A row of $n_1$ balls, all coloured with colour #1, is laid out before me. I do a set of $k$ replacements (where $k<n_1$) in steps:
Step 1: Choose a (uniform) random number $n_2$ on the range $[1,n_1]$. Repaint some $n_2$ balls with colour #2. There are now $n_1-n_2$ balls of colour #1, and $n_2$ of colour #2.
Step $i$ (for $2\leq i\leq k$): Now there are $n'_{j}$ balls of colour #$j$, for each $j:1\leq j\leq i$. Some of the $n'_{j}$ could be possibly zero.
I pick a ball at random, and note its colour $j$ and count the number of like-coloured balls to be $n'_j$. Then I generate a (uniform) random number $n_{i+1}$ in the range $[1,n'_j]$. I repaint some $n_{i+1}$ balls of my chosen colour #$j$ with a new colour, # $(i+1)$.
Consequently, the set of $n'_j$ balls of colour # $j$ has split into $n'_j - n_{i+1}$ balls of colour # $j$ and $n_{i+1}$ of the new colour # $(i+1)$. The total number of balls is still $n_1$, and balls with colours other than #$j$ were left unaffected.
I'd like some references for answering the question:
What is the distribution of the number of balls of each colour, after Step $k$?
My question seems to be a variant of Polya's Urn scheme, with a random number of replacements per step. At first glance, it seems to be different from the ones in Mahmoud.
If you use a new colour each time, and you are not worried about the names of the colours but just the partition into colours, you have a Markov chain on the partitions of $n_1$ with an absorbing state at $1+1+\cdots+1$.
If you write the partitions lexicographically like the following for $n_1=5$, then each step takes you (weakly) upwards
The analysis gets easier if you simply count how many colours you have at each step: either you stay with the same partition or you increase the number of colours by 1. If you have $c$ colours at any particular point then you have probability $\frac{c}{n_1}$ of staying with the same partition (swapping an existing colour completely for a new colour) and probability $\frac{n_1-c}{n_1}$ of adding a new colour without extinguishing an old colour.
You are now in the realm of coupon collectors and occupancy problems. For example, the expected number of steps to reach $n_1$ different colours is $\displaystyle n_1 \sum_{j=1}^{n_1} \frac{1}{j} -1$ (subtracting $1$ since you start at the the $0$th step with $1$ colour though more traditionally this happens after the first move).
The probability after $k$ steps of having $c$ colours is $\dfrac{S_2(k+1,c)\, n_1!}{ n_1^{k+1} (n_1-c)!}$ where $S_2(,)$ gives Stirling numbers of the second kind.