Euler s partition theorem and the combinatorics of l-sequences

Transcription

1 Euler s partition theorem and the combinatorics of l-sequences Carla D. Savage North Carolina State University June 26, 2008

2 Inspiration BME1 Mireille Bousquet-Mélou, and Kimmo Eriksson, Lecture hall partitions, Ramanujan. J., 1:1, 1997, BME2 Mireille Bousquet-Mélou, and Kimmo Eriksson, Lecture hall partitions II, Ramanujan. J., 1:2, 1997, Reference SY07 C. D. S. and Ae Ja Yee, Euler s partition theorem and the combinatorics of l-sequences, JCTA, 2007, to appear, available online. LS08 Nicholas Loehr, and C. D. S. Notes on l-nomials, in preparation.

3 Overview Euler s partition theorem

4 Overview 1, 2, 3,... l-sequences Euler s partition theorem

5 Overview 1, 2, 3,... l-sequences Euler s partition theorem The l-euler theorem

6 Overview 1, 2, 3,... l-sequences Euler s partition theorem The l-euler theorem Lecture hall partitions l-lecture hall partitions

7 Overview 1, 2, 3,... l-sequences Euler s partition theorem The l-euler theorem Lecture hall partitions l-lecture hall partitions Binomial coefficients l-nomial coefficients

8 Euler s Partition Theorem: The number of partitions of an integer N into odd parts is equal to the number of partitions of N into distinct parts. Example: N = 8 Odd parts: (7,1) (5,3) (5,1,1,1) (3,3,1,1) (3,1,1,1,1,1) (1,1,1,1,1,1,1,1) Distinct Parts: (8) (7,1) (6,2) (5,3) (5,2,1) (4,3,1)

9 Sylvester s Bijection

10 Sylvester s Bijection

11 l-sequences For integer l 2, define the sequence {a n (l) } n 1 by a (l) n = la (l) n 1 a(l) n 2, with initial conditions a (l) 1 = 1, a (l) 2 = l.

12 l-sequences For integer l 2, define the sequence {a n (l) } n 1 by a (l) n = la (l) n 1 a(l) n 2, with initial conditions a (l) 1 = 1, a (l) 2 = l. {a (3) n } = 1, 3, 8, 21, 55, 144, 377,... {a (2) n } = 1, 2, 3, 4, 5, 6, 7,...

13 l-sequences For integer l 2, define the sequence {a n (l) } n 1 by a (l) n = la (l) n 1 a(l) n 2, with initial conditions a (l) 1 = 1, a (l) 2 = l. {a (3) n } = 1, 3, 8, 21, 55, 144, 377,... {a (2) n } = 1, 2, 3, 4, 5, 6, 7,... (These are the (k, l) sequences in [BME2] with k = l = l.)

14 l 2 The l-euler theorem [BME2]: The number of partitions of an integer N into parts from the set {a (l) 0 + a (l) 1, a(l) 1 + a (l) 2, a(l) 2 + a (l) 3,...} is the same as the number of partitions of N in which the ratio of consecutive parts is greater than Proof: via lecture hall partitions. c l = l + l 2 4 2

15 l = 2 The l-euler theorem [BME2]: The number of partitions of an integer N into parts from the set {0 + 1, 1 + 2, 2 + 3,...} = {1, 3, 5,...} is the same as the number of partitions of N in which the ratio of consecutive parts is greater than c 2 = = 1

16 l = 3 The l-euler theorem [BME2]: The number of partitions of an integer N into parts from the set {0 + 1, 1 + 3, 3 + 8,...} = {1, 4, 11, 29,...} is the same as the number of partitions of N in which the ratio of consecutive parts is greater than c 3 = = (3 + 5)/2

17 l = 3 The l-euler theorem [BME2]: The number of partitions of an integer N into parts from the set {0 + 1, 1 + 3, 3 + 8,...} = {1, 4, 11, 29,...} is the same as the number of partitions of N in which the ratio of consecutive parts is greater than c 3 = = (3 + 5)/2 Stanley: bijection?

18 Θ (l) : Bijection for the l-euler Theorem [SY07] Given a partition µ into parts in {a 0 + a 1, a 1 + a 2, a 2 + a 3,...} construct λ = (λ 1, λ 2,...) by inserting the parts of µ in nonincreasing order as follows: To insert a k 1 + a k into (λ 1, λ 2,...): If k = 1, then add a 1 to λ 1 ; otherwise, if (λ 1 + a k a k 1 ) > c l (λ 2 + a k 1 a k 2 ), add a k a k 1 to λ 1, add a k 1 a k 2 to λ 2 ; recursively insert a k 2 + a k 1 into (λ 3, λ 4,...) otherwise, add a k to λ 1, and add a k 1 to λ 2.

19 The insertion step To insert a k + a k 1 into (λ 1, λ 2, λ 3, λ 4,...) either do (i) (λ 1 + a k, λ 2 + a k 1, λ 3, λ 4,...) or (ii) (λ 1 + (a k a k 1 ), λ 2 + (a k 1 a k 2 ), insert (a k 1 + a k 2 )into(λ 3, λ 4,...)) How to decide? Do (ii) if the ratio of first two parts is okay, otherwise do (i).

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21 Interpretation of a (l) n Define T n (l) : set of l-ary strings of length n that do not contain (l 1)(l 2) (l 1) T (3) 1 = {0, 1, 2} T (3) 1 = 3 T (3) 2 = {00, 01, 02, 11, 10, 12, 20, 21} T (3) 2 = 8 T (3) 3 : all 27 except 220, 221, 222, 022, 122, and 212. Theorem T (l) n = a(l) n+1. T (3) 3 = 27 6 = 21

22 Binary numeration system = 22 Ternary numeration system (unique representation) = 138 (unique representation)

23 Binary numeration system = 22 Ternary numeration system (unique representation) = 138 A Fraenkel numeration system: ternary, but... (unique representation) = 75

24 Binary numeration system = 22 Ternary numeration system (unique representation) = 138 A Fraenkel numeration system: ternary, but... (unique representation) = = = 21

25 Binary numeration system = 22 Ternary numeration system (unique representation) = 138 A Fraenkel numeration system: ternary, but... (unique representation) = = = 21 but unique representation by ternary strings with no 21*2

26 Theorem [Fraenkel 1985] Every nonnegative integer has a unique (up to leading zeroes) representation as an l-ary string which does not contain the pattern (l 1) (l 2) (l 1). Proof Show that is a bijection f : b n b n 1... b 1 n b i a i i=1 f : T (l) n {0, 1, 2,... a (l) n+1 1}. The l-representation of an integer x is [x] = f 1 (x).

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30 Can revise bijection To insert a k + a k+1 into (λ 1, λ 2, λ 3, λ 4,...) either do (i) (λ 1 + a k, λ 2 + a k 1, λ 3, λ 4,...) or (ii) (λ 1 + (a k a k 1 ), λ 2 + (a k 1 a k 2 ), How to decide? insert (a k 1 + a k 2 )into(λ 3, λ 4,...)) Do (i) if it does not create a carry in Fraenkel arithmetic; otherwise do (ii). So, insertion becomes a 2-d form of Fraenkel arithmetic.

31 Lecture Hall Partitions

32 The Lecture Hall Theorem [BME1] The generating function for integer sequences λ 1, λ 2,..., λ n satisfying: is L n : λ 1 n λ 2 n 1... λ n 1 0 L n (q) = n i=1 1 1 q 2i 1

33 The Lecture Hall Theorem [BME1] The generating function for integer sequences λ 1, λ 2,..., λ n satisfying: is L n : λ 1 n λ 2 n 1... λ n 1 0 L n (q) = n i=1 1 1 q 2i 1 lim n (Lecture Hall Theorem) = Euler s Theorem since λ 1 n λ 2 n 1... λ n 1 n i=1 0 partitions into distinct parts 1 partitions into odd parts 1 q2i 1

34 Let {a n } = {a (l) n }. The l-lecture Hall Theorem [BME2]: The generating function for integer sequences λ 1, λ 2,..., λ n satisfying: L (l) n : λ 1 a n λ 2 a n 1... λ n 1 a 2 λ n a 1 0 is L (l) n (q) = n i=1 1 (1 q a i 1 +a i ) lim n (l-lecture Hall Theorem) = l-euler Theorem since as n, a n /a n 1 c l

35 Θ (l) n : Bijection for the l-lecture Hall Theorem Given a partition µ into parts in {a 0 + a 1, a 1 + a 2, a 2 + a 3,..., a n 1 + a n } construct λ = (λ 1, λ 2,..., λ n ) by inserting the parts of µ in nonincreasing order as follows: To insert a k 1 + a k into (λ 1, λ 2,..., λ n ): If k = 1, then add a 1 to λ 1 ; otherwise, if (λ 1 + a k a k 1 ) (a n /a n 1 )(λ 2 + a k 1 a k 2 ), add a k a k 1 to λ 1, add a k 1 a k 2 to λ 2 ; recursively insert a k 2 + a k 1 into (λ 3, λ 4,..., λ n ) otherwise, add a k to λ 1, and add a k 1 to λ 2.

36 Relationship between bijections

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44 Truncated lecture hall partitions L (l) n,k : λ 1 n λ 2 n 1 λ k 1 n k + 2 λ k n k + 1 > 0 Theorem: [Corteel,S 2004] [ L (l) n,k (q) = q(k+1 2 ) n k ] q ( q n k+1 ; q) k (q 2n k+1 ; q) k, where (a; q) n = (1 a)(1 aq) (1 aq n 1 ) Analog for l-lecture hall partitions?

45 Truncated lecture hall partitions L (l) n,k,j : j λ 1 n λ 2 n 1 λ k 1 n k + 2 λ k n k + 1 > 0 Theorem: [Corteel,S 2004] [ L (l) n,k (q) = q(k+1 2 ) n k Theorem [Corteel,Lee,S 2005] ( ) L (l) n n,k,j (1) = jk k ] q ( q n k+1 ; q) k (q 2n k+1 ; q) k, where (a; q) n = (1 a)(1 aq) (1 aq n 1 ) Analog for l-lecture hall partitions?

46 Theorem [Corteel,S 2004] Given positive integers s 1,..., s n, the generating function for the sequences λ 1,..., λ n satisfying is λ 1 s 1 λ 2 s 2 λ n 1 s n 1 λ n s n 0 s2 1 s3 1 z 2 =0 z 3 =0 s n 1 z n=0 s1z2 + P n s q 2 i=2 z i n 1 n i=1 (1 qb i ) i=2 qb i zi+1 z i s i+1 s i where b 1 = 1 and for 2 i n, b i = s 1 + s s i. Corollary As q 1, (1 q) n this gf s 2 s 3 s n n i=1 b. i

47 Truncated l-lecture hall partitions l-nomials Corollary For l-sequence {a i }: L (l) n+k,k : λ 1 a n+k λ 2 a n+k 1 λ k a n+1 > 0 but L (l) n+k,k (q) =? ( n ) (l) lim ((1 q 1 q)k L (l) n+k,k (q)) = k (p 1 p 2 p k )(p n+k p n+k 1 p n+1 ) where p i = a i + a i 1 and ( ) n (l) is the l-nomial... k

48 The l-nomial coefficient Example ( n k ) (l) = a(l) n a (l) n 1 a(l) n k+1 a (l) k a(l) k 1 a(l) 1. ( ) 9 (3) = = 174, 715, 376. Theorem [Lucas 1878] ( n ) (l) k is an integer. like Fibonomials, e.g. Ron Knott s web page:

49 Pascal-like triangle for the l-nomial coefficient: Theorem ( ) n (l) ( = (a (l) n 1 k k+1 a(l) k ) k ) (l) + (a (l) n k a(l) n k 1 ) ( n 1 k 1 ) (l)

50 Let u l and v l be the roots of the polynomial x 2 lx + 1: Then u l = l + l 2 4 ; v l = l l u l + v l = l; u l v l = 1. a (l) n = un l v n l u l v l.

51 Let u l and v l be the roots of the polynomial x 2 lx + 1: Then u l = l + l 2 4 ; v l = l l u l + v l = l; u l v l = 1. a (l) n = un l v n l u l v l. For real r, define (l) r = ul r + v l r. Then for integer n (l) n = a (l) n+1 a(l) n 1. ( is the l analog of 2.)

52 Let u l and v l be the roots of the polynomial x 2 lx + 1: Then u l = l + l 2 4 ; v l = l l u l + v l = l; u l v l = 1. a (l) n = un l v n l u l v l. For real r, define (l) r = ul r + v l r. Then for integer n (l) n = a (l) n+1 a(l) n 1. ( is the l analog of 2.) ( (l) n/2 )2 = u n l + 2(u lv l ) n/2 + v n l = (l) n + 2. (l) r = u r l + v r l = vl r + ur l = (l) r

53 A 3-term recurrence for the l-nomial [LS] ( ) n (l) = k ( n 2 k ) (l) + (l) n 1 ( ) n 2 (l) + k 1 ( ) n 2 (l) k 2 Proof: Use identities: a (l) n+k a(l) n k = (l) n a (l) k. a n (l) a (l) n 1 a(l) k a(l) k 1 = a(l) n k a(l) n+k 1.

54 An l-nomial theorem [LS]: An analog of is n k=0 ( ) n z k = (1 + z) n k n k=0 ( ) n (l) z k = k n 1 i=0 (u i (n 1)/2 l + v i (n 1)/2 l z) = (1 + (l) 1 z + z2 )(1 + (l) 3 z + z2 ) (1 + (l) n 1 z + z2 ) (1 + z)(1 + (l) 2 z + z2 )(1 + (l) 4 z + z2 ) (1 + (l) n 1 z + z2 ) n even n odd

55 A coin-flipping interpretation of the l-nomial For real r, let C r (l) be a weighted coin for which the probability of tails is ul r / (l) r and the probability of heads is vl r / (l) r. The probability of getting exactly k heads when tossing the n coins C (l) (n 1)/2, C (l) 1 (n 1)/2, C (l) 2 (n 1)/2,..., C (l) (n 1)/2 :

56 A coin-flipping interpretation of the l-nomial For real r, let C r (l) be a weighted coin for which the probability of tails is ul r / (l) r and the probability of heads is vl r / (l) r. The probability of getting exactly k heads when tossing the n coins C (l) (n 1)/2, C (l) 1 (n 1)/2, C (l) 2 (n 1)/2,..., C (l) (n 1)/2 : ( ) (l) n k (l) (n 1)/2 (l) 1 (n 1)/2 (l) (n 1)/2 1 (l) (n 1)/2 (l) j/2 may not be an integer, but (l) j/2 (l) j/2 = (l) j + 2 is.

57 Figure: The 8 possible results of tossing the weighted coins C (l) 1, C (l) 0, C (l) 1. Heads (pink) are indicated with their v weight and tails (blue) with their u weight. The weight of each toss is shown.

58 Define a q-analog of the l-nomial: a n (l) (q) = un l v l nqn u l v l q ; (l) n (q) = ul n + v l n qn Then [ n k ] (l) q = a(l) n (q) a (l) n 1 (q) a(l) n k+1 (q) a (l) (q) a(l) (q) a(l) 1 (q). k k 1 [ n k ] (l) q = q k [ n 2 k ] (l) q [ ] + (l) n 2 (l) n 1 (q) k 1 q [ ] n 2 ( + q n k k 2 q and n k=0 [ n k ] (l) q q k(k+1)/2 z k = α n (q, z) n/2 i=1 (1 + zq i (l) n 2i+1 (q) + z2 q n+1 ). where α n (q, z) = 1 if n is even; (1 + zq (n+1)/2 ) if n is odd.

59 Proofs can be derived from this partition-theoretic interpretation:

60 Another q-analog of the l-nomial Let a n (q) = (1 q an )/(1 q). Then [ n k ] (l) q = (µ,f ) q shapeweight(µ) q fillweight(µ,f ) where the sum is over all pairs (µ, f ) such µ is a partition in [k (n k)] and f is a filling f (i, j) of the cells of [k (n k)] with elements of {0, 1,... l 1} so that (i) no row of µ or column of µ c contains (l 1)(l 2) (l 1) and... (a bit more) Indexing cells of k (n k) bottom to top, left to right: cell (i, j) has a shape weight (a i a i 1 )(a j a j 1 ) shapeweight(µ) is sum of shape weights of cells in µ cell (i, j) has a fill weight a i a j fillweight(µ, f ) is i,j f (i, j)a ia j. (Now starting to get something related to lecture hall partitions.)

61 Question: Is there an analytic proof of the l-euler theorem? When l = 2, the standard approach is to show the equivalence of the generating functions for the set of partitions into odd parts and the set of partitions into distinct parts: 1 1 q 2k 1 = (1 + q k ). k=1 k=1 The sum/product form of Euler s theorem is k=1 1 1 q 2k 1 = k=0 q k(k+1) (q; q) k Is there an analog for the l-euler theorem? What is the sum-side generating function for partitions in which the ratio of consecutive parts is greater than c l?

62 Question: When l = 2, several refinements of Euler s theorem follow from Sylvester s bijection. What refinements of the l-euler theorem can be obtained from the bijection? We have some partial answers, but here s one we can t answer: Sylvester showed that if, in his bijection, the partition µ into odd parts maps to λ (distinct parts), then the number of distinct part sizes occurring in µ is the same as the number of maximal chains in λ. (A chain is a sequence of consecutive integers.) Is there an analog for l > 2?

63 Question: Can we enumerate either of these finite sets: The integer sequences λ 1,..., λ n in which the ratio of consecutive parts is greater than c l and λ 1 < a (l) n+1? We have a combinatorial characterization: these can be viewed as fillings of a staircase of shape (n, n 1,..., 1) such that (i) the filling of each row is an l-ary string with no (l 1)(l 2) (l 1) and (ii) the rows are weakly decreasing, lexicographically. For l = 2 the answer is 2 n. The integer sequences λ 1,..., λ n satisfying 1 λ 1 a n λ 2 a n 1... λ n 1 a 2 λ n a 1 0 For l = 2 this is the same set as above and the answer is 2 n.

64 Question: What is the generating function for truncated l-lecture hall partitions? Question: What is the right q-analog of the l-nomial (and the right interpretation) to explain and generalize lecture hall partitions? (We know what lecture hall partitions look like in the l-world. What about ordinary partitions? partitions into distinct parts?) Question: There are several q-series identities related to Euler s theorem, such as Lebesgue s identity the Roger s-fine identity Cauchy s identity Are there l-analogs?

65 Sincere thanks to... You, the audience The FPSAC08 Organizing Committee The FPSAC08 Program Committee Luc Lapointe

66 CanaDAM nd Canadian Discrete and Algorithmic Mathematics Conference May 25-28, 2009 CRM, Montreal, Quebec, Canada

67 AMS 2009 Spring Southeastern Section Meeting North Carolina State University Raleigh, NC April 4-5, 2009 (Saturday - Sunday) Some of the special sessions: Applications of Algebraic and Geometric Combinatorics (Sullivant, Savage) Rings, Algebras, and Varieties in Combinatorics (Hersh, Lenart, Reading) Recent Advances in Symbolic Algebra and Analysis (Singer, Szanto) Kac-Moody Algebras, Vertex Algebras, Quantum Groups, and Applications (Bakalov, Misra, Jing) Enumerative Geometry and Related Topics (Rimayi, Mihalcea) Homotopical Algebra with Applications to Mathematical Physics (Lada, Stasheff) Low Dimensional Topology and Geometry (Dunfield, Etnyre, Ng)