Can irrational tilings give Catalan numbers? Igor Pak, UCLA (joint - PDF document

Can irrational tilings give Catalan numbers? Igor Pak, UCLA (joint work with Scott Garrabrant) Stanley’s 70th Birthday Conference, MIT June 24, 2014 1

Biographical tidbit My best investment ever: 2.38 roubles = ⇒ [new world].

Extraterrestrial research: a modest proposal Carl Sagan: We should communicate with aliens using prime numbers. SETI: Systematically sends prime number sequence to outer space. My proposal: Start sending Catalan numbers! Hey, You Never Know!

Difficult Question: Can there be a world without Catalan numbers? In other words, maybe there is a model of computation which is powerful enough, yet is unable to count any Catalan objects? Answer: Maybe! Consider the world of 1-dimensional irrational tilings!

Tilings of [1 × n ] rectangles Fix a finite set T = { τ 1 , . . . , τ k } of rational tiles of height 1. Let a n = a n ( T ) the number of tilings of [1 × n ] with T . n a n t n = P ( t ) /Q ( t ), where P, Q ∈ Z [ t ]. Transfer-matrix Method: A T ( t ) = � Therefore, NO Catalan numbers! a n = F n 1 A ( t ) = 1 − t − t 2 n 1 2 � n − 2 � a n = 2 t 4 A ( t ) = n (1 − t ) 3

Irrational Tilings of [1 × ( n + ε )] rectangles Fix ε > 0 and a finite set T = { τ 1 , . . . , τ k } of irrational tiles of height 1. Let a n = a n ( T, ε ) the number of tilings of [1 × ( n + ε )] with T . Observe: we can get algebraic g.f.’s A T ( t ). � 2 n 1 � α / ∈ Q a n = A ( t ) = √ 1 − 4 t n 1 1 2 − α 2 + α [1 × n ]

Main Conjecture: Let F denote the class of g.f. A T ( t ) enumerating irrational tilings. Then: C ( t ) = 1 − √ 1 − 4 t C ( t ) / ∈ F , . where 2 t In other words, there is no set T of irrational tiles and ε > 0, s.t. 1 � 2 n � a n ( T, ε ) = C n for all n ≥ 1 , where C n = . n + 1 n

Diagonals of Rational Functions n b n t n , where Let G ∈ Z [[ x 1 , . . . , x k ]]. A diagonal is a g.f. B ( t ) = � x n 1 , . . . , x n � � b n = G ( x 1 , . . . , x k ) . k Theorem 1. Every A ( t ) ∈ F is a diagonal of a rational function P/Q , for some polynomials P, Q ∈ Z [ x 1 , . . . , x k ] . For example, � 2 n � 1 = [ x n y n ] 1 − x − y . n Proof idea: Say, τ i = [1 × α i ], α i ∈ R . Let V = Q � α 1 , . . . , α k � , d = dim( V ). ε �→ ( c 1 , . . . , c d ), α i �→ v i ∈ Z d ⊂ V . We have natural maps Interpret irrational tilings as walks O → ( n + c 1 , . . . , n + c d ) with steps { v 1 , . . . , v k } .

Properties of Diagonals of Rational Functions (1) must be D -finite , see [Stanley, 1980], [Gessel, 1981]. (2) when k = 2, must be algebraic , and (2 ′ ) every algebraic B ( t ) is a diagonal of P ( x, y ) /Q ( x, y ), see [Furstenberg, 1967]. No surprise now that Catalan g.f. C ( t ), tC ( t ) 2 − C ( t ) + 1 = 0, is a diagonal: C n = [ x n y n ] y (1 − 2 xy − 2 xy 2 ) 1 − x − 2 xy − xy 2 , see e.g. [Rowland–Yassawi, 2014]. Moral: Theorem 1 is not strong enough to prove the Main Conjecture.

N -Rational Functions R k Definition: Let R k be the smallest set of functions F ( x 1 , . . . , x k ) which satisfies (1) 1 , x 1 , . . . , x k ∈ R k , (2) F, G ∈ R k = ⇒ F + G, F · G ∈ R k , (3) F ∈ R k , F (0) = 0 = ⇒ 1 / (1 − F ) ∈ R k . Note that all F ∈ R k satisfy: F ∈ N [[ x 1 , . . . , x k ]], and F = P/Q , for some P, Q ∈ Z [ x 1 , . . . , x k ]. Let N be a class of diagonals of F ∈ R k , for some k ≥ 1. For example, � 2 n � 1 t n ∈ N � 1 − x − y ∈ R 2 . because n n

N -rational functions of one variable: Word of caution: R 1 is already quite complicated, see [Gessel, 2003]. For example, take the following F, G ∈ N [[ t ]] : t + 5 t 2 1 + t F ( t ) = 1 + t − 5 t 2 − 125 t 3 , G ( t ) = 1 + t − 2 t 2 − 3 t 3 . Then F / ∈ R 1 and G ∈ R 1 ; neither of these are obvious. The proof follows from results in [Berstel, 1971] and [Soittola, 1976] , see also [Katayama–Okamoto–Enomoto, 1978].

Main Theorem: F = N . In other words, every tile counting function A T ∈ F is a diagonal of an N -rational function F ∈ R k , k ≥ 1 , and vice versa. n f ( n ) t n , Mail Lemma: Both F and N coincide with a class of g.f. F ( t ) = � where f : N → N is given as finite sums f = � g j , and each g j is of the form r j  � α ij ( v, n ) � � �  if m = p j n + k j ,   β ij ( v, n ) g j ( m ) = i =1 v ∈ Z dj   0 otherwise,  for some α ij = a ij v + a ′ ij n + a ′′ ij , β ij = b ij v + b ′ ij n + b ′′ ij , and p j , k j , r j , d j ∈ N .

Asymptotic applications Corollary 2. There exist � n f n , � n g n ∈ F , s.t. √ π � 3 � 3 Γ 2 π 5 / 2 n − 3 / 2 384 n � 128 n , 4 f n ∼ g n ∼ √ � 5 � 7 � Γ Γ 3 8 8 Proof idea: Take n � 4 k �� 3 k � � 128 n − k f n := . k k k =0 Note: We have b n ∼ B n β γ n , where β ∈ N , and B , γ ∈ A , for all � n b n t n = P/Q . n f n ∈ F , we have f n ∼ B n β γ n , where β ∈ Z / 2 , γ ∈ A , Conjecture 3. For every � and B is spanned by values of p Φ q ( · ) at rational points, cf. [Kontsevich–Zagier, 2001].

Back to Catalan numbers Recall 1 √ π n − 3 / 2 4 n . C n ∼ √ n f n t n ∈ F , s.t. f n ∼ 3 3 Corollary 4. There exists � π C n . Furthermore, ∀ ǫ > 0 , n f n t n ∈ F , s.t. f n ∼ λC n for some λ ∈ [1 − ǫ, 1 + ǫ ] . there exists � Moral: Main Conjecture cannot be proved via rough asymptotics. However: n f n t n ∈ F , s.t. f n ∼ C n . Conjecture 5. There is no � Note: Conj. 5 does not follow from Conj. 3; probably involves deep number theory.

Bonus applications n f n t n ∈ F , s.t. Proposition 6: For every m ≥ 2 , there is � f n = C n mod m, for all n ≥ 1 . n g n t n ∈ F , s.t. Proposition 7: For every prime p ≥ 2 , there is � ord p ( g n ) = ord p ( C n ) , for all n ≥ 1 , where ord p ( N ) is the largest power of p which divides N . Moral: Elementary number theory doesn’t help either to prove the Main Conjecture. Note: For ord p ( C n ), see [Kummer, 1852], [Deutsch–Sagan, 2006]. Proof idea: Take � 2 n � 2 n � � f n = + ( m − 1) . n − 1 n

In summary: As promised, we created a rich world of tile counting functions, which may have Catalan objects, but probably not!

Happy Birthday, Richard!