Is there a function that draws a triangle with rounded edges? Gabriel Lamé formula allows to convert a circle to a rounded rectangle and finally to a rectangle:
$$
x^n + y^n=1
$$
Is there a formula from which to connect the triangle to a rounded triangle to a circle?
I found the Reuleaux triangle, but it gets rounder by inflating the sides instead of rounding the edges.
 A: We can regard $|x|^n+|y|^n$ as the sum of four "barrier functions" $|x|^n$, $|y|^n$, $\lvert-x|^n$, $\lvert-y|^n$ at $90^\circ$ to each other. (Well, it's true up to a multiplicative factor.) To obtain a triangle, we can try constructing three barrier functions at $120^\circ$.
Define $g(t)=\bigl| t-\frac13\bigr|^n$ and $\theta_i=2\pi\frac i3$, and let
$$\begin{align}
f(x,y) &= \sum_{i=1}^3g(x\cos\theta_i+y\sin\theta_i) \\
&= \sum_{i=1}^3\left|x\cos\frac{2\pi i}3+y\sin\frac{2\pi i}3-\frac13\right|^n.
\end{align}$$
Here are plots of $f(x,y)=1$ for different $n$.

The $-\frac13$ is necessary in the definition of $g$ because otherwise you obtain a hexagon.
A: Let's consider your problem from the point of view of polar coordinates.
An equilateral triangle may be expressed as 
$$\tag{1}\rho(\theta)=\frac {-1}{2\;\cos\,g(\theta)}$$
with $g$ defined by 
$\quad g(\theta)=\begin{cases}
\theta-\frac{2\pi}3,&-\frac{2\pi}3<\theta<0\\
\theta+\frac{2\pi}3,&0\le\theta<\frac{2\pi}3\\
\theta,&\frac{2\pi}3\le\theta\le\frac{4\pi}3\\
\end{cases}$
getting this picture :  
Polar coordinates allow an easy transition to the unit circle (as $n\to\infty$) for example using 
$$\tag{2}\rho_n(\theta)=\left(\frac {-1}{2\;\cos\,g(\theta)}\right)^{1/n}$$
Result for $n=3$ :

To get a triangle rounded at the corners a little more work on $g(\theta)$ produced :
(note that I replaced $1/n$ in $(2)$ by $1/n^p$ with the parameter $p=2$ here)
$$\tag{3}\rho_n(\theta)=\left(\frac {-1}{2\;\cos\,g_n(\theta)}\right)^{1/n^2}$$
with : $\quad g_n(\theta)=\begin{cases}
a\left(h_n\left(\frac{\theta}a\right)+\operatorname{sgn}(\theta)\right),&-a<\theta<a\\
a\left(h_n\left(\frac{\theta}a-1\right)+1\right),&a\le\theta\le 2a\\
\end{cases}$ 
where $a:=\dfrac{2\pi}3\;$ and $\;\displaystyle h_n(x):=\operatorname{sgn}(x)\,\frac{1+\operatorname{sgn}(2\,|x|-1)\,\left[1-(1-\left|2\,|x|-1\right|)^n\right]}2$
($\operatorname{sgn}$ is the "sign function" and $|x|$ the absolute value)
Result for $n=1.7$ 
Hoping this helped,
