Who can solve this?

hold on let me verify this

Thanks for having an actually interesting thread, OP

Hmm some corrections:

[eqn]2/3+\sum_{j=2}^{s} (j(1-\frac{a_j}{3*2^j}) \prod_{k=1}^{j-1}\frac{a_k}{3*2^k})[/eqn]
[eqn]a_n = 2 a_{n-1}-(-1)^{n-1}[/eqn]
[eqn]a_0 = 2[/eqn]

Fucking hell, last line should be

[eqn]a_1 = 2[/eqn]

I don't think you know what you're doing.

Here's what I have so far:

Let [math]d_t[/math] denote the distance from the origin after [math]t[/math] steps. Then [math]d_0 = 1[/math] and [math]d_{t+1} = \lvert d_t + \exp i \theta \rvert = \sqrt{d_t^2 + 2 d_t \cos \theta + 1}[/math] where [math]\theta \sim \mathcal{U}(0, 2\pi)[/math] and thus [math]\cos \theta[/math] has an arcsine distribution.

I meant [math]\theta_t[/math] since the angle depends on [math]t[/math].

ehh if I run this with s = 10, I get 46.28

Circle has no edges.

I must be writing it wrong, because on s = 8 I have 1.5534