$n=2$

**Circle:** $x^2+y^2=c^2$. We all know the role of $\pi$ in the circle and the trigonometric functions,

$$\pi_2 = \color{brown}{B\big(\tfrac12,\tfrac12\big)}=3.1415\dots$$

with the *beta function* $B(x,y)$ and circumference for $D=1$

$$C=\pi_2 = 3.1415\dots$$

$n=4$

**Lemniscate:** $(x^2+y^2)^2=2c^2(x^2-y^2)$ (or $r^2=2c^2\cos2\theta$ in polar form):

The lemniscate constant $L$ plays an analogous role to $\pi$,

$$L =\frac{\color{brown}{B\big(\tfrac14,\tfrac14\big)}}{2\sqrt2}= \frac{B\big(\tfrac12,\tfrac14\big)}{2}= \frac{\sqrt\pi\,\Gamma\big(\tfrac14\big)}{2\Gamma\big(\tfrac34\big)}=\frac{\Gamma^2\big(\tfrac14\big)}{2\sqrt{2\pi}}=2.622057\dots$$

and if $c=1$,

$$\text{Arclength} = 2L = 5.2441\dots$$

The minimal half period $\omega_1$ of the *lemniscatic elliptic functions* is,

$$\omega_1=\frac{L}{\sqrt2}=\frac{\Gamma^2\big(\tfrac14\big)}{4\sqrt{\pi}}$$

$n=3$

**Object?:** ($3$rd or $6$th deg poly?) However, it seems we missed a step,

$$\pi_3 = \color{brown}{B\big(\tfrac13,\tfrac13\big)}=2^{1/3}\,B\big(\tfrac12,\tfrac13\big) =\frac{\Gamma^2\big(\tfrac13\big)}{\Gamma\big(\tfrac23\big)}=\frac{\sqrt{3}\,\Gamma^3\big(\tfrac13\big)}{2\pi}=5.29991\dots$$

which conveniently is the fundamental constant of the *Dixonian elliptic functions*.

Q:So does the constant $\pi_3$ play an analogous role for some geometric object akin to the circle and lemniscate?

**Edit:** To clarify (courtesy of reuns' comments): to know the arc length of a circle you need $\pi_2$. For a lemniscate, you need $L$. So is there a simple geometric object where to know its arc length you need $\pi_3$?