Concentric rings on a snub square tiling

Question

This challenge takes place on the snub square tiling.

Start by choosing any triangle, and color it \$c_1\$. Next, find all tiles which touch this triangle at any vertex, and color them \$c_2\$. Next, find all tiles which share a vertex with any \$c_2\$-colored tile, and color these \$c_3\$. Continue this process ad infinitum.

Illustration

Initial terms

The sequence begins

  a(1) = 1
  a(2) = 9
  a(3) = 21
  a(4) = 35

Notice:

• a(1) = 1 corresponds to the red triangle;
• a(2) = 9 corresponds to the number of tiles in the second, orange layer;
• a(3) = 21 corresponds to the number of tiles in the third, green layer; and so on.

(Note, this sequence is now in the OEIS; OEIS sequence A296368 is closely related.)

Challenge

Your goal is to write a program that takes in a positive integer n and returns the number of tiles that are colored \$c_n\$, (i.e. the number of tiles in the \$n\$-th layer.) This is a code-golf challenge, so the shortest code in bytes wins.

Answer 1

Ruby, 26 bytes

->n{~-n*12-496/4**n%4+1/n}

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Revised version adding 1/n and subtracting 496/4**n%4 to get the +1,-3,-3,-1 offset for the first 4 terms.

Ruby, 32 bytes

->n{n>4?~-n*12:[0,1,9,21,35][n]}

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After 4, the sequence settles down to (n-1)*12. See diagram below (the equilateral triangles have been distorted into 45 degree isosceles triangles and the entire diagram rotated 45 degrees, but it remains topologically equivalent.)

Answer 2

JavaScript (ES6), 23 bytes

Based on Level River St's answer.

n=>[1,5,13,7][--n]^n*12

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How?

We compute \$(n-1)\times12\$ and adjust the first 4 values with a XOR.

$$\begin{array}{c|c} n&1&2&3&4&5&6&7&8&9&10\\ \hline (n-1)\times12&0&12&24&36&48&60&72&84&96&108\\ \hline \text{XOR}&1&5&13&7&\color{grey}0&\color{grey}0&\color{grey}0&\color{grey}0&\color{grey}0&\color{grey}0\\ \hline a(n)&1&9&21&35&48&60&72&84&96&108 \end{array}$$

Answer 3

05AB1E, 9 bytes

<©12*3®cα

Try it online! or try a test suite.

<           # input - 1
 ©          # save to register
  12*       # multiply by 12
      ®     # push the register
     3 c    # binomial coefficient(3, input - 1)
        α   # absolute difference

With 0-indexing, this would be 7 bytes:

12*3Icα

Answer 4

Jelly, 8 bytes

’3cạ×ʋ12

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How?

’3cạ×ʋ12 - Link: integer, n
’        - decrement              (n-1)                1   2   3   4   5   6   7  ...
      12 - twelve                 12                  12  12  12  12  12  12 12  ...
     ʋ   - dyad:
 3       -   three                3                    3   3   3   3   3   3   3  ...
  c      -   choose               3C(n-1)              1   3   3   1   0   0   0  ...
    ×    -   multiply             (n-1)*12             0  12  24  36  48  60  72  ...
   ạ     -   absolute difference  |3C(n-1)-(n-1)*12|   1   9  21  35  48  60  72  ...

Answer 5

Python, 31 bytes

lambda n:n*12-11-(n>4or 5%-n%5)

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Answer 6

APL (Dyalog Unicode), 14 bytes^SBCS

12(|×-3!⍨⊢)-∘1

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Direct translation of Jonathan Allan's Jelly answer. Even the code structure is the same. Jelly is a golfy descendant of APL; if you want to learn Jelly, learn APL first!

How it works

12(|×-3!⍨⊢)-∘1  ⍝ Monadic train, input: n
12(       )-∘1  ⍝ Pass on to the inner function with left←12 and right←n-1
    ×             ⍝ left × right
     -            ⍝ minus
      3!⍨⊢        ⍝ binomial(3, right)
   |              ⍝ absolute value of the above

Answer 7

Jelly, 15 9 bytes

’3cạ×12$Ʋ

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A monadic link taking \$n\$ as its argument and returning \$a(n)\$.

Based on @LevelRiverSt’s clever Ruby answer so be sure to upvote that one too!

Thanks to @Grimmy for saving 6 bytes!

Explanation

 ’       | Subtract 1
       Ʋ | Following as a monad
3c       | - Number of ways of picking (n-1) items from 3
  ạ   $  | - Absolute difference from:
   ×12   |   - Multiply (n-1) × 12

Answer 8

brainfuck, 50 bytes

>>>>>>+<++<<----<+<,-[>++++++++++++[[>+<-]<]>>-]>.

Does i/o as raw byte values, like the others here.

Answer 9

Perl 6, 26 bytes

{$_*12-[-1,3,3,1][$_]}o*-1

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If this could be zero-indexed the o*-1 at the end can be removed. Returns (n-1)*12, offsetting the first 4 values.

Answer 10

Python 3, 40 36 bytes

lambda n:~-n*12-(*n*[0],1,3,3,-1)[4]

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Answer 11

Python 3, 39 bytes

lambda n:n>4and~-n*12or[1,9,21,35][n-1]

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Total rip of Level River St's answer so upvote him.

Answer 12

Python, 38 bytes

lambda n:n*12-11-([1]*n+[2,4,4,0])[-n]

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Answer 13

J, 14 bytes

12(*|@-3!~])<:

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A J port of Bubbler's APL answer, so it's also a port of Jonathan Allan's Jelly answer.

Answer 14

C (gcc), ^{34 33} 32 bytes

Saved a byte thanks to Grimmy!!!

f(n){n=--n<4?"!)5C"[n]-32:n*12;}

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Answer 15

brainfuck, 78 bytes

++++[->+++<]>>>+>--->--->-<<<<<<,-[->>>[+]<[->+<]<[->+>+<<]<[->+<]>]>>[->+<]>.

You can try it online over at TIO (the input is a line-feed (ascii 10) and the output is an l (ascii 108))

You can also try the verbose code at this online interpreter where input can be inserted as decimals, e.g. \6 gives 6 as input. After running, you can hit the "view memory" button and check the value of the output cell in bold, to ensure the result is right.

Answer 16

brainfuck, 61 bytes

,-[>++<-[>+++<-[>+++>++<<-[>+++>+<<-[>+++<-]]]]]>[>++++<-]>+.

Input/output as character codes (meta).

Try it online! or try it with decimal I/O.

Answer 17

Stax, 10 bytes

Ų@8Ω@╩vä3

Run and debug it

Answer 18

Rust - 52 bytes

let m=|n|(match n{1=>1,2=>9,3=>21,4=>35,_=>n}-1)*12;

oh gosh.. it loses even to Brainf***