A sign-reversing involution to count labeled lone-child-avoiding trees
Abstract
We use a sign-reversing involution to show that trees on the vertex set [n], considered to be rooted at 1, in which no vertex has exactly one child are counted by 1/n sumk=1n (-1)(n-k) n-choose-k (n-1)!/(k-1)! k(k-1). This result corrects a persistent misprint in the Encyclopedia of Integer Sequences.
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