Concentric rings on a snub square tiling

Ruby, 26 bytes


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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


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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.)

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JavaScript (ES6), 23 bytes

Based on Level River St's answer.


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

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

05AB1E, 9 bytes


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<           # 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: