1. Permutations, sequences, and partially ordered sets Sergey Kitaev Reykjavik University Joint work with Mireille Bousquet-MélouAnders Claesson Mark Dukes

2. Overview of results Bijections (respecting several statistics) between the following objects unlabeled (2+2)-free posets on n elements pattern-avoiding permutations of length n ascent sequences of length n linearized chord diagrams with n chords = certain involutions Closed form for the generating function for these classes of objects Pudwell’s conjecture (on permutations avoiding 31524) is settled using modified ascent sequences __

3. Ascent sequences Number of ascents in a word: asc(0, 0, 2, 1, 1, 0, 3, 1, 2, 3) = 4 (0,0,2,1,1,0,3,1,2,3) is not an ascent sequence, whereas (0,0,1,0,1,3,0) is.

4. Unlabeled (2+2)-free posets A partially ordered set is called (2+2)-free if it contains no induced sub-posets isomorphic to (2+2) = Such posets arise as interval orders (Fishburn): P. C. Fishburn, Intransitive indifference with unequal indifference intervals, J. Math. Psych. 7 (1970) 144–149. bad guy good guy

5. Unlabeled (2+2)-free posets Theorem. (Not ours!) A poset is (2+2)-free iff the collection of strict down-sets may be linearly ordered by inclusion.

6. Unlabeled (2+2)-free posets How can one decompose a (2+2)-free poset?

7. Unlabeled (2+2)-free posets 2

8. 113 101 Read labels backwards: (0, 1, 0, 1, 3, 1, 1, 2) – an ascent sequence! Removing last point gives one extra 0.

9. Theorem. There is a 1-1 correspondence between unlabeled (2+2)-free posets on n elements and ascent sequences of length n. (0, 1, 0, 1, 3, 1, 1, 2)

10. Permutations avoiding 31524 avoids 32541 contains How can one decompose such permutations?

11. Permutations avoiding 61832547 avoids. Remove the largest element, 8: 61 32 54 7 24103 8 corresponds to the position labeled 1. 61 32 54 2103 7 corresponds to the position labeled 3. Etc. Read obtained labels starting from the recent one, to get (0, 1, 1, 2, 2, 0, 3, 1) – an ascent sequence! Remove 7:

12. Theorem. There is a 1-1 correspondence between permutations avoiding on n elements and ascent sequences of length n. (0, 1, 1, 2, 2, 0, 3, 1)6 1 8 3 2 5 4 7

13. Restricted permutations versus (2+2)-free posets

14. Modified ascent sequences

16. Some statistics preserved under the bijections (0, 1, 0, 1, 3, 1, 1, 2) 3 1 7 6 4 8 2 5 min zerosleftmost decr. run min max level last element 012 label in fron of max element (0, 3, 0, 1, 4, 1, 1, 2) Level distri- bution letter distribution in modif. sequence element distribution between act. sites

17. Some statistics preserved under the bijections (0, 1, 0, 1, 3, 1, 1, 2) 3 1 7 6 4 8 2 5 highest level number of ascents in the inverse 2 7 1 5 8 4 3 6 (0, 3, 0, 1, 4, 1, 1, 2) right-to-left max in mod. sequence max compo- nents Components in modif. sequence components (0, 3, 0, 1, 4, 1, 1, 2)

19. RLCD versus unlabeled (2+2)-free posets

21. Sketch of a proof for surjectivity (2+2)-free poset interval order

22. Sketch of a proof for surjectivity

25. Posets avoiding and Ascent sequences are restricted as follows: m-1, where m is the max element here Catalan many

26. An open problem Find a bijection between (2+2)-free posets and permutations avoiding using a graphical way, like one suggested below.

27. Thank you for your attention! Any questions?