Sequences of Enumerative Geometry: Congruences and Asymptotics

Daniel B. Grünberg and Pieter Moree, with an appendix by Don Zagier

CONTENTS We study the integer sequence vn of numbers of lines in hyper- n 1. Introduction surfaces of degree 2n − 3ofP , n>1. We prove a number 2. Background of congruence properties of these numbers of several diﬀerent v 3. Congruences types. Furthermore, the asymptotics of the n are described (in an appendix by Don Zagier). Finally, an attempt is made 4. Proofs of the Theorems at carrying out a similar analysis for numbers of rational plane 5. Asymptotics curves. 6. Comparison with Another Sequence from Enumerative Geometry Appendix: Exact and Asymptotic Formulas for vn 1. INTRODUCTION References We study the sequence of numbers of lines in a hypersurface of degree D =2n − 3ofPn, n>1. The sequence is deﬁned by (see, e.g., [Fulton 84]) D vn := c2n−2(Sym Q), (1–1) G(2,n+1)

where G(2,n+ 1) is the Grassmannian of C2 subspaces of Cn+1 (i.e., projective lines in Pn) of dimension 2(n + 1 − 2) = 2n − 2, Q is the bundle of linear forms on the line (of rank r = 2, corresponding to a particular point D of the Grassmannian), and Sym is its Dth symmetric D+r−1 − − product, of rank r−1 = D 1=2n 2. The top Chern class (Euler class) c2n−2 is the class dual to the 0-chain (i.e., points) corresponding to the zeros of the bundle SymD(Q), i.e., to the vanishing of a degree-D equation in Pn; this is the geometric require- ment that the lines lie in a hypersurface. The integral (1–1) can actually be written as a sum: D − a=0(awi +(D a)wj) vn = , (1–2) ≤k≤n,k i,j (wi − wk)(wj − wk) 0≤i

where w0,...,wn are arbitrary complex variables. This is a consequence of a localization formula due to Atiyah 2000 AMS Subject Classification: 11N37, 11N69, 11R45 and Bott from equivariant cohomology, which says that only the (isolated) fixed points of the (C∗)n+1 action con- Keywords: Sequence, congruence, asymptotic growth, number of plane rational curves tribute to the defining integral of vn. Hence the sum.

c A K Peters, Ltd. 1058-6458/2008 $ 0.50 per page Experimental Mathematics 17:4, page 409 410 Experimental Mathematics, Vol. 17 (2008), No. 4

For the first few values of n, computation yields v2 =1, 2. BACKGROUND v3 = 27, v4 = 2875, v5 = 698005, v6 = 305093061. D. Zagier gave a simple proof that the right-hand side The motivating idea behind this paper is the expectation of (1–2) is independent of w0,...,wn (as it must be for that certain problems in enumerative geometry are cou- (1–2) to hold), and that in fact it can be replaced by the pled to modularity. This is a recurrent theme in string much simpler formula theory, where partition functions have often an enumerative interpretation as counting objects (instantons, etc.) 2n−3 and must satisfy the condition of modularity covariance vn = (1 − x) (2n − 3 − j + jx) , (1–3) xn−1 in order to obtain the same amplitude when two world- j=0 sheets have the same intrinsic geometry. n where the notation [ ]xn means the coefficient of x .In Modular forms, as is well known, have Fourier coeffi- fact, formula (1–2) was proved in a very different way cients satisfying many interesting congruences (think of using methods from Schubert calculus by B. L. van der Ramanujan’s congruences for partitions or for his func- Waerden, who established it in part 2 of his celebrated tion τ(n)). 20-part Zur algebraischen Geometrie series of papers [van The same can happen for the coefficients of expan- der Waerden 33, van der Waerden 83]. sions related to modular forms, for example, the expan- n The number of linear subspaces of dimension k con- sions y = Anx obtained by writing a modular form y n tained in a generic hypersurface of degree d in P ,when (locally) as a power series in a modular function x.For it is finite, can be likewise expressed as the coefficient of instance, the famous Apéry numbers related to Apéry’s a monomial in a certain polynomial in several variables; proofoftheirrationalityofζ(3) are obtainable in this see, e.g., [Manivel 01, Theorem 3.5.18]. way [Beukers 87], and they satisfy many interesting con- Zagier also gave the formula gruences [Stienstra and Beukers 85]. The numbers appearing in the context of mirror sym- 27 2n−7/2 vn ∼ (2n − 3) metry, Picard–Fuchs equations for Calabi–Yau mani- π folds, Gromov–Witten invariants, and similar problems × − 9 − 111 − 9999 ··· of enumerative geometry are sometimes related to modu- 1 2 3 + , (1–4) 8n 640n 25600n lar forms and sometimes not, so we can reasonably hope where the right-hand side is an asymptotic expansion in for interesting congruence properties in these contexts powers of n−1 with rational coefficients that can be ex- also. plicitly computed. The proof of this formula, as well as In the case of degree-d instantons, for example, there 4 the derivation of (1–3) from (1–2), can be found in the is the conjecture of Clemens that for the quintics in CP , 3 appendix. 5 | nd for all d; see, for example, [Lian and Yau 96]. The remaining results, summarized in Theorems 3.1 In Section 3 we shall find astonishingly many congru- and 3.2, are concerned with congruence properties of the ences for our sequence vn. We shall first draw a few numbers vn. In this context it turns out to be convenient tables for congruences modulo 2, 3, 4, 5, and 11, and then summarize the observed congruences. to define v1 = 1 (even though there is no such thing as a hypersurface in P1 of degree −1) and even more In Section 4 we prove those congruences by elementary means starting from (1–3), and a few conjectures will be remarkably, v0 = −1. We do not doubt that the congruence results presented formulated. Sequences of numbers coming from modular here form only the tip of an iceberg. For reasons of space, forms also often have interesting asymptotic properties, not all our proofs of the congruence results have been and we therefore wish to study this, too. given here. For those not given, the reader is referred to In Section 5 we find the asymptotic properties of the the arXiv version of this paper [Grünberg and Moree 08]. vn numerically using a clever empirical trick shown to A first version of this paper was written by the first us by Don Zagier, which we call the “asympk trick.” (A author. The present version is similar to the first one, rigorous proof of these asymptotics, as already mentioned except for Sections 3 and 4, which have been greatly above, was also provided by Zagier and is reproduced in revised and expanded by the second author. The con- the appendix.) jectures outside these two sections are due to the first Section 6 presents congruences and asymptotics for author alone. Sections 5 and 6 were revised by both the another enumerative sequence (without proofs): the se- second author and Don Zagier. quence of rational curves on the plane. Grünberg and Moree: Sequences of Enumerative Geometry: Congruences and Asymptotics 411

Zagier’s asymptotic shows that the vn themselves are 112220002... not Fourier coefficients of any modular form on a sub- 112220002... 000000000... group of SL(2, Z) (whose coefficients typically grow like 2k−1 n for weight 2k). Nevertheless, we cannot exclude, TABLE 2. Reduction of the integers vn modulo k =3. for example, that vn = f(n)a(n)+g(n), with f,g simple functions and a(n) modular. 1111111... 1111111... 3333333... 3333333... 3. CONGRUENCES TABLE 3. Reduction of the integers vn modulo k =4. { }∞ We will consider the sequences vn (mod k) k=1 for some small values of k; that is, we study the reduction of the 8. For k =2q, all rows are constant, and in a twofold integers vn modulo k. It turns out to be instructive to way, they sweep out all odd residues, i.e., for every order the vn modulo k in a table. Each table has k rows. odd integer a with 1 ≤ a ≤ 2q there are precisely two The lth entry in the ith row (1 ≤ i ≤ k)isvlk+i (mod k), where the reduction mod k is taken to be in the interval rows that have only a as entry. − [0,...,k 1]. 9. For k =2q > 2 the entries in the rows 1, 2, 2q−1, For instance, the first few tables for k =2, 3, 4,... are 2q−1 +1, 2q−1 +2, 2q − 1 equal, respectively, 1, 1, presented as Tables 1–4. 2q−1 − 1, 2q−1 +1, 2q−1 +1, 2q − 1. Table 1 (k = 2) tells us that all the vn are odd integers. q q− We shall be interested primarily in the tables for prime 10. For k =2 > 2 the entries in row a and row a +2 1 q− q k. Table 5 is a typical such table. differ by 2 1 (mod 2 ). Study of these and other tables led us to formulate a number of conjectures, most of which we were able to Proof: These ten claims are proved respectively in Lem- prove. An overview of these results is given in Theo- mas 4.7, 4.4, 4.4, 4.5, 4.8 and 4.9, 4.12, 4.14, 4.22 and rem 3.1. 4.25, 4.24, and 4.24.

Theorem 3.1. The following hold for the tables of On computing the reductions of v1,...,v32 modulo 32 vn (mod k): one finds by part 8 of this theorem that for k =4,the table has constant rows 1, 1, 3, 3; for k = 8, constant 1. All vn are odd. rows 1, 1, 3, 3, 5, 5, 7, 7; for k = 16, constant rows 1, 1, 11, 11, 5, 5, 7, 7, 9, 9, 3, 3, 13, 13, 15, 15; and for k = 32, 2. The first two rows of each table are equal. constant rows 1, 1, 27, 27, 21, 5, 7, 23, 9, 9, 19, 19, 29, 3. If k is even, then rows k/2+1 and k/2+2 are equal. 13, 31, 15, 17, 17, 11, 11, 5, 21, 23, 7, 25, 25, 3, 3, 13, 29, 15, 31. Thus, for modulus 2q with q ≤ 3weobserve 4. For k odd, row (k +3)/2 contains only zeros. that pairs of values occur and that these, moreover, are in ascending order. 5. For k prime, the first two rows start with 1, 1 followed For q = 4 the values still come in pairs, but the or- − by k occurrences of k 1. der is no longer ascending. For n ≥ 5 it turns out that pairs with equal values become sparser and sparser. No- 6. For k>2 prime, the last (k − 1)/2 entries of the first tice that in the above cases, for every modulus, all odd column vanish. values are assumed exactly twice. By part 8, this always 7. For k>2 prime, there is a block of zeros at the bottom happens. Thus, given an odd integer a and any integer (after (k−1)/2 columns), of height (k−1)/2 and width (k +3)/2. 114444423... 114444423... 200232141... 1111111... 000000000... 1111111... 010000400...

TABLE 1. Reduction of the integers vn modulo k =2. TABLE 4. Reduction of the integers vn modulo k =5. 412 Experimental Mathematics, Vol. 17 (2008), No. 4

1 1 10101010101010101010108 ... 1 1 10101010101010101010108 ... 59107861027861085... 41 5 8 6 7108 2 6 7108 8... 09322000000015... 93100 1 4 81076 2 8107... 00000000000000... 025110000000810... 022270000000101... 0100102000000083... 02893000000075...

TABLE 5. Reduction of the integers vn modulo k = 11. q ≥ 1, there are inﬁnitely many integers m such that q vm ≡ a (mod 2 ) (or put more succinctly, modulo pow- 103125 ers of two, the sequence vn is equidistributed over the c1 = −81,c2 = , 8 odd residue classes). ≡ 210171535 1308348857025 For k prime, it is often the case that vn 0(modk) c3 = − ,c4 = , for trivial reasons. It then makes sense to consider divis- 64 1024 ibility of vn by higher powers of k. Our deepest result in 11660783598520749 c5 = − . this direction is provided by the following theorem. 16384

Theorem 3.2. 4. PROOFS OF THE THEOREMS ≥ 1. If p 5 is a prime, then 4.1 Some Generalities

3 4 First recall from the elementary theory of ﬁnite ﬁelds of v p+3 ≡−2p (mod p ), 2 order p that 3 p−1 5 v p+3 ≡ 2p (1 − p)(p − 1)!4 (mod p ). 2 p−1 p−1 − ≡ − 2. Let r ≥ 1 and let p ≥ 2r +1 be a prime. Then x 1 (x j)(modp). j=1 2r+2 2r+3 v p+3 rp ≡ Crp (mod p ), (3–1) 2 + (Here and below, the letter x denotes a variable.) By where substituting x = 0 one obtains Wilson’s theorem:

r r 2r +1 2 2 (p − 1)! ≡−1(modp). Cr = bj,r((1 − 2j)) r− , (−4)r−1 r! 2 1 j=0 We also recall the elementary identity (a + b)p ≡ ap + bp (mod p), from which we infer that if f(x) ∈ Z[x], then the integers bj,r are deﬁned implicitly by f(x)p ≡ f(xp)(modp). These results will be freely used 2r 2r in the sequel without further reference. j (2r +1− a + ax)= bj,rx , a=1 j=0 2 Lemma 4.1. We have vn ≡ 0(mod(2n − 3) ). a and ((u))a := j (u +2j − 2). =1 Proof: The term with j = 0 in (1–3) equals 2n − 3. The term with j =2n − 3equals(2n − 3)x. Hence vn = 2 Remark 3.3. Note that b2r−j,r = bj,r. Numerical ex- (2n − 3) wn,where perimentation suggests that the numerator of cr always 2n−4 equals a power of 2 and that the congruence (3–1) holds wn = (1 − x) (2n − 3 − j + jx) for all odd primes. xn−2 j=1

We record here some values of cr: completing the proof. Grünberg and Moree: Sequences of Enumerative Geometry: Congruences and Asymptotics 413

The following result was ﬁrst noticed by D. Kerner. 2ln ≡ (1 − x) (1 − (2ln − j)+(2ln − j)x) An alternative, slightly longer, proof was given by M. xln j Vlasenko. =0 2ln ≡ − − 3 (1 x) (1 + j jx) We have vn ≡ 0(mod(2n − 3) ). xln Lemma 4.2. j=0 2ln−1 ≡ − Proof: It suﬃces to show that wn 0(mod2n 3). ≡ (1 − x) (−1)(−1 − j + jx) − xln Note that modulo 2n 3, we have that j=0 2ln−1 2n−4 ≡ (1 − x) (−1 − j + jx) xln wn ≡ (1 − x) (−j + jx) j xn−2 =0 j=1 ≡ vln+1. ≡−(2n − 4)! (x − 1)2n−3 n−2 x This concludes the proof. − − − 2n 3 = (2n 4)! − n 2 The next lemma generalizes Lemma 4.1. It implies 2n − 3 2n − 4 part 4 of Theorem 3.1. = −(2n − 4)! = −(2n − 3)!Cn− , n − 1 n − 2 2 Lemma 4.5. If k is odd, then 1 2m where Cm := m m is the mth Catalan number. The 2 2l+2 +1 vlk k / ≡ 0(mod(2l +1) k ). Catalan numbers are integers that arise in numerous +( +3) 2 counting problems. ≡ − (2l+1)k − Proof: We have vlk+(k+3)/2 [(1 x) j=0 ((2l+1)k Remark 4.3. It might be interesting to see whether the j + jx)] lk+(k+1)/2 . The terms in the product with j =0 − i x integers vn/(2n 3) with i =1,2,or3alsohavea and j =(2l +1)k lead to a factor of (2l +1)2k2.The geometric meaning. remaining terms in the product that are divisible by k lead to a factor k2l. The next result was obtained in collaboration with Alexander Blessing. It establishes parts 2 and 3 of The- ∞ 4.2 The Sequence {vn}n=1 Modulo Primes orem 3.1. The following lemma will be repeatedly used in this section. Lemma 4.4. For l ≥ 0 we have vln+1 ≡ vln+2 (mod 2n). Lemma 4.6. Let p be a prime and c an integer. Then, Proof: Since v = v = 1, the result is trivially true for 1 2 modulo p, l = 0, and thus we may assume l ≥ 1. We have, modulo 2n, p−1 (ix − i + c) 2ln−1 i=1 vln+1 ≡ (1 − x) (−1 − j + jx) . xln p−1 j=0 −(x − 1) if p | c, ≡ p − 2 ··· p−1 x −x (x + x + + x )= 1−x otherwise. Furthermore, we have, modulo 2n,

2ln+1 Proof: If p | c, then the result is trivial, so assume p c. vln+2 ≡ (1 − x) (1 − j + jx) xln+1 We can write j=0 p−1 p−1 p−1 p 2ln c x − x (ix − i + c) ≡ i x − 1+ ≡ , ≡ (1 − x)x (1 − j + jx) i 1 − x xln+1 i=1 i=1 i=1 j=0 2ln where we have used that as i runs over 1, 2,...,p− 1, ≡ (1 − x) (1 − j + jx) −1+c/i runs over all residues modulo p except for −1. xln j=0 414 Experimental Mathematics, Vol. 17 (2008), No. 4 a The second lemma in this section is part 1 of Theo- Recall that b =0ifb>a. For example, by direct rem 3.1. computation we ﬁnd that

≥ ≡ p2 − 2 (−2) ···(−2p +1) Lemma 4.7. For n 1 we have vn 1(mod2). ≡ (1 − p) 2p − 2 1 ···(2p − 2) Proof: Modulo 2, the 2n−2 terms in the product in (1–3) ≡−(2p − 1)(p − 1) are alternately 1 and x. It thus follows that ≡−1(modp)

n−1 vn ≡ [(1 − x)x ]xn−1 ≡ 1(mod2). (here [ ] means skipping multiples of p). By Lucas’s theorem we ﬁnd that This concludes the proof. p2 − 2 (p − 1)p + p − 2 = 2p − 2 1 · p + p − 2 The next lemma together with Lemma 4.9 establishes − − part 5 of Theorem 3.1. ≡ p 1 p 2 1 p − 2 ≡ ≡ Lemma 4.8. Let p be a prime. Then vp+1 vp+2 ≡−1(modp). 1(modp). Likewise, we immediately ﬁnd using Lucas’s theorem that Proof: Modulo p we have with r =1andp>3,

p−1 2p − 1 r 2 ≡ 1(modp ). vp+1 ≡ (1 − x) (−1 − j − jx) p − 1 xp j =1 (This identity with r = 2 was proved in 1819 by Charles xp − x 2 ≡ (1 − x) , Babbage. For r = 3, it follows from Wolstenholme’s the- 1 − x xp orem [Bauer 88].) p−1 At various points we use the easy result that j ≡ where in the derivation of the ﬁrst congruence we noted j (−1) (mod p). To see this, observe that modulo p,the that modulo p,thejth term in the product (1–3) is equal entries except for the two outermost ones in the (p+1)th to the (j + p)th, and in the second congruence, we used row of Pascal’s triangle are zero modulo p. Since each Lemma 4.6. Now note that of these entries arises as the sum of the two elements p 2 2 above it in the pth row, the entries in the pth alternate x − x x p−2 (1 − x) = (1 − x) between 1 and −1. Similarly, one infers that j ≡ 1 − x 1 − x j xp xp (−1) (j +1)(modp). k For a nice survey of arithmetic properties of binomial = x =1. p coeﬃcients, we refer the reader to [Granville 97]. k≥2 x

Let p be a prime and 2 ≤ l ≤ p +1.Then Finally, by Lemma 4.4, vp+2 satisﬁes the same congru- Lemma 4.9. vlp ≡ vlp ≡−1(modp). ence as vp+1 modulo p. +1 +2

Proof: We require a few case distinctions, making the The proof of the next lemma involves congruences for proof rather longwinded. For the details see [Gr¨unberg binomial coeﬃcients. In all cases these can be found and Moree 08]. by direct computation, but often it is more convenient to invoke a classical result of E. Lucas. Let n ≥ m be 2 s The cases l = p and l = p + 1 in the proof of Lemma natural numbers and write n = a0 +a1p+a2p +···+asp 2 s 4.9 can be proved more succinctly, as is done in the proofs and m = b +b p+b p +···+bsp with 0 ≤ ai,bi ≤ p−1. 0 1 2 of Lemmas 4.10 and Lemma 4.11. Then Lucas’s theorem states that

Lemma 4.10. Let p be a prime. Then vp2+1 ≡ vp2+2 ≡ n a0 a1 as ≡ ··· (mod p). −1(modp). m b0 b1 bs Grünberg and Moree: Sequences of Enumerative Geometry: Congruences and Asymptotics 415

Proof: Note that modulo p,theintegervp2+1 is congruent Proof: In the case i = 0, the result follows by Lemma to 4.1, so assume that i ≥ 1. On using that modulo p,the jth term equals the (j + p)th term, we find that modulo p−1 (1 − x) (−1 − j − jx)2p p, xp2 j =1 p−1 2i p−1 2 v p+3 i ≡ 4i (1−x) (2i−j+jx) (2i−j+jx) . p 2 2 + x(p+1)/2+j ≡ (1 − x) (−1 − j − jx ) j j xp2 =1 =1 j=1 p p−1 +1 On invoking Lemma 4.6 and noting that p> 2 + i,we ≡ (−1 − j − jy)2 . yp infer that j=1 2i 2 p v p+3 i ≡ 4i (x − x) (2i − j + jx) On proceeding as in the previous proof, we find that 2 + x(p+1)/2+i j=1 yp − y 2 y 2 2i vp2+1 ≡ = 2 − p − p ≡ − 4i (2i − j + jx) . 1 y y 1 y y x(p−1)/2+j j=1 = kyk+1 ≡−1(modp). p y i p− k≥1 2 − 1 Since deg j=1(2i j + jx) =2i and 2i< 2 + i,the result follows. Finally, by Lemma 4.4, vp2+2 satisfies the same congruence as vp2+1 modulo p. The next lemma will be used in the proof of Lemma 4.14. Lemma 4.11. Let p be a prime. Then vp2+p+1 ≡ vp2 p ≡−1(modp). + +2 Lemma 4.13. Define Ar(x) and Br(x) recursively by

p−1 r Proof: We have the following congruence modulo p: A0(x)=0,Ar+1(x)=(x + ···+ x ) − Ar(x); p−1 r B (x)=0,Br (x)=−(x + ···+ x ) − Br(x). 2 p−1 2p+2 0 +1 2 ≡ − ··· vp +p+1 (1 x)(x + x + + x ) 2 xp +p Put ≡ (1 − x)(x + x2 + ···+ xp−1)2

− p ··· p(p−1) ··· p−1 r × p ··· p(p−1) 2 fr(x)=(x 1)(1 + x + + x )(x + + x ) . (x + + x ) 2 xp +p p p− Then ≡ (x − x )(x + x2 + ···+ x 1) ∞ − r − p ··· p(p−1) kp 2 fr(x)=( 1) (x 1)(1 + x + + x ) × x 2 xp2+p p + x Ar(x)+Br(x), k=1 ≡ (x2 + ···+ xp − xp+1 −···−x2p−1) where for r ≥ 1, the degree of Br(x) equals (r −1)(p−1). ∞ × (k +1)xkp xp2−p Proof: The result follows easily on noting that k=0 ∞ p−1 fr (x)=(1+···+ x )fr(x) − fr(x) ≡ (k+1)p ≡− +1 (k +1)x 2 1. xp −p p2 p−1 r k=0 =(x − 1)(x + ···+ x ) − fr(x).

Finally, by Lemma 4.4, vp2+p+2 satisﬁes the same con- This completes the proof. gruence as vp2+p+1 modulo p. The next lemma is part 7 of Theorem 3.1. The next lemma establishes part 6 of Theorem 3.1. Lemma 4.14. Let p be an odd prime. Suppose that 0 ≤ Lemma 4.12. If p is an odd prime, then v p+3 ≡ ≤ − − ≤ ≤ lp p / i ≡ 2 +i i (p 3)/2 and (p 1)/2 l p.Thenv +( +3) 2+ 0(modp) for i =0,...,(p − 3)/2. 0(modp). 416 Experimental Mathematics, Vol. 17 (2008), No. 4

Proof: See [Gr¨unberg and Moree 08]. Proof of Theorem 3.2.: Part 1: We have the following congruence modulo p5.

A further question concerning the distribution of vn p−1 2 modulo primes is how frequently certain residues appear. v p+3 = p (1 − x) (p − j + jx) p−1 2 x 2 For example, is it true that the zeros have density 1? Is it j=1 true that the nonzero entries are equidistributed? Ques- p−1 ≡ 2 − − − tions like this can be answered for the middle binomial p (1 x) ( j + jx)+(1 x)p 2k j=1 coeﬃcient k ; see, for example, [Berend and Harmse 98, Moshe 03, Moshe 05]. The following lemma suggests p−1 p−1 × − − 2 that perhaps techniques from those papers can be used ( j + jx)+(1 x)p k=1 j=1 to investigate this issue. j=k p−1 1 2k × − k ≡ k ≡ ( j + jx) p−1 Lemma 4.15. We have v1+3 v2+3 k+1 k (mod 3) x 2 1≤kCarl Pomerance (an independent proof was given ≡ xp−1 − 1 xp−3 by Paolo Dominici). =0(modp).

l p −1 − l Lemma 4.16. The polynomial i=0 (ix i+j)(modp ), Now it is an easy consequence of Eisenstein’s congru- as a polynomial in x, depends only on the class j (mod p) ence (1859), see [Hardy and Wright 79, Theorem 132], (i.e., replacing j by j + kp wouldyieldthesameresult). which states that

r p− p −1 | 2 1 − 1 1 1 1 Proof: Let fj(x)= i=0 (ix + j). If p j, then there ≡ 1+ + + ···+ (mod p), r− r− − are p 1 factors divisible by p and p 1 ≥ r,sothat p 3 5 p 2 r fj(x) ≡ 0(modp ). So assume p j.Letk be the r r that [Hardy and Wright 79, Theorem 133] ≡ p p−1 p− inverse of j,sojk 1modp . Then modulo p ,we − 2 ≡ 1 2 pr (p−1)/2 ( 1) 4 (mod p ). (Indeed, by Morley’s have fj(x) ≡ j f1(x) (since the expression ik runs over r congruence (1895), see [Cai 02], this congruence is valid a complete residue system modulo p as i does). Now even modulo p3.) ≡ suppose j j1 (mod p), with, say, j1 = j + kp.Using We thus ﬁnally infer that induction with respect to r, one then easily sees that pr r r p ≡ p r 3 p−1 5 j1 =(j + kp) j (mod p ), and we are done. v p+3 ≡ 2p (1 − p){(p − 1)!}4 (mod p ), 2 Grünberg and Moree: Sequences of Enumerative Geometry: Congruences and Asymptotics 417

3 4 ∞ which of course implies that v p+3 ≡−2p (mod p ). 4.4 The Sequence {vn} Modulo Powers of Two 2 n=1 Part 2: We have the formal series identity Before we can consider the sequence modulo powers of ∞ two, we need some preparatory lemmas. 1 k +2r − 1 k 2r = x . q (1 − x) 2r − 1 2 −1 2 k=0 Lemma 4.17. If j is odd, then i (ix − i + j) ≡ q =0 x2 (mod 2q). Note that − −2r (1 2s)2r−1 − p−1 ≡ Proof: The proof is by induction with respect to q.For [(x 1) ] −s+rp r− (mod p). x 2 22 1(2r − 1)! q = 1 the result is obvious. Assume that the result has ≤ ≤ Using the latter congruence, we find that modulo p2r+3, been established for 1 q q1.Wewrite we have 2q1+1−1 2q1 −1 2q1+1−1 (ix − i + j)2 = (ix − i + j)2 (ix − i + j)2 v q p+3 +rp i=0 i=0 i=2 1 2 (2r+1)p = P1(x)P2(x), = (1 − x) ((2r +1)p − j + jx) p+1 +rp x 2 q1 j=0 say. Note that P1(x) ≡ P2(x)(mod2 ). The induction q 2r+1 2 1 2r+2 hypothesis thus implies that we can write P1(x)=x + ≡ p (1 − x) (2r +1− j + jx) q1 q1 2 q1 j=0 2 f1(x)andP2(x)=x +2 f2(x). − 2 ≡ q1 − q1 (2r+1)p Since (ix i + j) ((i +2 )x (i +2 )+ 2 q1+1 × (−j + jx) p+1 ≡ +rp j) (mod 2 ), it even follows that P1(x) x 2 j=0 q1+1 pj P2(x)(mod2 ), from which we infer that f1(x) ≡ 2r f2(x) (mod 2) and hence f1(x)+f2(x) ≡ 0(mod2).It 2 2r+2 j (2r+1)p−2r ≡ (2r +1) p bj,r x (x − 1) q1+1 x(p−1)/2+rp follows that modulo 2 , the product under considera- j=0 tion equals ≡ (2r +1)2p2r+2 q1 q1 2r ∞ 2 q1 2 q1 j (2r+1)p k +2r − 1 k P1(x)P2(x)=(x +2 f1(x))(x +2 f2(x)) × bj,r x (x − 1) x 2r − 1 x(p−1)/2+rp q1+1 q j=0 k=0 = x2 (mod 2 1+1). ≡−(2r +1)2p2r+2 2r r ∞ This concludes the proof. j lp l k +2r − 1 k × bj,r x x (−1) x 2r − 1 x(p−1)/2+rp j=0 l=0 k=0 In the course of the above proof we have shown that r 2r +1 ≡−(2r +1)2p2r+2 (−1)l 2q −1 2q+1−1 l l=0 (ix − i + j)2 ≡ (ix − i + j)2 (mod 2q+1). 2r p−1 − j +(r − l)p +2r − 1 i=0 i=2q × b 2 j,r 2r − 1 j=0 The next result shows that the same identity holds for (2r +1)2p2r+2 r 2r +1 2r the “square roots.” Using this identity, the “square root” ≡− (−1)l b ((1 − 2j)) 22r−1(2r − 1)! l j,r 2r−1 l=0 j=0 on the left-hand side of Lemma 4.17 can be computed (−1)r−1(2r +1)2 2r 2r (Lemma 4.19). ≡ p2r+2 b ((1 − 2j)) 22r−1(2r − 1)! r j,r 2r−1 j=0 ≥ 2r+2 Lemma 4.18. Let j be odd and q 2.Then = Crp , 2q −1 2q+1−1 (ix − i + j) ≡ (ix − i + j)(mod2q+1). where in the next-to-last step we used the identity i=0 i=2q r 2r 2r +1 (−1)r = (−1)l, r l Proof: It is an easy observation that modulo 2, we have l=0 for 0 ≤ k ≤ 2q − 1that r which is obtained by comparing the coefficient of x on q − q−1 both sides of the identity (1−x)−1(1−x)2r+1 =(1−x)2r. 21 x2 −1 if k is odd, (j − a + ax) ≡ q−1 This finishes the proof. x2 if k is even. a=0,a= k 418 Experimental Mathematics, Vol. 17 (2008), No. 4

Using this identity we ﬁnd that modulo 2q+1, from which we infer that ∞ 2q −q k k Pq(x) ≡ x (1 − x) (−2) x q+1 2 −1 k=0 (ix − i + j) q−1 i=2q × (−2)q−1−rxr (mod 2q) 2q −1 = (ix − i + j +2q (x − 1)) r=0 q− i=0 22 2q −q m 2q −q 2q −1 2q −1 2q −1 ≡ x (1 − x) amx ≡ x q ≡ (ix − i + j)+2 (x − 1) (ix − i + j) m=0 i=0 k=0 i=0,i=k 2q−1 2q −1 m q × bmx (mod 2 ), ≡ (ix − i + j)+2q (x − 1) m=0 i=0 2q −2 2q −1 q−1 q−1 where x2 1+x2 −1 1 2|k 2k − − q−1−m ≤ ≤ − ≡ ( 2) /3if0m q 1; 2q −1 am −q m q−1 q−1 − − +1+ ≤ ≤ − ≡ (ix − i + j)+2q (x − 1)(x2 2q−1 + x2 −12q−1) ( 2) /3ifq m 2q 2 i=0 q and 2−1 ≡ (ix − i + j). − − q−1−m ≤ ≤ − i=0 ( 2) if 0 m q 1; bm ≡ (−2)m−q if q ≤ m ≤ 2q − 1.

q q ≡ ≡− This finishes the proof. Thus v2 bq−1 1(mod2 ). Recall that we defined v = −1. The reason for this Lemma 4.19. Let j be odd and q ≥ 3.Modulo2q,we 0 is that this definition allows us to formulate the next have lemma, which together with Lemma 4.25 is part 8 of Theorem 3.1, with j =0. 2q −1 − ≡ 2q−1 −2 q−1 4 3 2 (ix i + j) x 2 (x + x + x +1)+x . Lemma 4.22. (Periodicity.) Suppose that i, k ≥ 0.We i=0 q have vk2q +i ≡ vi (mod 2 ).

Proof: First assume that i ≥ 2. Note that

Proof: The proof is similar to that of Lemma 4.17, but 2q −1 ≡ 2k with the diﬀerence that instead of the equality P1(x) vk2q +i ≡ (1 − x) (2i − 3 − j + jx) q+1 P2(x)(mod2 ), we use Lemma 4.18. j=0 2i−3 q Remark 4.20. By Lemma 4.16, it suﬃces to work in the × (2i − 3 − j + jx) (mod 2 ). xk2q +i−1 proofs of Lemma 4.17, 4.18, and 4.19 with j =1. j=0

q By lemma 4.17, the ﬁrst product equals xk2 (mod 2q). The next result establishes a part of parts 8 and 9 of Thus Theorem 3.1. 2i−3 q q vk2q +i ≡ (1−x) (2i−3−j +jx) ≡ vi (mod 2 ). Lemma 4.21. For q ≥ 1 we have v2q ≡−1(mod2 ). xi−1 j=0

q+1 − 2 −3 − − In order to deal with the case i = 1, we note that using Proof: Put Pq(x)=(1 x) j=0 ( 3 j + jx). We q q 2 −1 q ≡ q ≡ ≡ want to compute the coefficient of x in Pq(x) modulo Lemma 4.4, vk2 +1 vk2 +2 v2 v1 (mod 2 ). In the 2q. On invoking Lemma 4.17, one finds that case i = 0, one finds, proceeding as above, that for k ≥ 1, q vk2q ≡ v2q (mod 2 ). On invoking Lemma 4.21, it then follows that vk2q ≡ 2q q q Pq(x)(1 + 2x)(2 + x) ≡ x (1 − x)(mod2 ), v2q ≡ v0 (mod 2 ). Grünberg and Moree: Sequences of Enumerative Geometry: Congruences and Asymptotics 419

The next result yields a part of part 9 of Theorem 3.1. Paolo Dominici [Dominici 98] states the following result for vn without reference. q−1 Lemma 4.23. Suppose that q ≥ 1.Thenv2q−1 ≡ 2 − q 1(mod2 ). Theorem 4.26. For 1 ≤ i ≤ 2n − 4 we put yi = i/(2n − 3 − i).Then Proof: The proof is similar to that of Lemma 4.21. For − 2 − q ≤ 3 one verifies the claim numerically. So assume q ≥ 4. vn =(2n 3) (2n 4)! 2q−1 −1 ×{ − } We want to compute the coefficient of x in Pq−1(x) Sn−2(y1,...,y2n−4) Sn−1(y1,...,y2n−4) . modulo 2q. On invoking Lemma 4.19, one finds that

Pq− (x)(1 + 2x)(2 + x) We will now derive this result from (1–3). We need 1 q−1 two lemmas. ≡ x2 −2 2q−1(x4 + x3 + x +1)+x2 (mod 2q),

Lemma 4.27. Let L1(x),...,Lr(x) be linear polynomials, whence then 2q−1 m q−1 1 d 2 −q−2 m { ··· } Pq− (x) ≡ x bmx m L1(x) Lr(x) 1 m! dx m=0 L1(x) Lr(x) ··· × q−1 4 3 2 q = Sm ,..., L1(x) Lr(x). 2 (x + x + x +1)+x (mod 2 ). L1(x) Lr(x)

(Note that the assumption q ≥ 4 implies that 2q−1 − q − 2 ≥ 0.) Another observation that we need is the following. q−1 Thus modulo 2q, the coeﬃcient of x2 −1,thatis, Lemma 4.28. Let x1,...,xr be distinct nonzero ele- v2q−1 ,equals ments such that x1 · x2 ···xr =1and {x1,...,xr} = q−1 1 1 q−1 ≡ { ,..., }.ThenSr−k(x ,...,xr)=Sk(x ,...,xr), v2 bq−1 +2 (bq−3 + bq−2 + bq + bq+1) x1 xr 1 1 ≤ ≤ ≡−1+2q−1(−4+2− 2+1)≡ 2q−1 − 1. with 1 r k.

This completes the proof. Proof: Note that

1 1 The next result with i = 0, 1, and 2 yields a part of Sr−k(x1,...,xr)=Sk ,..., x1 ···xr x xr part 9 of Theorem 3.1. It also yields part 10 of Theo- 1 rem 3.1. = Sk(x1,...,xr), where in the derivation of the ﬁrst equality we used the Lemma 4.24. For i ≥ 0 and q ≥ 2 we have v q−1 i ≡ 2 + assumption that xi = 0, and in that of the second the q−1 q vi +2 (mod 2 ). remaining assumptions.

Proof: See [Gr¨unberg and Moree 08]. Corollary 4.29. For 1 ≤ k ≤ 2n − 5 we have Sk(y ,...,y n− )=S n− −k(y ,...,y n− ). Using induction and Lemma 4.24, one easily infers the 1 2 4 2 4 1 2 4 following result, which, together with Lemma 4.22, gives Proof of Theorem 4.26.: Put part 8 of Theorem 3.1. 2n−4 Lemma 4.25. (Equidistribution.) Let q ≥ 1. For every odd Pn(x)= (2n − 3 − j + jx). q j=1 integer a there are precisely two integers 1 ≤ j1

2n−3 4.5 On a Result of Paolo Dominici vn = (1 − x) (2n − 3 − j + jx) xn−1 Let Sk(x ,...,xr)denotethekth elementary symmetric j 1 =0 function in r variables, i.e., S1(x1,...,xr)=x1 +···+xr, 2 =(2n − 3) (1 − x)Pn(x) . n−2 S2(x1,...,xr)=x1x2 + x1x3 + ···+ xr−1xr,etc. x 420 Experimental Mathematics, Vol. 17 (2008), No. 4

Thus, where 2 ≤ r ≤ 3p − 1. In the symmetric function Sr(z1,...,z3p−1)thereare 2 vn =(2n − 3) {[Pn(x)]xn−2 − [Pn(x)]xn−3 }. (4–1) 3p−3 r terms containing neither z1 nor z2.Thereare p− 3 3 terms containing z but not z . Finally, there are On noting that r−1 1 2 3p−3 terms containing both z1 and z2. m r−2 1 d It follows that modulo p, [Pn(x)]xm = m Pn(x) , m! dx x=0 r (−1) Sr(y ,...,y p− ) we obtain on invoking Lemma 4.27 that 1 3 1 1 ≡ Sr − , −2, 1,...,1 [Pn(x)]xm =(2n − 4)!Sm(y ,...,yi,...,y n− ). (4–2) 1 2 4 2 3p − 3 1 3p − 3 3p − 3 Combining (4–2) with (4–1) yields that = − (2 + ) + . r 2 r − 1 r − 2 2 vn =(2n − 3) (2n − 4)! Modulo p we have × Sn−2(y1,...,y2n−4) − Sn−3(y1,...,y2n−4) , n (−1) {Sn−2(y1,...,y2n−4) − Sn−1(y1,...,y2n−4)} or, on invoking Corollary 4.29, 3p − 2 5 3p − 2 3p − 2 ≡ − + 2 − − − vn =(2n − 3) (2n − 4)! n 1 2 n 2 n 3 p − 2 p − 2 p − 2 × Sn− (y ,...,y n− ) − Sn− (y ,...,y n− ) . ≡ − 2 1 2 4 1 1 2 4 2 − − 5 − − +2 − − n p 1 n p 2 n p 3 This concludes the proof. p − 2 p − 2 p − 2 ≡ 2 +4 +2 n − p − 1 n − p − 2 n − p − 3 From the above proof we infer that we may alterna- p − 2 tively define vn by − 9 − − n p 2 2n−3 p p − 2 vn = (x − 1) (2n − 3 − j + jx) . (4–3) ≡ 2 − 9 xn n − p − 1 n − p − 2 j=0 p − 2 9 ≡−9 ≡ (−1)n9(n − p − 1) ≡ (−1)n . We leave it to the reader to use the observation that n − p − 2 2 2n−4 − − P (x):= j=1 (2n 3 j +jx) is self-reciprocal, i.e., satisfies P (1/x)x2m−4 = P (x), to infer (4–3) directly from This completes the proof. (1–3). Many of the congruences can be also proved using The- Remark 4.30. In addition to Zagier’s proof for van der orem 4.26. As an example, we will show that if p is an Waerden’s formula (1–4) and the derivation of it from 4 5 Theorem 4.26, formula (1–4) can also be found using the odd prime, then v3(p+1)/2 ≡−81p (mod p ). This is the case r = 1 of part 2 of Theorem 3.2. theory of Chern classes. This was kindly pointed out to the second author by Professor Friedrich Hirzebruch, Proof of part 2 of Theorem 3.2 in case r =1.: Set n = who also gave a sketch of the proof. An inspiration for 3(p +1)/2. Note that the proof was a lecture he had given on 22 September 2005 in Klagenfurt on the Catalan number arising in the − 2 − ≡− 4 5 (2n 3) (2n 4)! 18p (mod p ). context of the Schubert calculus. It thus remains to be proven that the expression in braces in Theorem 4.26 equals 9/2 modulo p. 5. ASYMPTOTICS It turns out to be a little easier to work with wi = −yi. r Given a sequence of coefficients, there are many things Note that (−1) Sr(w ,...,w p− )=Sr(y ,...,y p− ). 1 3 1 1 3 1 we would like to know about it. Apart from the search We have wi = i/(i − 3p)for1≤ i ≤ 3p − 1. Thus for a generating function and for a recurrence formula, wp = −1/2, w p = −2, and the remaining wi satisfy 2 an interesting question is the asymptotic behavior. We wi ≡ 1(modp). Hence remind the reader that candidate Fourier series for mod-

1 ular forms of weight 2k for SL(2, Z)musthavecoeﬃcients Sr(w ,...,w p− ) ≡ Sr − , −2, 1, 1,...,1 (mod p), 2k−1 k 1 3 1 2 growing like n (and n for cusp forms). Grünberg and Moree: Sequences of Enumerative Geometry: Congruences and Asymptotics 421

In our case, without prior knowledge of the alterna- The crucial point in the success of asympk is that the tive definition (1–3), we managed to compute only the operator Δk sends nk to k! and kills polynomials of degree first 80 values of vn using the Schubert package for in- less than k, so that all the intermediate terms of the −k tersection theory [Katz and Stromme 92] and the first expansion of sn between c0 and ckn disappear. 225 values using the rational function (with the dummy Variants of asympk allow one to deal, for example, variables wi). Numerically, it is readily seen that the with asymptotic expansions of the form 2n log n leading term for the vn is e . Since this is strongly ∼ 2 ··· reminiscent of the behavior of (2n)! = exp(2n log(2n) − (I) sn A log n + c0 + c1/n + c2/n + , 1 1 1 2n + log 2n + log 2π + O( )), we study instead the 2 2 2 n (II) sn ∼ Bn + A log n + c + c /n + c /n + ···, vn 0 1 2 behavior of log (2n)! and now find the leading term to be λ 2 2n. Subtracting it, we find the next-to-leading term to (III) sn ∼ An (1 + c1/n + c2/n + ···). be −4logn, easily verified by applying n∂n (i.e., taking subsequent differences and multiplying by n). The next In case (I) we can apply asympk to the sequence − 2 ··· term is a constant, C = −5.62 ..., which we find diffi- n(sn+1 sn), which has the form A + c1/n + c2/n + , cult to recognize. We have learned from Don Zagier a to obtain A to high precision, after which we apply the clever technique that makes it possible to determine a original method to {sn − A log n}. large number of digits of C; we present it below under In case (II) we apply asympk twice to Δs to get B and the name “asympk trick.” A, and then subtract (our approximation for) B log n+A from sn and apply the standard version. { } 5.1 The asympk Trick For case (III) we can either look at log sn and apply variant (I) or else apply asympk to {n(sn+1/sn − 1)} to Assume that we are given numerically a few hundred λ get λ and then apply the standard method to {sn/n }. terms of a sequence s = {sn}n∈N that we believe has an asymptotic expansion in inverse powers of n, i.e., Remark 5.1. Applying the operation asympk with suit- k c1 c2 ably chosen k gives a rapidly convergent sequence s( ). sn ∼ c0 + + + ··· . n n2 To estimate how many decimals are probably correct, we

Goal: determine the coefficients ci numerically. look at some relatively widely spaced elements of this se- (k) Trick: Choose some moderate value of k (say k =8) quence (e.g., the terms sn with n = 300, 400, 500 if we (k) 1 k k know 500 terms of the sequence s)andseehowmanyof and define a new sequence s as k! Δ N s,whereΔis their digits agree. the difference operator (Δu)n = un − un−1 and N the multiplication operator (Nu)n = nun, i.e., Remark 5.2. One also has to experiment to find the op- k − j timal choice of k. Typically, one uses k =5ifoneknows (k) ( 1) − k sn = − (n j) sn−j. 200 terms of s,andk = 8 if one knows 1000 terms. This j j!(k j)! =0 suggests that perhaps k ≈ log N is a good choice for a For n large we have (assuming the above asymptotic ex- generic sequence with N computed terms. pansion for s itself) Remark 5.3. The asympk trick was first described in [Za- (k) − k ck+1 gier 01]. Here Zagier considers the Stoimenov numbers sn = c0 +( 1) k+1 n ξD, which bound the number V (D) of linearly indepen- k+1 (k +1)ck − ck − k +2 2 +1 ··· dent Vassiliev invariants of degree D. Stoimenov himself +( 1) k+2 + . D n thought that ξD behaves “something like D!/1.5 .” Cal- culating the values up to D = 200 and applying a vari- Thus, while sn approximates c0 only to within an ac- − (k) curacy O(n 1), sn approximates it to the much better ation of asympk suggested an asymptotic formula of the accuracy O(n−k), so we obtain a very good approxima- form √ tion for c0. Call this operation asympk. The further co- D! D C1 C2 ξD ∼ C + + + ··· , efficients ci are then obtained inductively: if c0,...,ci−1 (π2/6)D 0 D D2 are known to high precision, we get ci by applying i i−1 asympk to the sequence n sn − c0 −···−ci−1/n )= with C0 ≈ 2.704332490062429595, C1 ≈−1.52707, and ci + ci+1/n + ···. C2 ≈−0.269009. 422 Experimental Mathematics, Vol. 17 (2008), No. 4

Subsequently, Zagier was able to prove this with ex- • k = 2: Both rows vanish (except for the ﬁrst two plicitly√ computable constants Ci.Inparticular,C0 = values), i.e., all nd are even. −5/2 π2/12 12 3π e , which agrees to the accuracy given l l • k =2: All rows are 0, i.e., nd ≡ 0(mod2)for above with the empirically obtained value. n>l+1.

vn 5.2 Application to the Asymptotics of • k =3: n3d ≡ 0(mod3),n3d+2 ≡ 1(mod3), In our case of the sequence vn of lines in a hypersurface n3d+1 ≡ alternating 1 or 2 (mod 3) because n6d+2 ≡ n of P , the coefficient c0 =: C is difficult to recognize, but 4(mod6). all other coefficients, c1,c2,..., are rational numbers that • k =5: nd ≡ 0(mod5),ford>8. The same for we easily recognize from a sufficient number of digits. k =25(d>23). Once the first few rational coefficients have been found and the corresponding terms subtracted from Because of this high degree of symmetry for low the sequence s, the constant term C can be ob- primes, most nonprimes will yield constant or regular tained with 30 digits, say. This is enough to rows (i.e., rows repeating when shifted horizontally). The feed to the PARI software and apply the function only nonobvious case is k = 26, where there is a shift by 8 lindep([C,1,log(Pi),log(2),log(3)]) to find a ra- (because k = 13 shifts by 16) and rows 4 and 6 alternate tional linear combination of C in terms of a given basis with 0. (educated guess). The result, equivalent to (1–4), is Further, we found only three primes with regular fea- tures: vn − 11 141 ··· log =2n 4logn + C + + 2 + , (5–1) (2n)! 6n 160n • k = 7: All rows are regular (repeat when shifted − − − 3 8 horizontally by 4); rows 5 and 7 are 0. where C := 3 log π 2 log 3 . In the appendix Don Zagier presents a proof of this asymptotic formula. • k = 13: The same, shift by 16, no 0 row.

• k = 19: The same, shift by 12, no 0 row. 6. COMPARISON WITH ANOTHER SEQUENCE • FROM ENUMERATIVE GEOMETRY k =5, 11, 17, 23, 29: These primes give almost-0 rows (i.e., nd ≡ 0 except for a ﬁnite number of d). As a matter of curiosity, we now compare our results so far with another result for a sequence of enumerative We have not attempted to prove these observations. geometry. 6.1.1 Asymptotics We now turn to the asymptotics of the sequence nd for d →∞. Di Francesco and Itzykson 6.1 Numbers of Plane Rational Curves proved [Di Francesco and Itzykson 94, Proposition 3] that One sequence of integers from enumerative geometry is d nd, the number of plane rational curves of degree d nd A 1 = / B +O through 3d − 1pointsinP2. Kontsevich’s recurrence (3d − 1)! d7 2 d formula [Kontsevich and Manin 94] reads as d tends to inﬁnity, and found the approximate values ≈ ≈ d−1 A 0.138 and B 6.1 for the constants A and B. nd = nknd−k Assuming a full asymptotic expansion

k=1 d nd A B B 3d − 4 3d − 4 ∼ B + 1 + 2 + ··· × k2(d − k)2 − k3(d − k) , (3d − 1)! d7/2 0 d d2 3k − 2 3k − 1

and applying variant (II) of the asympk trick to with n1 = 1. The result is n1 =1,n2 =1,n3 = 12, etc. log(nd/(3d − 1)!), we obtain the much more accurate ap- That is, there is one line through 2 points of the plane, proximations one conic through 5 points of the plane, twelve cubics through 8 points of the plane, etc. A ≈ 0.138009346634518656829562628891755541716 We can similarly draw tables of nd mod k for any inte- 014121072, ger k. The results (with the same convention as before) ≈ are as follows: B0 6.0358078488159024106383768720948935, Grünberg and Moree: Sequences of Enumerative Geometry: Congruences and Asymptotics 423

as well as the further values B1 ≈−2.2352424409362074, monic polynomials of degrees n1 +1,...,nk +1withno B2 ≈ 0.054313787925. multiple roots, then Unfortunately, we are not able to recognize any of G(α ,...,αk) these apparently irrational numbers, e.g., PARI does not 1 ··· P1(α1) Pk(αk) see in log A and log B0 a linear combination of simple P1(α1)=···=Pk (αk)=0 numbers like 1, log 2, log 3, log π, π,andπ2. is independent of all the Pi and is equal to the coeffi- n1 ··· nk cient of x1 xk in G(x1,...,xk). In fact, G need not A. APPENDIX: EXACT AND ASYMPTOTIC even be homogeneous, but can be any polynomial in k FORMULAS FOR vn variables of degree less than or equal to n1 + ···+ nk. (by Don Zagier) In this appendix we prove the alternative definition (1–3) Corollary A.3. Let F (x, y) be a symmetric homoge- and the asymptotic formulas (1–4) and (5–1) for the num- neous polynomial of degree 2n − 2 in two variables and bers vn defined in (1–2). w0,...,wn distinct complex numbers. Then the expression F (wi,wj ) A.1 Exact Formulas 0≤k≤n (wi − wk)(wj − wk) 0≤i

r r α x dx v(n) = coeff(prod(j=0,2*n-3,2*n-3-j+j*x,1-x),n-1) = Resx α P (α) = P (x) P (α)=0 α∈C and it takes less than 2 seconds to compute vn up to xr dx = −Resx=∞ , n = 100, and 46 seconds to compute up to n = 224. P (x) We can rewrite (1–3) in several other forms using − and this equals 0 if 0 ≤ r

−cDt2 while simply making the substitution z → D−z in (A–3) It is easy to see that φD(0) = 1 and φD(t) ≤ e gives the similar expression for some absolute constant c>0(amuchmoreprecise formula will be given in a moment), so the main contri- D − 2n j=0(z j) bution to the integral comes from small t.Fort small vn = D Resz dz , (A–5) =0 zn(z − D)n+1 and D large we have (uniformly in both variables) and adding these two last expressions gives yet a third ∞ − j−1 2j form: ( 1) r 2j D − log φ (t)= 2j 1 t D j j r , ,...,D D (z − j) =1 =1 3 1 2n+1 j=0 vn = D Resz=0 dz . (A–6) D 1 D 1 2 2 zn+1(z − D)n+1 − 2 − 4 = + t + + 3 t 3 3D 5 3 D 15D Each of the formulas (A–4)–(A–6) expresses vn as the D 1 4 1 + − + − + O t6 constant term at z = 0 of the Laurent expansion of a 7 3D 9D3 D5 rational function; for instance, (A–4) says that D − 1 14 1 8 + + 3 + O 5 t n 2n n−1 9 3D 15D D vn =(−1) D · coeﬃcient of z D 1 1 (1 − z)(2 − z) ···(D − 1 − z) + − + + O t10 in as z → 0. (A–7) 11 3D D3 (D − z)n−1 D − 1 1 12 + + O 3 t Substituting z = Du,wecanwritethisas 13 3D D D 1 14 n 2 n−1 + − + O t vn =(−1) D · coeﬃcient of u (A–8) 15 D

(1 − Du)(2 − Du) ···(D − 1 − Du) → D 1 16 18 in n− as u 0, + + O t + O Dt , (1 − u) 1 17 D

2 from which we see again that D | vn (Lemma 4.1). By expanding (D − z)1−n by the binomial theorem, we also and hence obtain closed formulas for vn; for instance, (A–7) gives 2 x x n− φD √ 1 − − 2 2 n−1−m 2n 2 m m+1 D (1 + x /D) D n − v = ( 1) D , (A–9) 6 n − 1 m 2 x m=0 = e−x /3 x2 + − 2x4 D−1 5 D where , the coeﬃcient of zm in z(z+1)···(z+D−1), x10 19x8 x4 m + − +3x6 + D−2 50 35 3 is a Stirling number of the ﬁrst kind. x14 12x12 314x10 59x8 + − + − − x6 D−3 A.2 Asymptotics 750 175 315 15

To obtain the asymptotic expansion of vn, we write the x30 11x28 + ···+ − + ··· residue in (A–2) as 1 and we make the substitu- 393750000 19687500 2πi |x|=1 tion x =(1+it)/(1 − it) to obtain, after a short calcula- 10 8 355x 2x −7 tion, + + D 162 45 ∞ 2 D2 + r2t2 t2dt −8 n + O D . v = 2 2 2 (A–10) π −∞ r , ,...,D 1+t (1 + t ) =1 3 ∞ 2 2 D+1 t dt D = D φ (t) 2 2 , Substituting this expansion (with the 34 omitted π −∞ (1 + t ) terms√ included) into equation (A–10) with t replaced by where φD(t) denotes the rational function x/ D and using the standard evaluation

2 −2 2 1+r D t ∞ n√ φD(t)= . −x2/3 2n (2n)! 3 1+t2 e x dx = 3π, r=1,3,...,D −∞ n! 4 Grünberg and Moree: Sequences of Enumerative Geometry: Congruences and Asymptotics 425 we obtain REFERENCES 27 9 969 61479 [Bauer 88] F. L. Bauer. “For All Primes Greater Than 3, D−1/2 − −1 −2 − −3 2p−1 ≡ p3 vn = D 1 D + D D ( p−1 ) 1 (mod )Holds.”Math. Intelligencer 10:3 π 4 160 3200 (1988), 42. 25225773 −4 10092025737 −5 + D − D [Berend and Harmse 98] D. Berend and J. E. Harmse. “On 358400 35840000 Some Arithmetical Properties of Middle Binomial Coeffi- 2271842858513 4442983688169 + D−6 − D−7 cients.” Acta Arith. 84 (1998), 31–41. 2007040000 1146880000 [Beukers 87] F. Beukers. “Irrationality Proofs Using Modular + O D−8 . Forms.” Astérisque 147–148 (1987), 271–283. [Cai 02] T. Cai. “A Congruence Involving the Quotients of This asymptotic formula can of course be written in many Euler and Its Applications, I.” Acta Arith. 103 (2002), 313– other ways, e.g., 320. [Di Francesco and Itzykson 94] P. Di Francesco and C. Itzyk- 27 2n−7/2 son. “Quantum Intersection Rings.” In The Moduli Space vn = (2n − 3) π of Curves (Texel Island, 1994), pp. 81–148, Progr. Math. 9 111 9999 87261 129. Boston: Birkhäuser, 1995. × 1 − − − + −··· 8n 640n2 25600n3 5734400n4 [Dominici 98] P. Dominici. “Sequence A027363.” On-Line Encyclopedia of Integer Sequences. Available online (http: or //www.research.att.com/∼njas/sequences/), 1998. −3 27 2n−7/2 [Fulton 84] W. Fulton. Intersection Theory.NewYork: vn = e (2n) π Springer, 1984. 15 1689 79281 19691853 [Granville 97] A. Granville. “Arithmetic Properties of Bi- × 1+ + + + + ··· 8n 640n2 25600n3 5734400n4 nomial Coefficients, I.” In Binomial Coefficients Modulo Prime Powers, Organic mathematics (Burnaby, BC, 1995), or pp. 253–276, CMS Conf. Proc. 20. Providence: Amer. Math. Soc., 1997. vn 11 141 9973 log =2n − 4logn + C + + + (2n)! 6n 160n2 28800n3 [Grünberg and Moree 08] D. Grünberg and P. Moree (with 59673 appendix by D. Zagier). Version of the present paper with + + ··· all proofs given. arXiv:math.NT/0610286, 2008. 179200n4 [Hardy and Wright 79] G. H. Hardy and E. M. Wright. An 3 8 with C = −3 − log π − 2 log 3 . Of course, more terms Introduction to the Theory of Numbers,fifthedition.New could be obtained in any of these expansions if desired. York: Oxford University Press, 1979. [Katz and Stromme 92] S. Katz and S. A. Stromme. “Schu- bert: A Maple Package for Intersection Theory and Enu- ACKNOWLEDGMENTS merative Geometry.” Available online (http://www.mi.uib. The first author, who initiated this paper, is deeply grateful no/stromme/schubert/), 1992. to Don Zagier for his help throughout all the stages of this [Kontsevich and Manin 94] M. Kontsevich and Y. Manin. project. Not only did Don simplify matters considerably with “Gromov–Witten Classes, Quantum Cohomology, and his formula (1–3), he also provided the asympk trick. The first Enumerative Geometry.” Comm. Math. Phys. 164 (1994), author is also thankful for fruitful discussions with Robert 525–562. Osburn. The second author would like to thank Daniel Berend, [Lian and Yau 96] B. H. Lian and S.-T. Yau. “Arithmetic David Cox, Paolo Dominici, and Yossi Moshe for fruitful e- Properties of Mirror Map and Quantum Coupling.” Comm. mail correspondence and Carl Pomerance and Paolo Dominici Math. Phys. 176 (1996), 163–191. for providing us with Lemma 4.16. He would like especially [Manivel 01] L. Manivel. Symmetric Functions, Schubert to thank Don Zagier for his help in improving the exposition Polynomials and Degeneracy Loci,SMF/AMSTextsand of the paper and Professor F. Hirzebruch for several hours of Monographs 6. Providence: American Mathematical Soci- discussion concerning formula (1–4) and Chern classes. ety, 2001. We thank Dmitry Kerner, Masha Vlasenko, and Wadim [Moshe 03] Y. Moshe. “The Density of 0’s in Recurrence Dou- Zudilin for several helpful discussions. Several inaccuracies in ble Sequences.” J. Number Theory 103 (2003), 109–121. an earlier version were detected by Alexander Blessing during a two-week practicum he held with the second author. [Moshe 05] Y. Moshe. “The Distribution of Elements in Au- This project was made possible thanks to the support of tomatic Double Sequences.” Discrete Math. 297 (2005), 91– the MPIM in Bonn. 103. 426 Experimental Mathematics, Vol. 17 (2008), No. 4

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D. Gr¨unberg, 49 rue Fondary, 75015 Paris, France ([email protected])

P. Moree, Max-Planck-Institut f¨ur Mathematik, Vivatsgasse 7, D-53111 Bonn, Germany ([email protected])

D. Zagier, Max-Planck-Institut f¨ur Mathematik, Vivatsgasse 7, D-53111 Bonn, Germany ([email protected])

Received October 9, 2006; accepted in revised form March 18, 2008.