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I've been diving into combinatorial problems related to cardinal trees and am trying to ascertain the count of such trees that have a degree k with n nodes. I stumbled upon some leads in "Concrete Mathematics", but couldn't find a definitive proof.

By a cardinal tree (or trie) of degree k, we mean a rooted tree in which each node has k positions for an edge to a child.

The answer is $C^k_n\equiv\binom{kn+1}{n}/(kn+1)$ according to this paper, but I cannot figure out why, even dive into the text book I mentioned.

Any guidance or suggestions would be greatly appreciated!

grey bear
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  • The full solution is hard, but we can quickly prove $C_n^k \le \binom{kn}{n-1}$ from the generalized Jacobson encoding mentioned on p. 282 (p. 8 of the PDF): it is a sequence of $kn$ bits, $n-1$ of which are 1's, and there are $\binom{kn}{n-1}$ ways to choose the positions of the 1's. (But there are further constraints, so this is only an upper bound.) Note that $\frac1{kn+1}\binom{kn+1}{n} = \frac1n \binom{kn}{n-1}$, so this upper bound overcounts exactly by a factor of $n$. – Misha Lavrov Sep 11 '23 at 16:30
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    This has been asked several times on this site. See vonbrand's answer for a proof via generating functions and Lagrange inversion. See Brian's for an a great presentation of a bijective proof from Concrete Mathematics. – Mike Earnest Sep 11 '23 at 16:53
  • ... and see Mike Earnest's new answer for a solution to this problem that follows the Concrete Mathematics approach. – Misha Lavrov Sep 11 '23 at 18:32

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