers, squares and Fibonacci numbers. As an example, using the fact that squares are just sums of the first n odd integers, we can match a square input like this: (^.|..\1)+$


On the first iteration, ..\1 can't work, because \1 doesn't have a value yet. So we start with ^., capturing a single character into group 1. On subsequent iterations, ^. no longer matches due to the anchor, but now ..\1 is valid. It matches two more characters than the previous iteration and updates the capture. This way we match increasing odd numbers, getting a square after each iteration.

Now unfortunately, we can't use this technique as is. After matching i*i, we need to get i as well, so that we can multiply it by j. A simple (but long) way to do this is to make use of the fact that matching i*i takes i iterations, so that we've captured i things in group 1. We could now use balancing groups to extract this i, but like I said that's expensive.

Instead, I figured out a different way to write this "sum of consecutive odd integers" that also yields i in a capturing group at the end. Of course the ith odd number is just 2i-1. This gives us a way to increment the forward reference only by 1 on each iteration. That's this part:

^()(\1(?<1>.\1))+


This () just pushes an empty capture onto group 1 (initialising i to 0). This is pretty much equivalent to the ^.| in the simple solution above, but using | in this case would be a bit trickier.

Then we have the main loop (\1(?<1>.\1)). \1 matches the previous i, (?<1>.\1) then updates group 1 with i+1. In terms of the new i, we've just matched 2i-1 characters. Exactly what we need.

When we're done, we've matched some square i*i and group 1 still holds i characters.

The second part is closer to the simple square matching I showed above. Let's ignore the backreference to 1 for now:

(.(?(4).\1))*


This is basically the same as (^.|..\4)*, except that we can't make use of ^ because we're not at the start of the string. Instead we make use of a conditional, to match the additional .\1 only when we've already used group 4. But in effect this is exactly the same. This gives us j*j.

The only thing that's missing is the j*i term. We combine this with the j*j by making use of the fact that the j*j computation still takes j iterations. So for each iteration we also advance the cursor by i with \1. We just need to make sure not to write that into group 4, because that would mess with matching consecutive odd numbers. That's how we arrive at the:

(\1(.(?(4).\1)))*


## CJam (16 15 bytes)

{mF{~\3%2=&},!}


Online demo

This is a block (an "anonymous function") which takes input on the stack and leaves 0 or 1 on the stack. It uses the characterisation that a number is Loeschian iff it has no prime factor equal to 2 mod 3 with odd multiplicity.

Thanks to Dennis for a one-byte saving.

• Wow, nice characterization! – Luis Mendo Aug 4 '16 at 16:14

# Python 2, 56 bytes

lambda n:any(n==i*i%n+i/n*(i/n+i%n)for i in range(2*n*n))


f k=or[k==i*i+j*j+i*j|i<-[0..k],j<-[0..i]]


Usage example: f 501 -> False.

Tries all combinations of i from 0 to k and j from 0 to i . or returns True if the equality k==i*i+j*j+i*j holds for at least one of the combinations.

@flawr found a slightly different version with the same byte count:

f k|v<-[0..k]=or[(i+j)^2==k+i*j|i<-v,j<-v]

• I didn't know about or, cool=) Perhaps you have an idea how to golf this alternative phrasing: f k|v<-[0..k]=or[(i+j)^2==k+i*j|i<-v,j<-v]? – flawr Aug 4 '16 at 18:59
• @flawr: no, no idea how to golf your version further down. If you don't mind, I'l add it to my answer as an alternative version. – nimi Aug 5 '16 at 16:01

## Java 8, 81 bytes

k->{for(int i=0,j;i<=k;i++)for(j=0;j<=k;)if(i*i+j*j+i*j++==k)return 1;return 0;};


simple, naïve implementation. coincidentally same code as C# but uses -> rather than =>.

• Three less bytes because you can omit the curly braces and ending ;. DAMN! – TheLethalCoder Aug 4 '16 at 16:22
• @TheLethalCoder I actually can't, I made a mistake - same byte count as C#. – Justin Aug 4 '16 at 16:28
• Makes me feel better anyway :) – TheLethalCoder Aug 4 '16 at 17:18
• This doesn´t seem to test negative i or j. – Titus Jan 6 '17 at 10:04

# MATL, 14 13 bytes

t:0hU&+HM&*+m


Outputs 1 or 0.

### Explanation

t:    % Implicitly input number k. Duplicate. Generate vector [1 2 ...k]
0h    % Concatenate a 0. Gives [1 2 ... k 0]
U     % Square, element-wise. Gives [1 4 ... k^2 0]
&+    % Sum of all pairs from this vector. Gives a (k+1)×(k+1) matrix
HM    % Push [1 2 ... k 0] again
&*    % Product of all pairs from this vector. Gives a (k+1)×(k+1) matrix
+     % Add the two matrices
m     % True if k is a member of the resulting matrix. Implicitly display

• Did you just out-golf Jelly? – Leaky Nun Aug 4 '16 at 15:46
• @LeakyNun Let's see how long it lasts. Maybe I'll delay the code explanation a bit :-P – Luis Mendo Aug 4 '16 at 15:48
• Nope. – – – – – – Leaky Nun Aug 4 '16 at 15:49
• Your turn – – – – Leaky Nun Aug 4 '16 at 16:02
• @LeakyNun Aw :-( Now I can add the explanation :-) – Luis Mendo Aug 4 '16 at 16:03

## Python, 67 bytes

lambda k,r=range:any(i*i+j*j+i*j==k for i in r(k+1)for j in r(k+1))


https://repl.it/Cj6x

# Jelly, 151413 12 bytes

1 byte thanks to miles.

²S+P
‘ṗ2’Ç€i


Try it online!

A word of advice when testing for large numbers (bigger than 50): don't.

Truthy is a positive number. Falsey is zero.

### Explanation

‘ṗ2’Ç€i   main chain, argument: z
‘ṗ2’      generate all pairs of numbers between 0 and z inclusive
Ç€    apply the helper link to each pair
i   find the index of z in the result

²S+P   helper link, argument: [x,y] (a pair of numbers)
²      compute [x*x, y*y]
S     x*x+y*y
+P   x*x+y*y+x*y

• Tied (for) now... :-) – Luis Mendo Aug 4 '16 at 16:13
• Should we exploit Peter's characterization...? – Luis Mendo Aug 4 '16 at 16:15
• @LuisMendo That seems interesting, but it seems that it would be longer – Leaky Nun Aug 4 '16 at 16:17
• I don't think you need to flatten it. Your helper link already maps from tuples to integers. – miles Aug 4 '16 at 17:37
• @miles That's clever, thanks. – Leaky Nun Aug 4 '16 at 17:54

## Python, 49 bytes

lambda n:0in[(n-3*i*i+0j)**.5%1for i in range(n)]


Uses the equivalent quadratic form given on OEIS of n == 3*i*i+j*j. Check whether n-3*i*i is a perfect square for any i by taking its square root and checking if it's an integer, i.e. equals 0 modulo 1. Note that Python computes square roots of perfect squares exactly, without floating point error. The +0j makes it a complex number to avoid an error on the square root of a negative.

# Jellyfish, 56434129 28 bytes

2 bytes thanks to Zgarb

p
n    <
+/
1*
/
+
&*r&;>i


Try it online!

A fork of my Jelly answer.

# C (gcc), 71 69 bytes

i,j,r;f(n){for(r=i=n+1;i--;)for(j=n;j--;)r*=n!=i*i+j*j+i*j;return!r;}

• 69 bytes: i,j,r;f(n){for(r=i=n+1;i--;)for(j=n;j--;)r*=n!=i*i+j*j+i*j;return!r;}. – owacoder Aug 5 '16 at 19:00
• This doesn´t seem to test negative i or j. – Titus Jan 6 '17 at 10:02
• @Titus The question dictates non-negative i and j. – orlp Jan 6 '17 at 11:06
• positive k, but not i and j. Take a closer look at the examples. – Titus Jan 6 '17 at 11:11
• @Titus Quoting from the challenge: "k can be expressed as i*i + j*j + i*j for i, j non-negative integers." You take a closer look. – orlp Jan 6 '17 at 11:13

## C#, 8482 81 bytes

k=>{for(int i=0,j;i<=k;++i)for(j=0;j<=k;)if(i*i+j*j+i*j++==k)return 1;return 0;};


A naïve solution. 1 = true, 0 = false

## VBA, 68 67 bytes

Function L(N):For a=0To N:For b=0To a:L=L+(N=a^2+a*b+b^2):Next b,a


Naive search, starting to slow down slightly for n=1000. Excel recognizes zero return as falsy, all other returns as truthy.

Note that investigation of negative i and j is not needed, since given i>j>=0 :

(-i)2 + (-i)(-j) + (-j)2 = i2 + ij + j2

(the same result as for i and j)

(-i)2 + (-i)j + j2 = i2 - ij + j2

i2 + i(-j) + (-j)2 = i2 - ij + j2

(if one is negative, it doesn't matter which one), and then

(i-j)2 + (i-j)j + j2 = (i2 - 2ij + j2) + (ij - j2) + j2 = i2 - ij + j2

And since both (i-j) and j are non-negative, any generation of Loeschian numbers involving a negative number can be achieved using non-negative numbers.

Saved a byte, Next:Next -> Next b,a thanks to Taylor Scott.

• This doesn´t seem to test negative i or j. – Titus Jan 6 '17 at 10:03
• See the first point under "Other equivalent characterizations". Note that all test cases come up correctly. I'll add the mathematical justification to my answer (if I can). – Joffan Jan 6 '17 at 13:24
• Sorry, my fault. Overread that that´s not necessary. – Titus Jan 6 '17 at 14:22
• @Joffan you can condense Next:Next to Next b,a – Taylor Scott Nov 1 '17 at 12:05
• @Joffan looking at your solution again maybe that is because of a missing :End Functionat the end of your solution – Taylor Scott Nov 2 '17 at 3:11

# Javascript (using external library - Enumerable) (63 bytes)

k=>_.Range(0,k+1).Any(i=>_.Range(0,k+1).Any(j=>i*i+j*j+i*j==k))


Link to library: https://github.com/mvegh1/Enumerable Code explanation: Create a range of integers from 0 to k (call this the "i" range), and test if any "i" satisfies a certain predicate. That predicate creates a range from 0 to k (call this the "j" range), and tests if any "j" satisfies a certain predicate. That predicate is the loeschian formula # Perl 6,  52 51  50 bytes

->\k{?first ->(\i,\j){k==i*i+j*j+i*j},(0..k X 0..k)}
->\k{?grep ->(\i,\j){k==i*i+j*j+i*j},(0..k X 0..k)}

{?grep ->(\i,\j){$_==i*i+j*j+i*j},(0..$_ X 0..$_)}  ### Explanation: { # Turn the following into a Bool # ( Technically not necessary as a list of 1 or more values is truthy ) ? # find all where the code block returns a truthy value grep # pointy block that takes one value (list of 2 values) # and gives each of the values in it a name ->$ ( \i, \j )
{
# return true if the definition matches
$_ == i*i + j*j + i*j }, # a list of 2 element lists (possible i and j values) ( 0..$_ X 0..$_ ) }  ### Test: use v6.c; use Test; my @true = 0, 1, 4, 7, 12, 13, 108, 109, 192, 516, 999; my @false = 2, 5, 10, 42, 101, 102, 128, 150, 501, 1000; plan (@true + @false) * 2; my &is-loeschian = {?grep ->(\i,\j){$_==i*i+j*j+i*j},(0..$_ X 0..$_)}

for |(@true X True), |(@false X False) -> ( $input,$expected ) {
my ($result,$seconds) = $input.&time-it; is$result, $expected, ~$input;
cmp-ok $seconds, &[<], 60, "in$seconds seconds"
}

sub time-it ( $input ) { my$start = now;
my $result =$input.&is-loeschian;
my $finish = now; return ($result, $finish -$start )
}

1..42
ok 1 - 0
ok 2 - in 0.00111763 seconds
ok 3 - 1
ok 4 - in 0.00076766 seconds
...
ok 19 - 516
ok 20 - in 0.19629727 seconds
ok 21 - 999
ok 22 - in 0.1126715 seconds
ok 23 - 2
ok 24 - in 0.0013301 seconds
ok 25 - 5
ok 26 - in 0.00186610 seconds
...
ok 37 - 150
ok 38 - in 0.83877554 seconds
ok 39 - 501
ok 40 - in 9.2968558 seconds
ok 41 - 1000
ok 42 - in 37.31434146 seconds

• This doesn´t seem to test negative i or j. – Titus Jan 6 '17 at 10:04
• @Titus the (0..$_ X 0..$_) produces an empty list if $_ is less than 0, so there is no need to check for negative i and j because they will never be negative. Since it is only supposed to produce True for a positive Loeschian number, I don't have to do anything special for the negative case. – Brad Gilbert b2gills Jan 6 '17 at 17:03 • 9 = (3*3)+(-3*-3)+(3*-3) is a positive Loeschian with i=3, j=-3; BUT I overread that every Loeschian number has non-negative i and j. So looking for negative numbers is not necessary. So actually we could delete those comments. Sorry for bugging; my fault. – Titus Jan 6 '17 at 19:04 • @Titus modifying the code to {grep ->(\i,\j){$_==i*i+j*j+i*j},(-$_..$_ X -$_..$_)}(9) results in ((-3,0),(-3,3),(0,-3),(0,3),(3,-3),(3,0)). Honestly I probably just adapted it from other answers. – Brad Gilbert b2gills Jan 6 '17 at 22:11

## PowerShell v2+, 6356 55 bytes

param($k)(0..$k|%{0..($i=$_)|%{$i*($i+$_)+$_*$_}})-eq$k


Takes input $k, loops upwards twice (outer loop $i = 0 to $k, inner loop $j = 0 to $i), each iteration generates the result of i*i + j*j + i*j (shortened to i*(i+j) + j*j). Those results are encapsulated in parens, and passed as an array to -eq$k. This acts as a filter to select only elements that equal the input. Outputs a nonzero (the number back) for truthy, or nothing (empty) for falsey. Processes 1000 in about 15 seconds on my machine.

### Test Cases

PS C:\Tools\Scripts\golfing> (1,4,7,12,13,108,109,192,516,999|%{.\loeschian-numbers.ps1 $_})-join',' 1,4,7,12,13,108,109,192,516,999 PS C:\Tools\Scripts\golfing> (2,5,10,42,101,102,128,150,501,1000|%{.\loeschian-numbers.ps1$_})-join','

PS C:\Tools\Scripts\golfing>


# Perl, 54 + 1 (-n flag) = 55 bytes

for$i(0..$_){for$j(0..$_){$i*$i+$j*$j+$i*$j-$_?1:say}}  Needs -n and -M5.010 flags to run : perl -nE 'for$i(0..$_){for$j(0..$_){$i*$i+$j*$j+$i*$j-$_?1:say}}'


Outputs some stuffs if the number is a Loeschian number, and nothing otherwise.

This implementation is quite boring, so here is another one, for 87 bytes, regex-based, just for the eyes :

perl -pE '$_=(1 x$_)=~/^(.*)(??{$1x(-1+length$1)})(.*)(??{$2x(-1+length$2)})(??{$1x length$2})$/'  Carefull with this one, as the backtracking will use a lot of memory, so don't try to test numbers too big! (especially numbers that aren't Loeschians) # Dyalog APL, 19 bytes ⊢∊(∘.(×-⍨2*⍨+)⍨0,⍳)  Checks if k ∊ (i + j)² – ij, for any 0 ≤ i, jk. ⊢ is k ∊ a member of ∘. all combinations of × i times j -⍨ subtracted from 2*⍨ the square of + i plus j ⍨ for all i and j in 0, zero prepended to ⍳ the integers 1 through k 1000 takes 3.3 seconds on my M540 and even less on TryAPL. ## Matlab, 53 52 bytes n=input('');[a b]=ndgrid(0:n);find((a+b).^2-a.*b==n)  Simple search over all possibilities. Outputs empty array as falsy and a non-empty vector as truthy value. Considering all-zeros matrix as falsy and not-all-zeros matrix as truthy we can get rid of the find function resulting in 47 46 bytes solution: n=input('');[a b]=ndgrid(0:n);(a+b).^2-a.*b==n  One byte saved thanks to @flawr • (a+b).^2-a.*b==n is shorter. – flawr Aug 4 '16 at 18:49 # C, 66 bytes Call f() with the number to test. The function returns the number of solutions it found. q,r;f(n){for(r=q=0;q++<n*n;r+=n==q%n*(q%n+q/n)+q/n*q/n);return r;}  # Mathematica, 44 bytes MemberQ[(+##)^2-##&@@@0~Range~#~Tuples~2,#]&  Unnamed function taking an integer as input and returning True or False. The command 0~Range~#~Tuples~2 creates all ordered pairs of integers both between 0 and the input #. The function (+##)^2-##& computes the square of the sum of its arguments minus the product of its arguments; when called on two arguments i and j, this is exactly i^2+j^2+ij as desired. So that function is called on all the tuples, and then MemberQ[...,#] checks whether the input is one of the resulting values. # ASP, 39 + 4 = 43 bytes o:-k=I*I+J*J+I*J;I=1..k;J=1..k.:-not o.  Output: the problem is satisfiable iff k is Loeschian. Answer Set Programming is a logical language, similar to prolog. I use here the Potassco implementation, clingo. Input is taken from parameters (-ck= is 4 bytes long). Call example: clingo -ck=999  Output sample: SATISFIABLE  Tried with 1000: clingo -ck=1000  Output sample: UNSATISFIABLE  You can try it in your browser ; unfortunately, this method doesn't handle call flags, so you need to add the line #const k=999 in order to make it work. Ungolfed & explained code: v(1..k). % predicate v(X) holds for any X in [1..k] o:- k=I*I+J*J+I*J ; v(I) ; v(J). % o holds if k is Loeschian. :- not o. % discard models where o doesn't holds (make problem unsatisfiable)  # PHP, 70 bytes for(;$i++<$k=$argv;)for($j=$i+1;$j--;)$i*$i+$j*$j+$i*$j-$k?:die(1);


takes input from command line argument; exits with 1 for Loeschian number, with 0 else.
Run with -nr.

breakdown

for(;$i++<$k=$argv;) # loop$i from 1 to $k for($j=$i+1;$j--;)      # loop $j from$i to 0
$i*$i+$j*$j+$i*$j-$k? # if$i,$j,$k do not satisfy the equation, do nothing
:die(1);                # else exit with return code 1
# implicit: exit with code 0