Concentric rings on a snub square tiling

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.)

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.

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

05AB1E, 9 bytes

<©12*3®cα

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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:

12*3Icα