# Background

A triangular grid is a grid formed by tiling the plane regularly with equilateral triangles of side length 1. The picture below is an example of a triangular grid.

A triangular lattice point is a vertex of a triangle forming the triangular grid.

The origin is a fixed point on the plane, which is one of the triangular lattice points.

# Challenge

Given a non-negative integer n, find the number of triangular lattice points whose Euclidean distance from the origin is less than or equal to n.

# Example

The following figure is an example for n = 7 (showing only 60-degree area for convenience, with point A being the origin):

# Test Cases

Input | Output
---------------
0 |       1
1 |       7
2 |      19
3 |      37
4 |      61
5 |      91
6 |     127
7 |     187
8 |     241
9 |     301
10 |     367
11 |     439
12 |     517
13 |     613
14 |     721
15 |     823
16 |     931
17 |    1045
18 |    1165
19 |    1303
20 |    1459
40 |    5815
60 |   13057
80 |   23233
100 |   36295
200 |  145051
500 |  906901
1000 | 3627559


Hint: This sequence is not OEIS A003215.

# Rules

Standard rules for apply. The shortest submission wins.

Please include how you solved the challenge in your submission.

• OEIS A053416 is the sequence of the number of points contained in a circle of diameter rather than radius n, so has twice as many terms as you want. – Neil Apr 29 '18 at 10:18
• Relevant Wikipedia and Mathworld. Contains xnor's formula and not proof. – user202729 Apr 30 '18 at 2:23
• It is the sum of the first n^2+1 terms of OEIS A004016. – alephalpha May 1 '18 at 3:55

# Python 2, 43 bytes

f=lambda n,a=1:n*n<a/3or n*n/a*6-f(n,a+a%3)


Try it online!

This is black magic.

Offering 250 rep for a written-up proof. See Lynn's answer for a proof and explanation.

• How does this work? I've been wondering for a good 30 minutes... It looks so simple but I can't find a relationship between that recursion and circles... – JungHwan Min Apr 29 '18 at 9:24
• @JungHwanMin My proof is an epic journey through plane geometry, Eisenstein integers, factorization over number fields, quadratic reciprocity, arithmetic progressions, and interchanging summations -- all for such a simple expression. Writing it all would be a major undertaking that I don't have time for now, so I'm hoping someone else will give a proof, likely a simpler one that mine that makes the connection clearer. – xnor May 1 '18 at 0:44
• Proof. This is longer than Lynn's but more self-contained: it makes no use of unproven assertions about factorisation over the Eisenstein integers. – Peter Taylor May 12 '18 at 17:14
• @PeterTaylor Cheddar Monk? As in Darths & Droids? – Neil Feb 7 at 20:29
• @Neil, congratulations on being the first person ever to ask! I registered the domain to use it as a bargaining chip for Negotiation, Level 1 in the Academy. – Peter Taylor Feb 7 at 20:44

# Haskell, 48 bytes

f n=1+6*sum[(mod(i+1)3-1)*div(n^2)i|i<-[1..n^2]]


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Uses xnor's "black magic" formula:

$$f(n)=1+6\sum_{a=0}^\infty \left\lfloor \frac{n^2}{3a+1}\right\rfloor - \left\lfloor \frac{n^2}{3a+2}\right\rfloor$$

A proof of its correctness, and an explanation of how xnor managed to express it in 43 bytes of Python, can be found here.

Long story short: we count Eisenstein integers of norm $1 \le N \le n^2$, by factoring $N = (x+y\omega)(x+y\omega^*)$ into Eisenstein primes and counting how many solutions for $(x,y)$ come out of the factorization. We recognize the number of solutions as being equal to

$$6 \times ((\text{# of divisors of }N \equiv 1\space(\text{mod }3)) - (\text{# of divisors of }N \equiv 2\space(\text{mod }3)))$$

and apply a clever trick to make that really easy to compute for all integers between $1$ and $n^2$ at once. This yields the formula above. Finally, we apply some Python golf magic to end up with the really tiny solution xnor found.

• I certainly didn't expect this when xnor said "there's some deep mathematical insights behind golfing the problem". – Bubbler May 10 '18 at 23:18

# Wolfram Language (Mathematica), 5351 50 bytes

-1 byte thanks to @miles

Sum[Boole[x(x+y)+y^2<=#^2],{x,-2#,2#},{y,-2#,2#}]&


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## How?

Instead of thinking in this:

Think of it like this:

So we apply the tranformation matrix [[sqrt(3)/2, 0], [1/2, 1]] to transform the second figure to the first one.

Then, we must find the circle in the triangular grid in terms of Cartesian coordinates.

(sqrt(3)/2 x)^2 + (1/2 x + y)^2 = x^2 + x y + y^2


So we find lattice points x, y such that x^2 + x y + y^2 <= r^2

For example, with r = 3:

• FYI, the formula x^2+x y+y^2 can also be derived from the Law of Cosines with 120 degrees. – Bubbler Apr 29 '18 at 16:38
• x^2+x y+y^2 -> x(x+y)+y^2 saves a byte – miles Apr 30 '18 at 7:31
• The formula x^2 + xy + y^2 can also be derived from the norm of an Eistenstein integer, which is a^2 - ab + b^2. Note that the sign of a and b is irrelevant except in the term ab so it has the same amount of solutions. – orlp May 1 '18 at 13:07

# Wolfram Language (Mathematica), 48 bytes

Based on OEIS A004016.

1+6Sum[DivisorSum[i,#~JacobiSymbol~3&],{i,#^2}]&


Try it online!

### CJam (24 bytes)

{_*_,f{)_)3%(@@/*}1b6*)}


This is an anonymous block (function) which takes one argument on the stack and leaves the result on the stack. Online test suite. Note that the two largest cases are too slow.

### Explanation

alephalpha noted in a comment on the question that

It is the sum of the first n^2+1 terms of OEIS A004016

and xnor's answer implements this sum (although I'm not sure whether their unposted proof uses it explicitly) as $$f(n) = 1 + 6 \sum_{a=0}^\infty \left\lfloor\frac{n^2}{3a+1}\right\rfloor - \left\lfloor\frac{n^2}{3a+2}\right\rfloor$$

My proof of correctness of that formula is based on some information gleaned from alephalpha's OEIS link:

G.f.: 1 + 6*Sum_{n>=1} x^(3*n-2)/(1-x^(3*n-2)) - x^(3*n-1)/(1-x^(3*n-1)). - Paul D. Hanna, Jul 03 2011

for which the relevant reference is the paper by Hirschhorn. An elementary proof is possible using nothing more than a basic understanding of complex numbers (cube roots of unity, magnitude), the concept of generating functions, the derivative of $x^a$, and the chain rule of differentiation. In summary, we first prove from first principles the Jacobi triple-product identity $$\prod_{k=0}^\infty (1-q^{k+1})(1 + xq^{k+1})(1 + x^{-1}q^k) = \sum_{k\in \mathbb{Z}} q^{k(k+1)/2}x^k$$ That then bootstraps a proof that $$\sum_{m,n \in \mathbb{Z}} \omega^{m-n} q^{m^2+mn+n^2} = \prod_{k=1}^\infty \frac{(1-q^k)^3}{1-q^{3k}}$$ where $\omega$ is a primitive cube root of unity. The final big step is to use this to show that $$\sum_{m,n \in \mathbb{Z}} q^{m^2+mn+n^2} = 1 + 6 \sum_{k \ge 0} \left(\frac{q^{3k+1}}{1-q^{3k+1}} - \frac{q^{3k+2}}{1-q^{3k+2}} \right)$$

### Code dissection

{          e# Define a block. Stack: ... r
_*       e#   Square it
_,f{     e#   Map with parameter: invokes block for (r^2, 0), (r^2, 1), ... (r^2, r^2-1)
)      e#     Increment second parameter. Stack: ... r^2 x with 1 <= x <= r^2
_)3%(  e#     Duplicate x and map to whichever of 0, 1, -1 is equal to it (mod 3)
@@/*   e#     Evaluate (r^2 / x) * (x mod 3)
}
1b6*     e#   Sum and multiply by 6
)        e#   Increment to count the point at the origin
}


# J, 27 bytes

[:+/@,*:>:(*++&*:)"{~@i:@+:


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Based on JungHwan Min's method.

## Explanation

[:+/@,*:>:(*++&*:)"{~@i:@+:  Input: n
+:  Double
i:     Range [-2n .. 2n]
"{~        For each pair (x, y)
*:             Square both x and y
+                Add x^2 and y^2
+                 Plus
*                  Product of x and y
>:                   Less than or equal to
*:                     Square of n
,                       Flatten
+/                         Reduce by addition


# APL (Dyalog Classic), 23 bytes

{1+6×+/-/⌊⍵÷1+3⊥¨⍳⍵2}×⍨


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tribute to xnor's and lynn's answers

the last test is commented because it needs more memory, e.g. MAXWS=200M in the env

# Jelly, 14 bytes

ḤŒR+²_×ʋþF½»ċ


Try it online!

# Jelly,  15  13 bytes

-2 thanks to Dennis (just increment the square to avoid concatenation of a zero; avoid head by using a post-difference modulo-slice rather than a pre-difference slice)

Uses the "black magic" method of honing in on the answer that was exposed by xnor in their Python answer, but uses iteration rather than recursion (and a little less calculation)

²:Ð€‘$Im3S×6C  A monadic link accepting a non-negative integer and returning a positive integer. Try it online! Or see the test-suite. ### How? ²:Ð€‘$Im3S×6C - Main Link: non-negative integer, n     e.g. 7
²             - square                                     49
\$        - last two links as a monad:
‘         -   increment                                50
Ð€          -   map across (implicit range of) right with:
:            -     integer division                       [49,24,16,12,9,8,7,6,5,4,4,4,3,3,3,3,2,2,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0]
I       - incremental differences                    [-25,-8,-4,-3,-1,-1,-1,-1,-1,0,0,-1,0,0,0,-1,0,0,0,0,0,0,0,-1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1]
m3     - every third element                        [-25,-3,-1,0,0,-1,0,0,0,0,0,0,0,0,0,0,-1]
S    - sum (vectorises)                           -31
×6  - multiply by six                           -186
C - complement (1-X)                           187


# JavaScript (ES6), 65 bytes

This is a port of @JungHwanMin's solution.

f=(n,y=x=w=n*2)=>y-~w&&(x*x+x*y+y*y<=n*n)+f(n,y-=--x<-w&&(x=w,1))


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# Original answer (ES7), 70 bytes

Simply walks through the grid and counts the matching points.

f=(n,y=x=n*=2)=>y+n+2&&(x*x*3+(y-x%2)**2<=n*n)+f(n,y-=--x<-n&&(x=n,2))


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• Porting xnor's answer is shorter: 42 bytes (outputs true instead of 1; 46 if we also integer-divide it). And I don't know JavaScript well enough to golf the integer-divisions ~~(a/b), but I'm sure there is a shorter way for those as well.. – Kevin Cruijssen Aug 30 at 12:09

# Java 8, 65 bytes

n->f(n,1)int f(int n,int a){return n*n<a/3?1:n*n/a*6-f(n,a+a%3);}


Port of @xnor's Python 2 answer.

Try it online.

# Pari/GP, 42 bytes

Using the built-in qfrep.

n->1+2*vecsum(Vec(qfrep([2,1;1,2],n^2,1)))


qfrep(q,B,{flag=0}): vector of (half) the number of vectors of norms from 1 to B for the integral and definite quadratic form q. If flag is 1, count vectors of even norm from 1 to 2B.

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# C# (Visual C# Interactive Compiler), 68 bytes

n=>{int g(int x,int y)=>x*x<y/3?1:x*x/y*6-g(x,y+y%3);return g(n,1);}


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Same as everyone else, unfortunately. I know there's probably a better way of writing this, but declaring and calling a lambda at the same time in c# is not exactly something I do, well, ever. Though in my defense, I can't think of a good reason (outside code golf, of course) to do so. Still, if someone knows how you can do this, let me know and/or steal the credit, I guess.

# Wolfram Language (Mathematica), 39 bytes

Length@Solve[x(x+y)+y^2<=#^2,Integers]&


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Using JungHwan Min's coordinate transformation and simply counting the solutions over the integers.

# 05AB1E, 15 bytes

nD>L÷¥ā3%ÏO6*±Ì


Port of @JonathanAllans Jelly answer, which in turn is a derivative from @xnor's 'black magic' formula.

Explanation:

n                # Square the (implicit) input-integer
D>              # Duplicate it, and increase the copy by 1
L             # Create a list in the range [1, input^2+1]
÷            # Integer divide input^2 by each of these integers
¥           # Take the deltas
ā          # Push a list in the range [1, length] without popping the deltas itself
3%        # Modulo each by 3
Ï       # Only leave the values at the truthy (==1) indices
O      # Take the sum of this list
6*    # Multiply it by 6
±   # Take the bitwise NOT (-n-1)
Ì  # And increase it by 2
# (after which the result is output implicitly)
`