2 Bounded denominators of $h_{s, i}$ ’s

This section studies several families of functions in v and k that take rational values with bounded denominators at the parameters of nontrivial tight designs.

Let ${\mathbb {Z}\mathrm {Val}}[v, k]$ be the ring of integer-valued rational functions in v and k, ${\mathcal {D}}_s$ the collection of all $(v_0, k_0)$ ’s such that there exists a nontrivial tight $2 s$ - $(v_0, k_0, \lambda _0)$ design, and ${\mathcal {A}}_s$ the ring of integer-valued functions on ${\mathcal {D}}_s$ . Then, ${\mathbb {Q}} \otimes _{{\mathbb {Z}}} {\mathcal {A}}_s$ is the ring of rational-valued functions on ${\mathcal {D}}_s$ with bounded denominators. It is clear that ${\mathbb {Z}\mathrm {Val}}[v, k] \subseteq {\mathcal {A}}_s \subseteq {\mathbb {Q}} \otimes _{{\mathbb {Z}}} {\mathcal {A}}_s$ .

For designs, there are three families of important functions. For each $i \in [0, s]_{{\mathbb {Z}}}$ , let

(2.1) $$ \begin{align} \lambda_{s, i} &:= \frac{1}{s!} \frac{k^{\underline{s + i}}}{(v - s)^{\underline{i}}} \in {\mathbb{Q}}(v, k), \end{align} $$

(2.2) $$ \begin{align} \alpha_{s, i} &:= {s \choose i} \frac{(k - s)^{\overline{i}}(k - s + 1)^{\overline{i}}}{(v - 2 s + 1)^{\overline{i}}} \in {\mathbb{Q}}(v, k), \end{align} $$

(2.3) $$ \begin{align} h_{s, i} &:= \frac{(k - s)^{\overline{i + 1}}}{(v - 2 s + 1)^{\overline{i}}} \in {\mathbb{Q}}(v, k). \end{align} $$

It is well known that $\lambda _{s, i} \in {\mathcal {A}}_s$ , since every tight $2 s$ - $(v, k, \lambda )$ design is an $(s + i)$ - $(v, k, \lambda _{s, i})$ design. It was proved in [Reference Xiang20, Corollary 2.2] that $\alpha _{s, i} \in {\mathcal {A}}_s$ . More precisely, the zeros of the Wilson polynomial in z,

$$\begin{align*}\Phi_s(z) := \sum_{i = 0}^s (-1)^{s - i} \frac{{v - s \choose i} {k - i \choose s - i} {k - i - 1 \choose s - i}}{{s \choose i}} {z \choose i} = \frac{{v - s \choose s}}{s!} \left(\sum_{i = 0}^s (-1)^i \alpha_{s, i} z^{\underline{s - i}}\right), \end{align*}$$

are exactly the s intersection numbers of tight $2 s$ -designs by [Reference Delsarte5, Reference Ray-Chaudhuri and Wilson13, Reference Bannai2]. Therefore, $\alpha _{s, i}$ must take integer values at $(v_0, k_0) \in {\mathcal {D}}_s$ .

The goal of this section is to prove $h_{s, i} \in {\mathbb {Q}} \otimes _{{\mathbb {Z}}} {\mathcal {A}}_s$ . In other words, to prove that $h_{s, i}(v_0, k_0)$ has bounded denominator for $(v_0, k_0) \in {\mathcal {D}}_s$ . The author proved such a result in [Reference Xiang20, Lemma 4.1] that the denominators of $h_{s, i}(v_0, k_0)$ are bounded above by $\exp (\frac {1}{2} s^2 \ln s + O(s^2))$ . The main result of this section, Proposition 2.2, gives a much better estimate. It shows that the denominators are bounded above by $\exp (s \ln s + O(s))$ .

Note that such an improvement is not necessary to prove Theorem 1.1 for sufficiently large s. However, for small s, this improvement can reduce the amount of computation needed in Section 6 by a factor of $10^{50}$ . So, it is an essential step in the proof of Theorem 1.1.

2.1 Linear combinations of $\alpha _{s, i}$ and $\lambda _{s, i}$

For a natural number n, let

$$\begin{align*}\ell_n := \mathrm{lcm} \left\{ {i \choose j} :\ 0 \leq j \leq i \leq n \right\}. \end{align*}$$

The number $\ell _n$ has many alternative expressions, for example, $\ell _n = \mathrm {lcm} \{1, \dots , n\}$ . Moreover, the statement $\ell _n = \exp (n + o(n))$ is equivalent to the prime number theorem. This is a technical number we need in the estimates in this section.

We first express $h_{s, s}$ in terms of $\lambda _{s, i}$ ’s and $\alpha _{s, i}$ ’s. The quotients $\frac {\lambda _{s, 1}}{h_{s, s}}, \dots , \frac {\lambda _{s, s}}{h_{s, s}}$ and $\frac {\alpha _{s, 1}}{h_{s, s}}, \dots , \frac {\alpha _{s, s}}{h_{s, s}}$ are all rational functions in v and k. In Lemma 2.1, we express $1$ as an explicit rational linear combination of these rational functions, from which we conclude that $h_{s, s}$ is a rational linear combination of $\lambda _{s, 1}, \dots , \lambda _{s, s}$ and $\alpha _{s, 1}, \dots , \alpha _{s, s}$ .

Lemma 2.1. Let s be a natural number with $s \geq 2$ . Then, $s! \ell _{s - 2} h_{s, s}$ is a ${\mathbb {Z}\mathrm {Val}}[v, k]$ -linear combination of $\alpha _{s, 1}, \dots , \alpha _{s, s}$ and $\lambda _{s, 1}, \dots , \lambda _{s, s}$ .

Proof. As polynomials in ${\mathbb {Q}}[v, k]$ , we have the identity

(2.4) $$ \begin{align} \begin{aligned} 1 & = \sum_{i = 0}^{s - 1} \left( \sum_{j = i + 1}^{s - 1} \frac{ (-1)^{i + j} (j - i - 1)! }{ (s - i - 1)! (j - 1)! } {k - j - 1 \choose s - j - 1} {k - i - 1 \choose j - i - 1} {v - s \choose i} \right) (v - 2 s + 1)^{\overline{s - i - 1}} (k - 1)^{\underline{i}} \\ &\quad + \sum_{i = 0}^{s - 1} \left( \sum_{j = 1}^{i} \frac{ (-1)^{i + j} (i - j)! }{ i! (s - j - 1)! } {k - j - 1 \choose i - j} {k - 1 \choose j - 1} {v - s - i - 1 \choose s - i - 1} \right) (v - s)^{\underline{i}} (k - s + 1)^{\overline{s - i - 1}}. \end{aligned} \end{align} $$

This is because the right side of equation (2.4) is a polynomial in v of degree $\leq s - 1$ and a polynomial in k of degree $\leq s - 2$ , and a direct calculation shows that equality holds for $(v, k) \in \{s, \dots , 2 s - 1\} \times \{1, \dots , s - 1\}$ .

By equations (2.1) to (2.3),

(2.5) $$ \begin{align} s! \frac{\lambda_{s, i + 1}}{h_{s, s}} & = (v - 2 s + 1)^{\overline{s - i - 1}} (k - 1)^{\underline{i}}, \end{align} $$

(2.6) $$ \begin{align} \frac{ i! i! (s - i)! }{ s! } {k \choose i} \frac{\alpha_{s, s - i}}{h_{s, s}} & = (v - s)^{\underline{i}} (k - s + 1)^{\overline{s - i - 1}}. \end{align} $$

Substituting equations (2.5) and (2.6) into equation (2.4) gives

$$ \begin{align*} s! \ell_{s - 2} & = \sum_{i = 0}^{s - 1} \left( \sum_{j = i + 1}^{s - 1} \frac{ (-1)^{i + j} s! s! (j - i - 1)! \ell_{s - 2}}{ (s - i - 1)! (j - 1)! } {k - j - 1 \choose s - j - 1} {k - i - 1 \choose j - i - 1} {v - s \choose i} \right) \frac{\lambda_{s, i + 1}}{h_{s, s}} \\ &\quad + \sum_{i = 0}^{s - 1} \left( \sum_{j = 1}^{i} (-1)^{i + j} i! (s - i) \frac{\ell_{s - 2}}{ {s - j - 1 \choose i - j} } {k - j - 1 \choose i - j} {k - 1 \choose j - 1} {v - s - i - 1 \choose s - i - 1} {k \choose i} \right) \frac{\alpha_{s, s - i}}{h_{s, s}}, \end{align*} $$

from which the result follows.

We then express $h_{s, i}$ in terms of $\alpha _{s, i}$ and $h_{s, s}$ . We use a similar idea as in Lemma 2.1, except that in this case we consider the quotients $\frac {\alpha _{s, i}}{h_{s, i}}$ and $\frac {h_{s, s}}{h_{s, i}}$ .

Proposition 2.2. Let s be a natural number with $s \geq 2$ . For every $i \in [0, s]_{{\mathbb {Z}}}$ , $s! \ell _{s - 1} \ell _{s - 2} h_{s, i}$ is a ${\mathbb {Z}\mathrm {Val}}[v, k]$ -linear combination of $\alpha _{s, 0}, \dots , \alpha _{s, s}$ and $\lambda _{s, 1}, \dots , \lambda _{s, s}$ .

Proof. If $i = 0$ , then $h_{s, i} = k - s = (k - s) \alpha _{s, 0}$ by equations (2.2) and (2.3); hence, the result holds trivially. From now on, we assume that $i \geq 1$ .

As polynomials in ${\mathbb {Q}}[k]$ , we have the identity

(2.7) $$ \begin{align} \begin{aligned} 1 & = \left( \sum_{j = 0}^{s - i - 1} (-1)^{s - j} \frac{ (s - i - j)! }{ (s - j - 1)! } {k \choose j} {k - j - 1 \choose s - i - j - 1} \right) (k - s + 1)^{\overline{i - 1}} \\ &\quad + \left( \sum_{j = s - i + 1}^{s - 1} (-1)^{s - j - 1} \frac{ (i + j - s)! }{ j! } {k - j - 1 \choose s - j - 1} {k - s + i - 1 \choose i + j - s - 1} \right) k^{\underline{s - i}}. \end{aligned} \end{align} $$

This is because the right side of equation (2.7) is a polynomial in k of degree $\leq s - i$ , and a direct calculation shows that equality holds for $k \in \{0, \dots , s - i - 1\} \cup \{s - i + 1, \dots , s - 1\}$ .

By equations (2.2) and (2.3),

(2.8) $$ \begin{align} \frac{ i! (s - i)! }{ s! } \frac{\alpha_{s, i}}{h_{s, i}} & = (k - s + 1)^{\overline{i - 1}}, \end{align} $$

(2.9) $$ \begin{align} (s - i)! {v - s \choose s - i} \frac{h_{s, s}}{h_{s, i}} & = k^{\underline{s - i}}. \end{align} $$

Substituting equations (2.8) and (2.9) into equation (2.7) gives

$$ \begin{align*} s! \ell_{s - 1} \ell_{s - 2} & = \left( \sum_{j = 0}^{s - i - 1} (-1)^{s - j} i! (s - i - j) \ell_{s - 2} \frac{ \ell_{s - 1} }{ {s - j - 1 \choose s - i} } {k \choose j} {k - j - 1 \choose s - i - j - 1} \right) \frac{\alpha_{s, i}}{h_{s, i}} \\& \quad + \left( \sum_{j = s - i + 1}^{s - 1} (-1)^{s - j - 1} \frac{\ell_{s - 1}}{{j \choose s - i}} {k - j - 1 \choose s - j - 1} {k - s + i - 1 \choose i + j - s - 1} \right) {v - s \choose s - i} \frac{s! \ell_{s - 2} h_{s, s}}{h_{s, i}}. \end{align*} $$

Therefore, $s! \ell _{s - 1} \ell _{s - 2} h_{s, i}$ is a ${\mathbb {Z}\mathrm {Val}}[v, k]$ -linear combination of $\alpha _{s, i}$ and $s! \ell _{s - 2} h_{s, s}$ . Then, the result follows from Lemma 2.1 .

Since ${\mathbb {Z}\mathrm {Val}}[v, k] \subseteq {\mathcal {A}}_s$ , any ${\mathbb {Z}\mathrm {Val}}[v, k]$ -linear combination of elements in ${\mathbb {Q}} \otimes _{{\mathbb {Z}}} {\mathcal {A}}_s$ is also in ${\mathbb {Q}} \otimes _{{\mathbb {Z}}} {\mathcal {A}}_s$ . Applying this to Proposition 2.2 gives Corollary 2.3.

Corollary 2.3. For every $i \in [0, s]_{{\mathbb {Z}}}$ , $s! \ell _{s - 1} \ell _{s - 2} h_{s, i} \in {\mathcal {A}}_s$ .

Proof. The result follows from Proposition 2.2 and from the fact that $\alpha _{s, 0}, \dots , \alpha _{s, s} \in {\mathcal {A}}_s$ and $\lambda _{s, 0}, \dots , \lambda _{s, s} \in {\mathcal {A}}_s $ .

3 Auxiliary function $H_{s, r}$

Let s be a natural number, and r an even natural number with $r \leq s$ . We construct an auxiliary function $H_{s, r}$ in $h_{s, i}$ ’s as follows:

(3.1) $$ \begin{align} H_{s, r} := \sum_{i = 0}^{r} (-1)^i {r \choose i} h_{s, s - r + i} h_{s, s - i}. \end{align} $$

The goal of this section is to prove that this $H_{s, r}$ satisfies conditions (i), (ii) and (iii) in Section 1. First, this complicated sum $H_{s, r}$ has a simple closed-form formula as shown in Theorem 3.3, which is proved in Section 3.1. The conditions (ii) and (iii) follow directly from the explicit closed-form formula. Second, since Corollary 2.3 proves that $h_{s, 0}, \dots , h_{s, s}$ satisfy the condition (i), namely taking rational values with bounded denominators, so does their rational quadratic combination $H_{s, r}$ . Proposition 3.4 is an explicit version of this statement and is proved in Section 3.2.

The auxiliary function $H_{s, r}$ was discovered in our attempts to give a proof of Theorem 1.1. It is quite mysterious, and, at this moment, we do not have any combinatorial interpretations of $H_{s, r}$ .

3.1 Simplifying $H_{s, r}$

Dixon’s identity is an identity on the hypergeometric function

$$\begin{align*}{}_{3}F_{2}\bigg[\begin{matrix}a & b & c\\ & d & e\end{matrix}\,;{z}\bigg] := \sum_{i = 0}^\infty \frac{a^{\overline{i}} b^{\overline{i}} c^{\overline{i}}}{d^{\overline{i}} e^{\overline{i}}} \frac{z^i}{i!}. \end{align*}$$

Theorem 3.1 (Dixon’s identity).

Let $a, b, c$ be complex variables. In the region ${\mathrm {Re}} (1 + a / 2 - b - c) > 0$ ,

$$\begin{align*}{}_{3}F_{2}\bigg[\begin{matrix}a & b & c\\& 1 + a - b & 1 + a - c\end{matrix}\,;1\bigg] = \frac{\Gamma(1 + a / 2) \Gamma(1 + a / 2 - b - c) \Gamma(1 + a - b) \Gamma(1 + a - c)}{\Gamma(1 + a) \Gamma(1 + a - b - c) \Gamma(1 + a / 2 - b) \Gamma(1 + a / 2 - c)}, \end{align*}$$

where $\Gamma $ is the Gamma function, and both sides should be understood as their analytic continuations.

When we choose $a := -r$ for some natural number r, $a^{\overline {i}} = 0$ for $i \geq r + 1$ . In this case, the hypergeometric function reduces to a polynomial in z with coefficients in ${\mathbb {Q}}(b, c, d, e)$ :

(3.2) $$ \begin{align} {}_{3}F_{2}\bigg[\begin{matrix}-r & b & c\\ & d & e\end{matrix}\,;{z}\bigg] = \sum_{i = 0}^{r} (-1)^i {r \choose i} \frac{b^{\overline{i}} c^{\overline{i}}}{d^{\overline{i}} e^{\overline{i}}} z^i. \end{align} $$

We have the following specialization of Dixon’s identity.

Corollary 3.2. Let s be a natural number, and r an even natural number. Let v and k be real variables. If $v \geq k + s$ and $k \geq r$ , then

$$\begin{align*}{}_{3}F_{2}\bigg[\begin{matrix}-r & k - r + 1 & - v + s\\& - k & v - r - s + 1\end{matrix}\,;{1}\bigg] = \frac{r^{\underline{r / 2}} (v - k - s)^{\overline{r / 2}}}{k^{\underline{r / 2}} (v - s - r + 1)^{\overline{r / 2}} }, \end{align*}$$

where both sides are regarded as rational functions in v and k.

We could prove Corollary 3.2 by setting $b := k - r + 1$ , $c := - v + s$ and then taking the limit of Theorem 3.1 as $a \to - r$ . The inequalities $v \geq k + s$ and $k \geq r$ are to make sure that the summands in the hypergeometric series $_3F_2$ have nonzero denominators and ${\mathrm {Re}}(1 + a / 2 - b - c) > 0$ . The details of the proof are omitted here.

It turns out that when we substitute $h_{s, i}$ ’s into $H_{s, r}$ in equation (3.1), the complicated sum $H_{s, r}$ is just the left side of Dixon’s identity in disguise. This means that we could use Dixon’s identity to give a closed-form formula for $H_{s, r}$ .

Theorem 3.3. Let s be a natural number, and r an even natural number with $r \leq s$ . Then,

$$\begin{align*}H_{s, r} = \frac{r!}{(r / 2)!} G_{s, r}, \end{align*}$$

where

(3.3) $$ \begin{align} G_{s, r} := \frac{ (v - k - s)^{\overline{r / 2}} (k - s)^{\overline{s - r / 2 + 1}} (k - s)^{\overline{s - r + 1}} }{ (v - 2 s + 1)^{\overline{s}} (v - 2 s + 1)^{\overline{s - r / 2}} }. \end{align} $$

Proof. We proceed with the following calculation.

$$ \begin{align*} H_{s, r} & = \sum_{i = 0}^{r} (-1)^i {r \choose i} h_{s, s - r + i} h_{s, s - i} & (\text{by equation~}(3.1)) \\& = \sum_{i = 0}^{r} (-1)^i {r \choose i} \frac{(k - s)^{\overline{s - r + i + 1}}}{(v - 2 s + 1)^{\overline{s - r + i}}} \frac{(k - s)^{\overline{s - i + 1}}}{(v - 2 s + 1)^{\overline{s - i}}} & (\text{by equation~}(2.3)) \\& = \frac{ (k - s)^{\overline{s + 1}} (k - s)^{\overline{s - r + 1}} }{ (v - 2 s + 1)^{\overline{s}} (v - 2 s + 1)^{\overline{s - r}} } \cdot \sum_{i = 0}^{r} (-1)^i {r \choose i} \frac{(k - r + 1)^{\overline{i}} (v - s)^{\underline{i}}}{k^{\underline{i}} (v - s - r + 1)^{\overline{i}}} \\& = \frac{ (k - s)^{\overline{s + 1}} (k - s)^{\overline{s - r + 1}} }{ (v - 2 s + 1)^{\overline{s}} (v - 2 s + 1)^{\overline{s - r}} } \cdot \sum_{i = 0}^{r} (-1)^i {r \choose i} \frac{(k - r + 1)^{\overline{i}} (- v + s)^{\overline{i}}}{(-k)^{\overline{i}} (v - s - r + 1)^{\overline{i}}} & ( x^{\underline{n}} = (-1)^n (-x)^{\overline{n}} ) \\& = \frac{(k - s)^{\overline{s + 1}} (k - s)^{\overline{s - r + 1}} }{ (v - 2 s + 1)^{\overline{s}} (v - 2 s + 1)^{\overline{s - r}} } \cdot {}_{3}F_{2}\bigg[\begin{matrix}-r & k - r + 1 & - v + s\\&\quad -k & v - r - s + 1\end{matrix}\,;1\bigg] & (\text{by equation~}(3.2)) \\& = \frac{ (k - s)^{\overline{s + 1}} (k - s)^{\overline{s - r + 1}} }{ (v - 2 s + 1)^{\overline{s}} (v - 2 s + 1)^{\overline{s - r}} } \cdot \frac{r^{\underline{r / 2}} (v - k - s)^{\overline{r / 2}} }{ k^{\underline{r / 2}} (v - s - r + 1)^{\overline{r / 2}}} & (\text{by Corollary }3.2) \\&= \frac{r!}{(r / 2)!} G_{s, r}. & (\text{by equation~}(3.3)) \end{align*} $$

3.2 Denominators of $G_{s, r}$

Since $h_{s, i} \in {\mathbb {Q}} \otimes _{{\mathbb {Z}}} {\mathcal {A}}_s$ and $H_{s, r}$ is a rational quadratic polynomial in $h_{s, i}$ , we have $H_{s, r} \in {\mathbb {Q}} \otimes _{{\mathbb {Z}}} {\mathcal {A}}_s$ . Using the estimate of denominators of $h_{s, i}$ ’s in Corollary 2.3, we could estimate the denominator of $H_{s, r}$ . Combined with Theorem 3.3, we could estimate the denominator of $G_{s, r}$ .

Proposition 3.4. Let s be a natural number, and r an even natural number with $r \leq s$ . Let

(3.4) $$ \begin{align} F_{s, r} := s!^2 \ell_{s - 1}^2 \ell_{s - 2}^2 \frac{r!}{(r / 2)!} \in {\mathbb{Z}}. \end{align} $$

Then, $F_{s, r} G_{s, r} \in {\mathcal {A}}_s$ .

Proof. We first rewrite $F_{s, r} G_{s, r}$ as follows.

$$ \begin{align*} F_{s, r} G_{s, r} & = \frac{(r / 2)!}{r!} F_{s, r} H_{s, r} & (\text{by Theorem }3.3) \\ & = \sum_{i = 0}^r (-1)^i {r \choose i} \left(s! \ell_{s - 1} \ell_{s - 2} h_{s, s - r + i}\right) \left(s! \ell_{s - 1} \ell_{s - 2} h_{s, s - i}\right). & (\text{by equations {(3.1)} and {(3.4)}}) \end{align*} $$

Then, the result follows from Corollary 2.3.

4 Upper bounds for v

The goal of this section is to give upper bounds for v in terms of s and some other parameters that could be chosen.

Assume that $(v, k)$ are parameters of a nontrivial tight $2 s$ -design. Recall that in Proposition 3.4 we proved that $F_{s, r} G_{s, r} \in {\mathcal {A}}_s$ . Moreover, $F_{s, r} G_{s, r}$ is positive by equations (3.3) and (3.4). In other words, $F_{s, r} G_{s, r}$ takes a positive integer value at $(v, k)$ . It is possible to obtain an upper bound for v by analyzing the inequality $\ln F_{s, r} + \ln G_{s, r} \geq 0$ . However, if we write down the complete arguments, which we will not do in this paper, it would only allow us to prove $v \ll s^{5}$ as $s \to \infty $ .

We can use this idea to obtain a much better bound, as long as we apply a small twist. Let

(4.1) $$ \begin{align} \tilde{G}_{s, r} := \frac{ {v - k - s + r / 2 - 1\choose r / 2} {k - r / 2 \choose s - r / 2 + 1} {k - r \choose s - r + 1}}{ {v - s \choose s} {v - s - r / 2 \choose s - r / 2} } = G_{s, r} \frac{s!}{(r / 2)! (s - r + 1)! (s - r / 2 + 1)} \in {\mathbb{Q}}(v, k), \end{align} $$

and

(4.2) $$ \begin{align} \tilde{F}_{s, r} := s! r! \ell_{s - 1}^2 \ell_{s - 2}^2 (s - r + 1)! (s - r / 2 + 1) = F_{s, r} \frac{(r / 2)! (s - r + 1)! (s - r / 2 + 1)}{s!} \in {\mathbb{Z}}. \end{align} $$

Clearly, $\tilde {F}_{s, r} \tilde {G}_{s, r} = F_{s, r} G_{s, r} \in {\mathcal {A}}_s$ by equations (4.1) and (4.2).

The constant $\tilde {F}_{s, r}$ contains a lot of small primes, which are usually more than enough to make $\tilde {F}_{s, r} \tilde {G}_{s, r}$ an element in ${\mathbb {Z}}_{(p)}$ . This suggests that we carefully choose an $\tilde {F}^b_{s, r}$ , which is much smaller than $\tilde {F}_{s, r}$ , and prove that $\tilde {F}^b_{s, r} \tilde {G}_{s, r} \in {\mathcal {A}}_s$ .

We first construct such an $\tilde {F}^b_{s, r}$ in Section 4.1. In Section 4.2, by examining the inequality $\ln \tilde {F}^b_{s, r} + \ln \tilde {G}_{s, r} \geq 0$ , Theorem 4.2 gives an upper bound for v in terms of s, r and b. If we choose some special parameters r and b in Theorem 4.2, then it gives $v \ll s$ as $s \to \infty $ . In Section 4.3, we give an explicit version of this asymptotic result that $v \leq 2{,}000{,}000 s$ for $s \geq 627$ .

4.1 Construction of $\tilde {F}^b_{s, r}$

We construct $\tilde {F}^b_{s, r}$ by artificially reducing the valuation of primes $p \leq b$ in $\tilde {F}_{s, r}$ . Let

(4.3) $$ \begin{align} \tilde{F}^b_{s, r} := \prod_{\text{prime } p \in [2, b]_{{\mathbb{Z}}}} p^{2 \lfloor \log_p (v - s) \rfloor} \prod_{\text{prime } p \in (b, s]_{{\mathbb{Z}}}} p^{\mathrm{val}_p(\tilde{F}_{s, r})}. \end{align} $$

Proposition 4.1. Let r be an even natural number with $r \leq s$ . If there exists a nontrivial tight $2s$ - $(v, k, \lambda )$ design, then $\tilde {F}^b_{s, r} \tilde {G}_{s, r}$ is a positive integer.

Proof. Since the design is nontrivial, $v - k \geq 2 s + 1$ and $k \geq 2 s + 1$ . Then, the factorization of $G_{s, r}$ in equation (3.3) shows that $G_{s, r}$ is a positive rational number; hence, so is $\tilde {F}^b_{s, r} \tilde {G}_{s, r}$ by equations (4.1) and (4.3).

For primes $p \leq b$ ,

$$ \begin{align*} & \mathrm{val}_p(\tilde{F}^b_{s, r} \tilde{G}_{s, r}) \\&\quad \geq 2 \lfloor \log_p (v - s) \rfloor - \mathrm{val}_p({v - s \choose s}) - \mathrm{val}_p({v - s - r / 2 \choose s - r / 2}) & (\text{by equations }(4.1)\text{ and }(4.3)) \\& \quad \geq 2 \lfloor \log_p (v - s) \rfloor - \lfloor \log_p (v - s) \rfloor - \lfloor \log_p (v - s - r / 2) \rfloor & (\text{since } \mathrm{val}_p({n \choose m}) \leq \lfloor \log_p n \rfloor) \\&\quad \geq 0. \end{align*} $$

For primes $p > b$ ,

$$ \begin{align*} \mathrm{val}_p(\tilde{F}^b_{s, r} \tilde{G}_{s, r}) & = \mathrm{val}_p(\tilde{F}_{s, r} \tilde{G}_{s, r}) & (\text{by equation~}(4.3)) \\& = \mathrm{val}_p(F_{s, r} G_{s, r}) & (\text{by equations }(4.1)\text{ and }(4.2)) \\& \geq 0. & (\text{by Proposition }3.4) \end{align*} $$

Therefore, $\tilde {F}^b_{s, r} \tilde {G}_{s, r}$ is a positive integer.

4.2 An upper bound for v

Since $\tilde {F}^b_{s, r} \tilde {G}_{s, r}$ is a positive integer by Proposition 4.1, it follows that $\ln \tilde {F}^b_{s, r} + \ln \tilde {G}_{s, r} \geq 0$ . Analyzing this inequality gives an upper bound for v.

Theorem 4.2. Let s be a natural number, r an even natural number with $6 \leq r \leq s$ and b a real number with $b \leq s$ . Let

(4.4) $$ \begin{align} \psi^b_{s, r} := r - 4 - 4 \pi(b), \end{align} $$

where $\pi (x)$ is the prime counting function that counts the number of primes no greater than x. Let

(4.5) $$ \begin{align} \begin{aligned} \kappa^b_{s, r} & := 2 \left( \ln \frac{(2 s - 3 r / 2 + 2)! s!}{(2 s - r + 2)! (s - r + 1)! (s - r / 2 + 1)} + (s - r) \ln 2 \right) \\ &\quad + 2 \left( \sum_{\text{prime } p \in (b, s]_{{\mathbb{Z}}}} \mathrm{val}_p\left(\tilde{F}_{s, r}\right) \ln p \right) + (4 \ln \frac{3}{2}) \pi(b). \end{aligned} \end{align} $$

Suppose that $\psi ^b_{s, r} > 0$ . If a nontrivial tight $2 s$ - $(v, k, \lambda )$ design exists, then

$$\begin{align*}v - 2 s + 1 \leq \exp(\kappa^b_{s, r} / \psi^b_{s, r}). \end{align*}$$

Proof. By Proposition 4.1, $\tilde {F}^b_{s, r} \tilde {G}_{s, r}$ is a positive integer; hence,

(4.6) $$ \begin{align} 0 \leq 2 \ln \left( \tilde{F}^b_{s, r} \tilde{G}_{s, r} \right) = 2 \ln \frac{\tilde{G}_{s, r}}{G_{s, r}} + 2 \ln \tilde{F}^b_{s, r} + 2 \ln G_{s, r}. \end{align} $$

We use the properties of the falling factorials and the rising factorials to estimate $G_{s, r}$ as follows.

$$ \begin{align*} G_{s, r} &= \frac{ (v - k - s)^{\overline{r / 2}} (k - s)^{\overline{s - r / 2 + 1}} (k - s)^{\overline{s - r + 1}} }{ (v - 2 s + 1)^{\overline{s}} (v - 2 s + 1)^{\overline{s - r / 2}} } & (\text{by equation~}(3.3)) \\ & \leq \frac{ (v - k - s)^{\overline{r / 2}} (k - s)^{\overline{2 s - 3 r / 2 + 2}} }{ (v - 2 s + 1)^{\overline{s}} (v - 2 s + 1)^{\overline{s - r / 2}} } & \text{(since } x^{\overline{a}} x^{\overline{b}} \leq x^{\overline{a + b}}) \\ & = \frac{ (v - k - s + r / 2 - 1)^{\underline{r / 2}} (k + s - 3 r / 2 + 1)^{\underline{2 s - 3 r / 2 + 2}} }{ (v - s)^{\underline{s}} (v - 2 s + 1)^{\overline{s - r / 2}} } & \text{(since } x^{\overline{a}} = (x + a - 1)^{\underline{a}}) \\ & \leq \frac{1}{ {2 s - r + 2 \choose r / 2} } \frac{ (v - r)^{\underline{2 s - r + 2}} }{ (v - s)^{\underline{s}} (v - 2 s + 1)^{\overline{s - r / 2}} } & \text{(since } {a + b \choose a} x^{\underline{a}} y^{\underline{b}} \leq (x + y)^{\underline{a + b}}) \\ & = \frac{1}{ {2 s - r + 2 \choose r / 2} } \frac{ (v - 2 s)^{\underline{2}} (v - r)^{\underline{s - r}} }{ (v - 2 s + 1)^{\overline{s - r / 2}} } \\ & \leq \frac{1}{ {2 s - r + 2 \choose r / 2} } \frac{v^{s - r}}{ (v - 2 s + 1)^{s - r / 2 - 2}} & \text{(since } x^{\underline{a}} \leq x^a \leq x^{\overline{a}}) \\ & \leq \frac{2^{s - r}}{ {2 s - r + 2 \choose r / 2} } \frac{1}{ (v - 2 s + 1)^{r / 2 - 2}}. & \text{(since } v \leq 2 (v - 2 s - 1)) \end{align*} $$

Taking logarithms on both sides gives the following estimate on $2 \ln G_{s, r}$ .

(4.7) $$ \begin{align} 2 \ln G_{s, r} \leq - (r - 4) \ln (v - 2 s + 1) - 2 \ln {2 s - r + 2 \choose r / 2} + 2 (s - r) \ln 2. \end{align} $$

Substituting equations (4.3) and (4.7) into equation (4.6),

$$ \begin{align*} 0 &\leq 2 \ln \frac{\tilde{G}_{s, r}}{G_{s, r}} + 4 \pi(b) \ln (v - s) + 2 \left(\sum_{\text{prime } p \in (b, s]_{{\mathbb{Z}}}} \mathrm{val}_p\left(\tilde{F}_{s, r}\right) \ln p \right) \\&\quad - (r - 4) \ln (v - 2 s + 1) - 2 \ln {2 s - r + 2 \choose r / 2} + 2 (s - r) \ln 2 \\& = - (r - 4) \ln (v - 2 s + 1) + 4 \pi(b) (\ln (v - s) - \ln \frac{3}{2}) + \kappa^b_{s, r} & \\& \qquad (\text{by equations }(4.1)\text{ and }(4.5)) \\& \leq - (r - 4) \ln (v - 2 s + 1) + 4 \pi(b) \ln (v - 2 s + 1) + \kappa^b_{s, r} & \\& \qquad (\text{since } v \geq 2 (2 s + 1) \text{ and } 6 \leq r \leq s) \\& = - \psi^b_{s, r} \ln (v - 2 s + 1) + \kappa^b_{s, r}. & (\text{by equation~}(4.4)) \end{align*} $$

Since $\psi ^b_{s, r} > 0$ , we have $\ln (v - 2 s + 1) \leq \kappa ^b_{s, r} / \psi ^b_{s, r}$ , from which the result follows.

4.3 Explicit upper bound for v when $s \geq 627$

From Theorem 4.2, it is easy to get an asymptotic upper bound for v. Choose $b := s$ and $r := 2 \lfloor s / 2 \rfloor $ . As $s \to \infty $ , $\psi ^b_{s, r} \asymp s$ and $\kappa ^b_{s, r} \asymp s \ln s$ . Therefore, $v < \exp (\kappa ^b_{s, r} / \psi ^b_{s, r}) + 2 s \ll s$ if there exists a nontrivial tight $2s$ - $(v, k, \lambda )$ design. We provide an explicit version of this asymptotic bound in Proposition 4.3.

Proposition 4.3. Let $b := s$ and $r := 2 \lfloor s / 2 \rfloor $ . If $s \geq 627$ , then $\psi ^b_{s, r} > 0$ and

$$\begin{align*}\exp(\kappa^b_{s, r} / \psi^b_{s, r}) < 2{,}000{,}000 s. \end{align*}$$

Proof. Dusart gave in [Reference Dusart7, Corollary 5.2] an explicit upper bound on the prime counting function,

(4.8) $$ \begin{align} \pi(x) \leq \frac{x}{\ln x} \left(1 + \frac{1.2762}{\ln x}\right). \end{align} $$

With this, we estimate $\psi ^b_{s, r}$ .

$$ \begin{align*} \psi^b_{s, r} & \geq s - 5 - 4 \pi(s) & (\text{by equation~}(4.4)\text{ and } r \in \{s - 1, s\}) \\& > s - 5 - 4 \frac{s}{\ln s} \left(1 + \frac{1.2762}{\ln s}\right) & (\text{by equation~}(4.8)) \\& > 0. & (\text{since } s \geq 172) \end{align*} $$

Stirling’s formula says that for $n \geq 1$ ,

(4.9) $$ \begin{align} f(n) < \ln n! \leq f(n) + 1, \end{align} $$

where $f(n) := n \ln n - n + \frac {1}{2} \ln n$ . With this and Dusart’s bound, we estimate $\kappa ^b_{s, r}$ .

$$ \begin{align*} \kappa^b_{s, r} & \leq 2 \ln (s / 2 + 7 / 2)! + 2 \ln s! - 2 \ln (s + 2)! \\&\quad - 2 \ln (s / 2 + 1) + 4 (\ln \frac{3}{2}) \pi(s) & (\text{by equation~}(4.5)\text{ and } r \in \{s - 1, s\}) \\& \leq 2 f(s / 2 + 7 / 2) + 2 f(s) - 2 f(s + 2) + 4 \\&\quad - 2 \ln (s / 2 + 1) + 4 (\ln \frac{3}{2}) \frac{s}{\ln s} \left(1 + \frac{1.2762}{\ln s}\right) & (\text{by equations }(4.8)\text{ and }(4.9)) \\& < \left(\ln s - \frac{5}{4}\right) s. & (\text{since } s \geq 186) \end{align*} $$

Therefore, combining the estimates of $\psi ^b_{s, r}$ and $\kappa ^b_{s, r}$ ,

$$ \begin{align*} & \kappa^b_{s, r} / \psi^s_{s, r} - \ln s \\ & \leq \left(\left(\ln s - \frac{5}{4}\right) s\right) \left(s - 5 - 4 \frac{s}{\ln s} \left(1 + \frac{1.2762}{\ln s}\right)\right)^{-1} - \ln s \\ & < \ln 2{,}000{,}000, & (\text{since } s \geq 627) \end{align*} $$

from which the result follows.

5 Lower bounds for v

For a nontrivial tight $2s$ - $(v, k, \lambda )$ design, the parameter $\lambda $ is just $\lambda _{s, s}$ in equation (2.1). Moreover,

(5.1) $$ \begin{align} s! \lambda_{s, s} = \frac{k \cdots (k - 2 s + 1)}{(v - s) \cdots (v - 2 s + 1)} \end{align} $$

takes integer values at parameters of nontrivial tight designs. This means that the denominator of the right side should not contain any prime factor larger than k. It indicates that we can use number-theoretic properties to give a lower bound for v.

Theorem 5.1. Let $\rho _s$ be the smallest positive integer n such that there are no primes in $(n, n + s - 1]_{{\mathbb {Z}}}$ . If there exists a nontrivial tight $2 s$ - $(v, k, \lambda )$ design, then

$$\begin{align*}v - 2 s \geq \rho_{s + 1}. \end{align*}$$

Remark 5.2. The number $\rho _s$ can be rewritten as

$$\begin{align*}\rho_s = \min \{p_n :\ p_{n + 1} - p_n \geq s\}, \end{align*}$$

where $p_n$ is the n-th prime. Due to its connection with prime gaps, the exact values of $\rho _s$ are known for $s \leq 1{,}550$ , which is the largest known maximal prime gap. In particular, $\rho _{288} = 1{,}294{,}268{,}491$ .

An upper bound on prime gaps will give us a lower bound for $\rho _s$ . We use one upper bound given by Dusart. Dusart’s result is not asymptotically the best known one, but it is explicit and has very nice constants.

Proposition 5.3. For $s \geq 288$ ,

$$\begin{align*}\rho_s > 5{,}000 s (14.6 + \ln s)^2. \end{align*}$$

In particular,

$$\begin{align*}\rho_s > 2{,}000{,}000 s. \end{align*}$$

Proof. Dusart showed in [Reference Dusart7, Corollary 5.5] that for $x \geq 468{,}991{,}632$ , there exists a prime p in the interval

$$\begin{align*}\left(x, x + \frac{x}{5000 (\ln x)^2}\right]. \end{align*}$$

Since $\rho _s \geq \rho _{288} > 468{,}991{,}632$ by Remark 5.2, $\left (\rho _s, \rho _s + \frac {\rho _s}{5000 (\ln \rho _s)^2}\right ]$ contains a prime. On the other hand, by definition, $\left (\rho _s, \rho _s + s - 1\right ]$ contains no primes. Therefore,

$$\begin{align*}\frac{\rho_s}{5000 (\ln \rho_s)^2} \geq s. \end{align*}$$

Then,

$$ \begin{align*} \rho_s & \geq 5{,}000 s (\ln \rho_s)^2 \geq 5{,}000 s (\ln (5{,}000 s (\ln \rho_s)^2))^2 \geq 5{,}000 s (\ln (5{,}000 s (\ln \rho_{288})^2))^2 \\ & > 5{,}000 s (14.6 + \ln s)^2 > 2{,}000{,}000 s. \end{align*} $$

6 Proof of the main result

Section 4 gives upper bounds for v, and Section 5 gives lower bounds for v. Now, we have all the ingredients needed to prove the main result, Theorem 1.1.

Proof of Theorem 1.1.

Through a series of previous works [Reference Bremner3, Reference Peterson11, Reference Xiang20, Reference Dukes and Short-Gershman6], it has been proved that the only nontrivial tight $2s$ -designs with $s \in [2, 9]_{{\mathbb {Z}}}$ are the Witt $4$ - $(23, 7, 1)$ design and its complement, the Witt $4$ - $(23, 16, 52)$ design. From now on, we assume that $s \geq 10$ and that there exists a nontrivial tight $2 s$ - $(v, k, \lambda )$ design.

Since the design is nontrivial, $k \geq 2 s + 1$ and $v - k \geq 2 s + 1$ . Its complementary design is a nontrivial tight $2 s$ - $(v, v - k, \overline {\lambda })$ design for some parameter $\overline {\lambda }$ . Since the complementary design has the same s and v as the original $2 s$ - $(v, k, \lambda )$ design, we may assume without loss of generality that $v \leq 2 k$ .

Choose $r := 2 \lfloor s / 2 \rfloor $ . Combining Theorem 4.2 and Theorem 5.1,

(6.1) $$ \begin{align} \rho_{s + 1} \leq v - 2 s < \exp(\kappa^b_{s, r} / \psi^b_{s, r}), \end{align} $$

whenever $\psi ^b_{s, r} > 0$ for some $b \leq s$ .

We divide the situation into three cases depending on the size of s and show contradictions in each case.

Case 1: $s \geq 627$ .

Choose $b := s$ . Proposition 4.3 shows $\psi ^b_{s, r} > 0$ and gives an upper bound

$$\begin{align*}\exp(\kappa^b_{s, r} / \psi^b_{s, r}) < 2{,}000{,}000 s. \end{align*}$$

On the other hand, Proposition 5.3 gives a lower bound

$$\begin{align*}\rho_s > 2{,}000{,}000 s. \end{align*}$$

These two bounds contradict equation (6.1).

Case 2: $s \in [288, 626]_{{\mathbb {Z}}}$ .

Choose $b := s / 3$ . Direct calculations show that $\psi ^b_{s, r} > 0$ and

$$\begin{align*}\exp(\kappa^b_{s, r} / \psi^b_{s, r}) < 1{,}000{,}000{,}000. \end{align*}$$

On the other hand,

$$\begin{align*}\rho_{s + 1} \geq \rho_{288} > 1{,}000{,}000{,}000 \end{align*}$$

by Remark 5.2. These two bounds contradict equation (6.1).

Case 3: $s \in [10, 287]_{{\mathbb {Z}}}$ .

For each $s \in [10, 287]_{{\mathbb {Z}}}$ , calculations over $b \in [1, s]_{{\mathbb {Z}}}$ show that we can choose a suitable b such that $\psi ^b_{s, r} > 0$ and $\exp (\kappa ^b_{s, r} / \psi ^b_{s, r}) < 15{,}000{,}000{,}000$ . Therefore, by equation (6.1), $v - 2 s \leq 15{,}000{,}000{,}000$ .

Let $x := k - s$ and $y := v - 2 s + 1$ . Then, $\alpha _{s, i}$ in equation (2.2) can be rewritten as

(6.2) $$ \begin{align} \alpha_{s, i} = {s \choose i} \frac{x^{\overline{i}}(x + 1)^{\overline{i}}}{y^{\overline{i}}}. \end{align} $$

The $\alpha _{s, i}$ ’s are integers by [Reference Xiang20, Corollary 2.2]. Moreover, $y \in [x + s + 2, 2 x + 1]_{{\mathbb {Z}}}$ and $x < k \leq v - 2 s - 1 \leq 15{,}000{,}000{,}000$ .

However, a computer search shows that there are no integer triples $(s, x, y)$ with $s \in [10, 287]_{{\mathbb {Z}}}$ , $x \in [1, 15{,}000{,}000{,}000]_{{\mathbb {Z}}}$ and $y \in [x + s + 2, 2 x + 1]_{{\mathbb {Z}}}$ such that $\alpha _{s, 1}, \dots , \alpha _{s, 6}$ in equation (6.2) are all integers.