### Recursive construction

Consider the equilateral triangle, $J_0$, in $\mathbb{R}^2$ with vertices $F_0=(a_0,a_1,a_2)$ where

(1)

Inductively, define the vertices

(2)
\begin{align} F_{n+1}=F_n \cup (2^n a_1+F_n) \cup (2^n a_2 +F_n)\ , \end{align}

where by $y+A=(y+x~|~x\in A)$ and $\lambda A=(\lambda x~|~x\in A)$. Thus, it is easy to see that $F_n\subset 2^n J_0$. For example, the first iteration gives the set of points:

(3)
\begin{align} F_1=\Big[(0,0),(1,0),(\tfrac12,\tfrac{\sqrt{3}}2), \mathbf{(1,0)},(2,0),(\tfrac32,\tfrac{\sqrt{3}}2), \mathbf{(\tfrac12,\tfrac{\sqrt{3}}2),(\tfrac32,\tfrac{\sqrt{3}}2)},(1,\sqrt3)\Big] \end{align}

The three points in boldface are duplicates and must be dropped giving us six distinct points in $F_1$. Thus the number of points in $F_n$ is three less than the number of points in $F_{n-1}$.

### The spectrum of the Laplacian for low generations

#### Generation zero

The vertices are ordered as follows:

(4)
\begin{align} F_0=\Big[(0,0),(1,0),(\tfrac12,\tfrac{\sqrt{3}}2)\Big] \end{align}

The (negative of the) discrete Laplacian is

(5)
\begin{align} \Delta_0=\begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{pmatrix}\ . \end{align}

Its eigenvalues are

(6)
\begin{align} \Big(3,3,0\Big)\ . \end{align}

#### Generation one

The vertices are ordered as follows:

(7)
\begin{align} F_1=\Big[(0,0),(1,0),(\tfrac12,\tfrac{\sqrt{3}}2), (2,0),(\tfrac32,\tfrac{\sqrt{3}}2),(1,\sqrt3)\Big]\ , \end{align}

and the negative of the discrete Laplacian is

(8)
\begin{align} \Delta_1=\begin{pmatrix} 2 & -1 & -1 & 0 & 0 & 0 \\ -1 & 4 & -1 & -1 & -1 & 0 \\ -1 & -1 & 4 & 0 & -1 & -1 \\ 0 & -1 & 0 & 2 & -1 & 0 \\ 0 & -1 & -1 & -1 & 4 & -1 \\ 0 & 0 & -1 & 0 & -1 & 2 \end{pmatrix}\ . \end{align}

Its eigenvalues are

(9)
\begin{align} \Big(\tfrac{1}{2} \left(7+\sqrt{13}\right),\tfrac{1}{2} \left(7+\sqrt{13}\right),4,\tfrac{1}{2} \left(7-\sqrt{13}\right),\tfrac{1}{2} \left(7-\sqrt{13}\right),0\Big) \end{align}

#### Generation two

(10)
\begin{align} \Delta_2=\begin{pmatrix} 2 & -1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ -1 & 4 & -1 & -1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ -1 & -1 & 4 & 0 & -1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 4 & -1 & 0 & -1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & -1 & -1 & 4 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & -1 & 4 & 0 & 0 & -1 & -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & 4 & -1 & 0 & 0 & -1 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & -1 & 4 & 0 & 0 & 0 & -1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 4 & -1 & 0 & 0 & -1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & -1 & 4 & 0 & 0 & 0 & -1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 2 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & -1 & 0 & 0 & -1 & 4 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & -1 & 0 & 0 & -1 & 4 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & -1 & 0 & 0 & -1 & 4 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & -1 & 2 \end{pmatrix}\ . \end{align}

Its eigenvalues are

(11)
\begin{align} \Big(6,6,6,\lambda_4,\lambda_4,5,5,\lambda_3,\lambda_3,\lambda_2,\lambda_2,2,\lambda_1,\lambda_1,0\Big)\ , \end{align}

where $\lambda_i$ are solutions to the equation $x^4 -12 x^3+48x^2-71x+27=0$ arranged in increasing order. Numerically solving, the eigenvalues are approximately

(12)
\begin{align} \Big(6, 6, 6, 5.22337, 5.22337, 5, 5, 3.85548, 3.85548, 2.35084, 2.35084, 2, 0.57031, 0.57031, 0\Big) \end{align}

### References

1. Martin T. Barlow and Edwin A. Perkins, Brownian Motion on the Sierpinski Gasket, Probab. Th. Rel. Fields 79 (1988) 543-623.