2019
DOI: 10.48550/arxiv.1909.06518
|View full text |Cite
Preprint
|
Sign up to set email alerts
|

Counting Labeled Threshold Graphs with Eulerian Numbers

Abstract: A threshold graph is any graph which can be constructed from the empty graph by repeatedly adding a new vertex that is either adjacent to every vertex or to no vertices. The Eulerian number n k counts the number of permutations of size n with exactly k ascents. Implicitly Beissinger and Peled proved that the number of labelled threshold graphs on n ≥ 2 vertices isTheir proof used generating functions. We give a direct combinatorial proof of this result.

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
3

Citation Types

0
3
0

Year Published

2019
2019
2019
2019

Publication Types

Select...
1

Relationship

1
0

Authors

Journals

citations
Cited by 1 publication
(3 citation statements)
references
References 5 publications
0
3
0
Order By: Relevance
“…Each term contributes 2 asc(π)−1 , so we conclude the result by Lemma 3.4 after noting that n−1 −1 = 0. Corollary 1.6 shows that, for n ≥ 2, IPF 2 (n) is equal to the number of connected threshold graphs on n vertices [3]. This can be proven bijectively from essentially the same proof as in [3], but for brevity we omit the details.…”
Section: Interval Resultsmentioning
confidence: 84%
See 2 more Smart Citations
“…Each term contributes 2 asc(π)−1 , so we conclude the result by Lemma 3.4 after noting that n−1 −1 = 0. Corollary 1.6 shows that, for n ≥ 2, IPF 2 (n) is equal to the number of connected threshold graphs on n vertices [3]. This can be proven bijectively from essentially the same proof as in [3], but for brevity we omit the details.…”
Section: Interval Resultsmentioning
confidence: 84%
“…Corollary 1.6 shows that, for n ≥ 2, IPF 2 (n) is equal to the number of connected threshold graphs on n vertices [3]. This can be proven bijectively from essentially the same proof as in [3], but for brevity we omit the details. The formulas for k = 3 and k = n − 2 seem complicated (though the latter can be put into a closed form), and as of this writing neither sequence appears in the OEIS.…”
Section: Interval Resultsmentioning
confidence: 84%
See 1 more Smart Citation