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Register a new userA sign-reversing involution to count labeled lone-child-avoiding trees
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David Callan
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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 sum_{k=1}^{n} (-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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