THE LEAST COMMON MULTIPLE AND

LATTICE POINTS ON HYPERBOLAS

Andrew Granville and Jorge Jiménez-Urroz

Abstract. We bound, from below, the least common multiple of k integers from a short interval. This is used to bound the length of an arc Γ of the hyperbola, xy = N, containing k integer lattice points.

1. Introduction

The problem of finding lattice points on curves has been studied by many authors, most famously Gauss’s investigation on lattice points inside the circle. If we restrict our attention to lattice points on short arcs, then there are several bounds known; for example, for conics see [1, 2, 3, 4]. We consider the hyperbola xy = N so that each lattice point is a divisor of N. Therefore counting lattice points on a small arc of the hyperbola, is the same as counting divisors of N in a short interval. We prove the following result:

Theorem 1. Given k integers X ≤ a_{1} < ⋯ < a_{k} ≤ X + L, we have

(1)  LCM[a_{1}, ⋯, a_{k}] ≥ C_{k} \frac{X^{k}}{L^{\binom{k}{2}}} ,

where C_{k} is positive and

(2)  C_{k}^{2} = \frac{(\frac{2k}{k−1})^{k} \binom{2k−2}{k}^{2k−1} ∏_{m=1}^{k−2} m!^{2}}{2(k!) ∏_{m=1}^{k−1} \binom{2m}{m}^{2}} .

Using Stirling’s formula (see section 5) one can show that

(3)  C_{k} = (4e^{−3/2}k)^{k^{2}/2−3k/2} (16πe^{−5/2})^{k/2} k^{7/24}e^{α+O(1/k)}

where

α = \frac{log 4}{3} + \frac{log π}{8} − \frac{53}{24} + \frac{5γ}{12} + \frac{5}{4} ∑_{j≥2} \frac{ζ(j) − 1}{j + 2},

γ is the Euler constant, and ζ(j) := ∑_{n≥1} 1/n^{j} is the Riemann zeta function.

Typeset by AMS-TEX

The first author is a Presidential Faculty Fellow and is supported, in part, by the National Science Foundation.

We give the first few values of C_{k} in the following table.

\begin{matrix}k & C_{k} \\ 2 & 1 \\ 3 & 2^{2} \\ 4 & 2 · 5^{2}\sqrt{5} \\ 5 & \frac{2^{6} · 7^{3}\sqrt{3 · 7}}{3} \\ 6 & 2^{7} · 3^{6} · 7^{3}\sqrt{3 · 7} \\ 7 & \frac{2^{16} · 3^{7} · 11^{5} · \sqrt{3 · 5 · 11}}{5^{2}} \\ 8 & \frac{2^{12} · 3^{6} · 11^{5} · 13^{6} · \sqrt{3 · 5 · 11 · 13}}{5} \\ 9 & \frac{2^{24} · 3^{7} · 5^{8} · 11^{5} · 13^{6} · \sqrt{7 · 11 · 13}}{7^{3}} \\ 10 & \frac{2^{23} · 3^{9} · 5^{4} · 11^{5} · 13^{6} · 17^{8} · \sqrt{7 · 11 · 13 · 17}}{7^{2}} \end{matrix}

If p is a prime in the interval (k, 2k − 1) then p divides C_{k}^{2} but p^{2} does not, as may be seen by examining (2). Thus, by Bertrand’s postulate, C_{k} is irrational for all k ≥ 4.

Write N := LCM[a_{1}, ⋯, a_{k}] and A_{i} = N/a_{i} for each i. Then Theorem 1 gives, since N/(X + L) ≤ A_{1} < ⋯ < A_{k} ≤ N/X that N^{\binom{k−1}{2}} ≥ C_{k}X^{k(k−2)}/L^{\binom{k}{2}}. Thus

Corollary 1. Given k ≥ 3 integers X ≤ a_{1} < ⋯ < a_{k} ≤ X + L, we have

(4)  LCM[a_{1}, ⋯, a_{k}] ≥ C_{k}^{1/\binom{k−1}{2}}X^{2k/(k−1)}/L^{k/(k−2)}.

Notice that the bound in (1) is sharper than that of (4) exactly when C_{k}X^{k−2} ≥ L^{\binom{k}{2}} for k ≥ 4 (the bounds are identical for k = 3).

To prove Theorem 1 we write a_{i} = X + δ_{i}L for each i, so that 0 ≤ δ_{1} < δ_{2} < ⋯ < δ_{k} ≤ 1. Note that

Λ : = LCM[a_{1}, ⋯, a_{k}] = (a_{1}a_{2} ⋯ a_{k}/L^{\binom{k}{2}}) · (Λ_{2}/Λ_{1})

≥ (X^{k}/L^{\binom{k}{2}}) · (Λ_{2}/Λ_{1}) (4a)

where

Λ_{1} = Λ_{1}(a_{1}, ⋯, a_{k}; L) : = ∏_{1≤i<j≤k} \frac{(a_{j} − a_{i})}{L} = ∏_{1≤i<j≤k}(δ_{j} − δ_{i}),

and Λ_{2} = Λ_{2}(a_{1}, ⋯, a_{k}) : = \frac{LCM[a_{1}, ⋯, a_{k}]}{∏_{i=1}^{k} a_{i}} ∏_{1≤i<j≤k}(a_{j} − a_{i}),

In Proposition 1 we give a sharp upper bound on Λ_{1} by analytic methods, and in Proposition 2 we give a sharp lower bound on Λ_{2} by elementary methods. Combining these two results implies Theorem 1, using (4a).

Hilbert [10], Stieltjes [16], and others [14,15], have all shown how to maximize (Λ_{1} =) ∏_{1≤i<j≤k}(δ_{j} − δ_{i}) when 0 ≤ δ_{1} < δ_{2} < ⋯ < δ_{k} ≤ 1. We present our own proof in section 2, though like most previous proofs, it involves computing discriminants of certain classical polynomials from the theory of orthogonal polynomials.

Proposition 1. We have

max_{0≤δ_{1}<δ_{2}<···<δ_{k}≤1} ( ∏_{1≤i<j≤k}(δ_{j} − δ_{i}))^{2} = \frac{2(k − 1)^{k}k! ∏_{m=1}^{k−1} \binom{2m}{m}^{2}}{(2k)^{k}\binom{2k−2}{k}^{2k−1}} .

We bound Λ_{2} from below, by more elementary methods, in section 3.

Proposition 2. Given integers a_{1} < a_{2} < ⋯ < a_{k}, the number Λ_{2}(a_{1}, ⋯, a_{k}), defined above, is an integer and is divisible by ∏_{m=1}^{k−2} m!.

Remark. In section 4a we exhibit many examples with Λ_{2}(a_{1}, ⋯, a_{k}) = ∏_{m=1}^{k−2} m!, so Proposition 2 cannot be improved.

The lower bounds (1) and (4) are trivial for sufficiently large L, so we want to find the largest L for which these inequalities are sharp.

Theorem 2. Given integer k ≥ 2 and sufficiently large L, there exists X ≪ L^{\binom{k}{2}}, such that

M_{k} := min_{X≤a_{1}<a_{2}<···<a_{k}≤X+L} LCM[a_{1}, ⋯, a_{k}] = (1 + O (\frac{1}{\sqrt{L}})) C_{k} \frac{X^{k}}{L^{\binom{k}{2}}} .

Remark 1. Our proof gives many such k-tuples of integers with X ≫ L^{\binom{k}{2}}. With a better understanding of the distribution of congruence classes we could give such a result in a wider range.

Remark 2. One can give examples in Theorem 2 described by polynomials. For example, if k = 2 let X = a_{1} = nL and a_{2} = (n + 1)L for a given integer L. Since (a_{2}, a_{1}) = L = a_{2} − a_{1}, we see that

[a_{1}, a_{2}] = \frac{a_{1}a_{2}}{(a_{1}, a_{2})} = (n + 1)X = (1 + \frac{1}{n}) \frac{X^{2}}{L} .

For k = 3 we take, for any integers n and t, with L = 4n,

X = a_{1} = (n + 1 + (2n + 1)t)4n(2n − 1),
a_{2} = a_{1} + (2n − 1) = ((2n + 1) + 4nt)(2n + 1)(2n − 1)
a_{3} = a_{1} + 4n = (n + (2n − 1)t)4n(2n + 1)

Again note that (a_{j}, a_{i}) = a_{j} − a_{i} and that (a_{1}, a_{2}, a_{3}) = 1. Therefore

[a_{1}, a_{2}, a_{3}] = \frac{a_{1}a_{2}a_{3}}{∏_{1≤i<j≤3}(a_{j}, a_{i})} = \frac{a_{1}a_{2}a_{3}}{4n(2n + 1)(2n − 1)}

= 4\frac{X^{3}}{L^{3}} \{1 + O (\frac{n}{X} + \frac{1}{n^{2}} )\}

In both of these cases we have given polynomial examples which are as good as possible (in that they give examples with lcm∼ C_{k}X^{k}/L^{\binom{k}{2}}). This is not possible for k ≥ 4: To see this notice that if it were then every f_{j}(t) = a_{j} − a_{1} (j ≥ 2) would have the same degree and would have integer coefficients. The ratios of the leading coefficients of the f_{j}(t) would thus all be rational and so Λ_{1} would be a rational number. However if lcm∼ C_{k}X^{k}/L^{\binom{k}{2}} then the value of Λ_{1}^{2} is given in Proposition 1, and this is not the square of a rational for any k ≥ 4.

On the other hand we will show, in section 4c, how to construct such polynomials for each k which lead to examples with lcm≪_{k} X^{k}/L^{\binom{k}{2}}.

We may re-interpret our results to give results about lattice points on the hyperbola xy = N in as small an arc as possible. As a consequence of (1), (4) and Theorem 2 we have (taking X = a_{1} and L = a_{k} − a_{1}):

Theorem 3. If there are k distinct lattice points (a_{i}, b_{i}) on the hyperbola ab = N with a_{1} < a_{2} < ⋯ < a_{k} then

(4b)  a_{k} − a_{1} ≥ c_{k} max \{\frac{a_{1}^{2− \frac{2}{k−1}}}{N^{1−2/k}} , \frac{a1^{\frac{2}{k−1}}}{N^{1/\binom{k}{2}}} \} .

where c_{k} := C_{k}^{1/\binom{k}{2}}. On the other hand one can find integer N, together with k distinct lattice points (a_{i}, b_{i}) on the hyperbola ab = N with

either a_{1} ≫_{k} N^{1− \frac{1}{k−1}} and a_{k} − a_{1} ≤ \{c_{k} + o(1)\}a_{1}^{2− \frac{2}{k−1}} /N^{1−2/k},
or a_{1} ≪_{k} N^{\frac{1}{k−1}} and a_{k} − a_{1} ≤ \{c_{k} + o(1)\}a1^{\frac{2}{k−1}} /N^{1/\binom{k}{2}}.

Remarks. The first of the bounds in (4b) is larger if and only if X ≥ N^{1/2}, when k ≥ 4. When k = 3 they are equal. Note that if a = a_{1} ≪_{k} N^{1/k} then we can have a_{k} − a_{1} ≪ 1 simply by taking each a_{j} = a + j and N = (a + 1)(a + 2) ⋯ (a + k).

Taking k = 3 in Theorem 3, we find that one always has a_{3} − a_{1} ≥ 2^{2/3}a_{1}/N^{1/3}. By making an analagous remark about the b_{i}s, we find that the Euclidean distance between (a_{1}, b_{1}) and (a_{3}, b_{3}) is always ≫ N^{1/6} (however one can have two lattice points on the hyperbola a bounded distance apart, for example (m, m + 1) and (m + 1, m) when N = m(m + 1)). We expect that by the methods of [5b] this lower bound can be improved to ≫ N^{1/4}.

Acknowledgment.. This work is part of the second author’ s Ph. D. Thesis. He would like to thank his advisor, Javier Cilleruelo, for suggesting the problem and for his interest. The second author wishes to thank as well Andrew Granville for his invitation to work with him in the University of Georgia at Athens and University of Michigan at Ann Arbor, and for his constant encouragement during this period. Both universities were very kindly in their hospitality. Finally he would like to thank Trevor Wooley for his attention in the University of Michigan, Henryk Iwaniec for his invitation to go to Rutgers University, Fernando Chamizo for checking some calculations concerning the appendix, and Christian Ballot for remembering some properties of the resultant.

2. The transfinite diameter of [0, 1]

The transfinite diameter of the interval [0, 1] is defined as

D = lim_{k→∞} max ∏_{i<j} (δ_{j} − δ_{i})^{2/k(k−1)} ,

and it is known that D = 1/4 (see [6] and chapter 11 of [13] and [17]). This also follows from Proposition 1 and Stirling’s formula.

The proof of Proposition 1. Let us suppose that

F (δ_{1}, δ_{2}, ⋯, δ_{k}) := ∏_{1≤i<j≤k}(δ_{j} − δ_{i})

attains its maximum in this range at η_{1}, η_{2}, ⋯, η_{k}. Define

H(x) := ∏_{i=1}^{k}(x − η_{i}).

so that our maximum equals Δ(H), the discriminant of H. (H(x) is a certain Jacobi polynomial, and its roots are known as the Fekete numbers of order k.)

First note that η_{1} = 0 and η_{k} = 1, else η_{k} − η_{1} < 1 and so taking δ_{i} := (η_{i} − η_{1})/(η_{k} − η_{1}) we have δ_{j} − δ_{i} = (η_{j} − η_{i})/(η_{k} − η_{1}) > η_{j} − η_{i} implying that F (δ_{1}, ⋯, δ_{k}) > F (η_{1}, ⋯, η_{k}), a contradiction.

F is differentiable function, and so attains its maximum in the closed set 0 ≤ δ_{2} ≤ ⋯ ≤ δ_{k−1} ≤ 1 (evidently η_{i} ≠ η_{j} else F = 0). Therefore the maximum occurs at a critical point, so for 2 ≤ i ≤ k − 1 we have

0 = \frac{1}{F} \frac{∂F}{∂δ_{i}} (η_{2}, ⋯, η_{k−1}) = ∑_{j≠i} \frac{1}{η_{i} − η_{j}} = \frac{1}{2} \frac{H′′(η_{i})}{H′(η_{i})},

and thus H′′(η_{i}) = 0. Since H′′(x) is a polynomial of degree k − 2, its roots are exactly η_{2}, η_{3}, ⋯, η_{k−1}, which are exactly the roots of H(x), other than 0 and 1. Therefore x(x−1)H′′(x) = CH(x) for some constant C: Since H is monic of degree k, the leading coefficient of H′′(x) is C = k(k − 1), and so

(5)  x(x − 1)H′′(x) = k(k − 1)H(x).

(This equation can be used to show that the roots of H are symmetric about 1/2.) By comparing coefficients of both sides in (5), we find that

H(x) = ∑_{j=1}^{k}(−1)^{k−j} \begin{matrix}\binom{k}{j} & \binom{k−1}{j−1} \\ ( & {2k−2 \atop k−j} & ) \end{matrix}x^{j}.

which may be verified by substituting into (5).

Given polynomial f of degree n and polynomial g of degree m ≤ n, define R(f, g) to be the absolute value of the resultant of f and g. To determine Δ(H), we will use the fact that Δ(H)^{2} = R(H, H′). We need several facts about resultants (see [11]); for example, R(f, cg) = c^{n}R(f, g) for any constant c. Also, if g has leading coefficient b, and h ≡ f (mod g) where h has degree r ≤ n, then R(f, g) = b^{n−r}R(g, h).

We will define a sequence of polynomials H_{2k} = H(x), H_{2k−1} = H′(x) and

H_{2m−i+1} = (−1)^{k−1} \frac{(k − 1)!(m)^{i}}{\binom{2k−2}{k}m!} ∑_{j=i}^{m}(−1)^{i−j}\binom{m + j − i}{m − j}\binom{2j − i}{j}x^{j}

for 0 ≤ 2m − i + 1 ≤ 2k − 2, where i = 0 or 1, and m is an integer. By comparing coefficients one shows that these polynomials satisfy

H_{2k−2} = kH_{2k} − xH_{2k−1},
H_{2k−3} = H_{2k−1} + 2H_{2k−2},
H_{2k−4} = (k − 1)H_{2k−2} − (2k − 3)xH_{2k−3}, and
H_{2m−i−1} = mH_{2m−i+1} + (−1)^{i}2^{1−i}(2m − 1)^{i}H_{2m−i}x^{i}

for i = 0 or 1, and 1 ≤ m ≤ k − 3. We will write these relationships as H_{l−2} = β_{l}H_{l} + γ_{l}H_{l−1} where β_{l} = [l/2] except for β_{2k−1} = 1. Note also that the degree of H_{l} is [l/2], and write A_{l} for the leading coefficient of H_{l}. Therefore

R(H_{l+1}, H_{l}) = A_{l}β_{l+1}^{−[l/2]}R(H_{l}, H_{l−1}),

for each l ≥ 1, and since R(H_{2}, H_{1}) = (k − 1)!/\binom{2k−2}{k}, we deduce that

Δ^{2}(H) = R(H_{2k}, H_{2k−1}) = \frac{(k − 1)!}{\binom{2k−2}{k}} ∏_{l=2}^{2k−1} A_{l} ∏_{j=2}^{k}(β_{2j}β_{2j−1})^{−(j−1)}

= \frac{(k − 1)^{k}(k − 1)! ∏_{m=1}^{k−1} \binom{2m}{m}^{2}}{(2k)^{k−1}\binom{2k−2}{k}^{2k−1}}

which gives our result after some re-arrangement.

3. Combinatorial number theory

Our next objective is to prove Proposition 2. To do this we first prove the following well-known result, giving a proof which arose in discussion with Konyagin:

Lemma 1. For any set of integers b_{1}, b_{2}, ⋯, b_{n}, we have that ∏_{m≤n−1} m! = ∏_{1≤i<j≤n}(j − i) divides ∏_{1≤i<j≤n}(b_{j} − b_{i}).

Proof. Let D be the determinant of the n-by-n matrix with (i, j)th entry \binom{b_{j}}{i−1}. Each entry of this matrix is an integer, and so D is an integer (as can be seen from expanding minors to compute D). The entries of the ith row of the matrix are all the same polynomial of degree i−1 in b_{j}; thus we can subtract appropriate rational multiples of rows 1, 2, …, I − 1 from the Ith row (these rational multiples being independent of the b_{j}) to get a matrix with the same determinant, but whose (i, j)th entry is b_{j}^{i−1}/(i − 1)!. Multiplying through the i row by (i − 1)! we are left with a Vandermonde matrix whose determinant is D ∏_{i≤n}(i − 1)! = ∏_{1≤i<j≤n}(b_{j} − b_{i}). The result follows from this equation.

Proof of Proposition 2. Define v_{p}(r) to be the exact power of prime p which divides integer r, and let v_{p}(r/s) = v_{p}(r) − v_{p}(s).

For a given prime p select ℓ = ℓ_{p} so that v_{p}(a_{ℓ}) is maximal. By definition v_{p}(a_{ℓ}) = v_{p}(LCM[a_{1}, ⋯, a_{k}]). Moreover if i ≠ ℓ then v_{p}(a_{ℓ} − a_{i}) ≥ v_{p}(a_{i}), and so

v_{p} (\frac{LCM[a_{1}, ⋯, a_{k}]}{a_{ℓ}} · ∏_{\begin{matrix}1≤i≤k \\ i≠ℓ \end{matrix}} \frac{(a_{ℓ} − a_{i})}{a_{i}} ) ≥ 0.

Taking {b_{1}, ⋯, b_{k−1}} = {a_{i} : i ≠ ℓ} in Lemma 1, we see that

v_{p} ( ∏_{\begin{matrix}1≤i<j≤k \\ i,j≠ℓ \end{matrix}}(a_{j} − a_{i})) ≥ v_{p} ( ∏_{m≤k−2} m!) .

Adding these two results we find that v_{p}(Λ_{2}) ≥ v_{p}(∏_{m≤k−2} m!) for all primes p, and so the result follows.

4a. Proposition 2 is “best possible”

Given k ≥ 2, let P := LCM[1, ⋯, k] and α_{1} = 0. We shall select 0 < α_{2} < α_{3} < ⋯ < α_{k}, then determine a large positive integer z, and finally take each a_{i} = z + α_{i}. So, given α_{1}, α_{2}, …, α_{r−1}, let Q be the set of primes > k which divide ∏_{1≤i<j≤r−1}(α_{j} − α_{i}), and define ω(p) to be the number of distinct residue classes of α_{1}, α_{2}, …, α_{r−1} (mod p) for each p ∈ Q.

The sieve of Eratosthenes-Legendre gives that for any y and N, the number of integers α in the range y < α ≤ y + NP^{2} for which α ≡ r − 1 (mod P^{2}), and α ≢ α_{1}, α_{2}, …, α_{r−1} (mod p) for any prime p ∈ Q is

(6)  N ∏_{p∈Q} (1 − \frac{ω(p)}{p}) + O(r^{|Q|}) ≫ \frac{N}{log^{r} |Q|} + O(r^{|Q|}).

We take y = α_{r−1} here, and see that for sufficiently large N, there must exist such an integer α, which we denote α_{r}.

After we have determined α_{1}, α_{2}, …, α_{k} we let z be a positive integer for which z ≡ 0 (mod P^{2}), and z ≡ −α_{j} (mod p^{s}) for any prime power p^{s} dividing α_{j} − α_{i} with p > k, for any 1 ≤ i < j ≤ k. We claim that such an integer exists, by the Chinese Remainder Theorem, since any common prime divisors of α_{j} − α_{i} and α_{J} − α_{I}, with {i, j} ≠ {I, J}, must be ≤ k, by construction.

Since each a_{i} ≡ i − 1 (mod P^{2}), it is easy to show that for any prime p ≤ k we have v_{p}(Λ_{2}) = v_{p}(∏_{m≤k−2} m!). On the other hand if prime p > k divides some difference a_{j} − a_{i} = α_{j} − α_{i} then this pair is unique by our construction. Also a_{i} ≡ a_{j} = z + α_{j} ≡ 0 (mod p^{s}), where p^{s} is the exact power of p dividing a_{j} − a_{i}. Thus

v_{p}(Λ_{2}) = v_{p}(LCM[a_{i}, a_{j}]) + v_{p}(a_{i} − a_{j}) − v_{p}(a_{i}) − v_{p}(a_{j}) = 0.

Therefore, this construction gives examples with Λ_{2}(a_{1}, ⋯, a_{k}) = ∏_{m≤k−2} m!.

4b. Proof of Theorem 2

We proceed, more-or-less, with the construction above, though we change the range in which we look for our α_{i}. In particular, given L, we select α_{i}, for each i ≥ 2 to be the largest number < η_{i}L − 2\sqrt{L} which satisfies the given congruences.

Each α_{j}−α_{i} is ≤ L, and so has ≪ log L/ log log L prime factors. Taking N = \sqrt{L}, we find that (6) is ≫ N/(log log L)^{k} − L^{O(1/ log log L)} > 0, and so we can find α_{i} in the interval [η_{i}L−2\sqrt{L}, η_{i}L−\sqrt{L}]. Therefore Λ_{1}, and thus Λ, is within 1+O(1/\sqrt{L}) of the maximum possible (since the above construction gives Λ_{2} = ∏_{m≤k−2} m!).

Finally we take z satisfying all the given congruences. The modulus for these congruences is ≪_{k} L^{\binom{k}{2}}, so guarantee finding such an X(= a_{1}), with X ≪_{k} L^{\binom{k}{2}}.

4c. A construction with polynomials

Consider the integers b_{j} = 2kj^{3}+j^{2} for j ≤ k which verify that (b_{j}−b_{i})/(j−i) are all different. We will take a_{j} = a_{1}+(j −1)+P^{2}b_{j}t (where P is as above). Thus each a_{j} −a_{i} equals j−i+P^{2}(b_{j} −b_{i})t, and these polynomials are all distinct by our choices of the b_{j}. Select polynomial a_{1} of minimal degree satisfying a_{1} ≡ −((j −1)+P^{2}b_{j}t) (mod j − i + P^{2}(b_{j} − b_{i})t) in ℤ[t] for all 1 ≤ i < j ≤ k. As t → ∞ running through integer values, one finds that lcm∼ τ_{k}a_{1}^{k}/t^{\binom{k}{2}} for some constant τ_{k}, justifying the remarks made shortly before the statement of Theorem 3.

5. Asymptotics for C_{k}.

The estimate (3) for the asymptotic behaviour of the constant C_{k} defined in (2) is easily deduced from the estimate

(7)  ∏_{m=1}^{l−1} m! = η^{1/2} (\sqrt{2π})^{l} (e^{−3/2}l)^{l^{2}/2−1/12} e^{O(1/l)},

where η := \sqrt{2π} exp(−23/12 + γ/3 + ∑_{j≥2}(ζ(j) − 1)/(j + 2)), by noting that ∏_{m=1}^{l}(2m)!^{2} = 2^{l}l! ∏_{m=1}^{2l} m! (since (2m)! = 2m(2m − 1)!), and using Stirling’s formula, m! = \sqrt{2πm}(m/e)^{m}e^{1/12m+O(1/m^{2})}, and the estimate (k−1)^{k} = k^{k}e^{−1+O(1/k)}.

To establish (7) take f(x) = x^{2}/2 in the identity

l^{f(l)} ∏_{m=1}^{l−1} m^{f(m)−f(m+1)} = ∏_{m=2}^{l} (\frac{m}{m − 1})^{f(m)} ,

so that f(m) − f(m + 1) = −m − 1/2, to obtain

∏_{m=2}^{l} (\frac{m}{m − 1})^{m^{2}/2} = l^{l^{2}/2} ∏_{m=1}^{l−1} m^{−m−1/2}

= \frac{l^{l^{2}/2}}{(l − 1)!^{l+1/2}} ∏_{m=1}^{l−1} m^{l−m} = \frac{l^{(l+1)^{2}/2}}{l!^{l+1/2}} ∏_{m=1}^{l−1} m! (8)

Now

∑_{m=2}^{l} m^{2} log (\frac{m}{m − 1}) = − ∑_{m=2}^{l} m^{2} log (1 − \frac{1}{m}) = ∑_{m=2}^{l} m^{2} ∑_{j≥1} \frac{1}{jm^{j}} ,

∑_{m=2}^{l} \frac{1}{m} = log l + γ − 1 + O (\frac{1}{l}) ,

and ∑_{j≥4} ∑_{m≥l+1} 1/jm^{j−2} ≪ ∑_{m>l} 1/m^{2} ≪ 1. Thus exponentiating gives

∏_{m=2}^{l} (\frac{m}{m − 1})^{m^{2}} = l^{1/3} exp (l^{2}/2 + l − 11/6 + γ/3 + ∑_{j≥2} \frac{ζ(j) − 1}{j + 2} + O (\frac{1}{l})) .

Combining this with (8) gives (7), using Stirling’s formula.

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Department of Mathematics, University of Georgia, Athens, Georgia 30602, USA

E-mail address: andrew@math.uga.edu

Departamento de Matemáticas, Facultad de Ciencias, Universidad Autónoma de Madrid, 28049 Madrid, España

E-mail address: jorge.jimenez@uam.es