2. Auxiliary functions and linear forms

Let $r = {\textrm {num}{(r)}}/{\textrm {den}{(r)}}$ be a positive rational number, where $\textrm {num}{(r)}$ and $\textrm {den}{(r)}$ are the numerator and denominator of $r$ in reduced form, respectively. We refer to $r$ as the Ball–Rivoal length parameter [Reference Ball and RivoalBR01]. Eventually we will take the rational number $r$ arbitrarily close to $r_0 = {(\sqrt {4e^2+1}-1)}/{2} \approx 2.26388$ in order to maximize a certain quantity.

Let $s$ be a positive odd integer and $B$ be a positive real number. We will always assume that:

1. (1) both $s$ and $B$ are larger than some absolute constant;

2. (2) $s \geqslant 10(2r+1)B^2$.

Eventually we will take $B = cs^{1/2}/\log ^{1/2} s$ for some constant $c$.

The zero set $\mathcal {F}_B$ consists of the zeros in the interval $(0,1]$ of our auxiliary rational functions, and the denominator set ${\Psi }_B$ consists of different denominators of the zeros. We collect some properties of these two sets. The first property is known in the topic about the inverse totient problem.

Proof. For the first proposition, we refer the reader to [Reference DresslerDre70] or [Reference BatemanBat72]. Since

\[ \left|\mathcal{F}_{B}\right| = \sum_{b \in {\Psi}_B} \varphi(b) = \sum_{m=1}^{\lfloor B \rfloor } m\left( \left| {\Psi}_m \right| - \left| {\Psi}_{m-1} \right| \right) = \lfloor B \rfloor \left| {\Psi}_{\lfloor B \rfloor} \right| - \sum_{m=1}^{\lfloor B \rfloor -1} \left| {\Psi}_m \right|, \]

the first proposition implies that $|\mathcal {F}_{B}| =(\frac {1}{2}({\zeta (2)\zeta (3)}/{\zeta (6)})+o(1))B^2$. Now, ${\zeta (2)\zeta (3)}/{\zeta (6)} = 1.94\ldots < 2$ and the third proposition follows. For the second proposition, note that if $b \in {\Psi }_B$ and $b'$ is any divisor of $b$, then $\varphi (b') \leqslant \varphi (b) \leqslant B$, so $b' \in {\Psi }_B$. Therefore, for any $k \in \{ 1,2,\ldots ,b \}$, we have ${k}/{b} = ({k/\gcd (k,b)})/({b/\gcd (k,b)}) \in \mathcal {F}_{B}$.

We define the integer

(2.1)\begin{equation} P_{B,\textrm{den}(r)} = 2\,\textrm{den}(r) \cdot \text{LCM}_{\substack{b \in {\Psi}_B \\ p \mid b}}\{ p-1 \}, \end{equation}

where $\text {LCM}$ means taking the least common multiple. As a convention, the letter $p$ always denotes prime numbers.

For given $r$, $s$ and $B$, we define the following auxiliary rational functions.

Definition 2.3 (FSZ constructions) For any positive integer $n$ which is a multiple of $P_{B,\textrm {den}(r)}$, we define the rational function

\[ R_n(t) = A_1(B)^n A_2(B)^n \frac{n!^{s+1}}{( {n}/{\textrm{den}(r)})!~^{\textrm{den}(r)(2r+1)|\mathcal{F}_B|}} \frac{(t-rn)\prod_{\theta \in \mathcal{F}_{B}} \prod_{j=0}^{(2r+1)n-1} \left( t-rn+j+\theta \right) }{\prod_{j=0}^{n}(t+j)^{s+1}}, \]

where

\[ A_1(B) = \prod_{b \in {\Psi}_B} b^{(2r+1)\varphi(b)}, \]

we refer to $A_1(B)^n$ the major arithmetic (wasting) factor, and

\[ A_2(B) = \prod_{b \in {\Psi}_B} \prod_{p \mid b} p^{{(2r+1)\varphi(b)}/({p-1})}, \]

we refer to $A_2(B)^n$ the minor arithmetic (wasting) factor.

Notice that by (2.1), both $A_1(B)^n$ and $A_2(B)^n$ are integers; also, ${n}/{\textrm {den}(r)}$, $rn$ and $(2r+1)n$ are integers. The factors $A_1(B)^n$ and $A_2(B)^n$ are technical: approximately speaking, they are designed to remedy the arithmetic loss from the denominators of rational zeros (it will be clear later in § 3). In the following lemma we estimate $A_1(B)$ and $A_2(B)$.

Lemma 2.4 We have

\[ A_1(B) = \exp\left( \left( \frac{1}{2} \frac{\zeta(2)\zeta(3)}{\zeta(6)} + o_{B \rightarrow + \infty}(1) \right)\left(2r+1\right) B^2 \log B \right) \]

and, for any $B$ larger than some absolute constant,

\[ A_2(B) \leqslant \exp\left( 10(2r+1)B^2(\log\log B)^2 \right). \]

Proof. We start by $\log A_1(B) = (2r+1)\sum _{b \in {\Psi }_B} \varphi (b)\log b$. Firstly,

\begin{align*} \log A_1(B) &\geqslant (2r+1)\sum_{b \in {\Psi}_B} \varphi(b)\log \varphi(b) \\ &= (2r+1) \int_{1^{-}}^{B} x\log x \,{d}|{\Psi}_x|; \end{align*}

an integration by parts argument with the fact that $| {\Psi }_x| = (({\zeta (2)\zeta (3)}/{\zeta (6)})+o_{x \rightarrow +\infty }(1)) x$ (see Proposition 2.2(1)) gives $\log A_1(B) \geqslant (2r+1)(\frac {1}{2} ({\zeta (2)\zeta (3)}/{\zeta (6)}) + o_{B \rightarrow + \infty }(1) ) B^2 \log B$. On the other hand, it is well known (see, for instance, [Reference Montgomery and VaughanMV06, Theorem 2.9]) that

\[ \varphi(m) \geqslant ( e^{-\gamma} + o_{m \rightarrow +\infty}(1) )\frac{m}{\log\log m}, \]

where $\gamma = 0.577\ldots$ is Euler's constant. For any $b \in {\Psi }_B$, since $\varphi (b) \leqslant B$, we derive that

(2.2)\begin{equation} b \leqslant ( e^{\gamma}+o_{B \rightarrow +\infty}(1) ) B \log\log B; \end{equation}

thus, $\log A_1(B) \leqslant (2r+1)(1+o_{B\rightarrow +\infty }(1))\log B \sum _{b \in {\Psi }_B} \varphi (b)$ and a summation by parts argument as above gives $\log A_1(B) \leqslant (2r+1)( \frac {1}{2} ({\zeta (2)\zeta (3)}/{\zeta (6)}) + o_{B \rightarrow + \infty }(1) ) B^2 \log B$. Combining the two parts, we obtain the estimate for $A_1(B)$.

Now, for $A_2(B)$, by (2.2) and $e^{\gamma } = 1.78\ldots < 2$, when $B$ is larger than some absolute constant, we have

\begin{align*} \log A_2(B) &\leqslant (2r+1) \sum_{b \leqslant 2B\log\log B} \varphi(b) \sum_{p \mid b} \frac{\log p}{p-1} \\ &= (2r+1) \sum_{p \leqslant 2B\log\log B} \frac{\log p}{p-1} \sum_{\substack{b \leqslant 2B\log\log B \\ p \mid b}} \varphi(b). \end{align*}

Since $\sum _{\substack {b \leqslant 2B\log \log B \\ p \mid b}} \varphi (b) \leqslant \sum _{\substack {b \leqslant 2B\log \log B \\ p \mid b}} b \leqslant {4B^2 (\log \log B)^2}/{p}$, it implies that

\[ \log A_2(B) \leqslant 4(2r+1) B^2 (\log\log B)^2 \sum_{p} \frac{\log p}{p(p-1)}. \]

Because $\sum _{p} (({\log p)}/{p(p-1)}) \leqslant \sum _{p \leqslant 5} (({\log p})/{p(p-1)}) + \sum _{p \geqslant 7} ({1}/{p^{1.5}}) \leqslant 2$, the estimate for $A_2(B)$ follows.

We proceed to construct linear forms in Hurwitz zeta values. Since the numerator and denominator of $R_{n}(t)$ have a common factor $\prod _{j=0}^{n} (t+j)$, it can be rewritten as $R_{n}(t) = {Q_{n}(t)}/{\prod _{j=0}^{n} (t+j)^{s}}$, where $Q_{n}(t)$ is a polynomial in $t$ with rational coefficients. Since $\deg R_n <0$ (see below), we know that $R_n(t)$ has a (unique) partial fraction expansion

(2.3)\begin{equation} R_n(t) = \sum_{i=1}^{s}\sum_{k=0}^{n} \frac{a_{i,k}}{(t+k)^i} \end{equation}

with coefficients $a_{i,k} \in \mathbb {Q}$. Note that these coefficients $a_{i,k}$ also depend on $n$, $r$, $s$ and $B$.

We list two properties of $R_n(t)$ and $a_{i,k}$ which will be used later in Lemma 2.5.

(1) As a rational function, the degree of $R_{n}(t)$ is

\[ \deg R_n = 1 + (2r+1)|\mathcal{F}_B|n - (s+1)(n+1) \leqslant -2. \]

This is due to $|\mathcal {F}_B| \leqslant B^2$ and $s \geqslant 10(2r+1)B^2$.

(2) The auxiliary function $R_{n}(t)$ has the following symmetry:

\[ R_{n}(-t-n) = - R_{n}(t). \]

In fact, if we substitute $t$ by $-t-n$ in the definition of $R_n(t)$ (see Definition 2.3), the factor

\[ \frac{(t-rn)\prod_{\theta \in \mathcal{F}_{B}} \prod_{j=0}^{(2r+1)n-1} \left( t-rn+j+\theta \right) }{\prod_{j=0}^{n}(t+j)^{s+1}} \]

becomes

\begin{align*} & \frac{(-t-(r+1)n)\prod_{\theta \in \mathcal{F}_{B}} \prod_{j=0}^{(2r+1)n-1} \left( -t-(r+1)n+j+\theta \right) }{\prod_{j=0}^{n}(-t-n+j)^{s+1}} \\ &\quad = (-1)^{1+(2r+1)n|\mathcal{F}_B|-(s+1)(n+1)} \times \frac{(t+(r+1)n)\prod_{\theta \in \mathcal{F}_{B}} \prod_{j=0}^{(2r+1)n-1} \left( t+(r+1)n-j-\theta \right) }{\prod_{j=0}^{n}(t+n-j)^{s+1}} \\ &\quad = (-1) \times \prod_{j=-rn}^{(r+1)n}(t+j) \times \frac{\prod_{\theta \in \mathcal{F}_{B} \setminus \{ 1\}} \prod_{j=0}^{(2r+1)n-1} \left( t- rn +j + 1-\theta \right) }{\prod_{j=0}^{n}(t+j)^{s+1}}. \end{align*}

We have used the facts that $(2r+1)n$ is even and $s$ is odd in the above computation. By observing that

\[ \left\{ 1-\theta \mid \theta \in \mathcal{F}_{B} \setminus \{1 \} \right\} = \mathcal{F}_{B} \setminus \{1 \}, \]

we obtain $R_{n}(-t-n) = - R_{n}(t)$. In particular, since the partial fraction expansion for $R_n(t)$ is unique, we derive that

\[ (-1)^i a_{i,k} =- a_{i,n-k} \]

for all $1\leqslant i\leqslant s$, $0 \leqslant k \leqslant n$.

For all $\theta \in \mathcal {F}_B$, we define the following quantities:

(2.4)\begin{equation} r_{n,\theta} = \sum_{m=1}^{\infty} R_{n}(m+\theta). \end{equation}

The notation $r_{n,\theta }$ is adopted to keep pace with that in [Reference Fischler, Sprang and ZudilinFSZ19]. There is no risk for it to be confused with the Ball–Rivoal length parameter $r$.

We recall the definition of the Hurwitz zeta values:

\[ \zeta(i,\alpha) = \sum_{m=0}^{\infty} \frac{1}{(m+\alpha)^i}, \]

where $i\geqslant 2$ is an integer and $\alpha$ is a positive real number.

The following lemma is the same as [Reference Fischler, Sprang and ZudilinFSZ19, Lemma 1] (for a proof, see therein).

Lemma 2.5 (Linear forms) For all $\theta \in \mathcal {F}_B$, we have

\[ r_{n,\theta} = \rho_{0,\theta} + \sum_{\substack{3 \leqslant i \leqslant s \\ i \ \textrm{odd}}} \rho_{i} \zeta\left(i,\theta \right), \]

where the rational coefficient

\[ \rho_{i} = \sum_{k=0}^{n} a_{i,k} \quad \text{for } 3 \leqslant i \leqslant s, i \ {\rm odd} , \]

does not depend on $\theta \in \mathcal {F}_B$, and

\[ \rho_{0,\theta} = - \sum_{k=0}^{n}\sum_{\ell=0}^{k}\sum_{i=1}^{s} \frac{a_{i,k}}{\left( \ell+\theta \right)^i}. \]

3. Arithmetic lemma

The following proposition is elementary; we omit the proof.

Proposition 3.1 Let $L \in \mathbb {N} \cup \{ 0 \}$. Suppose that $x_1,x_2,\ldots ,x_L$ are any $L$ consecutive terms in an integer arithmetic progression with common difference $b \in \mathbb {N}$; then, for any prime $q \nmid b$, we have

\[ v_q(x_1x_2\ldots x_L) \geqslant \sum_{i=1}^{\infty} \left\lfloor \frac{L}{q^i} \right\rfloor . \]

In the degenerate case of $L=0$, we view $x_1x_2\ldots x_L = 1$.

For any ${a}/{b} \in \mathcal {F}_{B}$ with $\gcd (a,b)=1$, we define the following polynomials:

(3.1)\begin{align} F_{b,a}(t) &= \frac{\prod_{p \mid b} p^{{(2r+1)n}/({p-1})}}{( {n}/{\textrm{den}(r)})!^{\,\textrm{den}(r)(2r+1)}}\cdot b^{(2r+1)n} \prod_{j=0}^{(2r+1)n -1} \left( t-rn + j + \frac{a}{b} \right) \nonumber\\ &= \frac{\prod_{p \mid b} p^{{(2r+1)n}/({p-1})}}{( {n}/{\textrm{den}(r)})!^{\,\textrm{den}(r)(2r+1)}} \prod_{j=0}^{(2r+1)n -1} \left( bt-brn + a + bj \right). \end{align}

Then we define

\[ \widetilde{F}_{b,a}(t) = \begin{cases} F_{b,a}(t) & \text{if}\ {a}/{b} \neq 1, \\ (t-rn)F_{1,1}(t) & \text{if}\ {a}/{b} = 1. \end{cases} \]

Notice that since $n \in P_{B,\textrm {den}(r)}\mathbb {N}$, by (2.1), all of ${(2r+1)n}/({p-1})$, ${n}/{\textrm {den}(r)}$, $rn$ and $(2r+1)n$ are integers. By Definition 2.3, we have

(3.2)\begin{equation} R_n(t) = n!^{s+1} \frac{\prod_{{a}/{b} \in \mathcal{F}_{B}}\widetilde{F}_{b,a}(t)}{\prod_{j=0}^{n}(t+j)^{s+1}}. \end{equation}

For a formal series $U(t) = \sum _{\ell =0}^{\infty } u_{\ell } t^{\ell } \in \mathbb {Q}[[t]]$, we denote by $[t^{\ell }](U(t))$ the $\ell$th coefficient of $U(t)$, i.e., $[t^{\ell }](U(t)) = u_{\ell }$.

As usual, we denote by $d_n=\text {LCM}\{1,2,\ldots ,n\}$ the least common multiple of the first $n$ positive integers. By the prime number theorem, we have $\lim _{n \rightarrow +\infty } d_n^{1/n} = e$. We first establish the following arithmetic property of $\widetilde {F}_{b,a}(t)$.

Proposition 3.2 For any non-negative integers $\ell$ and $k$, we have

\[ d_{n}^{\ell} \cdot [t^{\ell}](\widetilde{F}_{b,a}(t-k)) \in \mathbb{Z}. \]

Proof. Note that we only need to prove the proposition with $\widetilde {F}_{b,a}$ replaced by $F_{b,a}$. If $\ell > \deg F_{b,a} =(2r+1)n$, the proposition trivially holds. In the rest of the proof, we assume that $\ell \leqslant \deg F_{b,a}$.

For a prime $q \mid b$, the $q$-adic order of the factor ${\prod _{p \mid b} p^{{(2r+1)n}/({p-1})}}/{({n}/{\textrm {den}(r)} )!^{\,\textrm {den}(r)(2r+1)}}$ is non-negative (recall that $v_q(m!) \leqslant {m}/({q-1})$). So, by (3.1), the $q$-adic order of every coefficient of $F_{b,a}(t-k)$ is non-negative. Therefore, for any prime $q \mid b$, we have $v_q( d_{n}^{\ell } \cdot [t^{\ell }](F_{b,a}(t-k))) \geqslant 0$.

Now consider a prime $q \nmid b$. Notice that $[t^{\ell }](\prod _{j=0}^{(2r+1)n -1} ( b(t-k)-brn + a + bj ))$ is a sum of finitely many terms all of the form

(3.3)\begin{equation} b^{\ell} \prod_{i=1}^{\ell + 1}\prod_{j \in J_i} \left( -bk-brn + a + bj \right), \end{equation}

where $J_i$ is a set consisting of $L_i \in \mathbb {N} \cup \{ 0 \}$ consecutive integers such that $L_1 + L_2 + \cdots + L_{\ell + 1} = (2r+1)n - \ell$. By Proposition 3.1, we derive that the $q$-adic order of the expression (3.3) is

\[ v_q(\hbox{Equation (3.3)}) \geqslant \sum_{i=1}^{\infty} \sum_{j=1}^{\ell + 1} \left\lfloor \frac{L_j}{q^i} \right\rfloor. \]

For a fixed $i \geqslant 1$, we have

\begin{align*} \sum_{j=1}^{\ell + 1} \biggl\lfloor \frac{L_j}{q^i} \biggr\rfloor \geqslant \sum_{j=1}^{\ell + 1} \frac{L_j - (q^i - 1)}{q^i} = \frac{(2r+1)n+1}{q^i} - \ell - 1 > \biggl\lfloor \frac{(2r+1)n}{q^i} \biggr\rfloor - \ell - 1, \end{align*}

but the left-hand side is a non-negative integer, so we obtain that $\sum _{j=1}^{\ell + 1} \lfloor {L_j}/{q^i} \rfloor \geqslant \max (0, \lfloor {(2r+1)n}/{q^i} \rfloor - \ell )$. Therefore,

(3.4)\begin{align} v_q(\hbox{Equation (3.3)}) &\geqslant \sum_{i=1}^{\lfloor \log_q n \rfloor} \left( \left\lfloor \frac{(2r+1)n}{q^i} \right\rfloor - \ell \right) \nonumber\\ &\geqslant \sum_{i=1}^{\lfloor \log_q n \rfloor} \left( \textrm{den}(r)(2r+1) \left\lfloor \frac{n/\textrm{den}(r)}{q^i} \right\rfloor - \ell \right) \nonumber\\ &= v_q\left( \left( \frac{n}{\textrm{den}(r)} \right)!^{\,\textrm{den}(r)(2r+1)} \right) - \ell v_q(d_n). \end{align}

(The non-trivial part is for cases $q\leqslant n$; for $q>n$, the above derivation is also valid but degenerates to trivial results.) In conclusion, for any prime $q \nmid b$, by (3.1) and inequality (3.4), we find that $d_{n}^{\ell } \cdot [t^{\ell }](F_{b,a}(t-k))$ is a sum of finitely many terms, each of these terms having non-negative $q$-adic order; this completes the proof of Proposition 3.2.

We prove the following arithmetic lemma, which corresponds to [Reference Fischler, Sprang and ZudilinFSZ19, Lemma 2]. In our situation, the Ball–Rivoal length parameter $r$ is just a rational number (not necessarily an integer or a half integer), so we have to modify the proof for $d_n^{s+1-i} a_{i,k} \in \mathbb {Z}$, but the rest of proof is the same as [Reference Fischler, Sprang and ZudilinFSZ19, Lemma 2].

Lemma 3.3 (Arithmetic lemma) We have

\[ d_n^{s+1-i} \rho_{i} \in \mathbb{Z} \]

for all odd integers $i$ with $3 \leqslant i \leqslant s$, and we have

\[ d_{n+1}^{s+1} \rho_{0,\theta} \in \mathbb{Z} \]

for all $\theta \in \mathcal {F}_B$.

Proof. For any $k \in \{ 0,1,\ldots ,n \}$ and any $i \in \{ 1,2,\ldots ,s \}$, by comparing (3.2) with the partial fraction expansion (2.3) of $R_n(t)$, and by viewing $t^{s+1}R_n(t-k) \in \mathbb {Q}[[t]]$ as a formal series, we have

\begin{align*} a_{i,k} &= [t^{s+1-i}]\big( t^{s+1}R_{n}(t-k) \big) \\ &= (-1)^{(s+1)k}\frac{n!^{s+1}}{k!^{s+1}(n-k)!^{s+1}} [t^{s+1-i}]\biggl( \prod_{{a}/{b} \in \mathcal{F}_{B}} \widetilde{F}_{b,a}(t-k) \prod_{\substack{0 \leqslant j \leqslant n \\ j \neq k}} (1+ \frac{t}{j-k} )^{-s-1} \biggr) \\ &= \binom{n}{k}^{s+1} \sum_{\substack{\underline{\ell} \\ \textrm{sum}(\underline{\ell}) = s+1-i}} \prod_{{a}/{b} \in \mathcal{F}_{B}} [t^{\ell_{b,a}}]( \widetilde{F}_{b,a}(t-k) ) \prod_{\substack{0 \leqslant j \leqslant n \\ j \neq k}} \frac{(-1)^{\ell_j}\binom{s+\ell_j}{\ell_j}}{(j-k)^{\ell_j}}, \end{align*}

where the sum is taken for all tuples $\underline {\ell }$ consisting of non-negative integers $\ell _{b,a}$ and $\ell _j$ such that

\[ \textrm{sum}(\underline{\ell}) = \sum_{{a}/{b} \in \mathcal{F}_{B}} \ell_{b,a} + \sum_{\substack{0 \leqslant j \leqslant n \\ j \neq k}} \ell_j = s+1-i. \]

By Proposition 3.2 and the fact that $d_n^{\ell _j} ({1}/{(j-k)^{\ell _j}}) \in \mathbb {Z}$, we derive that

\[ d_{n}^{s+1-i} a_{i,k} \in \mathbb{Z}. \]

Once $d_{n}^{s+1-i} a_{i,k} \in \mathbb {Z}$ is established, the rest of the arguments are the same as [Reference Fischler, Sprang and ZudilinFSZ19, Lemma 2]. We only outline the proof as follows: $d_n^{s+1-i}\rho _i \in \mathbb {Z}$ follows immediately from the expression of $\rho _i$. The proof for $d_{n+1}^{s+1} \rho _{0,\theta } \in \mathbb {Z}$ is more involved; it is proved by showing that

\[ \sum_{i=1}^{s} \frac{d_{n+1}^{s+1} a_{i,k}}{(\ell + \theta)^{i}} \]

is an integer for any $0 \leqslant \ell \leqslant k \leqslant n$ and $\theta \in \mathcal {F}_B$. In order to do this, we will use the fact that $R_n(t)$ has zeros $-n+\theta , -n+1+\theta , -n+2 +\theta ,\ldots , \theta$ for $\theta \in \mathcal {F}_B \setminus \{1\}$.

4. Analysis lemma

Under our assumptions that $s \geqslant 10(2r+1)B^2$ and $B$ is larger than some absolute constant, we have the following result.

Lemma 4.1 (Analysis lemma) We have

\[ \lim_{n \rightarrow +\infty} \left(r_{n,1}\right)^{{1}/{n}} = g(x_0), \]

where

\[ g(X) = A_1(B)A_2(B) \textrm{den}(r)^{(2r+1)|\mathcal{F}_B|} (X+2r+1)^{(2r+1)|\mathcal{F}_B|} \left( \frac{(X+r)^r}{(X+r+1)^{r+1}} \right)^{s+1} \]

and $x_0$ is the unique positive real solution of the equation

\[ f(X) = \left( \frac{X+2r+1}{X} \right)^{|\mathcal{F}_B|} \left( \frac{X+r}{X+r+1} \right)^{s+1} = 1. \]

Moreover, for any $\theta \in \mathcal {F}_B$, we have

\[ \lim_{n \rightarrow +\infty} \frac{r_{n,1}}{r_{n,\theta}} = 1. \]

Before proving the analysis lemma, we first collect some properties of the functions $f$ and $g$. Note that these two functions depend only on $r,s$ and $B$.

Proof. For the first proposition, by calculating ${f'(x)}/{f(x)}$, we find that $f'(x) = 0$ has a unique positive solution $x_1$ which satisfies

(4.1)\begin{equation} (s+1-(2r+1)|\mathcal{F}_B|)x_1^2 + (2r+1)(s+1-(2r+1)|\mathcal{F}_B|)x_1 - r(r+1)(2r+1)|\mathcal{F}_B| = 0 \end{equation}

and $f$ is decreasing on $(0,x_1)$, increasing on $(x_1,+\infty )$. Since $f(0^{+}) = +\infty$ and $f(+\infty ) = 1$, there exists a unique $x_0$ satisfying all the requirements. The last (very weak) bound for $x_0$ comes from $x_0 < x_1$ and (4.1).

The second proposition follows from the estimates for $A_1(B)$, $A_2(B)$, $|\mathcal {F}_B|$ (see Proposition 2.2 and Lemma 2.4) and $x_0 \rightarrow 0$.

Now we prove Lemma 4.1. We claim that it can be proved by the same strategy in [Reference Fischler, Sprang and ZudilinFSZ19, Lemma 3], but we give a slightly modified proof.

Proof Proof of Lemma 4.1 For any $\theta \in \mathcal {F}_B$, since $R_n(m+\theta ) = 0$ for $m=1,2,\ldots ,rn-1$, we define the shift version of the auxiliary rational functions:

\[ \widehat{R}_{n}(t) = R_{n}(t+rn); \]

then, by (2.4), we have

(4.2)\begin{equation} r_{n,\theta} = \sum_{k=0}^{\infty} \widehat{R}_{n}(k+\theta). \end{equation}

We have the following two expressions for $\widehat {R}_{n}(t)$:

(4.3)\begin{align} \widehat{R}_{n}(t) &= A_1(B)^n A_2(B)^n \frac{n!^{s+1}}{( {n}/{\textrm{den}(r)})!^{\,\textrm{den}(r)(2r+1)|\mathcal{F}_B|}} \frac{ t\prod_{\theta' \in \mathcal{F}_B} \prod_{j=0}^{(2r+1)n-1} \left(t+j+\theta' \right)}{ \prod_{j=0}^{n} (t+rn+j)^{s+1}} \end{align}

(4.4)\begin{align} &= A_1(B)^n A_2(B)^n \frac{n!^{s+1}}{( {n}/{\textrm{den}(r)})!^{\,\textrm{den}(r)(2r+1)|\mathcal{F}_B|}} \nonumber\\ &\quad \times t \biggl( \prod_{\theta' \in \mathcal{F}_B } \frac{\Gamma(t+(2r+1)n+\theta')}{\Gamma(t+\theta')} \biggr) \left( \frac{\Gamma(t+rn)}{\Gamma(t+(r+1)n+1)} \right)^{s+1} . \end{align}

We define $c_1 = \min (\frac {1}{2}e^{-10s/r},{x_0}/{2})$, which is independent of $n$. To estimate the series (4.2) for $r_{n,\theta }$, we divide it into three parts:

\[ r_{n,\theta} = \biggl( \sum_{0 \leqslant k < c_1n} + \sum_{c_1 n \leqslant k \leqslant n^{10}} + \sum_{k>n^{10}} \biggr) ( \widehat{R}_{n}(k+\theta)). \]

For the first part, by (4.3), for all $t \in (0,2c_1 n]$, we have

\begin{align*} \frac{\widehat{R}_{n}^{\prime}(t)}{\widehat{R}_{n}(t)} &> \sum_{j=0}^{(2r+1)n} \frac{1}{t+j} - (s+1) \sum_{j=0}^{n} \frac{1}{t+rn+j} \\ &> \log\left( \frac{t +(2r+1)n}{t} \right) - (s+1) \frac{n+1}{rn} \\ &> \log\left(\frac{1}{2c_1}\right) - 4s/r \\ &> 0. \end{align*}

So, $\widehat {R}_{n}(t)$ is increasing on $t \in (0,2c_1 n]$; we derive that (when $n > c_1^{-1}$)

(4.5)\begin{equation} \sum_{0 \leqslant k < c_1n} \widehat{R}_{n}(k+\theta) < \left( c_{1}n+1 \right)\widehat{R}_{n}([c_{1}n]+\theta). \end{equation}

To deal with the middle part, for all $c_1 n \leqslant k \leqslant n^{10}$, we denote $\kappa = \kappa (k,n) = {k}/{n} \in [c_1, +\infty )$. By applying Stirling's formula in the weak form

\[ \Gamma(x) = x^{O_{x \rightarrow +\infty}(1)}\left( \frac{x}{e} \right)^x \]

for the equation (4.4), a calculation shows that as $n \rightarrow +\infty$,

(4.6)\begin{align} \widehat{R}_{n}(k+\theta) &= n^{O(1)} \cdot A_1(B)^n A_2(B)^n \frac{(n/e)^{(s+1)n}}{(( {n}/{\textrm{den}(r)})/e)^{(2r+1)|\mathcal{F}_B|n}} \nonumber\\ &\quad \times \biggl( \frac{( (k+(2r+1)n)/e )^{k+(2r+1)n}}{(k/e)^{k}} \biggr)^{\left| \mathcal{F}_B \right|} \frac{\left( (k+rn)/e \right)^{(s+1)(k+rn)}}{\left( (k+(r+1)n)/e \right)^{(s+1)(k+(r+1)n)}} \nonumber \\ &= n^{O(1)} \cdot A_1(B)^n A_2(B)^n \textrm{den}(r)^{(2r+1)|\mathcal{F}_B|n} \nonumber\\ &\quad \times \left( \frac{(\kappa + 2r + 1)^{\kappa + 2r + 1}}{\kappa^\kappa} \right)^{|\mathcal{F}_B|n} \left( \frac{(\kappa + r)^{\kappa + r}}{(\kappa + r + 1)^{\kappa + r + 1}} \right)^{(s+1)n} \nonumber\\ &= n^{O(1)} \cdot \left( f(\kappa)^{\kappa} g(\kappa) \right)^n \nonumber\\ &= n^{O(1)} \cdot h(\kappa)^n \end{align}

uniformly for any $k \in [c_1n,n^{10}]$ and any $\theta \in \mathcal {F}_B$ (the absolute bound for $O(1)$ depends only on $s,B,r$ and $\textrm {den}(r)$); here the function $h(x)$ is defined for $x>0$ as $h(x)= f(x)^x g(x)$ and a direct computation shows that ${h'(x)}/{h(x)} = \log f(x)$. Hence, $h(x)$ achieves its maximum only at $x = x_0$ with maximal value $h(x_0) = g(x_0)$.

In particular, we have the following bound for each $k \in [c_1 n , n^{10}]$:

(4.7)\begin{equation} \widehat{R}_{n}(k+\theta) \leqslant n^{O(1)} \cdot g(x_0)^n. \end{equation}

Finally, we treat the tail part. For any $k > n^{10}$, when $n$ is sufficiently large in terms of $r,s$ and $B$ (more precisely, when $n \geqslant \max (10(2r+1), 10A_1(B)A_2(B), {10}/{g(x_0)})$), by (4.3) and our assumption that $s \geqslant 10(2r+1)B^2$, we have

\begin{align*} \widehat{R}_{n}(k+\theta) &< A_1(B)^n A_2(B)^n n^{(s+1)n} \cdot \frac{(2k)\prod_{\theta' \in \mathcal{F}_B}\prod_{j=0}^{(2r+1)n-1} (2k)}{\prod_{j=0}^{n}(k)^{s+1}} \\ &< \frac{(2A_1(B)A_2(B)n)^{(s+1)n}}{k^{({9}/{10})(s+1)n + 2}} \\ &< \left( \frac{2A_1(B)A_2(B)}{n^8}\right)^{(s+1)n}\frac{1}{k^2} \\ &< \left( \frac{g(x_0)}{2} \right)^n \frac{1}{k^2}. \end{align*}

As a conclusion, we obtain the following bound for the tail part for all sufficiently large $n$:

(4.8)\begin{equation} \sum_{k> n^{10}} \widehat{R}_{n}(k+\theta) \leqslant \left( \frac{g(x_0)}{2} \right)^n. \end{equation}

Now, in view of the estimates (4.5), (4.7) and (4.8), we have $r_{n,1} \leqslant n^{O(1)} g(x_0)^n$. On the other hand, (4.6) implies that $r_{n,1} \geqslant \widehat {R}_{n}(\lfloor x_0n \rfloor ) = n^{O(1)}h(x_0 + o(1))^n$. Therefore,

\[ \lim_{n \rightarrow +\infty} \left(r_{n,1}\right)^{{1}/{n}} = g(x_0). \]

To prove the last statement in the lemma, we first fix an arbitrary (sufficiently) small $\varepsilon _0 > 0$. For all $\theta \in \mathcal {F}_B$, we have

(4.9)\begin{equation} r_{n,\theta} \geqslant \sum_{(x_0 - \varepsilon_0)n \leqslant k \leqslant (x_0 + \varepsilon_0)n} \widehat{R}_{n}(k+\theta). \end{equation}

In view of the estimates (4.5), (4.6) and (4.8), we also have

(4.10)\begin{align} r_{n,\theta} &\leqslant n^{O(1)}\max ( h(x_0 - \varepsilon_0), h(x_0 + \varepsilon_0))^n \quad + \quad \sum_{(x_0 - \varepsilon_0)n \leqslant k \leqslant (x_0 + \varepsilon_0)n} \widehat{R}_{n}(k+\theta) \nonumber\\ &< (1+\varepsilon_0) \sum_{(x_0 - \varepsilon_0)n \leqslant k \leqslant (x_0 + \varepsilon_0)n} \widehat{R}_{n}(k+\theta), \end{align}

provided $n$ is sufficiently large with respect to $\varepsilon _0$, i.e., $n \geqslant n_0(\varepsilon _0)$. For all $k$ with $(x_0 - \varepsilon _0)n \leqslant k \leqslant (x_0 + \varepsilon _0)n$, let $\kappa = \kappa (n,k) = {k}/{n}$ as before. We now use the fact that, for any fixed real number $\tau$,

(4.11)\begin{equation} \dfrac{\Gamma(x+\tau)}{\Gamma(x)} = \left( 1+o_{x \rightarrow +\infty}(1) \right) x^{\tau} . \end{equation}

Applying (4.11) to (4.4), we derive that

(4.12)\begin{align} \frac{\widehat{R}_{n}(k+1)}{\widehat{R}_{n}(k+\theta)} &= \left( 1+o(1) \right) \left( \frac{\kappa + 2r + 1}{\kappa} \right)^{|\mathcal{F}_B| (1-\theta)}\left( \frac{\kappa + r}{\kappa + r+1} \right)^{(s+1)(1-\theta)} \nonumber\\ &= \left( 1+o(1) \right) f(\kappa)^{1 - \theta} \end{align}

uniformly for $k \in [(x_0 - \varepsilon _0)n , (x_0 + \varepsilon _0)n]$ as $n \rightarrow +\infty$. By (4.9), (4.10) and (4.12), we find that

\[ \left( 1+o(1) \right)\frac{1}{1+\varepsilon_0} f(x_0 + \varepsilon_0)^{1 - \theta} \leqslant \frac{r_{n,1}}{r_{n,\theta}} \leqslant \left( 1+o(1) \right)(1+\varepsilon_0)f(x_0 - \varepsilon_0)^{1-\theta}; \]

thus,

\[ \frac{1}{1+\varepsilon_0} f(x_0 + \varepsilon_0)^{1 - \theta} \leqslant \liminf_{n \rightarrow +\infty} \frac{r_{n,1}}{r_{n,\theta}} \leqslant \limsup_{n \rightarrow +\infty} \frac{r_{n,1}}{r_{n,\theta}} \leqslant (1+\varepsilon_0) f(x_0 - \varepsilon_0)^{1-\theta}. \]

It is true for all sufficiently small $\varepsilon _0>0$. Letting $\varepsilon _0 \rightarrow 0^+$, we deduce that

\[ \lim_{n \rightarrow +\infty} \frac{r_{n,1}}{r_{n,\theta}} = 1. \]

This completes the proof of Lemma 4.1.

5. Elimination procedure and proof of the theorem

We prove Theorem 1.1 in this section. We will use the same strategy as [Reference Fischler, Sprang and ZudilinFSZ19, § 5], namely, an elimination procedure. So, we only give an outline of this elimination procedure.

We denote $I_s = \{ 3,5,7,\ldots ,s \}$. For any subset $J \subset I_s$ with $|J| = |{\Psi }_B| -1$, since the following general Vandermonde matrix (see, for instance, [Reference Gantmacher and KreinGK02, pp. 76–77]):

\[ \big[b^{j}\big]_{b \in {\Psi}_B, \ j \in \{1\} \cup J } \]

is invertible, there exist integers $w_b \in \mathbb {Z}$ for all $b \in {\Psi }_B$ such that $\sum _{b \in {\Psi }_B} w_{b} b^{j} = 0$ for any $j \in J$ and $\sum _{b \in {\Psi }_B} w_{b} b \neq 0$. (Note that these $w_b$ depend only on $J$ and ${\Psi }_B$.) Since

(5.1)\begin{equation} \sum_{k=1}^{b} \zeta\left(i,\frac{k}{b}\right) = \sum_{k=1}^{b} \sum_{m=0}^{\infty} \frac{b^i}{(mb+k)^i} = b^i \zeta(i), \end{equation}

we derive that (recall Proposition 2.2(2), ${k}/{b} \in \mathcal {F}_B$)

\[ \widehat{r}_{n,b} := \sum_{k=1}^{b} r_{n,{k}/{b}} = \sum_{k=1}^{b} \rho_{0,{k}/{b}} + \sum_{i \in I_s} \rho_{i} b^i \zeta(i) \]

is a linear combination of $1$ and odd zeta values. By Lemma 4.1, we have $\widehat {r}_{n,b} = (b+o(1))r_{n,1}$ as $n \rightarrow +\infty$. Let

\[ \widetilde{r}_{n} := \sum_{b \in {\Psi}_B} w_b \widehat{r}_{n,b}; \]

then

(5.2)\begin{equation} \widetilde{r}_{n} = \sum_{b \in {\Psi}_B} w_b \sum_{k=1}^{b} \rho_{0,{k}/{b}} + \sum_{i \in I_s \setminus J} \biggl( \sum_{b \in {\Psi}_B} w_b b^i \biggr) \rho_{i} \zeta(i) \end{equation}

and, as $n \rightarrow +\infty$,

(5.3)\begin{equation} \widetilde{r}_{n} = \biggl( \sum_{b \in {\Psi}_B} w_{b} b + o(1)\biggr) r_{n,1} \quad \text{with}\ \sum_{b \in {\Psi}_B} w_{b} b \neq 0. \end{equation}

Equation (5.2) shows that we can eliminate any $|{\Psi }_B|-1$ odd zeta values.

Proof. We argue by contradiction. Suppose that the number of irrationals in $\{ \zeta (i) \}_{i \in I_s}$ is less than $|{\Psi }_B|$; then we can take a subset $J \subset I_s$ with $|J| = |{\Psi }_B| -1$ such that $\zeta (i) \in \mathbb {Q}$ for all $I_s \setminus J$; let $A$ be the common denominator of these rational zeta values. Define $\widetilde {r}_{n}$ as above for this $J$; then, by (5.2) and Lemma 3.3, for all $n \in P_{B,\textrm {den}(r)}\mathbb {N}$, we derive that

\[ Ad_{n+1}^{s+1} \widetilde{r}_{n} \in \mathbb{Z}. \]

But by (5.3), Lemma 4.1 and the hypothesis $g(x_0) < e^{-(s+1)}$, we have

\[ 0 < \lim_{n \rightarrow +\infty} \left| Ad_{n+1}^{s+1}\widetilde{r}_{n} \right|^{{1}/{n}} = e^{s+1}g(x_0) < 1. \]

This is a contradiction.

So, we seek parameters $r$, $s$ and $B$ to meet the requirement that $g(x_0) < e^{-(s+1)}$, and at the same time to make $|{\Psi }_B| \sim ({\zeta (2)\zeta (3)}/{\zeta (6)}) B$ as large as possible. By Proposition 4.2(2), for a fixed $r$ (such that ${r^r}/{(r+1)^{r+1}} < e^{-1}$), if we take $B = cs^{1/2} / \log ^{1/2} s$ for some constant $c$, then $\lim _{s \rightarrow +\infty } g(x_0)^{{1}/({s+1})} < e^{-1}$ if and only if

\[ c < \sqrt{ \frac{4\zeta(6)}{\zeta(2)\zeta(3)} \frac{(r+1)\log(r+1)-r\log(r)-1}{2r+1}}. \]

The maximum point of the function $r \mapsto ({(r+1)\log (r+1)-r\log (r)-1})/({2r+1})$ is

\[ r_0 = \frac{\sqrt{4e^2+1}-1}{2} \approx 2.26388, \]

with maximal value $1 - \log r_0$. The constant $c_0$ in Theorem 1.1 is designed by

\[ c_0 = \sqrt{\frac{4\zeta(2)\zeta(3)}{\zeta(6)}\left( 1 -\log r_0 \right)}. \]

This leads to the following proof.

Proof Proof of Theorem 1.1 Given any small $\varepsilon > 0$, we first fix a rational number $r = r(\varepsilon )$ sufficiently close to $r_0$ such that

\[ \frac{c_0 - \varepsilon/10}{\zeta(2)\zeta(3)/\zeta(6)} < \sqrt{ \frac{4\zeta(6)}{\zeta(2)\zeta(3)} \frac{(r+1)\log(r+1)-r\log(r)-1}{2r+1}}. \]

Take $B = cs^{1/2}/\log ^{1/2} s$ with constant $c = ({c_0 - \varepsilon /10})/{(\zeta (2)\zeta (3)/\zeta (6))}$; by Propositions 4.2(2) and 2.2(1), there exists $s_0(r,\varepsilon )$ such that for any odd integer $s \geqslant s_0(r,\varepsilon )$, we have $g(x_0) < e^{-(s+1)}$ and $|{\Psi }_B| > (\zeta (2)\zeta (3)/\zeta (6) - \varepsilon /10)B$. Therefore, by Proposition 5.1, the number of irrationals among $\zeta (3),\zeta (5),\ldots , \zeta (s)$ is at least

\[ |{\Psi}_B| > (c_0 - \varepsilon) \frac{s^{1/2}}{\log^{1/2} s}. \]

6. Remarks on FSZ constructions

If we choose a general finite set $\mathcal {F} \subset (0,1]$ of rational numbers to be the zero set of the auxiliary function $R(t)$, like [Reference Fischler, Sprang and ZudilinFSZ19] and this note, we design the factor

(6.1)\begin{equation} A_1(\mathcal{F})^n = \prod_{\theta \in \mathcal{F}} \textrm{den}(\theta)^{(2r+1)n} \end{equation}

to remedy the arithmetic loss from the denominators of rational zeros. Suppose that our goal is to prove that there exist $D$ irrational numbers among $\zeta (3), \zeta (5), \ldots , \zeta (s)$. In order to eliminate $D-1$ zeta values, in view of (5.1), we assume that there exist $D$ pairwise different positive integers $b_1, b_2 , \ldots , b_D$ such that

(6.2)\begin{equation} \mathcal{F} \supset \left\{ \frac{1}{b_i} ,\frac{2}{b_i} , \ldots, \frac{b_i}{b_i} \right\} \end{equation}

for any $i=1,2,\ldots ,D$. Then $\mathcal {F}$ contains the following disjoint union:

\[ \mathcal{F} \supset \bigcup_{i=1}^{D} \left\{ \left.\frac{a}{b_i} \,\right| 1 \leqslant a \leqslant b_i, \gcd(a,b_i) = 1 \right\}. \]

Hence, we have

\[ A_1(\mathcal{F}) \geqslant \prod_{i=1}^{D} b_i^{(2r+1)\varphi(b_i)}. \]

Now we consider the minimal magnitude of $A_1(\mathcal {F})$.

Proposition 6.1 We have

\[ \min \prod_{i=1}^{D} b_i^{\varphi(b_i)} = \exp\left( \left( \frac{1}{2}\frac{\zeta(6)}{\zeta(2)\zeta(3)} +o_{D \rightarrow +\infty}(1)\right) D^2\log D \right), \]

where the minimum is taken for all $D$-tuples $(b_1,\ldots ,b_D)$ such that $b_1,b_2,\ldots ,b_D$ are $D$ pairwise distinct positive integers.

Proof. We have

\begin{align*} \log \prod_{i=1}^{D} b_i^{\varphi(b_i)} &\geqslant \sum_{i=1}^{D} \varphi(b_i)\log \varphi(b_i) \\ &\geqslant \sum_{i=1}^{D} \varphi(b'_i)\log \varphi(b'_i), \end{align*}

where $b'_1,b'_2,\ldots ,b'_D$ are the $D$ smallest positive integers in the linear order $\prec$ defined by

\[ m_1 \prec m_2 \Leftrightarrow \left( \varphi(m_1)<\varphi(m_2)\quad \text{or}\quad \left( \varphi(m_1) = \varphi(m_2)\ \text{and}\ m_1 < m_2 \right) \right). \]

Recall that for any positive real number $x$, we define ${\Psi }_x = \{ b \in \mathbb {N} \mid \varphi (b) \leqslant x \}$. Then there exists an integer $B$ such that ${\Psi }_{B-1} \subset \{ b'_1,\ldots ,b'_D \} \subset {\Psi }_{B}$. By Proposition 2.2(1), we have the asymptotical relation $B = ({\zeta (6)}/{\zeta (2)\zeta (3)} + o_{D \rightarrow +\infty }(1) ) D$. Following the first paragraph in the proof of Lemma 2.4, we derive that

\begin{align*} \sum_{i=1}^{D} \varphi(b'_i)\log \varphi(b'_i) &\geqslant \sum_{b \in {\Psi}_{B-1}} \varphi(b)\log \varphi(b) \\ &\geqslant \left( \frac{1}{2} \frac{\zeta(2)\zeta(3)}{\zeta(6)} +o_{B \rightarrow +\infty}(1) \right) B^2 \log B \\ &= \left( \frac{1}{2}\frac{\zeta(6)}{\zeta(2)\zeta(3)} +o_{D \rightarrow +\infty}(1)\right) D^2\log D. \end{align*}

This shows that

\[ \prod_{i=1}^{D} b_i^{\varphi(b_i)} \geqslant \exp\left( \left( \frac{1}{2}\frac{\zeta(6)}{\zeta(2)\zeta(3)} +o_{D \rightarrow +\infty}(1)\right) D^2\log D \right) \]

for any pairwise distinct positive integers $b_1,b_2,\ldots ,b_D$. On the other hand, if we take $(b'_1, \ldots , b'_D)$ as defined above, then Lemma 2.4 also implies that

\begin{align*} \prod_{i=1}^{D} b_i^{\prime\varphi(b'_i)} &\leqslant \prod_{b \in {\Psi}_{B}} b^{\varphi(b)} \\ &= \exp\left( \left( \frac{1}{2}\frac{\zeta(2)\zeta(3)}{\zeta(6)} +o_{B \rightarrow +\infty}(1)\right) B^2\log B \right) \\ &= \exp\left( \left( \frac{1}{2}\frac{\zeta(6)}{\zeta(2)\zeta(3)} +o_{D \rightarrow +\infty}(1)\right) D^2\log D \right). \end{align*}

We complete the proof of Proposition 6.1.

Proposition 6.1 shows that, under the constraints (6.1) and (6.2), the choice $\mathcal {F} = \mathcal {F}_B$ in Definition 2.1 is optimal up to an $o(1)$ term for minimizing the major arithmetic wasting factor $A_1(\mathcal {F})$. We do not need to consider the minor arithmetic wasting factor $A_2(\mathcal {F})$ for general $\mathcal {F}$, since $A_2(\mathcal {F}_B)$ is asymptotically negligible with respect to $A_1(\mathcal {F}_B)$ (see Lemma 2.4); we already have that

\[ A_1(\mathcal{F}_B) A_2(\mathcal{F}_B) = \exp\left( \left( \frac{1}{2}\frac{\zeta(2)\zeta(3)}{\zeta(6)} +o_{B \rightarrow +\infty}(1)\right) (2r+1)B^2\log B \right). \]

For the factorial factor

\[ \frac{n!^{s+1}}{( {n}/{\textrm{den}(r)} )!~^{\textrm{den}(r)(2r+1)|\mathcal{F}|}}, \]

comparing to the corresponding factor $n!^{s+1-(2r+1)|\mathcal {F}|}$ in [Reference Ball and RivoalBR01] or [Reference Fischler, Sprang and ZudilinFSZ19], we have an extra waste of

\[ \binom{n}{ \underbrace{\frac{n}{\textrm{den}(r)},\ldots,\frac{n}{\textrm{den}(r)}}_{\textrm{den}(r)\ \text{in number}}}^{(2r+1)|\mathcal{F}|} \leqslant \textrm{den}(r)^{(2r+1)|\mathcal{F}|n}, \]

which is asymptotically negligible with respect to $A_1(\mathcal {F})^n$ (at least for $\mathcal {F} = \mathcal {F}_B$). Usually the Ball–Rivoal length parameter $r$ is taken to be an integer in the literature, since in the case that $r$ is integral, the corresponding arithmetic lemmas (see [Reference Ball and RivoalBR01, Lemme 5] and [Reference Fischler, Sprang and ZudilinFSZ19, Lemma 2]) can be proved in a simpler way. We need to take non-integral rational $r$ in order to let it be arbitrarily close to $r_0$. The idea of taking non-integral $r$ is not new: it has appeared in the literature (for example, [Reference ZudilinZud01]), but in a different form.

There exist some arithmetic saving factors known as $\Phi _n$ factors. They are certain products over primes in the range $C_{B,r} \sqrt {n} \leqslant p \leqslant n$ (here we can take $C_{B,r} = \sqrt {2(r+1)B\log \log B}$). We mention that the saving from $\Phi _n^{-1}$ plays an important role in small cases for the odd zeta problem; see, for instance, [Reference ZudilinZud01, Reference Rivoal and ZudilinRZ20], [Reference ZudilinZud02, § 4] or [Reference Krattenthaler and RivoalKR07, Chapitre 11]. However, like [Reference Fischler, Sprang and ZudilinFSZ19, Remark 2], the known types of $\Phi _n$ factors have no effect on asymptotics. The reason is that, by Definition 2.3, (2.3) and (3.2), for any $k \in \{ 0,1,\ldots ,n \}$,

\begin{align*} a_{s,k} &= \left( (t+k)^s R_n(t) \right) |_{t=-k} \\ &= (-1)^{k}\binom{n}{k}^s \frac{n!(k+1)_{rn}(-k+n+1)_{rn}}{ ( {n}/{ \textrm{den}(r)})!^{\,\textrm{den}(r)(2r+1)} } \prod_{{a}/{b} \in \mathcal{F}_B \setminus \{1\} } F_{b,a}(-k), \end{align*}

where $(x)_{m} := x(x+1)\cdots (x+m-1)$. For any prime $p$ with $C_{B,r} \sqrt {n} \leqslant p \leqslant n$, the $p$-adic order of $a_{s,0}$ is relatively small. If we define

\[ \widetilde{\Phi}_n = \prod_{C_{B,r} \sqrt{n} \leqslant p \leqslant n}\ p^{v_p(\gcd\{ a_{s,k} \}_{k=0}^{n} )}, \]

then we can show that $\widetilde {\Phi }_n \leqslant A_2(B)^n \cdot d_n^{(\textrm {den}(r)(2r+1) + 1)|\mathcal {F}_B|}$, which is asymptotically negligible with respect to $A_1(\mathcal {F}_B)^n$. One may want to directly save the common divisor of $d_{n+1}^{s+1}\rho _{0, \theta }$ and $d_{n+1}^{s+1}\rho _i$, but it is out of current research. The small cases are more difficult to study; up to now, besides Apéry's theorem [Reference ApéryApé79] that $\zeta (3)$ is irrational, the most remarkable result is that at least one of $\zeta (5),\zeta (7),\zeta (9),\zeta (11)$ is irrational, due to Zudilin [Reference ZudilinZud01] in 2001. The question whether $\zeta (5)$ is irrational remains open.