A (semi-) automatic method for the determination of differentially algebraic integer sequences modulo powers of 2 and 3

Transcription

1 A (semi-) automatic method for the determination of differentially algebraic integer sequences modulo powers of 2 and 3 Universität Linz; Universität Wien; Queen Mary, University of London

2 Let (a n ) n 0 be a sequence of integers, where a n is the number of... of size n.

3 Let (a n ) n 0 be a sequence of integers, where a n is the number of... of size n. What can we say about modular properties of these numbers?

4 Let (a n ) n 0 be a sequence of integers, where a n is the number of... of size n. What can we say about modular properties of these numbers? In this talk: What can we say about the value of a n modulo 2 k?

5 Let (a n ) n 0 be a sequence of integers, where a n is the number of... of size n. What can we say about modular properties of these numbers? In this talk: What can we say about the value of a n modulo 2 k? What can we say about the value of a n modulo 3 k?

6 Congruences modulo powers of 2

7 Congruences modulo powers of 2 Example: Catalan numbers The Catalan numbers are the numbers Cat n = 1 ( ) 2n, n = 1, 2, 3,.... n + 1 n

8 Congruences modulo powers of 2 Example: Catalan numbers The Catalan numbers are the numbers Cat n = 1 ( ) 2n, n = 1, 2, 3,.... n + 1 n The first few numbers are 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, , , , , , , , , ,....

9 Congruences modulo powers of 2 Example: Catalan numbers The Catalan numbers are the numbers Cat n = 1 ( ) 2n, n = 1, 2, 3,.... n + 1 n The first few numbers are 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, , , , , , , , What can we say modulo 2? , ,....

10 Congruences modulo powers of 2 Example: Catalan numbers The Catalan numbers are the numbers Cat n = 1 ( ) 2n, n = 1, 2, 3,.... n + 1 n The first few numbers are 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, , , , , , , , , ,.... What can we say modulo 2? The Catalan numbers are odd for n = 1, 3, 7, 15,....

11 Congruences modulo powers of 2 Example: Catalan numbers The Catalan numbers are the numbers Cat n = 1 ( ) 2n, n = 1, 2, 3,.... n + 1 n The first few numbers are 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, , , , , , , , , ,.... What can we say modulo 2? The Catalan numbers are odd for n = 1, 3, 7, 15,.... Guess: The Catalan number Cat n is odd if and only if n = 2 s 1, for some non-negative integer s.

12 Guess: The Catalan number Cat n is odd if and only if n = 2 s 1, for some non-negative integer s.

13 Guess: The Catalan number Cat n is odd if and only if n = 2 s 1, for some non-negative integer s. A trivial c proof. Everybody knows that the generating function C(z) = n 0 Cat n z n for the Catalan numbers satisfies zc 2 (z) C(z) + 1 = 0. In terms of generating functions, our guess is equivalent to C(z) = z 1 Φ(z) modulo 2, where Φ(z) = s 0 z2s = z + z 2 + z 4 + z 8 + z c Doron Zeilberger: A computer can do this!

14 Guess: The Catalan number Cat n is odd if and only if n = 2 s 1, for some non-negative integer s. A trivial c proof. Everybody knows that the generating function C(z) = n 0 Cat n z n for the Catalan numbers satisfies zc 2 (z) C(z) + 1 = 0. In terms of generating functions, our guess is equivalent to C(z) = z 1 Φ(z) modulo 2, where Φ(z) = s 0 z2s = z + z 2 + z 4 + z 8 + z To prove the guess, we substitute in the equation and reduce: z(z 1 Φ(z)) 2 z 1 Φ(z) + 1 = z 1 Φ 2 (z) z 1 Φ(z) + 1 = z 1 (Φ(z) + z) z 1 Φ(z) + 1 = 0 modulo 2. c Doron Zeilberger: A computer can do this!

15 What about Catalan numbers modulo 4, 8, 16,...?

16 What about Catalan numbers modulo 4, 8, 16,...? Maybe, after reduction modulo 2 k, the generating function C(z) = n 0 Cat n z n is expressible as a polynomial in Φ(z), C(z) = d a i (z)φ i (z), i=0 where the a i (z) are suitable Laurent polynomials in z, and d is a suitable degree bound.

17 What about Catalan numbers modulo 4, 8, 16,...? Maybe, after reduction modulo 2 k, the generating function C(z) = n 0 Cat n z n is expressible as a polynomial in Φ(z), C(z) = d a i (z)φ i (z), i=0 where the a i (z) are suitable Laurent polynomials in z, and d is a suitable degree bound. Recall that Φ 2 (z) Φ(z) z = 0 modulo 2. Hence, (Φ 2 (z) Φ(z) z) k = 0 modulo 2 k. So, we may choose d = 2k 1.

18 What about Catalan numbers modulo 4, 8, 16,...? Maybe, after reduction modulo 2 k, the generating function C(z) = n 0 Cat n z n is expressible as a polynomial in Φ(z), C(z) = 2k 1 i=0 a i (z)φ i (z), where the a i (z) are suitable Laurent polynomials in z,

19 What about Catalan numbers modulo 4, 8, 16,...? Maybe, after reduction modulo 2 k, the generating function C(z) = n 0 Cat n z n is expressible as a polynomial in Φ(z), C(z) = 2k 1 i=0 a i (z)φ i (z), where the a i (z) are suitable Laurent polynomials in z, This Ansatz is then substituted for C(z) in the equation zc 2 (z) C(z) + 1 = 0 modulo 2. One reduces high powers of Φ(z) by the relation (Φ 2 (z) Φ(z) z) k = 0 modulo 2 k, compares coefficients of powers Φ j (z), j = 0, 1,,2k 1, obtains a system of (algebraic) equations for the unknowns a i (z) over Z/2 k Z, and....

20 What about congruences modulo higher powers of 2? Well...

21 What about congruences modulo higher powers of 2? Well... The proposed approach has two problems: 1 The relation (Φ 2 (z) Φ(z) z) k = 0 modulo 2 k. may not be the minimal one. In fact, we have Φ 4 (z)+6φ 3 (z)+(2z +3)Φ 2 (z)+(2z +6)Φ(z)+2z +5z 2 = 0 modulo 8.

22 What about congruences modulo higher powers of 2? Well... The proposed approach has two problems: 1 The relation (Φ 2 (z) Φ(z) z) k = 0 modulo 2 k. may not be the minimal one. In fact, we have Φ 4 (z)+6φ 3 (z)+(2z +3)Φ 2 (z)+(2z +6)Φ(z)+2z +5z 2 = 0 modulo 8. 2 Solving a system of (algebraic-differential) equations over Z/2 k Z is not a piece of a cake....

23 What about congruences modulo higher powers of 2? Re 1). In general, we are not able to provide a formula for a monic polynomial of minimal degree satisfied by Φ(z) modulo 2 k. (More on this later.) So, as a best compromise, we base our considerations on the congruence (Φ 4 (z)+6φ 3 (z)+(2z +3)Φ 2 (z)+(2z +6)Φ(z)+2z +5z 2 ) 2α = 0 This is a polynomial relation of degree 2 α+2. modulo 8 2α = 2 3 2α.

24 The method for proving congruences modulo 2 k Re 2). The general problem. Suppose we have a sequence (f n ) n 0 which we want to determine modulo a power of 2. We form the generating function F (z) = n 0 f nz n, and suppose that we know that it satisfies a differential equation of the form P(z; F (z), F (z), F (z),..., F (s) (z)) = 0, where P is a polynomial with integer coefficients, which has a unique formal power series solution.

25 The method for proving congruences modulo 2 k Re 2). The general problem. Suppose we have a sequence (f n ) n 0 which we want to determine modulo a power of 2. We form the generating function F (z) = n 0 f nz n, and suppose that we know that it satisfies a differential equation of the form P(z; F (z), F (z), F (z),..., F (s) (z)) = 0, where P is a polynomial with integer coefficients, which has a unique formal power series solution. Idea: Make the Ansatz F (z) = 2 α+2 1 i=0 a i (z)φ i (z) modulo 2 3 2α, where the a i (z) s are (at this point) undetermined Laurent polynomials in z. Then, gradually determine approximations a i,β (z) to a i (z) such that our differential equation holds modulo 2 β, for β = 1, 2,..., 3 2 α.

26 The method for proving congruences modulo 2 k The base step: Substitute F (z) = 2 α+2 1 i=0 a i,1 (z)φ i (z) modulo 2 into the differential equation, considered modulo 2, P(z; F (z), F (z), F (z),..., F (s) (z)) = 0 modulo 2, use Φ (z) = 1 modulo 2, reduce high powers of Φ(z) modulo the polynomial relation of degree 2 α+2 satisfied by Φ(z), and compare coefficients of powers Φ k (z), k = 0, 1,..., 2 α+2 1. This yields a system of 2 α+2 (algebraic differential) equations (modulo 2) for the unknown Laurent polynomials a i,1 (z), i = 0, 1,..., 2 α+2 1, which may or may not have a solution.

27 The method for proving congruences modulo 2 k The iteration: Provided we have already found a i,β (z), i = 0, 1,..., 2 α+2 1, such that F (z) = 2 α+2 1 i=0 a i,β (z)φ i (z) solves our differential equation modulo 2 β, we put a i,β+1 (z) := a i,β (z) + 2 β b i,β+1 (z), i = 0, 1,..., 2 α+2 1, where the b i,β+1 (z) s are (at this point) undetermined Laurent polynomials in z. Next we substitute F (z) = in the differential equation. 2 α+2 1 i=0 a i,β+1 (z)φ i (z)

28 The method for proving congruences modulo 2 k The iteration: One uses Φ (z) = β 2 n z 2n 1 n=0 modulo 2 β+1, one reduces high powers of Φ(z) using the polynomial relation satisfied by Φ(z), and one compares coefficients of powers Φ j (z), j = 0, 1,..., 2 α+2 1. After simplification, this yields a system of 2 α+2 (linear differential) equations (modulo 2) for the unknown Laurent polynomials b i,β+1 (z), i = 0, 1,..., 2 α+2 1, which may or may not have a solution.

29 Catalan numbers again

30 Catalan numbers again The Ansatz: C(z) = 2 α+2 1 i=0 a i (z)φ i (z) modulo 2 3 2α.

31 Catalan numbers again The Ansatz: The base step: We have C(z) = C(z) = 2 α+2 1 i=0 a i (z)φ i (z) modulo 2 3 2α. α z 2k 1 + z 1 Φ 2α+1 (z) modulo 2. k=0

32 Catalan numbers again The Ansatz: The base step: We have C(z) = C(z) = 2 α+2 1 i=0 a i (z)φ i (z) modulo 2 3 2α. α z 2k 1 + z 1 Φ 2α+1 (z) modulo 2. k=0 The iteration: works automatically without problems.

33 Theorem Let Φ(z) = n 0 z2n, and let α be some positive integer. Then the generating function C(z) for Catalan numbers, reduced modulo 2 3 2α, can be expressed as a polynomial in Φ(z) of degree at most 2 α+2 1 with coefficients that are Laurent polynomials in z. Moreover, for any given α, this polynomial can be found automatically.

34 Journal of Integer Sequences, Vol. 13 (2010), Article Catalan Numbers Modulo 2 k Shu-Chung Liu 1 Department of Applied Mathematics National Hsinchu University of Education Hsinchu, Taiwan liularry@mail.nhcue.edu.tw and Jean C.-C. Yeh Department of Mathematics Texas A & M University College Station, TX USA Abstract In this paper, we develop a systematic tool to calculate the congruences of some combinatorial numbers involving n!. Using this tool, we re-prove Kummer s and Lucas theorems in a unique concept, and classify the congruences of the Catalan numbers cn (mod 64). To achieve the second goal, cn (mod 8) and cn (mod 16) are also classified. Through the approach of these three congruence problems, we develop several general properties. For instance, a general formula with powers of 2 and 5 can evaluate cn (mod Manuel Kauers, Christian Krattenthaler 2 k ) for any k. and AnThomas equivalence W. Müller cn 2 k c n is derived, Congruences where for n isdifferentially the number obtained algebraic sequences

35 For those cn (mod 64) with ω2(cn) = 2, we can simply plug u16(cn) given in (47) into (32). Here we also show a precise classification by tables. Theorem 6.3. Let n N with d(α) =2. Then we have cn 64 ( 1) zr(α) 4 5 u16(cf2(cn)), where u16(cf2(cn)) is given in (47). Precisely, let [α]2 = 10 a 10 b 2, i.e., [n]2 = 10 a 10 b+1 1 β 2, and then we have cn (mod 64) shown in the following four tables. a =0 a =1 a =2 a 3 b = b = b = b when β =0 a =0 a =1 a =2 a 3 b = b = b = b when β =1 a =0 a =1 a =2 a 3 b = b = b = b when β =2 a =0 a =1 a =2 a 3 b = b = b = b when β 3 Proof. Notice that there are difference between a 3 and a = 3, and similarly for b and β. We split (47) into two parts as follows: A := χ(β = 0)(2 α1 α0 1) χ(β = 1) + 2χ(β = 2) α0 +2χ(β = 3)(1 α0), B := 2 [ c 2 ( α) + α0(1 α2) + #(S4( α), { , 1x00 2}) ] r1( α) zr1( α) + α0 α1 +1. Clearly, B is independent on β. We will only prove the first table of this theorem. The other three tables can be checked in the same way. With simple calculation we obtain the values of A as β = 0 and B as follows: a =0 a =1 a =2 a =3 b = Manuel Kauers, Christian Krattenthaler b =1 1 and Thomas 1 1W. Müller 1 a =0 a =1 a =2 a =3 b = Congruences b =1 0 for differentially 1 2 algebraic 0 sequences

36 Theorem (Liu and Yeh, compactly) Let Φ(z) = n 0 z2n. Then, modulo 64, we have Cat n z n = 32z 5 +16z 4 +6z 2 +13z +1+ ( 32z z z z + 40 ) Φ(z) n=0 + ( 16z z z z ( + 32z z z z ( + 48z ) Φ 6 (z) + z ) ( Φ 2 (z) + 32z z z ) Φ 4 (z) + ( 32z ) Φ 5 (z) ( 32z z ) Φ 3 (z) ) Φ 7 (z) modulo 64.

37 Theorem Let Φ(z) = n 0 z2n. Then, modulo 4096, we have Cat n z n = 2048z z z z z z z 8 n= z z z z z 3 + 2z 2 + z ( 2048z z z z z z z z z 2) Φ(z) + ( 2048z z z z z z z z z z z z ) Φ 2 (z) + ( 2048z z z z z z z z z z z ) Φ 3 (z) + ( 2048z z z z z z z z z z z z ) Φ 4 (z) z + ( 2048z z z z z z z z z z z ) Φ 5 (z)

38 + ( 2048z z z z z z z z z z ) Φ 6 (z) z + ( 2048z z z z z z z z ) Φ 7 (z) z + ( 2048z z z z z z z z z z ) Φ 8 (z) z + ( 2048z z z z z z z z + ( 2048z z z z z z z z z + ( 2048z z z z z 3 ) Φ 9 (z) ) Φ 10 (z) +2688z z z ) Φ 11 (z)

39 + ( 2048z z z z z z z z z ) Φ 12 (z) ( z z z z z z z ( z z z z z z z ( z z z z ) Φ 15 (z) z ) Φ 13 (z) ) Φ 14 (z) modulo 4096.

40 + ( 2048z z z z z z z z z ) Φ 12 (z) ( z z z z z z z ( z z z z z z z ( z z z z ) Φ 15 (z) z ) Φ 13 (z) ) Φ 14 (z) modulo 4096.

41 + ( 2048z z z z z z z z z ) Φ 12 (z) ( z z z z z z z ( z z z z z z z ( z z z z ) Φ 15 (z) z ) Φ 13 (z) ) Φ 14 (z) modulo We have also a procedure for extracting coefficients of powers of Φ(z).

42 Subgroup numbers modulo powers of 2

43 Subgroup numbers modulo powers of 2 Theorem (Stothers 1977) The number s n of index-n-subgroups in the inhomogeneous modular group PSL 2 (Z) is odd if, and only if, n is of the form 2 k 3 or 2 k+1 6, for some positive integer k 2.

44 Subgroup numbers modulo powers of 2 Theorem (Stothers 1977) The number s n of index-n-subgroups in the inhomogeneous modular group PSL 2 (Z) is odd if, and only if, n is of the form 2 k 3 or 2 k+1 6, for some positive integer k 2.

45 Subgroup numbers modulo powers of 2 Theorem (Stothers 1977) The number s n of index-n-subgroups in the inhomogeneous modular group PSL 2 (Z) is odd if, and only if, n is of the form 2 k 3 or 2 k+1 6, for some positive integer k 2. In terms of Φ(z): s n+1 z n = (z 7 + z 4 )Φ(z) + z 6 + z 5 + z 2 modulo 2. n 0

46 DIVISIBILITY PROPERTIES OF SUBGROUP NUMBERS FOR THE MODULAR GROUP Thomas W. Müller and Jan-Christoph Schlage-Puchta Abstract. Let Γ = PSL2(Z) be the classical modular group. It has been shown by Stothers (Proc. Royal Soc. Edinburgh 78A, ) that sn, the number of index n subgroups in Γ, is odd if and only if n+3 or n+6 is a 2-power. Moreover, Stothers loc. cit. also showed that fλ, the number of free subgroups of index 6λ in Γ, is odd if and only if λ + 1 is a 2-power. Here, these divisibility results for fλ and sn are generalized to congruences modulo higher powers of 2. We also determine the behaviour modulo 3 of fλ. Our results are naturally expressed in terms of the binary respectively ternary expansion of the index. 1. Introduction and results Let Γ = PSL 2(Z) be the classical modular group. We denote by s n the number of index n subgroups in Γ, and by f λ the number of free subgroups in Γ of index 6λ. These days, quite a lot is known concerning the subgroup arithmetic of Γ. Newman [5, Theorem 4] gave an asymptotic formula for s n; for a more general and more precise result see [3, Theorem 1]. Based on numerical computations of Newman, Johnson conjectured that s n is odd if and only if n =2 a 3,a 2 or n =2 a 6,a 3. This conjecture was first proved by Stothers [6]. He first used coset diagrams to establish a relation between s n and f λ for various λ in the range 1 λ n+4, and then showed that fλ is odd if and 6 only if λ =2 a 1,a 1. The parity pattern for f λ found by Stothers has been shown to hold for a larger class of virtually free groups, including free products Γ = G 1 S G 2 Manuel Kauers, Christian of two Krattenthaler finite groups andg Thomas i with anw. amalgamated Müller Congruences subgroup S of forodd differentially order, whose algebraic indicessequences

47

48 14 DIVISIBILITY PROPERTIES OF SUBGROUP NUMBERS In this way we may simplify the last displayed expression as follows. 2#{n =2 a +2 b, a > b 2} + 2#{n =2 a +2 b 3,a 3,b 2} + 2#{n =2 a +2 b 6, a > b 3} 2#{n =2 a +2 b +3, a > b} 2#{n =2 a +2 b,a 3,b 2} 2#{n =2 a +2 b 3, a > b 3} + 4#{n =2 b +2 c +4, b > c 2,b 4} + 4#{n =2 a +9,a 3} + 4#{n =2 b +2 c +1, b > c 2} + 4#{n =2 a +2 b +2 c 6, b > c 2,a 3} + 4#{n =2 a +2 b +2 c +3, b > c 2,a 2,b 4} + 4#{n =2 a +2 b +2 c, b > c 2,a 2} + 4#{n =2 a +2 b +2 c 3, b > c 2,a 3} + 4#{n =2 a +2 b +3,a 3,b 2} + 4#{n =2 a +2 b +9, a, b 2} Next consider for example the quantity 4#{n =2 a +2 b +6,a 3,b 2}. If (a, b) is a solution with a>b 3, then (b, a) is also a solution, that is, the number of solutions is even, unless n is of the form n =2 a + 10,a 3, or n is of the form 2 a + 6 with a 4. The same argument may be applied to several other terms as well, which allows us to simplify the expression further to obtain the following. 2#{n =2 a +2 b, a > b} + 4#{n =2 a +1,a 3} + 2#{n =2 a 3,a 4}} +4#{n =2 a +2 b 3, a > b 2}+2#{n =2 a +2 b 6, a > b 3} 2#{n =2 a +2 b +3, a > b} 2#{n =2 a +4,a 3} 2#{n =2 a,a 4} + 4#{n =2 a +2 b, a > b 2} 2#{n =2 a +2 b 3, a > b 3} + 4#{n =2 b +2 c +4, b > c 2,b 4} + 4#{n =2 b +2 c +1, b > c 2} + 4#{n =2 a +2 b +2 c 6, b > c 2,a 3} + 4#{n =2 a +2 b +2 c +3, b > c 2,a 2,b 4} + 4#{n =2 a +2 b +2 c, b > c 2,a 2} + 4#{n =2 a +2 b +2 c 3, b > c 2,a 3} + 4#{n =2 a +7,a 3} + 4#{n =2 a +3,a 4} Finally, consider the quantity #{n =2 a +2 b +2 c, b > c 2,a 2}. Let (a, b, c) be a solution counted. If all three components are distinct, there are no solutions with two of the variables equal to each other, and there are 3 possible orderings of a, b and c. In a b c

49 DIVISIBILITY PROPERTIES OF SUBGROUP NUMBERS 17 To ease further computations, we consider sets with one, two, and three parameters separately. Sets defined by one parameter contribute {4 n =2 a, 2 a 3,a 3} + {2 n =2 a 2, 2 a +1,a 3} + {1 n =2 a,a 3} + {4 n =3 2 a,a 3} + {4 n =2 a +9,a 3} + {6 n =2 a +1, 2 a +4,a 3} + {7 n =2 a +3,a 3} + {4 n =3 2 a +3,a 3} + {4 n =2 a + 12} + {1 n =2 a 6, 2 a } + {7 n =2 a 3, 2 a +3} + {4 n =2 a + 12, 2 a + 15, a > b 2} + {4 n =2 a +1,a 3} + {2 n =2 a 3,a 4}} {2 n =2 a +4,a 3} {2 n =2 a,a 4} + {4 n =2 a 2,a 5} + {4 n =2 a 6,a 5} + {4 n =3 2 a 6,a 5} + {4 n =2 a + 15,a 2} + {4 n =2 a +7,a 4} + {4 n =2 a +3,a 4} + {4 n =3 2 a +3,a 4} + {4 n =2 a +4,a 4} + {4 n =2 a,a 4} + {4 n =3 2 a,a 4} + {4 n =2 a +1,a 4} + {4 n =2 a 3,a 4} + {4 n =3 2 a 3,a 4} + {4 n =2 a +7,a 3} which is congruent to {5 n =2 a 6,a 5} + {1 n =2 a 3,a 3} + {6 n =2 a 2,a 3} + {6 n =2 a +3,a 3} + {4 n =2 a +9,a 3} Next, we collect all 2-parameter sets. These contribute + {4 n =2 a +3,a 4}, + {4 n =3 2 a 6,a 3} + {4 n =3 2 a 3,a 4}. {4 n =2 a +2 b +1, 2 a +2 b 2, a > b 2} + {2 n =2 a +2 b, a > b 2} + {4 n =2 a +2 b +4, 2 a +2 b +1, a > b 2} + {2 n =2 a +2 b +3, a > b 2} + {4 n =2 a +2 b, 2 a +2 b +3, 2 a +2 b 6, 2 a +2 b 3, a > b 2} + {2 n =2 a +2 b, a > b} + {4 n =2 a +2 b 3, a > b 2} + {2 n =2 a +2 b 6, a > b 3} { 2 n =2 a +2 b +3, a > b} + {4 n =2 a +2 b, a > b 2} { 2 n =2 a +2 b 3, a > b 3} Manuel Kauers, Christian Krattenthaler + {4 n and =2 a Thomas +2 b +1, W. a > Müller b 2} + {4 n Congruences =2 a +2 b for differentially 2, a > b 2} algebraic sequences

50 In terms of the series Φ(z), the result of Müller and Schlage Puchta can be compactly expressed in the form s n+1 z n = z 57 +4z 20 +4z 17 +4z 14 +4z 12 +4z 11 +4z 10 +4z 9 +2z 8 +4z 5 +2z 4 +4z 3 +2z 2 n 0 ( 6 z 6 + ) Φ(z) + 4z z z z z z z z 4 + 4z3 + 2 z 3 + 4z2 + 4 z ( + 4z z z z 5 + 4z4 + 3 z 4 + 4z3 + 6 z 3 + 2z2 + 2 z ) z + 4 Φ 2 (z) ( 6 + z z z z z 3 + 4z2 + 4 ) z 2 Φ 3 (z) modulo 8.

51 Subgroup numbers modulo powers of 2 Let S(z) = n 0 s n+1z n be the generating function for the subgroups numbers of PSL 2 (Z). Then Godsil, Imrich and Razen found the differential equation ( 1+4z 3 +2z 4 +4z 6 2z 7 4z 9 )S(z)+(z 7 z 10 )(S (z)+s 2 (z)) z + 4z 2 + 4z 3 z 4 + 4z 5 2z 6 2z 8 = 0.

52 Subgroup numbers modulo powers of 2 Let S(z) = n 0 s n+1z n be the generating function for the subgroups numbers of PSL 2 (Z). Then Godsil, Imrich and Razen found the differential equation ( 1+4z 3 +2z 4 +4z 6 2z 7 4z 9 )S(z)+(z 7 z 10 )(S (z)+s 2 (z)) z + 4z 2 + 4z 3 z 4 + 4z 5 2z 6 2z 8 = 0. Theorem Let Φ(z) = n 0 z2n, and let α be some positive integer. Then the generating function S(z) = S PSL2 (Z)(z), when reduced modulo 2 3 2α, can be expressed as a polynomial in Φ(z) of degree at most 2 α+2 1 with coefficients that are Laurent polynomials in z. Moreover, for any given α, this polynomial can be found automatically.

53 Theorem Let Φ(z) = n 0 z2n. Then, modulo 64, we have s n+1 (PSL 2 (Z)) z n n 0 = z z z z z z z z z z z z z z z z z z z z z z z z z z 9 + 2z z 7 + 4z z z z z z z + 36 z z z z ( z z z z z z z z z z z z z z z z 7 + 8z 5 + 8z z z + 32 ) Φ(z) + 20 z + 12 z z z z z 6 ( + 32z z z z z z z z z z z z z z z z z z z z z z z + 50 z z z z z ) z 7 Φ 2 (z)

54 ( + 32z z z z z z z z z z z z 9 + 8z z z 4 + 8z z z z + 20 z z z z z ) z 7 Φ 3 (z) ( + 32z z z z z z z z z z z z z z z z z z z z z 13 +8z z z z 9 +44z 8 +40z 7 +48z 6 +16z 5 +20z 4 +56z 3 +30z 2 ) Φ 4 (z) + 32z z + 34 z z z z z z 7 ( + 32z z z z z z z z z z z z z z z z z z z z z z z + 24 z z z z ) z 6 Φ 5 (z)

55 ( + 32z z z z z z z z z z z z z z z z z 5 + 8z z 3 + 4z z z + 52 z z z z z ) z 7 Φ 6 (z) ( + 32z z z z z z z z z z 9 +8z z z z 3 + 8z z z + 8 z z z z z ) z 7 Φ 7 (z) modulo 64.

56 Congruences modulo powers of 3

57 Journal of Number Theory 117 (2006) Congruences for Catalan and Motzkin numbers and related sequences Emeric Deutsch a,, Bruce E. Sagan b a Department of Mathematics, Polytechnic University, Brooklyn, NY 11201, USA b Department of Mathematics, Michigan State University, East Lansing, MI , USA Received 19 July 2004 Available online 22 August 2005 Communicated by D. Goss Abstract We prove various congruences for Catalan and Motzkin numbers as well as related sequences. The common thread is that all these sequences can be expressed in terms of binomial coefficients. Our techniques are combinatorial and algebraic: group actions, induction, and Lucas congruence for binomial coefficients come into play. A number of our results settle conjectures of Cloitre and Zumkeller. The Thue Morse sequence appears in several contexts Elsevier Inc. All rights reserved. 1. Introduction

58 Central trinomial coefficients modulo 3 Theorem (Deutsch and Sagan) Let T n denote the n-th central trinomial coefficient, that is, the coefficient of z n in (1 + z + z 2 ) n. Then { 1 (mod 3), if n T (01), T n 0 (mod 3), otherwise. Here, T (01) denotes the set of all positive integers n, which have only digits 0 and 1 in their ternary expansion.

59 Motzkin numbers modulo 3 Theorem (Deutsch and Sagan) The Motzkin numbers M n satisfy 1 (mod 3), if n 3T (01) or n 3T (01) 2, M n 1 (mod 3), if n 3T (01) 1, 0 (mod 3), otherwise. Here, T (01) denotes the set of all positive integers n, which have only digits 0 and 1 in their ternary expansion.

60 Central binomial coefficients modulo 3 Theorem (Deutsch and Sagan) The central binomial coefficients satisfy ( ) 2n n { ( 1) δ 3(n) (mod 3), if n T (01), 0 (mod 3), otherwise. Here, T (01) denotes the set of all positive integers n, which have only digits 0 and 1 in their ternary expansion, and δ 3 (n) denotes the number of 1s in the ternary expansion of n.

61 Catalan numbers modulo 3 Theorem (Deutsch and Sagan) The Catalan numbers C n satisfy { ( 1) δ 3 (n+1) (mod 3), if n T (01) 1, C n 0 (mod 3), otherwise. Here, T (01) denotes the set of all positive integers n, where all digits in their ternary expansion are 0 or 1 except for the right-most digit, and δ3 (n) denotes the number of 1s in the ternary expansion of n ignoring the right-most digit.

62 Central Eulerian numbers modulo 3 Let A(n, k) denote the number of permutations of {1, 2,..., n} with exactly k 1 descents. Theorem (Deutsch and Sagan) The central Eulerian numbers A(2n 1, n) and A(2n, n) satisfy { 1 (mod 3), if n T (01) + 1, A(2n 1, n) 0 (mod 3), otherwise. and 1 (mod 3), if n T (01) + 1, A(2n, n) 1 (mod 3), if n T (01) or n T (01) + 2, 0 (mod 3), otherwise. Here, T (01) denotes the set of all positive integers n, which have only digits 0 and 1 in their ternary expansion.

63 The paper by Deutsch and Sagan contains results of similar nature for Motzkin prefix numbers, Riordan numbers, sums of central binomial coefficients, central Delannoy numbers, Schröder numbers, and hex tree numbers.

64 Let us have another look at the central trinomial numbers theorem: Theorem (Deutsch and Sagan) Let T n denote the n-th central trinomial coefficient, that is, the coefficient of z n in (1 + z + z 2 ) n. Then { 1 (mod 3), if n T (01), T n 0 (mod 3), otherwise, where T (01) denotes the set of all positive integers n, which have only digits 0 and 1 in their ternary expansion.

65 Let us have another look at the central trinomial numbers theorem: Theorem (Deutsch and Sagan) Let T n denote the n-th central trinomial coefficient, that is, the coefficient of z n in (1 + z + z 2 ) n. Then { 1 (mod 3), if n T (01), T n 0 (mod 3), otherwise, where T (01) denotes the set of all positive integers n, which have only digits 0 and 1 in their ternary expansion.

66 Let us have another look at the central trinomial numbers theorem: Theorem (Deutsch and Sagan) Let T n denote the n-th central trinomial coefficient, that is, the coefficient of z n in (1 + z + z 2 ) n. Then { 1 (mod 3), if n T (01), T n 0 (mod 3), otherwise, where T (01) denotes the set of all positive integers n, which have only digits 0 and 1 in their ternary expansion. In other words: Let Then: Ψ(z) = k 0 n 1 > >n k 0 z 3n 1 +3 n nk = (1 + z 3j ) j=0 = 1 + z + z 3 + z 4 + z 9 + z 10 + z 12 + z n 0 T nz n = Ψ(z) modulo 3.

67 A functional equation modulo 3 satisfied by Ψ(z)

68 Lemma A functional equation modulo 3 satisfied by Ψ(z) The series Ψ(z) = j=0 (1 + z3j ) satisfies Ψ 2 (z) = z modulo 3.

69 Lemma A functional equation modulo 3 satisfied by Ψ(z) The series Ψ(z) = j=0 (1 + z3j ) satisfies Ψ 2 (z) = z modulo 3. Proof. We have Ψ 2 (z) = (1 + z 3j ) 2 = z (1 + z) (1 + z 3j ) 2 j=0 = 1 (1 + z)3 1 + z j=1 j=0 = z (1 + z 9 )Ψ 2 (z 9 ) modulo 3 = = z modulo 3. (1 + z 3j ) 2 = z (1 + z 3 )Ψ 2 (z 3 ) modulo 3

70 Lemma A functional equation modulo 3 satisfied by Ψ(z) The series Ψ(z) = j=0 (1 + z3j ) satisfies Ψ 2 (z) = z modulo 3.

71 Lemma A functional equation modulo 3 satisfied by Ψ(z) The series Ψ(z) = j=0 (1 + z3j ) satisfies Ψ 2 (z) = z modulo 3.

72 Lemma A functional equation modulo 3 satisfied by Ψ(z) The series Ψ(z) = j=0 (1 + z3j ) satisfies Ψ 2 (z) = z modulo 3. It is well-known that the generating function T (z) n 0 T nz n is given by T (z) = 1/ 1 2z 3z 2, or, phrased differently, (1 2z 3z 2 )T 2 (z) 1 = 0. Morover, this functional equation determines T (z) uniquely.

73 Lemma A functional equation modulo 3 satisfied by Ψ(z) The series Ψ(z) = j=0 (1 + z3j ) satisfies Ψ 2 (z) = z modulo 3. It is well-known that the generating function T (z) n 0 T nz n is given by T (z) = 1/ 1 2z 3z 2, or, phrased differently, (1 2z 3z 2 )T 2 (z) 1 = 0. Morover, this functional equation determines T (z) uniquely. Taken modulo 3, the above functional equation becomes: Consequently: (1 + z)t 2 (z) 1 = 0 modulo 3. T n z n = Ψ(z) modulo 3. n 0

74 What are the common features?

75 What are the common features? In each case, the generating function satisfied a quadratic equation (and, as a matter of fact, this applies as well for Motzkin prefix numbers, Riordan numbers, sums of central binomial coefficients, central Delannoy numbers, Schröder numbers, and hex tree numbers).

76 What are the common features? In each case, the generating function satisfied a quadratic equation (and, as a matter of fact, this applies as well for Motzkin prefix numbers, Riordan numbers, sums of central binomial coefficients, central Delannoy numbers, Schröder numbers, and hex tree numbers). In each case, one could express the generating function, after reduction of its coefficients modulo 3, as a linear expression in Ψ(±z).

77 What are the common features? In each case, the generating function satisfied a quadratic equation (and, as a matter of fact, this applies as well for Motzkin prefix numbers, Riordan numbers, sums of central binomial coefficients, central Delannoy numbers, Schröder numbers, and hex tree numbers). In each case, one could express the generating function, after reduction of its coefficients modulo 3, as a linear expression in Ψ(±z). Can this be so many accidents?

78 A meta-theorem Theorem Let F (z) be a formal power series with integer coefficients which satisfies a quadratic equation where c 2 (z)f 2 (z) + c 1 (z)f (z) + c 0 (z) = 0 modulo 3, Then F (z) = c 1(z) c 2 (z) ± zf1 (1 + εz γ ) f2+1 Ψ(εz γ ) modulo 3. c 2 (z)

79 A meta-theorem Theorem Let F (z) be a formal power series with integer coefficients which satisfies a quadratic equation where c 2 (z)f 2 (z) + c 1 (z)f (z) + c 0 (z) = 0 modulo 3, 1 c 2 (z) = z e 1 (1 + εz γ ) e 2 modulo 3, with non-negative integers e 1, e 2 and ε {1, 1}; Then F (z) = c 1(z) c 2 (z) ± zf1 (1 + εz γ ) f2+1 Ψ(εz γ ) modulo 3. c 2 (z)

80 A meta-theorem Theorem Let F (z) be a formal power series with integer coefficients which satisfies a quadratic equation where c 2 (z)f 2 (z) + c 1 (z)f (z) + c 0 (z) = 0 modulo 3, 1 c 2 (z) = z e 1 (1 + εz γ ) e 2 modulo 3, with non-negative integers e 1, e 2 and ε {1, 1}; 2 c 2 1 (z) c 0(z)c 2 (z) = z 2f 1 (1 + εz γ ) 2f 2+1 modulo 3, with non-negative integers f 1, f 2. Then F (z) = c 1(z) c 2 (z) ± zf1 (1 + εz γ ) f2+1 Ψ(εz γ ) modulo 3. c 2 (z)

81 Proof.

82 Proof.

83 Proof

84 Proof The corresponding choices of c 2 (z), c 1 (z), c 0 (z) are: c 2 (z) c 1 (z) c 0 (z) c1 2(z) c 0(z)c 2 (z) mod 3 trinomial 1 2z 3z z Motzkin z 2 z z cent.bin. 1 4z z Catalan z z Motz.pref. z 3z 2 1 3z z Riordan z + z z z Delannoy 1 6z + z z 2 Schröder z z z 2 hex tree z 2 3z z

85 Can we also do congruences modulo powers of 3?

86 Can we also do congruences modulo powers of 3? Yes! One follows the recipe that we developed for the series Φ(z) in order to find congruences modulo powers of 2: One expresses the generating function now as polynomial in Ψ(z) (or Ψ( z), or Ψ(z 2 ), or... ), with undetermined coefficients, which may be Laurent polynomials in z and 1 + z (respectively 1 z, 1 + z 2,... ). Again, high powers of Ψ(z) can be reduced, here by means of the relation ( Ψ 2 (z) 1 ) 3 α = 0 modulo 3 3α. 1 + z

87 Can we also do congruences modulo powers of 3? Yes! One follows the recipe that we developed for the series Φ(z) in order to find congruences modulo powers of 2: One expresses the generating function now as polynomial in Ψ(z) (or Ψ( z), or Ψ(z 2 ), or... ), with undetermined coefficients, which may be Laurent polynomials in z and 1 + z (respectively 1 z, 1 + z 2,... ). Again, high powers of Ψ(z) can be reduced, here by means of the relation ( Ψ 2 (z) 1 ) 3 α = 0 modulo 3 3α. 1 + z It is also possible to extract coefficients from powers of Ψ(z).

88 As it turns out, there is even a meta-theorem which refines all the modulo 3-results of Deutsch and Sagan to any power of 3. The corresponding results can be found automatically. Moreover, this meta-theorem produces as well several new congruence results.

89 Theorem Let α be some positive integer. Furthermore, suppose that the formal power series F (z) with integer coefficients satisfies the functional-differential equation c 2 (z)f 2 (z) + c 1 (z)f (z) + c 0 (z) where + 3Q(z; F (z), F (z), F (z),..., F (s) (z)) = 0, Then F (z), when coefficients are reduced modulo 3 3α, can be expressed as a polynomial in Ψ(εz γ ) of the form 2 3 α 1 F (z) = a 0 (z) + a i (z)ψ i (εz γ ) modulo 3 3α, i=0 where the coefficients a i (z), i = 0, 1,..., 2 3 α 1, are Laurent polynomials in z and 1 + εz γ.

90 Theorem Let α be some positive integer. Furthermore, suppose that the formal power series F (z) with integer coefficients satisfies the functional-differential equation c 2 (z)f 2 (z) + c 1 (z)f (z) + c 0 (z) where + 3Q(z; F (z), F (z), F (z),..., F (s) (z)) = 0, 1 c 2 (z) = z e 1 (1 + εz γ ) e 2 modulo 3, with non-negative integers e 1, e 2 and ε {1, 1}; Then F (z), when coefficients are reduced modulo 3 3α, can be expressed as a polynomial in Ψ(εz γ ) of the form 2 3 α 1 F (z) = a 0 (z) + a i (z)ψ i (εz γ ) modulo 3 3α, i=0 where the coefficients a i (z), i = 0, 1,..., 2 3 α 1, are Laurent polynomials in z and 1 + εz γ.

91 Theorem Let α be some positive integer. Furthermore, suppose that the formal power series F (z) with integer coefficients satisfies the functional-differential equation c 2 (z)f 2 (z) + c 1 (z)f (z) + c 0 (z) where + 3Q(z; F (z), F (z), F (z),..., F (s) (z)) = 0, 1 c 2 (z) = z e 1 (1 + εz γ ) e 2 modulo 3, with non-negative integers e 1, e 2 and ε {1, 1}; 2 c 2 1 (z) c 0(z)c 2 (z) = z 2f 1 (1 + εz γ ) 2f 2+1 modulo 3, with non-negative integers f 1, f 2 ; Then F (z), when coefficients are reduced modulo 3 3α, can be expressed as a polynomial in Ψ(εz γ ) of the form 2 3 α 1 F (z) = a 0 (z) + a i (z)ψ i (εz γ ) modulo 3 3α, i=0 where the coefficients a i (z), i = 0, 1,..., 2 3 α 1, are Laurent polynomials in z and 1 + εz γ.

92 Theorem Let α be some positive integer. Furthermore, suppose that the formal power series F (z) with integer coefficients satisfies the functional-differential equation c 2 (z)f 2 (z) + c 1 (z)f (z) + c 0 (z) where + 3Q(z; F (z), F (z), F (z),..., F (s) (z)) = 0, 1 c 2 (z) = z e 1 (1 + εz γ ) e 2 modulo 3, with non-negative integers e 1, e 2 and ε {1, 1}; 2 c 2 1 (z) c 0(z)c 2 (z) = z 2f 1 (1 + εz γ ) 2f 2+1 modulo 3, with non-negative integers f 1, f 2 ; 3 Q is a polynomial with integer coefficients. Then F (z), when coefficients are reduced modulo 3 3α, can be expressed as a polynomial in Ψ(εz γ ) of the form 2 3 α 1 F (z) = a 0 (z) + a i (z)ψ i (εz γ ) modulo 3 3α, i=0 where the coefficients a i (z), i = 0, 1,..., 2 3 α 1, are Laurent polynomials in z and 1 + εz γ.

93 Central trinomial numbers modulo 27 Theorem We have T n z n = ( 9z z + 15 ) Ψ(z 3 ) n 0 + ( 15z z 4 + 4z z z + 19 ) Ψ 3 (z 3 ) + ( 9z 8 + 6z 7 + 6z 6 + 9z z 4 + 3z z + 24 ) Ψ 5 (z 3 ) modulo 27.

94 Minimal polynomials for Φ(z) and Ψ(z) It would be desirable to know minimal relations for the series Φ(z) and Ψ(z).

95 Minimal polynomials for Φ(z) and Ψ(z) It would be desirable to know minimal relations for the series Φ(z) and Ψ(z). Definition A polynomial A(z, t) in z and t is minimal for the modulus 2 γ, if it is monic (as a polynomial in t), has coefficients which are Laurent polynomials in z, satisfies A(z, Φ(z)) = 0 modulo 2 γ, and there is no monic polynomial B(z, t) whose coefficients are Laurent polynomials in z, whose t-degree is less than that of A(z, t), and which satisfies B(z, Φ(z)) = 0 modulo 3 γ.

96 Minimal polynomials for Φ(z) and Ψ(z) It would be desirable to know minimal relations for the series Φ(z) and Ψ(z). Definition A polynomial A(z, t) in z and t is minimal for the modulus 2 γ, if it is monic (as a polynomial in t), has coefficients which are Laurent polynomials in z, satisfies A(z, Φ(z)) = 0 modulo 2 γ, and there is no monic polynomial B(z, t) whose coefficients are Laurent polynomials in z, whose t-degree is less than that of A(z, t), and which satisfies B(z, Φ(z)) = 0 modulo 3 γ.

97 Minimal polynomials for Φ(z) and Ψ(z) It would be desirable to know minimal relations for the series Φ(z) and Ψ(z). Definition A polynomial A(z, t) in z and t is minimal for the modulus 2 γ, if it is monic (as a polynomial in t), has coefficients which are Laurent polynomials in z, satisfies A(z, Φ(z)) = 0 modulo 2 γ, and there is no monic polynomial B(z, t) whose coefficients are Laurent polynomials in z, whose t-degree is less than that of A(z, t), and which satisfies B(z, Φ(z)) = 0 modulo 3 γ. Definition A polynomial A(z, t) in z and t is minimal for the modulus 3 γ, if it is monic (as a polynomial in t), has coefficients which are Laurent polynomials in z and 1 + z, satisfies A(z, Ψ(z)) = 0 modulo 3 γ, and there is no monic polynomial B(z, t) whose coefficients are Laurent polynomials in z and 1 + z, whose t-degree is less than that of A(z, t), and which satisfies B(z, Ψ(z)) = 0 modulo 3 γ.

98 Conjecture The degree of a minimal polynomial for the modulus 2 γ, γ 1, is the least d such that 2 γ d!.

99 Conjecture The degree of a minimal polynomial for the modulus 2 γ, γ 1, is the least d such that 2 γ d!. Proposition Minimal polynomials for the moduli 2, 4, 8, 16, 32, 64, 128 are A 1 (z, t) := t 2 + t + z modulo 2, A 1 (z, t) 2 modulo 4, A 2 (z, t) := A 1 (z, t) 2 + 4t 3 + 2t 2 + 6t + 2z + 4z 2 modulo 8, A 1 (z, t)a 2 (z, t) modulo 16, A 2 (z, t) 2 modulo 32, A 2 (z, t) 2 modulo 64, t t 7 + t 6 (68z + 18) + t 5 (124z + 24) + t 4 ( 62z z + 81 ) + t 3 ( 20z z + 28 ) + t 2 ( 116z z z + 92 ) + t ( 116z z 2 + 8z + 16 ) + 9z z z z modulo 128.

100 Conjecture The degree of a minimal polynomial for the modulus 3 γ, γ 1, is 2d, where d is the least positive integer such that 3 γ 3 d d!.

101 Conjecture The degree of a minimal polynomial for the modulus 3 γ, γ 1, is 2d, where d is the least positive integer such that 3 γ 3 d d!. Proposition Minimal polynomials for the moduli 3, 9, 27, 81, 243, 729, 2187,..., 3 13 are A 0 (z, t) := t z modulo 3, A 2 0 (z, t) modulo 9, A 3 0 (z, t) modulo 27, ( A 1 (z, t) := t 2 ) 1+z 1 3 ( 9 (1+z) 2 t 2 ) 1+z z (1+z) 5 modulo 81, A 0 (z, t)a 1 (z, t) modulo 243, A 2 0 (z, t)a 1(z, t) modulo 729, A 2 1 (z, t) modulo 2189, A 2 1 (z, t) modulo 38, A 0 (z, t)a 2 1 (z, t) modulo 39, A 2 0 (z, t)a2 1 (z, t) modulo 310, A 3 1 (z, t) modulo 311, Manuel Kauers, Christian A 3 1 (z, t) Krattenthaler and Thomas W. Müller Congruences for differentially modulo algebraic 312 sequences,