patterns Lara Pudwell faculty.valpo.edu/lpudwell

Transcription

1 faculty.valpo.edu/lpudwell joint work with Andrew Baxter Permutation 2014 East Tennessee State University July 7, 2014

2 Ascents Definition An ascent in the string x 1 x n is a position i such that x i < x i+1. Example:

3 Ascents Definition An ascent in the string x 1 x n is a position i such that x i < x i+1. Example: Definition asc(x 1 x n ) is the number of ascents of x 1 x n. Example: asc(01024) = 3

4 Ascent Sequences Definition An ascent sequence is a string x 1 x n of non-negative integers such that: x 1 = 0 x n 1 + asc(x 1 x n 1 ) for n 2 A n is the set of ascent sequences of length n A 2 = {00, 01} More examples: 01234, A 3 = {000, 001, 010, 011, 012} Non-example: 01024

5 Ascent Sequences Definition An ascent sequence is a string x 1 x n of non-negative integers such that: x 1 = 0 x n 1 + asc(x 1 x n 1 ) for n 2 A n is the set of ascent sequences of length n A 2 = {00, 01} More examples: 01234, A 3 = {000, 001, 010, 011, 012} Non-example: (Bousquet-Mélou, Claesson, Dukes, & Kitaev, 2010) A n is the nth Fishburn number (OEIS A022493). A n x n = n (1 (1 x) i ) n 0 n 0 i=1

6 Definition The reduction of x = x 1 x n, red(x), is the string obtained by replacing the ith smallest digits of x with i 1. Example: red(273772) =

7 Definition The reduction of x = x 1 x n, red(x), is the string obtained by replacing the ith smallest digits of x with i 1. Example: red(273772) = Pattern containment/avoidance a = a 1 a n contains σ = σ 1 σ m iff there exist 1 i 1 < i 2 < < i m n such that red(a i1 a i2 a im ) = σ. a B (n) = {a A n a avoids B} contains 012, 000, 1102; avoids 210.

8 Definition The reduction of x = x 1 x n, red(x), is the string obtained by replacing the ith smallest digits of x with i 1. Example: red(273772) = Pattern containment/avoidance a = a 1 a n contains σ = σ 1 σ m iff there exist 1 i 1 < i 2 < < i m n such that red(a i1 a i2 a im ) = σ. a B (n) = {a A n a avoids B} contains 012, 000, 1102; avoids 210. Goal Determine a B (n) for many of choices of B.

9 Previous Work Duncan & Steingrímsson (2011) Pattern σ {a σ (n)} n 1 OEIS 001, , n 1 A , 0112 (3 n 1 + 1)/2 A , n ) 0101 n+1( n A Mansour and Shattuck (2014) Callan, Mansour and Shattuck (2014) Pattern σ {a σ (n)} n 1 OEIS 1012 n 1 ( n 1 ) k=0 k Ck A x+3x 0123 ogf: 2 A x+6x 2 x 3 8 pairs of length 1 2n ) n+1( n A

10 Overview 13 length 3 6 ( permutations, 000, 001, 010, 100, 011, 101, ) 2 = 78 pairs at least 35 different sequences a σ,τ (n) 16 sequences in OEIS 3 sequences from Duncan/Steingrímsson 1 eventually zero 1 from pattern-avoiding set partitions 3 from pattern-avoiding permutations 1 sequence from Mansour/Shattuck (Duncan/Steingrímsson conjecture)

11 a 010,021 (n) = a 010 (n) = a 10 (n) = 2 n 1

12 a 010,021 (n) = a 010 (n) = a 10 (n) = 2 n 1 If σ contains 10, then a 010,σ = 2 n 1.

13 a 010,021 (n) = a 010 (n) = a 10 (n) = 2 n 1 If σ contains 10, then a 010,σ = 2 n 1. a 101,201 (n) = a 101 (n) = C n

14 a 010,021 (n) = a 010 (n) = a 10 (n) = 2 n 1 If σ contains 10, then a 010,σ = 2 n 1. a 101,201 (n) = a 101 (n) = C n 101-avoiders are restricted growth functions. If σ contains 201, then a 101,σ = C n.

15 a 010,021 (n) = a 010 (n) = a 10 (n) = 2 n 1 If σ contains 10, then a 010,σ = 2 n 1. a 101,201 (n) = a 101 (n) = C n 101-avoiders are restricted growth functions. If σ contains 201, then a 101,σ = C n. a 101,210 (n) = 3n

16 a 010,021 (n) = a 010 (n) = a 10 (n) = 2 n 1 If σ contains 10, then a 010,σ = 2 n 1. a 101,201 (n) = a 101 (n) = C n 101-avoiders are restricted growth functions. If σ contains 201, then a 101,σ = C n. a 101,210 (n) = 3n Proof sketch: bijection with ternary strings with even number of 2s (Duncan/Steingrímsson proof that a 102 (n) = 3n uses bijection with same strings.)

17 A n n 2 a 000,012 (n) = 3 n = 3 or n = 4 0 n 5 A 1 (000, 012) = {0} A 2 (000, 012) = {00, 01} A 3 (000, 012) = {001, 010, 011} A 4 (000, 012) = {0011, 0101, 0110}

18 a 0 a,012 b(n) = 0 for n (a 1) ((a 1)(b 2) + 2) + 1 Proof: largest letter preceeded by at most b 1 smaller values at most a 1 copies of each value How to maximize number of ascents: (a 1)(b 2) ascents before largest letter largest possible digit is (a 1)(b 2) + 1 Use all digits in {0,..., (a 1)(b 2) + 1} each a 1 times.

19 a 0 a,012 b(n) = 0 for n (a 1) ((a 1)(b 2) + 2) + 1 Proof: largest letter preceeded by at most b 1 smaller values at most a 1 copies of each value How to maximize number of ascents: (a 1)(b 2) ascents before largest letter largest possible digit is (a 1)(b 2) + 1 Use all digits in {0,..., (a 1)(b 2) + 1} each a 1 times. Maximum avoider example: (a=3, b=5)

20 OEIS Formula 000,011 A n 000,001 A F ( n+1 011,100 A n ) ,100 A F n+2 1 ( 001,210 A n ) 3 + n 000,101 A M n 100,101 A (Generalized Catalan) 021,102 A S n (123, 3241) 102,120 A S n (132, 4312) 101,120 A S n (231, 4123) 101,110 A F 2n 1 201,210 A n 1 k=0 ( n 1 k ) Ck

21 Avoiding 100 and 101 a 100,101 (n) = GC n, the nth generalized Catalan number a 100,101 (n) = a 0100,0101 (n) ascent sequences avoiding a subpattern of are restricted growth functions Mansour & Shattuck (2011): 1211, 1212-avoiding set partitions are counted by GC n used algebraic techniques known: GC n counts DDUU-avoiding New: bijective proof

22 Avoiding 100 and 101 Bijection from DDUU-avoiding to ascent sequences: Heights of left sides of up steps:

23 Avoiding 100 and 101 Bijection from DDUU-avoiding to ascent sequences: Heights of left sides of up steps:

24 Avoiding 100 and 101 Bijection from DDUU-avoiding to ascent sequences: Heights of left sides of up steps:

25 Avoiding 100 and 101 Bijection from DDUU-avoiding to ascent sequences: Heights of left sides of up steps:

26 Avoiding 100 and 101 Bijection from DDUU-avoiding to ascent sequences: Heights of left sides of up steps:

27 Avoiding 100 and 101 Bijection from DDUU-avoiding to ascent sequences: Heights of left sides of up steps:

28 Avoiding 100 and 101 Bijection from DDUU-avoiding to ascent sequences: Heights of left sides of up steps:

29 Avoiding 100 and 101 Bijection from DDUU-avoiding to ascent sequences: Heights of left sides of up steps:

30 A n :

31 A n (10):

32 A n (10): Root: (2) Rule: (2) (2)(2) A 10 (n) = 2 n 1

33 Permutations a 102,120 (n) = S n (132, 4312) Proof: Isomorphic generating tree

34 Permutations a 102,120 (n) = S n (132, 4312) Proof: Isomorphic generating tree a 101,120 (n) = S n (231, 4123) Proof: Isomorphic generating tree

35 Permutations a 102,120 (n) = S n (132, 4312) Proof: Isomorphic generating tree a 101,120 (n) = S n (231, 4123) Proof: Isomorphic generating tree a 021,102 (n) = S n (123, 3241) Proof:... 5 labels. Permutations 8 labels. (Vatter, FINLABEL, 2006) Transfer matrix method gives same enumeration, bijective proof open.

36 Avoiding 201 and 210 a 201,210 (n) = n 1 ( n 1 ) k Ck k=0 Proof scribble: generating tree recurrence system of functional equations experimental solution plug in for catalytic variables

37 Avoiding 201 and 210 a 201,210 (n) = n 1 ( n 1 ) k Ck k=0 Proof scribble: generating tree recurrence system of functional equations experimental solution plug in for catalytic variables Conjecture (Duncan & Steingrímsson) a 0021 (n) = a 1012 (n) = n 1 k=0 ( n 1 ) k Ck Note: Proving this would complete Wilf classification of 4.

38 A familiar sequence... Conjecture (Duncan & Steingrímsson) a 0021 (n) = a 1012 (n) = n 1 k=0 ( n 1 ) k Ck (Mansour & Shattuck) a 1012 (n) = n 1 k=0 ( n 1 ) k Ck

39 A familiar sequence... Conjecture (Duncan & Steingrímsson) a 0021 (n) = a 1012 (n) = n 1 k=0 ( n 1 ) k Ck (Mansour & Shattuck) a 1012 (n) = n 1 k=0 ( n 1 ) k Ck a 0021 (n) = n 1 ( n 1 ) k Ck k=0 Proof: Similar technique to a 201,210 (n).

40 Summary and Future work 16 pairs of 3- appear in OEIS. Erdős-Szekeres analog for ascent sequences. New bijective proof connecting 100,101-avoiders to. Completed Wilf classification of 4-. Open: 19 sequences from pairs of 3- not in OEIS. Bijective explanation that a021,102 (n) = S n (123, 3241).

41 Summary and Future work 16 pairs of 3- appear in OEIS. Erdős-Szekeres analog for ascent sequences. New bijective proof connecting 100,101-avoiders to. Completed Wilf classification of 4-. Open: 19 sequences from pairs of 3- not in OEIS. Bijective explanation that a021,102 (n) = S n (123, 3241). Forthcoming: Enumeration schemes for pattern-avoiding ascent sequences Details on a 201,210 (n) and a 0021 (n) More bijections with other combinatorial objects?

42 References A. Baxter and L. Pudwell,, arxiv: , submitted. M. Bousquet-Mélou, A. Claesson, M. Dukes, S. Kitaev, (2+2)-free posets, ascent sequences, and pattern avoiding permutations, J. Combin. Theory Ser. A 117 (2010), D. Callan, T. Mansour, and M. Shattuck. Restricted ascent sequences and Catalan numbers. arxiv: , March P. Duncan and E. Steingrímsson. Pattern avoidance in ascent sequences. Electronic J. Combin. 18(1) (2011), #P226 (17pp). T. Mansour and M. Shattuck. Restricted partitions and generalized Catalan numbers. Pure Math. Appl. (PU.M.A.) 22 (2011), no. 2, A18 (05A15) T. Mansour and M. Shattuck. Some enumerative results related to ascent sequences. Discrete Mathematics (2014), V. Vatter. Finitely labeled generating trees and restricted permutations, J. Symb. Comput. 41 (2006),

43 References A. Baxter and L. Pudwell,, arxiv: , submitted. M. Bousquet-Mélou, A. Claesson, M. Dukes, S. Kitaev, (2+2)-free posets, ascent sequences, and pattern avoiding permutations, J. Combin. Theory Ser. A 117 (2010), D. Callan, T. Mansour, and M. Shattuck. Restricted ascent sequences and Catalan numbers. arxiv: , March P. Duncan and E. Steingrímsson. Pattern avoidance in ascent sequences. Electronic J. Combin. 18(1) (2011), #P226 (17pp). T. Mansour and M. Shattuck. Restricted partitions and generalized Catalan numbers. Pure Math. Appl. (PU.M.A.) 22 (2011), no. 2, A18 (05A15) T. Mansour and M. Shattuck. Some enumerative results related to ascent sequences. Discrete Mathematics (2014), V. Vatter. Finitely labeled generating trees and restricted permutations, J. Symb. Comput. 41 (2006), Thanks for listening!