We consider a growing tree. Let $f(i,N)$ be the expected number of nodes of degree $i$ after $N$ nodes were attached to the tree.
Assume that nodes are degree-averse, i.e., they attach to nodes with probability inversely proportional to degrees.
Then,
\begin{align} f(1,N+1)&=\frac{f(1,N)}{\sum f(j,N)/j } f(1,N) + \left(1-\frac{f(1,N)}{\sum f(j,N)/j } \right)(f(1,N)+1) \\ f(i,N+1)&=\frac{f(i,N)/i}{\sum f(j,N)/j} (f(i,N)-1) +\frac{f(i-1,N)/(i-1)}{\sum f(j,N)/j} (f(i,N)+1) +\left(1-\frac{f(i,N)/i}{\sum f(j,N)/j} - \frac{f(i-1,N)/(i-1)}{\sum f(j,N)/j} \right)f(i,N) \\ \end{align} i.e., \begin{align} f(1,N+1)&=\left(1-\frac{1}{\sum f(j,N)/j } \right)f(1,N)+1 \\ f(i,N+1)&=\left(1-\frac{1/i}{\sum f(j,N)/j } \right) {f(i,N)} +\frac{f(i-1,N)/(i-1)}{\sum f(j,N)/j} \\ \end{align}
We also have \begin{align} f(1,2)&=2 \\ f(i,N)&=0, \textrm{ for } N\leq i \end{align}
Letting $g(i,N)=f(i,N)/i$,
\begin{align} g(1,N+1)&=\left(1-\frac{1}{\sum g(j,N) } \right)g(1,N)+1 \\ g(i,N+1)&=\left(1-\frac{1/i}{\sum g(j,N) } \right) { g(i,N)} +\frac{g(i-1,N)}{i \sum g(j,N)} \\ \end{align}
\begin{align} g(1,2)&=2 \\ g(i,N)&=0, \textrm{ for } N\leq i \end{align}
How to solve the above recursion, for $i=2, 3, \ldots, N$?
PS: this question is inspired by Distribution of nodes of a tree by degree Na, Hwa Sung; Rapoport, Anatol 1970 https://deepblue.lib.umich.edu/handle/2027.42/32846
Also, it seems that the average node quality, $(1/N)\sum f(j,N)/j = \sum g(j,N)/N$, converges to an asymptotic value. Which value is this? Here I'm assuming that the quality of a node is inversely proportional to its degree.
