Using standard assumptions (birthdays i.i.d. uniformly across $d$ days), then
- the probability that $k$ out of $n$ other people share your birthday is binomial:
$\mathbb P(K=k)={n \choose k}\frac{(d-1)^k}{d^n}$
- the probability at least $k$ share your birthday is $\mathbb P(K\ge k)=\frac{\sum_{i=k}^{n}{n \choose i}(d-1)^i}{d^n} = 1-\frac{\sum_{j=0}^{k-1}{n \choose j}(d-1)^{n-j}}{d^n}$
- which in the case $k=1$ gives $\mathbb P(K\ge 1) = 1-\frac{(d-1)^{n}}{d^n}$, as you already have
The expected number who share your birthday is $\mathbb E[K]=\frac{n}{d}$
If instead $k$ is fixed and you increase $n$ until $k$ matches occur, you have a version of the negative binomial distribution, the sum of $k$ geometric distributions each with expected value $d$, so $\mathbb E[N]=kd$
The Wikipedia article's same birthday as you variant in fact looks at the question of the minimum $n$ where $\mathbb P(K\ge 1)\ge \frac12$, and this would be $n = \Big\lceil\frac{\log(2)}{ \log(d)-\log(d-1)}\Big\rceil$. There is not such a simple calculation for the more general $\mathbb P(K\ge k)\ge \frac12$, but the number rises by close to $d$ each time $k$ increases by $1$. For example with $d=365$, you would get
k min(n: P(K>=k)>=1/2) difference
1 253
2 613 360
3 976 363
4 1340 364
5 1705 365
6 2070 365
7 2435 365
8 2799 364
9 3164 365
10 3529 365
