The University of Waterloo CEMC's Problem of the Month for November, the problem statement for which can be seen here, was authored by me. (As far as I know, it is original, though I have come to expect that such low hanging fruits have already been plucked. Forgive me if someone authored a similar problem earlier.)
In essence, I was inspired by the fact that I had $n$ pairs of matching socks in a laundry bag and I was pulling socks out one at a time and hoping that, at some point, I would obtain a sock that matched a previously drawn sock (this was before I had degenerated to the point that I did not care about my socks matching). The main part of the problem asks for the pulled sock number $k+1$ at which the probability is highest that the earliest match occurs. The official solution can be read here and it interestingly has to do with triangular numbers.
Given $2n$ total socks (so $n$ pairs of matching socks), the probability of the first match occurring upon removing the $(k+1)^{\text{th}}$ sock for $1\le k\le n$ is $$P(n,k) = \frac{k2^k n! (2n-k-1)!}{(2n)!(n-k)!}.$$ For those wondering, this probability is the same regardless of whether the socks within a single pair are distinguishable or indistinguishable, though in reality, they are of course indistinguishable.
This means the expected sock number at which the first match occurs is $$E(n)=\sum_{k=1}^{n}{k\cdot P(n,k)} = \sum_{k=1}^{n}{\frac{k^2 2^k n! (2n-k-1)!}{(2n)!(n-k)!}}.$$ According to Wolfram Alpha, the first $9$ values of this function are: $$1,\frac{5}{3},\frac{11}{5},\frac{93}{35},\frac{193}{63},\frac{793}{231},\frac{1619}{429},\frac{26333}{6435},\frac{53381}{12155}.$$ Does it seem to grow like a square root? I'm not sure.
My question is: Can this expected value function be written in a closed form, or be approximated by bounds or by an asymptotic formula?
First of all, your "expected sock number" is $1$ off, because $P(n,k)$ is the probability for the $(k+1)^{\text{th}}$ sock. I am making this correction because the formula when you add $1$ is nicer. So I will give a formula for the sequence $$2, \frac83, \frac{16}{5}, \frac{128}{35}, \dots$$ which is shifted by $1$ from your sequence. (Just looking at this might lead you to some guesses about what the formula is!)
It is often easier to compute the expected value of a (nonnegative integer-valued) random variable $\mathbf X$ by the formula $$ \mathbb E[\mathbf X] = \sum_{k \ge 0} \Pr[\mathbf X > k]. $$ (You can check that $\Pr[\mathbf X = k]$ is counted exactly $k$ times in this sum.) In this case, $\Pr[\mathbf X > k]$ is the probability that the first $k$ socks are all different. This has a nicer formula: $$\Pr[\mathbf X > k] = \frac{\binom{n}{k} 2^k}{\binom{2n}{k}} = \frac{n!(2n-k)!2^k}{(2n)!(n-k)!} = \frac{\binom{2n-k}{k}2^k}{\binom{2n}{n}}.$$ So we now want to evaluate the sum $$ \mathbb E[\mathbf X] = \frac{1}{\binom{2n}{n}} \sum_{k=0}^n \binom{2n-k}{n}2^{k}. $$ This sum has a combinatorial interpretation: $\binom{2n-k}{n}2^{k}$ is A276418, the number of paths of length $2n$ that start at $0$, go up or down at each step, and return to $0$ exactly $k$ times. (I don't know if there's an easy argument that counts these.) Summing these over all $k$ gives $4^n$: the number of paths of length $2n$ that start at $0$ and go up or down at each step.
Therefore $\mathbb E[\mathbf X] = 4^n/\binom{2n}{n}$. Asymptotically as $n \to \infty$, $\mathbb E[\mathbf X] \sim \sqrt{\pi n}$.