Yes, okay, but are there non-computable sets that are not extremely sparse?

The link is broken, could someone who remembers the quote edit the quote itself into the answer?

(I'm not the downvoter btw, my comment was just an answer)

Each $w_j$ is a linear combination of the $v_i$. Concretely if the entries in the $j$'th column of $S$ are $s_{1j}, \ldots s_{nj}$ then $w_j = s_{1j} v_1 + \ldots + s_{nj} v_n$. Taking a linear combination is a well known and well and well named operation on vector sequences; I thinks here you then would call it 'taking $n$ linear combinations of the $n$ vectors in the sequence'

Where does this sequence come from? Why is it interesting?

Maybe a stupid question, but suppose I can access this algorithm as a very fast black box, how would it help me find all subgroups of $SU(3)$, say? I can keep coming up with presentations of finite groups and from some point onwards the black box will only return 'no' in each instance, but when do I know that I can stop? Is there some a priori bound on the number of generators or relations?

What is the meaning of $\beta$?

Ehm, sometimes "odds of $X$" is also used in the meaning of "probability of $X$" rather than as "probability of $X$ divided by probability of not-$X$" as it usually used in statistics. This of course doesn't solve my problem, but it seems to be the interpretation that Iosif uses below. It is interesting that he does find the same sequence albeit for the total number of balls rather than the white balls

I'm probably fooling myself in some stupid way, but isn't the probability of drawing 2 red balls equal to the probability of drawing a red ball times the probability of drawing a red ball given that you already removed a red ball? To me it seems that the scenario with 4 white balls and 3 red balls gives a probability of $(3/7)*(2/6) = 1/7$ and so an odds of $(1/7)/(6/7) = 1/6$. How would you get to odds $1/2$??

w.r.t quesions (i) en (ii) of Iosif above: I think that, given that you already wrote $PR_0$ for the initial condition, it would make sense to replace the $PR$ in the LHS of the formula with $PR_{i+1}$ and the $PR$ in the RH with $PR_i$ for clarity. The question then asks for the relationship between $\lim_{i \to \infty} PR_i$ and the eigenvectors

Just checking: the curly U and V are the same as the ordinary U and V?

Right, they don't need to respect the grading on the ring. That's what I missed. Thanks!

Probably a stupid question, but why isn't it just GL(n, k)?

The first of these statements, by Fourier, seems to be exactly what the OP is looking for!

Thanks for asking this! I remember struggling with the same issue some 15 years ago. Back then someone explained it to me IRL but I forgot what the answer was. It is good to have it written somewhere on the internet in a findable place!