New representations for all sporadic Apéry-like sequences, with applications to congruences
aa r X i v : . [ m a t h . N T ] F e b New representations for all sporadic Ap´ery-like sequences, withapplications to congruences
Ofir Gorodetsky
Abstract
Sporadic Ap´ery-like sequences were discovered by Zagier, by Almkvist and Zudilin and by Cooper intheir searches for integral solutions for certain families of second- and third-order differential equations.We find new representations, in terms of constant terms of powers of Laurent polynomials, for all the 15sporadic sequences.The new representations in turn lead to binomial expressions for the sequences, which, as opposedto previous expressions, do not involve powers of 8 and powers of 3. We use these to establish thesupercongruence B np k ≡ B np k − mod p k for all primes p ≥ n, k ≥
1, where B n is asequence discovered by Zagier and known as Sequence B .Additionally, for 14 out of the 15 sequences, the Newton polytopes of the Laurent polynomials usedin our representations contain the origin as their only interior integral point. This property allows us toprove that these 14 sporadic sequences satisfy a strong form of the Lucas congruences, extending workof Malik and Straub. Moreover, we obtain lower bounds on the p -adic valuation of these 14 sequencesvia recent work of Delaygue. Roger Ap´ery’s ingenious proof of the irrationality of ζ (3) = P n ≥ /n and ζ (2) = P n ≥ /n [Ap´e79,Ap´e81] utilized two particular sequences: a n = n X k =0 (cid:18) nk (cid:19) (cid:18) n + kk (cid:19) , b n = n X k =0 (cid:18) nk (cid:19) (cid:18) n + kk (cid:19) (1.1)for n ≥
0. These sequences are referred to as Ap´ery sequences or Ap´ery numbers. Ap´ery constructed explicitrational approximations to ζ (3) and ζ (2), with denominators being small multiples of a n and b n , respectively.These rational approximations turn out to converge quickly enough (in a particular sense) to ζ (3) and ζ (2)as n → ∞ in order to conclude that ζ (3), ζ (2) / ∈ Q .Originally, Ap´ery defined a n and b n through 3-term recurrences with given initial values:( n + 1) a n +1 − (2 n + 1)(17 n + 17 n + 5) a n + n a n − = 0 , ( a = 1 , a = 5)( n + 1) b n +1 − (11 n + 11 n + 3) b n − n b n − = 0 , ( b = 1 , b = 3) (1.2)and showed that a n and b n can be expressed as the binomial sums in (1.1). The recurrences for a n and b n can be restated in terms of differential equations for the g.f.s (generating functions) F a ( z ) = X n ≥ a n z n and F b ( z ) = X n ≥ b n z n , namely, setting θ := z ddz , F a is a solution to the third-order equation( θ − z (2 θ + 1)(17 θ + 17 θ + 5) + z ( θ + 1) ) y = 0 , (1.3)while F b solves the second-order equation( θ − z (11 θ + 11 θ + 3) − z ( θ + 1) ) y = 0 . (1.4)1eukers [Beu87a, Beu87b] showed that F a , F b enjoy modular parametrization. Namely, there is a modularfunction t ( z ) ( z in the upper-half plane) with respect to the modular subgroup Γ (6) such that F a ◦ t is amodular form of weight 2 with respect to Γ (6). Similarly, there is a modular function t ( z ) with respect toΓ (5) such that F b ◦ t is a modular form of weight 1 with respect to Γ (5).This is not a coincidence, and in fact related to the shape of the specific differential equations. Equation(1.4) is a Picard-Fuchs equation, while (1.3) is a symmetric square of a Picard-Fuchs equation. We shall notdiscuss this geometric aspect here, and instead refer the reader to the papers [SB85, PS89, Ver01] and toZagier’s survey [Zag18]. There is no apparent reason for solutions of the recurrences (1.2) to be integers (or, equivalently, for solutionsof (1.3) or (1.4), which are regular at the origin, to have only integral coefficients in their Taylor expansionaround z = 0). It is a surprise that a n and b n are integers. For this reason, Zagier [Zag09] searched fortriples ( A, B, λ ) ∈ Z such that the recurrence( n + 1) u n +1 − ( An + An + λ ) u n + Bn u n − = 0 , n ≥ − , (1.5)has a solution in integers with u − = 0, u = 1. A solution u n gives rise to a power series y = P u n z n satisfying the second-order differential equation( θ − z ( Aθ + Aθ + λ ) + Bz ( θ + 1) ) y = 0 (1.6)and vice-versa. For ( A, B, λ ) = (11 , − ,
3) we already know that there is such a desired solution, namely b n . In his computer search, two well-understood families were found: Legendrian and hypergeometric. Inaddition, six sporadic solutions are found. The two families are easily explained, in the sense that the g.f.s P u n z n have particularly simple forms, which explain the integrality. The six sporadic sequences, whichinclude u n = b n , do not fall into either of these families and are not explained as easily. They are given inthe following table.( A, B, λ ) Name Formula Other names(7 , − , A u n = P nk =0 (cid:0) nk (cid:1) Franel numbers(9 , , B u n = P ⌊ n/ ⌋ k =0 ( − k n − k (cid:0) n k (cid:1)(cid:0) k k (cid:1)(cid:0) kk (cid:1) (10 , , C u n = P nk =0 (cid:0) nk (cid:1) (cid:0) kk (cid:1) (11 , − , D u n = P nk =0 (cid:0) nk (cid:1) (cid:0) n + kn (cid:1) Ap´ery numbers(12 , , E u n = P ⌊ n/ ⌋ k =0 n − k (cid:0) n k (cid:1)(cid:0) kk (cid:1) (17 , , F u n = P nk =0 ( − k n − k (cid:0) nk (cid:1) P kj =0 (cid:0) kj (cid:1) Zagier conjectured that his (finite) search yielded all the integral solutions to (1.5). This is suggested by thefact that so few sporadic solutions were found in a very large parameter domain. As in the case of b n , thedifferential equations associated with the sequences found by Zagier are Picard–Fuchs equations, and thesolutions to the differential equations have modular parametrizations [Zag09, § § a n . Almkvist and Zudilin [AZ06] searched for ( a, b, c ) ∈ Z such that the recurrence ( n + 1) u n +1 − (2 n + 1)( an + an + b ) u n + cn u n − = 0has integer solutions with u − = 0, u = 1. They found 6 sporadic solutions, including a n , which correspondsto ( a, b, c ) = (17 , , a, b, c ) Name Formula Other names(7 , ,
81) ( δ ) u n = P ⌊ n/ ⌋ k =0 ( − k n − k (cid:0) n k (cid:1)(cid:0) n + kn (cid:1)(cid:0) k k (cid:1)(cid:0) kk (cid:1) Almkvist-Zudilin numbers(11 , , η ) u n = P ⌊ n/ ⌋ k =0 ( − k (cid:0) nk (cid:1) (cid:16)(cid:0) n − k − n (cid:1) + (cid:0) n − k n (cid:1)(cid:17) (10 , ,
64) ( α ) u n = P nk =0 (cid:0) nk (cid:1) (cid:0) kk (cid:1)(cid:0) n − kn − k (cid:1) Domb numbers(12 , ,
16) ( ǫ ) u n = P nk = ⌈ n/ ⌉ (cid:0) nk (cid:1) (cid:0) kn (cid:1) (9 , , −
27) ( ζ ) u n = P k,l (cid:0) nk (cid:1) (cid:0) nl (cid:1)(cid:0) kl (cid:1)(cid:0) k + ln (cid:1) (17 , ,
1) ( γ ) u n = P nk =0 (cid:0) nk (cid:1) (cid:0) n + kk (cid:1) Ap´ery numbersLater, Cooper [Coo12] introduced an additional parameter d ∈ Z :( n + 1) u n +1 − (2 n + 1)( an + an + b ) u n + n ( cn + d ) u n − = 0 . A solution u n gives rise to a power series y = P u n z n satisfying the third-order differential equation( θ − z (2 θ + 1)( aθ + aθ + b ) + z ( c ( θ + 1) + d ( θ + 1))) y = 0 (1.7)and vice-versa. Cooper found 3 additional sporadic solutions, named s , s and s . He established theirmodular parametrization (the subscript is related to the level of the corresponding modular form). Formulasfor the terms of s , s and s were obtained by Zudilin:( a, b, c, d ) Name Formula Other names(13 , , − , s u n = P nk = ⌈ n/ ⌉ (cid:0) nk (cid:1) (cid:0) n + kk (cid:1)(cid:0) kn (cid:1) (6 , , − , s u n = P nk =0 (cid:0) nk (cid:1) Yang-Zudilin numbers(14 , , , − s u n = P ⌊ n/ ⌋ k =0 ( − k (cid:0) nk (cid:1)(cid:0) kk (cid:1)(cid:0) n − kn − k (cid:1) (cid:16)(cid:0) n − k − n (cid:1) + (cid:0) n − kn (cid:1)(cid:17) In total, we have 15 sporadic solutions, which are often referred to as Ap´ery-like sequences.
We say that a sequence { u n } n ≥ is the constant term sequence of a Laurent polynomial Λ ∈ C [ x , x − , x , x − , . . . , x d , x − d ]if u n is the constant term of Λ n for every n ≥
1. Our main result is
Theorem 1.1.
All 15 sporadic sequences are constant term sequences of Laurent polynomials with integercoefficients. If the sequence satisfies a k th order differential equation ( k ∈ { , } ), the polynomial may betaken to be k -variate.For all the sporadic sequences, the Laurent polynomials may be chosen in such a way that they factorizeinto a product of Laurent polynomials with coefficients in the set {− , , } only.For all but possibly ( η ) , the Laurent polynomials may be chosen in such a way that their Newton polytopescontain the origin as their only interior integral point. Specific Laurent polynomials are given in Proposition 3.3, from which we derive new binomial sumrepresentations for B , F and ( δ ). Proposition 1.2.
Consider the set S ( n ) = n ( a , b , c ) ∈ Z ≥ : a + a + a = n,b + b + b = n,c + c + c = n, ∀ ≤ i ≤ a i + b i + c i = n, | b +2 b +2 c + c o . The sequence B can be written as − n B n = 3 X ( a , b , c ) ∈ S ( n ) (cid:18) n a (cid:19)(cid:18) n b (cid:19)(cid:18) n c (cid:19) − (cid:18) nn, n, n (cid:19) . onsider the set T ( n ) = (cid:26) ( a , b , c , d , e ) ∈ Z ≥ : a + a + a = n,b + b + b = n,c + c + c = n, d + d + d = n,e + e + e = n, ∀ ≤ i ≤ a i + b i + c i + d i +2 e i =2 n (cid:27) . The sequence F can be written as F n = X ( a , b , c , d , e ) ∈ T ( n ) (cid:18) n a (cid:19)(cid:18) n b (cid:19)(cid:18) n c (cid:19)(cid:18) n d (cid:19)(cid:18) n e (cid:19) ( − a + b + c . (1.8) Consider the set U ( n ) = ( ( a , b , c , d ) ∈ Z ≥ : a + a + a = n,b + b + b = n,c + c + c = n,d + d + d = n, b + c + d = n,a + b + d = n,a + b + c = n ) . The sequence ( δ ) can be written as AZ n = X ( a , b , c , d ) ∈ U ( n ) (cid:18) n a (cid:19)(cid:18) n b (cid:19)(cid:18) n c (cid:19)(cid:18) n d (cid:19) ( − a + b + d . (1.9) Both a n and b n possess a wealth of arithmetic properties. We concentrate on Gauss congruences. Definition 1.3.
A sequence { u n } n ≥ of integers is said to satisfy the Gauss congruences of order r if, forall primes p ≥ r + 1 and all positive integers k, n we have u np k ≡ u np k − mod p rk . (1.10)The case r = 1 is simply called ‘Gauss congruences’, while the case r > supercongru-ences . A well-known example of a sequence satisfying the Gauss congruences is u n = a n for any choice of a ∈ Z . Another example of such a sequence is given by u n = (cid:0) nn (cid:1) [Rob00, Ch. 5.3.3]. In fact, (cid:0) nn (cid:1) satisfiesthe Gauss congruences of order 3, as shown by Jacobsthal [BSF +
52] (cf. [Rob00, Ch. 7.1.6]).Chowla, Cowles and Cowles [CCC80], based on numerical observations, made several conjectures on thevalues of a n modulo primes and prime powers. Gessel [Ges82] proved generalizations of their conjectures.In particular, in [Ges82, Thm. 3] he proved that a n satisfies the Gauss congruences of order 3, but only inthe special case where k = 1 in (1.10). This was established independently by Mimura [Mim83, Prop. 2].Finally, in his PhD thesis, Coster [Cos88] showed that the Gauss congruences of order 3 are satisfied by a n (without any restriction on k ), as well as by b n .Following numerical observations, it was conjectured that all sporadic sequences satisfy the Gauss con-gruences of order 2 or 3 (depending on the sequence), see [OSS16, Table 2]. The fact that they all satisfythe Gauss congruences of order 1 is explained by modular parametrization, see e.g. [Ver10, OS11a, Moy13].Establishing Gauss congruences of higher order is a harder task. In his thesis, Coster [Cos88] establishedsuch congruences for ‘generalized Ap´ery numbers’, that is u ( r,s,ε ) n = n X k =0 (cid:18) nk (cid:19) r (cid:18) n + kk (cid:19) s ε k for ε ∈ {± } and r, s ≥
0. Specifically, he proved that u ( r,s,ε ) satisfies the Gauss congruences of order 3 aslong as r ≥
2, or r = 1, s ≥ ε = −
1. The sequences u ( r,s,ε ) include, as special cases, the sequences( γ ) and D , as well as A and s . Coster’s method for proving these congruences was extended and used byvarious authors: • Osburn and Sahu [OS11b] proved that C satisfies the Gauss congruences of order 3. • Later, Osburn and Sahu [OS13] proved that ( α ) satisfies the Gauss congruences of order 3 as well,extending a result of Chan, Cooper and Sica [CCS10]. They also sketched the proof of a similar resultfor E (using the representation (3.3)). 4 Osburn, Sahu and Straub [OSS16] proved that s satisfies the Gauss congruences of order 3, as wellas a family that includes s , D and ( ǫ ). In [OSS16, Ex. 3.1] they sketch a proof of a similar result for( η ).All these sequences involved only one summation variable, k . Recently, the author established that thesequence ( ζ ), whose definition involves summation over two variables k and l , satisfies the Gauss congruencesof order 3 [Gor19, Cor. 1.4]. Although the proof uses q -congruences, this can be avoided, and one canformulate the proof using Coster’s method.This leaves three sequences for which little is known: B , F and ( δ ). It was proved by Amdeberhan andTauraso [AT16] that ( δ ) satisfies the Gauss congruences of order 3, but only in the special case where k = 1in (1.10). The main difficulty in dealing with B , F and ( δ ) has to do with the appearance of powers (of 3and 8) in their definition. As opposed to binomial coefficients, which behave nicely modulo high powers ofprimes, powers’ behavior is much worse. For instance, u n = 2 n satisfies the Gauss congruences of order 1but not of higher order, as opposed to u n = (cid:0) nn (cid:1) . In fact, it is not even known if there are infinitely manyprimes p such that 2 p ≡ mod p ; such primes are known as Wieferich primes. We prove the following. Theorem 1.4.
Sequence B satisfies the Gauss congruences of order 2. This confirms a conjecture of Osburn, Sahu and Straub [OSS16].
Remark 1.5.
Theorem 1.1 also gives a new proof for the fact that all the sporadic sequences are integraland satisfy the Gauss congruences (of order 1), since these congruences hold for the constant terms of powersof any integral Laurent series.
Acknowledgements
In addition to useful comments, corrections and suggestions, I would like to thank Armin Straub for crucialcomputational help in the proof of Proposition 3.3 for ( η ), and Wadin Zudilin for letting me know of thesurvey [Zag18]. This project has received funding from the European Research Council (ERC) under theEuropean Union’s Horizon 2020 research and innovation programme (grant agreement No 851318). Definition 2.1.
A sequence { u n } n ≥ of integers is said to satisfy the Lucas congruences if, for any prime p , and any positive integer n with base- p expansion P si =0 n i p i (0 ≤ n i < p ) we have u n ≡ s Y i =0 u n i mod p. The name of these congruences comes from a theorem of Lucas [Luc78], stating that if n = P si =0 n i p i and k = P si =0 k i p i are base- p expansions of n ≥ k ≥
0, then (cid:18) nk (cid:19) ≡ s Y i =0 (cid:18) n i k i (cid:19) mod p. Gessel [Ges82, Thm. 1], inspired by the previously mentioned conjectures of Chowla, Cowles and Cowles, hasshown that a n satisfies the Lucas congruences. Deutsch and Sagan [DS06, Thm. 5.9] showed more generallythat for any positive integers r and s , the sequence u n = n X k =0 (cid:18) nk (cid:19) r (cid:18) n + kk (cid:19) s satisfies the Lucas congruences, which implies that a n and b n satisfy the Lucas congruences. They alsoshowed that u n = (cid:0) nn (cid:1) satisfies Lucas congruences [DS06, Thm. 4.4]. A property stronger than Lucascongruences is the D3 property, named after Dwork [Dwo69].5 efinition 2.2. Let { u n } n ≥ be a sequence of integers, and set u := 1. We say that { u n } n ≥ satisfies theD3 congruences with respect to a prime p if u n + mp s u ⌊ np ⌋ ≡ u n u ⌊ n + mpsp ⌋ mod p s (2.1)for all s, m, n ≥
0. If this holds for all primes p , we simply say that { u n } n ≥ satisfies the D3 congruences.These congruences are implicit in the work of Dwork [Dwo69] and have significance in p -adic analysis. Werefer the reader to [SvS15, MV16] for further information, as well as definitions of related congruences, knownas D1 and D2, which together imply D3. One can show that (2.1) with s = 1 implies the Lucas congruences.Samol and van Straten [SvS15] proved that a certain class of sequences satisfies the D3 congruences. Adifferent proof of this was found by Mellit and Vlasenko [MV16]. Before we state their result, we recall thatthe Newton polytope of a Laurent polynomial f = P ( a ,...,a d ) ∈ Z d c a ,...,a d x a · · · x a d d is the convex hull of thesupport of f , that is, of { ( a , . . . , a d ) ∈ Z d : c a ,...,a d = 0 } . We use the notation CT( f ) for the constant term c , ,..., of f (or any Laurent series). Theorem 2.3. [SvS15, MV16] Let Λ ∈ Z [ x , x − , . . . , x d , x − d ] be a Laurent polynomial. If the Newtonpolytope of Λ contains the origin as its only interior integral point, then the sequence u n = CT(Λ n ) satisfiesthe D3 congruences. As observed e.g. by Mellit and Vlasenko [MV16, § b n is the constant term of the n th power ofΛ = (1 + x )(1 + x )(1 + x + x ) x x . (2.2)Straub [Str14, Rem. 1.4] observed that a n is the constant term of the n th power ofΛ = ( x + x )( x + 1)( x + x + x )( x + x + 1) x x x . (2.3)A short computation shows that the Newton polytopes of (2.2) and (2.3) satisfy the conditions of Theo-rem 2.3, and so a n and b n satisfy the D3 congruences. Malik and Straub [MS16, Thm. 3.1] proved that all 15 Ap´ery-like sequences satisfy the Lucas congruences.For all sequences except for ( η ) and s they did this by extending and applying a general method developedby McIntosh [McI92], while for ( η ) and s a much more delicate analysis was needed. From Theorems 2.3and 1.1 we immediately obtain Corollary 2.4.
All the sporadic sequences, except possibly ( η ) , satisfy the D3 congruences. p -adic valuation of sporadic sequences Given a sequence of integers u = { u n } n ≥ and a prime p , let Z p ( u ) = { ≤ i ≤ p − p | u i } . For anynon-negative integer n whose base- p expansion is n = P Ni =0 n i p i , let α p ( u , n ) = { ≤ i ≤ N : n i ∈ Z p ( u ) } .Let v p ( n ) be the p -adic valuation of an integer n . Recently, Delaygue proved that v p ( u n ) ≥ α p ( u , n )under two conditions on { u n } n [Del18, Thm. 2]. The first condition is that u n = CT(Λ n ) for some Laurentpolynomial Λ with integer coefficients, whose Newton polytope contains the origin as its only interior integralpoint. This holds for all the sporadic sequences except possibly ( η ). The second condition involves arestriction on the differential operator annihilating P n ≥ u n z n . According to (1.6), (1.7) and [MS16, p. 23],this condition holds for all 15 sequences. We thus obtain Corollary 2.5.
Every sporadic sequence { u n } n , except possibly ( η ) , satisfies v p ( u n ) ≥ α p ( u , n ) for every n ≥ and every prime p . Constant term and binomial representations for sporadic se-quences
Given a power series A ( x , . . . , x d ) = P ( a ,...,a d ) ∈ Z d ≥ c a ,a ,...,a d x a x a · · · x a d d ∈ C [[ x , . . . , x d ]], we define itsdiagonal as the power series ∆ d ( A ) := X i ≥ c i,i,...,i u i ∈ C [[ u ]] . Given a rational function A ( x , . . . , x d ) ∈ C ( x , . . . , x d ) whose denominator does not vanish at the origin,we can identify it with its Taylor expansion around ~
0. For such A we have the following result of Lipshitz. Theorem 3.1. [Lip88] Let A ∈ K ( x , . . . , x d ) ( K a field of characteristic ). If the denominator of A doesnot vanish at the origin, then the diagonal ∆ d ( A ) satisfies a homogeneous linear differential equation withcoefficients in K [ u ] . Given a Laurent series f ( x , . . . , x d ) = P ( a ,...,a d ) ∈ Z d c a ,a ,...,a d x a x a · · · x a d d ∈ C [[ x , . . . , x d , x − , . . . , x − d ]],we define its constant term series as the power seriesCTS( f ) := X i ≥ CT( f i ) u i ∈ C [[ u ]] . The constant term series of f is also known as its fundamental period . Suppose a Laurent polynomial A ( x )in d variables becomes a polynomial when multiplied by Q di =1 x i . Then it is known that A can be expressedas the diagonal of a rational function in d + 1 variables, namelyCTS( A ) = ∆ d +1 (cid:18) − x x . . . x d +1 A ( x , . . . , x d ) (cid:19) . (3.1)In general, the diagonal of a rational function cannot be expressed as the constant term series of a Laurentpolynomial. However, we have the following partial converse. This converse is a special case of the MacMahonMaster Theorem [Mac60]. Lemma 3.2.
Given a square matrix M = ( M i,j ) ≤ i,j ≤ n ∈ Mat d ( C ) , consider the rational function A ( x , . . . , x d ) := 1det( I d − M · Diag( x , . . . , x d )) . We have ∆ d ( A ) = CTS Q di =1 ( P dj =1 M i,j x j ) x x · · · x d ! . (3.2) As the rational function in the right-hand side of (3.2) is homogeneous of degree , we may substitute x i = 1 for some i without changing the constant term series. There are various benefits that come from representing a sequence as the diagonal of a rational function,some of which are described in [Str14, p. 1987]. For instance, it allows one to extract the asymptoticsof the sequence directly from the rational function. A much simpler benefit, which we make use of inProposition 1.2, is that it allows us to easily extract binomial sum representations for the sequence . In [Zag09, § An example of a binomial sum is (1.1). The interested reader may find a rigorous definition of ‘binomial sums’ in [BLS17]. A and ( γ ) can be found in [Str94, Sch95], for ( δ ), ( α ) and ( γ ) can be foundin [CZ10] and for ( α ) and ( γ ) can be found in [CTYZ11]. Additional representations for A , D , E and ( α )are found in their OEIS pages [Slo96], and we single out the following representation for E , u n = n X k =0 (cid:18) nk (cid:19)(cid:18) kk (cid:19)(cid:18) n − kn − k (cid:19) . (3.3)This formula for E was used in [OS13] in establishing Gauss congruences of order 2; the original representationof E , given by Zagier, involves powers of 4, which get in the way of establishing supercongruences. A proofof (3.3) is sketched in [Del18, p. 257]. Bostan, Lairez and Salvy [BLS17, Thm. 3.5] proved (algorithmically) that if { u n } n ≥ is a binomial sum thenthe g.f. of u n is a diagonal of a rational function. Hence, every sporadic sequence is the diagonal of somerational function. We are aware of the following explicit diagonal representations for the 10 sequences A , B , C , D , F , ( γ ), ( δ ), ( α ), ( ǫ ), s and s : • Zagier [Zag09] already explained how his Sporadic sequences have constant term representations (cf.Verrill [Ver99]). In particular, B is the constant term sequence of x + x + x − x x x − x x x (3.4)and so of ( x + x + 1 − x x ) / ( − x x ) as well, and F is the constant term sequence of x x + x x + x + x + x + x − x x − x x . • Various rational functions whose diagonal is the g.f. for ( γ ) are given in [Chr84, Chr85, BBC +
13, Str14].Of particular interest is the one found by Straub, which involves 4 variables, while others involvebetween 5 to 8 variables. His rational function is [Str14, Thm. 1.1]((1 − x − x )(1 − x − x ) − x x x x ) − . (3.5)Recently, Gheorghe Coserea added to the OEIS [Slo96] two rational functions, in 4 variables, whosediagonal is the g.f. of ( γ ): (1 − ( x x x x + x x x + x x + x x + x x + x + x )) − and (1 − x (1 + x )(1 + x )(1 + x )( x x x + x x + x + x + 1)) − . • Straub observed [Str14, Ex. 3.4] that the diagonal of((1 − x − x )(1 − x ) − x x x ) − (3.6)is the g.f. of D . Bostan, Boukraa, Maillard and Weil [BBMW15], as well as Coserea [Slo96], havefound very similar representations. • Straub observed that((1 − x )(1 − x )(1 − x ) − x x x ) − , (1 − x − x − x + 4 x x x ) − (3.7)are both diagonals for A [Str14, Ex. 3.6, 3.7]. In the OEIS, Coserea provides three more rationalfunctions, in 3 variables as well, whose diagonal is the g.f. of A :(1 − x x − x x − x x − x x x ) − , (1 + x + x + x + 2( x x + x x + x x ) + 4 x x x ) − (3.8)and (1 − ( x + x + x ) + x x + x x + x x − x x x ) − .8 Straub also observed [Str14, Eq. (32)] that the g.f. for ( δ ) coincides with the diagonal of(1 − ( x + x + x − x ) − x x x x ) − . • The diagonal of ((1 − x )(1 − x )(1 − x )(1 − x ) − x x x x ) − is the g.f. for s , see Straub [Str14,Ex. 3.6]. Coserea [Slo96] also provides the rational function (1 − ( x x x + x x x + x x x + x x x + x x + x x )) − . • Sequences C and ( α ) are the constant term sequences of the Laurent polynomial P d ( x , . . . , x d ) =( x + x + . . . + x d ) (cid:16) x + . . . + x d (cid:17) with d = 3 and d = 4, respectively (see [BSW13, Ex. 6], [BSV16,Ex. 3.17], [BNSW11, Eq. (8)]). By (3.1), this also gives diagonal representations for C and ( α ). • Coserea lists the functions (1 − ( x x + x x − x x + 3 x x x )) − , (1 − ( x + x − x + 3 x x x )) − and (1 + x + x + x − x x x ) − as rational functions whose diagonal is the g.f. for B [Slo96]. • Coserea has also found that the diagonal of (1 − ( x ( x x + x x + x x ) + x x + x x + x + x )) − is the g.f. for s . • In [SZ15, Ex. 4.2], Straub and Zudilin prove that the g.f. of ( ǫ ) is the diagonal of the Kauers–Zeilbergerrational function (1 − ( x + x + x + x ) + 2 e ( x , x , x , x ) + 4 x x x x ) − , where e is the thirdelementary symmetric polynomial.In the proof of [Str14, Thm. 1.1], Straub observes that the rational function (3.5) may be written as det( I − M · Diag( x , . . . , x )) − where M = , and so by Lemma 3.2 it gives rise to a constant term representation for ( γ ), the one defined by (2.3).Representations of the form det( I d − M · Diag( x , . . . , x d )) − are hidden in some of the other functions thatwe listed, as we now show. The rational function in (3.6) can be written as det( I − M · Diag( x , x , x )) − where M = , and by Lemma 3.2, we find that the constant term series of (2.2) is the g.f. of D . The rational functions in(3.7) and (3.8) can be written as det( I − M · Diag( x , x , x )) − for M = , − − , , − − − − − − . For Zagier’s sequences, he explains in [Zag09, §
7] why his six sequences have constant term representations.However, we are not aware of a reason why the associated Laurent polynomials should always have the originas the only interior point of their Newton polygon, and also not aware of a reason for Zagier’s argument,involving Picard-Fuchs equations, to generalize to the sequences of Almkvist and Zudilin and of Cooper.The discussion in § B , C , F , ( γ ) and( α ), and also that previously known diagonal representations for A and D in fact give rise to constant termrepresentations through Lemma 3.2.We call a Laurent polynomial good if it may be factored into a product of Laurent polynomial, each ofwhich has coefficients in the set {− , , } . We have conducted a computer search of the following form: weenumerated good Laurent polynomials f (in several variables) with bounded degree and coefficients. Foreach such polynomial we tested whether CTS( f ) is the g.f. of a sporadic sequence. For each of the 159equences, the search yielded more than one Laurent polynomial whose diagonal is its g.f.. The reason forlooking only at good Laurent polynomials is that they lead to binomial sum representations which do notinvolve powers of integers other than ± Proposition 3.3.
The constant term series of the Laurent polynomials in the second column are g.f.s of thesequences in the first column:Sequence Laurent polynomial A ( xy ) − ( x + 1)( y + 1)( x + y ) B ( − xy ) − ( x + y + 1)( x + y − xy − x − y + 1) C ( xy ) − ( x + y + 1)( xy + x + y ) D ( xy ) − ( x + 1)( y + 1)( x + y + 1) E ( − xy ) − ( xy + x + y − xy − x − y − F ( xy ) − ( − x + y + 1)( x − y + 1)( x + y − x + y + 1)( x + y + 1) Sequence Laurent polynomial ( δ ) ( xyz ) − ( y − z + 1)( − x + y + z )( x + z + 1)( x + y − xyz ) − ( xy + yz + zx )( x + y + z − xy − yz − zx + x + y + z + 1)( η ) ( xyz ) − ( zx + xy − yz − x − xy + yz − zx − y − yz + zx − xy − z − α ) ( xyz ) − ( − x − y − z + 1)( x − y )( x − y + z + 1)( x + y − z + 1)( xyz ) − ( x + y + z + 1)( xyz + xy + yz + zx )( ǫ ) ( xyz ) − ( x + 1)( y + 1)( z + 1)( x + y + z + 1)( ζ ) ( xyz ) − ( x + y + z )( x + y + z + xy + yz + zx + xyz )( γ ) ( xyz ) − ( y + z )( x + 1)( x + y + 1)( x + y + z ) Sequence Laurent polynomial s ( xyz ) − ( x − y + 1)( x + z )( y − x − z + 1) s ( xyz ) − ( x + 1)( y + 1)( z + 1)( xyz + 1) s ( − xyz ) − ( x + y + z − xy − yz − zx − x + y − z )( x + y + z + xy + yz − zx + x + y + z )As can be seen, we were always able to find good symmetric polynomials, except in the cases of ( γ ), s and s . For ( γ ), P = ( xyz ) − ( x + y + z + 1)( x y + xy + y z + yz + z x + zx + xy + yz + zx + 2 xyz ) is asymmetric polynomial whose constant term series is (empirically) the g.f. of ( γ ), but P is not good accordingto our definition. For s , P = ( xyz ) − ( xy + yz + zx + x + y + z )( x + y + z − xy − yz − zx − x − y − z + 1)is a symmetric polynomial whose constant term series is (empirically) the g.f. of s , and is good accordingto our definition, but we do not know how to prove it. Note that the polynomial for B was given by Zagier(see (3.4)), although it was not factored; its factorization is useful in the proofs of Proposition 1.2 andTheorem 1.4. Proof of Proposition 3.3.
For each of the Laurent polynomials in the proposition, we may easily computethe first three terms and see that they agree with the first three terms of the respective sequence. It nowsuffices to compute the differential operator annihilating the constant term series of the polynomial, andthat it coincides with (1.6) or (1.7).The proof of Theorem 3.1 is constructive and may give rise to such a differential operator (by appealingto (3.1)). However, the algorithm arising from Lipshitz’s theorem is often not efficient enough for practicalpurposes. However, Zeilberger’s Creative Telescoping algorithm [Zei91], as generalized by Chyzak [Chy00],gives a much faster way to compute the operator. See [BCvHP12] for more details.The algorithm is implemented in the package
Mgfun To find an operator annihilating the diagonal of f ( s, t, x ), we run the command Version 4.1, available from https://specfun.inria.fr/chyzak/mgfun.html . _, S_, T_ := op(op(Mgfun:-creative_telescoping(F, x::diff, [s::diff, t::diff]))) on F := ( st ) − f ( s, t/s, x/t ), and P will be the operator. So if we want to find the operator annihilatingCTS( f ( s, t )), we shall run it on F := ( st ) − (1 − xf ( s, t/s, x/t )) − . To find an operator annihilating thediagonal of f ( s, t, x, w ), we run the command P_, S_, T_, X_ := op(op(Mgfun:-creative_telescoping(F, w::diff, [s::diff, t::diff, x::diff]))) on F := ( stx ) − f ( s, t/s, x/t, w/x ), and P will be the operator. So if we want to find the operator annihilatingCTS( f ( s, t, x )), we shall run it on F := ( st ) − (1 − xf ( s, t/s, x/t )) − .In the two-variate case, the algorithm returns the expected operator in under a minute on an averagelaptop. For the three-variate case, the algorithm takes a few minutes when we ran it on the polynomialsfor ( ǫ ), ( ζ ), ( γ ), s and s . For ( δ ), ( η ), ( α ) and s we had to ‘massage’ the Laurent polynomials slightlybefore running the program, as we now explain.For ( δ ), we consider f = ( xyz ) − ( y − z + 1)( − x + y + z )( x + z + 1)( x + y − M = − − − , we have CTS ( f ) = ∆ ( 1det( I − M · Diag( x , . . . , x )) ) . Instead of running the algorithm on ( stx ) − (1 − wf ( s, t/s, x/t, w/x )) − , we ran it on ( stx ) − g ( s, t/s, x/t, w/x )where g := det( I − M · Diag( x , . . . , x )), and it proved to be much quicker. The same trick works for the firstpolynomial of ( α ). The second polynomial for α comes from the polynomial ( x + x + x + x ) (cid:16) x + . . . + x (cid:17) mentioned in § x = 1.For the second polynomial f for ( δ ), note that CT( f n ) = CT( g n ) for g = f ( xyz , yz , z ). We ran thealgorithm on ( stx ) − (1 − wg ( s, t/s, x/t, w/x )) − . For s , the same trick works.Finally, for ( η ) we use a different implementation of Creative Telescoping, which turned out to be fasterfor this example, namely the Mathematica package
HolonomicFunctions by Christoph Koutschan. To findthe recurrence satisfied by CT( f n ) for the polynomial of ( η ), we ran CreativeTelescoping[((z x + x y - y z - x - 1) (x y + y z - z x - y - 1)(y z + z x - x y - z - 1)/(x y z))^n/(x y z), Der[x], {Der[y], Der[z], S[n]}]CreativeTelescoping[First[%], Der[y], {Der[z], S[n]}]CreativeTelescoping[First[%], Der[z], {S[n]}] which outputs the expected recurrence.
We use bold Latin letters a , b , . . . as short for sequences ( a , a , a ) , ( b , b , b ) , . . . . If e.g. a sums to n wewrite (cid:0) n a (cid:1) for the multinomial coefficient (cid:0) na ,a ,a (cid:1) . For the first part, note that (cid:18) ( x + y + 1)( x + y − xy − x − y + 1) − xy (cid:19) n = ( x + y + 1) n ( x + yω + ω ) n ( x + yω + ω ) n ( − xy ) n = ( − xy ) − n X a + a + a = n x a y a (cid:18) n a (cid:19) X b + b + b = n x b y b ω b +2 b (cid:18) n b (cid:19)X c + c + c = n x c y c ω c + c (cid:18) n c (cid:19) , Version 1.7.3, available from . ω = e πi/ . By Proposition 3.3, CTS(( x + y + 1)( x + y − xy − x − y + 1) / ( − xy )) is the generatingseries for B n , and so B n = ( − n X a ,a ,a b ,b ,b c ,c ,c (cid:18) n a (cid:19)(cid:18) n b (cid:19)(cid:18) n c (cid:19) ω b +2 b +2 c + c , (3.9)where the sum is over tuples ( a , a , a , b , b , b , c , c , c ) of non-negative integers satisfying a + a + a = b + b + b = c + c + c = n and a i + b i + c i = n for i = 1 , ,
3. Observe that ℜ (2 ω k ) = 3 · | k −
1. Since B n is real, we may take the real part of (3.9), obtaining2( − n B n = 3 X a ,a ,a b ,b ,b c ,c ,c (cid:18) n a (cid:19)(cid:18) n b (cid:19)(cid:18) n c (cid:19) − CT (cid:18)(cid:18) ( x + y + 1) xy (cid:19) n (cid:19) , where we sum over tuples ( a , a , a , b , b , b , c , c , c ) in S ( n ). The second term evaluates to (cid:0) nn,n,n (cid:1) by themultinomial theorem, as needed. For the second part, note that (cid:18) ( − x + y + 1)( x − y + 1)( x + y − x + y + 1)( x + y + 1) x y (cid:19) n = ( − x + y + 1) n ( x − y + 1) n ( x + y − n ( x + y + 1) n ( x + y + 1) n ( xy ) n = ( xy ) − n X a + a + a = n x a y a ( − a (cid:18) n a (cid:19) X b + b + b = n x b y b ( − b (cid:18) n b (cid:19)X c + c + c = n x c y c ( − c (cid:18) n c (cid:19) X d + d + d = n x d y d (cid:18) n d (cid:19)X e + e + e = n x e y e (cid:18) n e (cid:19) . Since CTS(( − x + y + 1)( x − y + 1)( x + y − x + y + 1)( x + y + 1) / ( x y )) is the generating series for F n (by Proposition 3.3), one obtains (1.8). For the third part, note that (cid:18) ( y − z + 1)( − x + y + z )( x + z + 1)( x + y − xyz (cid:19) n = ( y − z + 1) n ( − x + y + z ) n ( x + z + 1) n ( x + y − n ( xyz ) n = ( xyz ) − n X a + a + a = n y a z a ( − a (cid:18) n a (cid:19) X b + b + b = n x b y b z b ( − b (cid:18) n b (cid:19)X c + c + c = n x c z c (cid:18) n c (cid:19) X d + d + d = n x d y d ( − d (cid:18) n d (cid:19) . Since CTS(( y − z + 1)( − x + y + z )( x + z + 1)( x + y − / ( xyz )) is the generating series for AZ n (byProposition 3.3), one obtains (1.9). For A - E , we can check that the polynomials in Proposition 3.3 satisfy the conditions of the theorem bydrawing the corresponding Newton polygon. For instance,12 y is the Newton polygon of the polynomial corresponding to A in Proposition 3.3. For F , the polynomial P ( x, y ) = ( xy ) − ( − x + y + 1)( x − y + 1)( x + y − x + y + 1)( x + y + 1) does not satisfy the conditions,but it turns out that P ( x, y ) = Q ( x , y ) for Q ( x, y ) = ( − xy ) − ( x + y + 1)( x + y − xy − x − y + 1).Since CT( P n ) = CT( Q n ), it suffices to check the Newton polygon of Q , which is given below: xy Indeed, (0 ,
0) is its only interior point. For ( δ )-( γ ) and s - s , drawing a picture is harder. In SageMath ,one can check whether a Laurent polynomial in 3 variables contains the origin as its only interior point usingthe following procedure: def checkInterior(poly):polyhedron = Polyhedron(vertices = poly.exponents())for pt in polyhedron.integral_points():if tuple(pt) != (0,0,0) and polyhedron.interior_contains(pt):return Falsereturn polyhedron.interior_contains((0,0,0))
For instance, the following code checks that the first polynomial of ( δ ) (from Proposition 3.3) satisfies theconditions of the theorem: R.
This allowed us to verify that the Newton polytopes contain the origin as their only interior point, except inthe case of ( η ), where (1 , ,
0) and (1 , ,
0) (as well as their permutations) are additional interior points.13
Proof of Theorem 1.4
The following lemma is a classical result of Jacobsthal [BSF + Lemma 4.1.
Let p ≥ be a prime and n ≥ m ≥ be integers. Then (cid:0) pnpm (cid:1)(cid:0) nm (cid:1) ≡ p v p ( nm ( n − m ))+3 − δ p, where δ p, = 1 if p = 3 and δ p, = 0 otherwise. The identity (cid:0) nm (cid:1) = (cid:0) n − m − (cid:1) n/m yields the following congruence. Lemma 4.2.
Let p be a prime and n ≥ m ≥ be integers. Suppose that v p ( n ) ≥ v p ( m ) . Then (cid:18) nm (cid:19) ≡ p v p ( n ) − v p ( m ) . As (cid:0) nn,n,n (cid:1) = (cid:0) n n (cid:1)(cid:0) nn (cid:1) satisfies the Gauss congruences of order 2 by Lemma 4.1, and so does ( − n , we dealwith ˜ B n = 2( − n B n + (cid:0) nn,n,n (cid:1) instead of B n .The vector ( a , b , c ) stands for ( a , a , . . . , c , c ) ∈ Z . Let k, n ≥ p ≥ p · ( a , b , c ) for ( p · a , p · b , p · c ), and p | ( a , b , c ) if all the entries of ( a , b , c ) are divisible by p . Given( a , b , c ) ∈ S ( m ) we write B ( a , b , c ) = (cid:18) m a (cid:19)(cid:18) m b (cid:19)(cid:18) m c (cid:19) . Let p be an odd prime and let n, k ≥
1. By Proposition 1.2 we may write ˜ B np k − ˜ B np k − as˜ B np k − ˜ B np k − = 3( S + S + S ) (4.1)where S = X ( a , b , c ) ∈ S ( np k ) p ∤ ( a , b , c ) B ( a , b , c ) ,S = X ( a , b , c ) ∈ S ( np k − ) ( B ( p · ( a , b , c )) − B ( a , b , c )) ,S = X p · ( a , b , c ) ∈ S ( np k )( a , b , c ) / ∈ S ( np k − ) B ( p · ( a , b , c ))Here we have used the following observation: if ( a , b , c ) ∈ S ( np k − ) then p · ( a , b , c ) ∈ S ( np k ).Let ( a , b , c ) ∈ S ( np k ) with p ∤ ( a , b , c ). Suppose without loss of generalization that p ∤ a . At least oneof b and c is indivisible by p as well, because of the condition a + b + c = np k . Suppose without lossof generalization that p ∤ b . The integer B ( a , b , c ) is divisible by (cid:0) np k a (cid:1)(cid:0) np k b (cid:1) , which by Lemma 4.2 impliesthat B ( a , b , c ) ≡ p k , and so S ≡ p k .We now deal with S . Let ( a , b , c ) ∈ S ( np k − ). We may write both B ( p · ( a , b , c )) and B ( a , b , c ) as aproduct of 6 binomial coefficients, namely B ( p · ( a , b , c )) = (cid:18) np k pa (cid:19)(cid:18) p ( a + a ) pa (cid:19)(cid:18) np k pb (cid:19)(cid:18) p ( b + b ) pb (cid:19)(cid:18) np k pc (cid:19)(cid:18) p ( c + c ) pc (cid:19) ,B ( a , b , c ) = (cid:18) np k − a (cid:19)(cid:18) a + a a (cid:19)(cid:18) np k − b (cid:19)(cid:18) b + b b (cid:19)(cid:18) np k − c (cid:19)(cid:18) c + c c (cid:19) . B ( p · ( a , b , c )) B ( a , b , c ) ≡ p d (4.2)where d = min { k + v p ( pa ) + v p ( p ( a + a )) , v p ( p a a ( a + a )) ,k + v p ( pb ) + v p ( p ( b + b )) , v p ( p b b ( b + b )) ,k + v p ( pc ) + v p ( p ( c + c )) , v p ( p c c ( c + c )) } − . Without loss of generalization, suppose that d = k + v p ( a ) + v p ( a + a ) + 1 or d = v p ( a a ( a + a )) + 2. Inthe first case, since (cid:0) np k − a (cid:1) | B ( a , b , c ), Lemma 4.2 implies that B ( a , b , c ) ≡ p max { k − − v p ( a ) , } andso B ( p · ( a , b , c )) ≡ B ( a , b , c ) mod p e , where e ≥ d + k − − v p ( a ) ≥ k + v p ( a + a ) ≥ k , and so the summand of S corresponding to( a , b , c ) vanishes modulo p k . Suppose now that we are in the second case, d = v p ( a a ( a + a )) + 2. Since a i + b i + c i = p k − n for each i ∈ { , , } , we have B ( a , b , c ) = Y i =1 (cid:18) np k − a i , b i , c i (cid:19) , and in particular (cid:0) np k − a (cid:1)(cid:0) np k − a (cid:1) | B ( a , b , c ), so that by Lemma 4.2 we have B ( a , b , c ) ≡ p max { k − − v p ( a ) , } +max { k − − v p ( a ) , } . (4.3)From (4.2) and (4.3) we obtain B ( p · ( a , b , c )) ≡ B ( a , b , c ) mod p e , where e ≥ d + ( k − − v p ( a )) + ( k − − v p ( a )) ≥ k + v p ( a + a ) ≥ k , and so the summand of S corresponding to ( a , b , c ) vanishes modulo p k in this case also. All in all, S ≡ p k .Finally, we deal with S . We have shown that S and S vanish modulo p k . By (4.1), it suffices to showthat p k | S . If p = 3 then clearly S = 0 since p | ( a , b , c ) forces ( a , b , c ) /p ∈ S ( np k − ). So from now onwe suppose that p = 3, and shall show 3 k − | S .We first demonstrate that each summand of S is divisible by 3 k − . As this is vacuous for k = 1,we assume k ≥
2. The condition ( a , b , c ) / ∈ S ( n k − ) implies 3 ∤ b + 2 b + 2 c + c , which forces oneof { b , b , c , c } not being divisible by 3, say 3 ∤ b . Next, at least one of a and c is also indivisibleby 3 (say, a ) since otherwise the condition n k − = a + b + c gives a contradiction modulo 3. Theequality (cid:0) k n a (cid:1)(cid:0) k n b (cid:1)(cid:0) k n c (cid:1) = k n b k n a (cid:0) k n − b , b − , b (cid:1)(cid:0) k n − a , a − , a (cid:1)(cid:0) k n c (cid:1) implies that indeed each summand of S is divisible by 3 k − . Assuming still that k ≥
2, we shall show that3 | X · ( a , b , c ) ∈ S ( n k )( a , b , c ) / ∈ S ( n k − ) (cid:18) k n a (cid:19)(cid:18) k n b (cid:19)(cid:18) k n c (cid:19) − (2 k − , (4.4)finishing the proof for this case. For each ( a , b , c ) in the summation range we define ∆ = a − a mod 3,∆ = b − b mod 3 and ∆ = c − c mod 3. The condition 3 ∤ b + 2 b + 2 c + c is equivalent to∆ = ∆ . Moreover, we have ∆ + ∆ + ∆ = 0. This forces { ∆ , ∆ , ∆ } being a permutation of { , , } , and so if ( a , b , c ) is in the summation range, then ( b , c , a ) and ( c , a , b ) are twoadditional distinct tuples in the range, and evidently cyclic permutations of ( b , c , a ) contribute the samevalue to the right-hand side of (4.4), proving the congruence (4.4) holds.It is left to deal with the case k = 1, which we deal with differently. We must prove that B n ≡ B n mod 9for 3 ∤ n . By the formula B n = P ⌊ n/ ⌋ k =0 ( − k n − k (cid:0) n k (cid:1)(cid:0) k k (cid:1)(cid:0) kk (cid:1) , dating back to Zagier’s paper [Zag09], wehave B n ≡ ( − n (cid:0) nn,n,n (cid:1) mod 9 and B n ≡ ( − ⌊ n ⌋ n − ⌊ n ⌋ (cid:0) n ⌊ n ⌋ (cid:1)(cid:0) ⌊ n ⌋⌊ n ⌋ , ⌊ n ⌋ , ⌊ n ⌋ (cid:1) mod 9. For n = 1, it is easyto check that both expression are 3 mod 9. For n > ∤ n ), both expressions are 0 modulo 9 by e.g.Kummer’s theorem [Kum52]. 15 Future directions
1. Suppose an integer sequence { u n } n ≥ is given in terms of a binomial sum, or as the diagonal of rationalfunction. Is there an algorithm deciding whether u n = CT(Λ n ) for some Laurent polynomial Λ, andfor finding such a Λ? Is there an efficient algorithm which produces Λ such that the only integral pointin its interior is the origin, in case such Λ exists?Positive answers to these questions would lead to an alternative proof of Theorem 1.1. The firstquestion is a special case of a question raised by Zagier [Zag18, p. 769, Q1], asking how to recognizethat a differential equation has geometric origin.2. Fix d ∈ { , } . Can one classify integer sequences { u n } n ≥ such that y = P n ≥ u n z n satisfies (1.6) (if d = 2) or (1.7) (if d = 3), and in addition u n = CT(Λ n ) for some Laurent polynomial Λ in d variables?Such a classification might give support to the conjecture that the 15 sporadic sequences that werediscovered are the only integral solutions to (1.6) and (1.7).3. Can one find new Calabi-Yau equations by considering the annihilating operator of CTS(Λ) for variousgood Laurent polynomials Λ? Here ‘good’ is used as in § u n = C nα n X k =0 ( − k (cid:18) − αn − k (cid:19) (cid:18) α − k (cid:19) , u n = C nα n X k =0 (cid:18) − αn − k (cid:19) (cid:18) α − k (cid:19) ,u n = C nα (cid:18) − αn (cid:19)(cid:18) α − n (cid:19) , u n = C nα (cid:18) − αn (cid:19)(cid:18) α − n (cid:19)(cid:18) nn (cid:19) for α ∈ { / , / , / , / } and C α = α − if α ∈ { / , / } and C α = 2 α − if α ∈ { / , / } . For thethird-order Legendrian sequence u n = 16 n P nk =0 (cid:0) − / n − k (cid:1) (cid:0) − / k (cid:1) = P nk =0 (cid:0) n − kn − k (cid:1) (cid:0) kk (cid:1) , we have foundthe representation u n = CT(Λ n ) withΛ = ( xyz ) − ( − x + y + z + 1)( x − y + z + 1)( x + y − z + 1)( x + y + z − . The only interior point of Λ’s Newton polytope is the origin.5. Straub generalized the third-order Gauss congruences for the Ap´ery nmbers ( γ ) as follows. He proved[Str14, Thm. 1.2] that for every n = ( n , n , n , n ) ∈ Z , A ( p r n ) ≡ A ( p r − n ) mod p r holds for all r ≥ p ≥
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Mathematical Institute, University of Oxford, Oxford, OX2 6GG, UK
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