Recent works on joint distributions involving the Poupard and Entringer statistics Guoniu Han IRMA, Strasbourg (Based on some joint papers with Dominique Foata) Shanghai Jiao Tong University June 25, 2018 1
• Enumerative Combinatorics • Generating Function A = ( A 0 , A 1 , A 2 , . . . ) : sequence of sets f : statistic on A • Problem ( A ; f ) : compute x f ( a ) ? � F n ( x ) = a ∈ A n 2
• “Each generating function problem ( A ; f ) produces a math- ematics research paper ?” • No. It depends the hardness of the problem: —— (easy) —— (interesting) —— (hard) —— > Only problems that are neither easy nor hard produce research papers. • It is difficult to find a interesting problem ( A ; f ) , since most generating function problems are easy, or hard. • A trick for finding interesting problem ? Yes. 3
• 2D - Generating Function (and 3D, 4D, ...) A = ( A 0 , A 1 , A 2 , . . . ) : sequence of sets f, g : statistics on A • Problem ( A ; f, g ) : compute x f ( a ) y g ( a ) ? � F n ( x, y ) = a ∈ A n 4
• If two 1D problems ( A , f ) and ( A , g ) are easy, then, the 2D problem ( A ; f, g ) is: (1) easy, (2) interesting, (3) hard? • Answer. Undefined: (1) or (2) or (3). • 2D-Principle . Although the answer is undefined, the prob- ability of “ ( A ; f, g ) is an interesting problem” is much bigger than a random problem ( B ; h ) . 5
Christiane Poupard Deux propri´ et´ es des arbres binaires ordonn´ es stricts, Europ. J. Combin. 1989 (TangentTree, eoc) (TangentTree, pom) R. C. Entringer A combinatorial interpretation of the Euler and Bernoulli num- bers, Nieuw. Arch. Wisk., 1966 (Alt, first) 6
Foata-Han Finite Difference Calculus for Alternating Permutations, Jour- nal of Difference Equations and Applications, 2013 (Alt; grn) • g.f. for n odd sec( x + y ) cos( x − y ) • g.f. for n even sec 2 ( x + y ) cos( x − y ) 7
Tree Calculus for Bivariate Difference Equations, Journal of Difference Equations and Applications, 2014 (TangentTree; eoc, pom) • g.f. for the lower triangle cos(2 x ) + cos(2 y ) cos(2 z ) 2 cos 2 � � x + y + z • g.f. for the upper triangle sin(2 x ) sin(2 z ) 2 cos 2 � � x + y + z 8
Seidel Triangle Sequences and Bi-Entringer Numbers, European Journal of Combinatorics, 2014 (Alt; first, last) • g.f. for n even, upper triangle cos x cos z cos( x + y + z ) • g.f. for n even, lower triangle sin x sin z cos( x + y + z ) • g.f. for n odd, lower and upper triangles sin x cos z cos( x + y + z ) 9
Andr´ e Permutation Calculus; a Twin Seidel Matrix Sequence, Sem. Loth. Comb. 2016, 54 pages (AndreI; first, nexttolast) (AndreII; last, grn) • A. n even, upper triangle cos x cos z sin( x + y + z ) cos 2 ( x + y + z ) • A. n even, lower triangle cos( x + y + z ) + sin x cos( x + y ) cos x sin z cos 2 ( x + y + z ) 10
• A. n odd, upper triangle cos x cos z cos 2 ( x + y + z ) • A. n odd, lower triangle cos( x + y ) cos( y + z ) cos 2 ( x + y + z ) • B. n even, upper triangle sin x cos z cos 2 ( x + y + z ) • B. n even, lower triangle cos( x + y ) sin( y + z ) cos 2 ( x + y + z ) 11
• B. n odd, upper triangle sin x cos z sin( x + y + z ) cos 2 ( x + y + z ) • B. n odd, lower triangle cos( x + y + z ) + cos x cos( x + y ) sin x sin z − cos 2 ( x + y + z ) 12
Entringer-Poupard Matrices, Submitted, 2016 (Alt; last, grn) • upper triangle (sin x + cos x ) sin(2 z ) cos 2 ( x + y + z ) • lower triangle (sin x + cos x ) cos( x + y − z ) cos 2 ( x + y + z ) 13
Secant Tree Calculus, Central European Journal of Mathe- matics, 2014 (SecantTree; eoc, pom) • g.f. for the upper triangle: cos(2 y ) + 2 cos(2( x − z )) − cos(2( z + x )) 2 cos 3 ( x + y + z ) • g.f. for the lower triangle : 14
Secant Tree Calculus, Central European Journal of Mathemat- ics, 2014 (SecantTree; eoc, pom) • g.f. for the upper triangle: cos(2 y ) + 2 cos(2( x − z )) − cos(2( z + x )) 2 cos 3 ( x + y + z ) • g.f. for the lower triangle : hard ? Open problem ! 15
Tangent numbers Taylor expansion of tan u : u 2 n +1 � tan u = (2 n + 1)! T 2 n +1 n ≥ 0 1!1 + u 3 3! 2 + u 5 5! 16 + u 7 7! 272 + u 9 = u 9! 7936 + · · · The coefficients T 2 n +1 ( n ≥ 0) are called the tangent numbers 16
Secant numbers Taylor expansion of sec u : u 2 n 1 � sec u = cos u = (2 n )! E 2 n n ≥ 0 = 1 + u 2 2! 1 + u 4 4! 5 + u 6 6! 61 + u 8 8! 1385 + u 10 10! 50521 + · · · The coefficients E 2 n ( n ≥ 0) are called the secant numbers 17
Alternating permutations D´ esir´ e Andr´ e’s (1879): A permutation σ = σ (1) σ (2) · · · σ ( n ) of 12 · · · n with the property that σ (1) > σ (2) , σ (2) < σ (3) , σ (3) > σ (4) , etc. in an alternating way is called alternating permutation . Let A n denote the set of all alternating permutations of 12 · · · n . Theorem: # A 2 n − 1 = T 2 n − 1 , # A 2 n = E 2 n . 18
Tangent tree 5 9 7 8 ❆ ✁✁ ❆ ✁✁ ❆ ❆ P P 3 P � P 4 6 P ✥� ❅✥✥✥✥✥✥✥ ❅ 2 1 2 n + 1 vertices, complete, binary, rooted, planar, labeled, increasing The set all off tangent trees : T 2 n +1 . # A 2 n +1 = # T 2 n +1 = T 2 n +1 19
Secant tree 5 8 7 ❆ ✁✁ ❆ ❆ ❆ P P 3 P � 4 P 6 P ✥� ❅✥✥✥✥✥✥✥ ❅ 2 1 2 n vertices, complete (execpt that the rightmost vertice is missing), binary, rooted, planar, labeled, increasing The set all off tangent trees : T 2 n . # A 2 n = # T 2 n = E 2 n . 20
Bijection 5 7 5 8 7 ❅ � ❅ � ❅ 3 ❅ � ❅ � ❅ t 1 = t 2 = ❍ ❍ ✟ ❍ ❍ ✘✟✟ � 4 4 3 6 6 ✘� ❍ ❍ ❅✘✘✘✘✘ ❅✘✘✘✘✘ ❅ ❅ 2 2 1 1 σ 1 = 6 1 5 4 7 2 3 σ 2 = 6 1 5 4 8 2 7 3 Tangent, secant trees and alternating permutations 21
Poupard statistic: eoc Poupard (1989) Let 1 = a 1 → a 2 → a 3 → · · · → a j − 1 → a j be the minimal chain of a tree t ∈ T n , the “end of the minimal chain” of t is defined to be eoc( t ) := a j . 5 9 7 8 ❆ ✁✁ ❆ ✁✁ ❆ ❆ P P 3 P � 4 P 6 ✥� P ❅✥✥✥✥✥✥✥ ❅ 2 1 For example, the minimal chain of the tree t is 1 → 2 → 3 → 7 , so that eoc( t ) = 7 22
Poupard statistic: pom If the leaf with the maximum label n is incident to a node labeled k , define its “parent of the maximum leaf” to be pom( t ) := k . 5 9 7 8 ❆ ✁✁ ❆ ✁✁ ❆ ❆ P P 3 P � P 4 6 ✥� P ❅✥✥✥✥✥✥✥ ❅ 2 1 The parent of its maximum leaf (equal to n = 9 ) is pom( t ) = 4 23
5 secant trees with 4 vertices 3 4 4 ◗◗◗◗ ◗◗◗◗ ✟ ✟ ❅ ❅ ✟ ✟ ❅ 4 3 4 ❅ 3 ❅ 3 4 2 ❅ ✟ ✟ ✟ 2 2 ❅✟✟ ❅✟✟ ❅✟✟ ❅ ❅ ❅ 2 2 3 ◗ 1 ◗ 1 1 1 1 4 1 3 2 3 1 4 2 4 2 3 1 3 2 4 1 2 1 4 3 eoc = 3 4 3 3 2 pom = 1 2 2 2 3 24
16 tangent trees with 5 vertices There 16 tangent trees from T 5 . Only 4 of them (reduced trees) are displayed, but each of them gives rise to three other tangent trees, having the same “eoc” and “pom” statistics, by pivoting each pair of subtrees. 4 5 4 5 3 5 3 5 ❅ � ❅ � ❅ � ❅ � 2 ❅ � 3 ❅ � 4 ❅ � 5 ❅ � ✟ ✟ ✟ ✟ ❅✟✟ ❅✟✟ ❅✟✟ ❅✟✟ ❅ ❅ ❅ ❅ 3 2 2 2 1 1 1 1 2 1 4 3 5 3 1 4 2 5 4 1 3 2 5 5 1 3 2 4 eoc = 2 4 3 3 pom = 3 2 2 1 25
Equidistribution Theorem. The statistics “ eoc − 1 ” and “ pom ” are equidistributed on each set T n . The tangent tree case was obtained by Poupard (1989). Her original proof, not of combinatorial nature, makes use of a clever finite difference analysis argument. Our proof: Bijection inspired from the classical “jeu de taquin” on directed acyclic graphs, (Sch¨ utzenberger, 1972) 26
Proof Let 1 = a 1 → a 2 → a 3 → · · · → a j − 1 → a j be the minimal chain of t . (i) for i = 1 , 2 , . . . , j − 1 replace each node label a i of the minimal chain by a i +1 − 1 ; (ii) replace the node label a j by n ; (iii) replace each other node label b by b − 1 . 6 7 9 6 9 8 5 4 8 7 4 5 3 2 2 3 1 1 eoc=6 pom=5 27
Poupard numbers for tagent trees: g n ( k ) g n ( k ) := # { t ∈ T 2 n − 1 :pom( t ) = k } = # { t ∈ T 2 n − 1 :eoc( t ) = k + 1 } k = 1 2 3 4 5 6 7 Sum n = 1 1 1 = T 1 2 0 2 0 2 = T 3 3 0 4 8 4 0 16 = T 5 4 0 32 64 80 64 32 0 272 = T 7 Theorem (Poupard, 1989). x 2 n +1 − k y k − 1 ( k − 1)! = cos( x − y ) � � 1+ g n +1 ( k ) (2 n + 1 − k )! cos( x + y ) n ≥ 1 1 ≤ k ≤ 2 n +1 28
Poupard numbers for secant trees: h n ( k ) h n ( k ) := # { t ∈ T 2 n :pom( t ) = k } = # { t ∈ T 2 n :eoc( t ) = k + 1 } k = 1 2 3 4 5 6 7 Sum n = 1 1 1 = E 2 2 1 3 1 5 = E 4 3 5 15 21 15 5 61 = E 6 4 61 183 285 327 285 183 61 1385 = E 8 Theorem (Foata-H., 2013). x 2 n +1 − k y k − 1 ( k − 1)! = cos( x − y ) � � 1+ h n +1 ( k ) cos 2 ( x + y ) (2 n + 1 − k )! n ≥ 1 1 ≤ k ≤ 2 n +1 29
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