Proof. (Theorem 2) The weight \(a_{n+1}\) is odd, as we showed in [6] (since our example with \(B=0\) included the same weight \(a_{n+1}\) as here). For the reader’s convenience, here is the argument: \(s_n\) is \(7\pmod{8}\) if \(n\geq 2\) is even and \(3\pmod{8}\) if \(n\geq 3\) is odd. This is immediate by induction from the recurrence \(s_{n+1}=s_n(s_n-1)+1\). It follows that \(s_n^2-s_n+2\) is \(4\pmod{8}\) if \(n\geq 2\) is even, and that \(s_n^2-3s_n+4\) is \(4\pmod{8}\) if \(n\geq 3\) is odd. So \(a_{n+1}\) is odd in both cases.

Next, we show that the weighted projective space \(Y=\text{\boldsymbol{P}}^{n+1}(a_0,\ldots,a_{n+1})\) is well-formed. That is, we have to show that \(\gcd(a_0,\ldots,\widehat{a_j}, \ldots,a_{n+1})=1\) for each \(j\). We have \(s_{n+1}-1=s_0\cdots s_n\) with \(s_0,\ldots,s_n\) pairwise coprime, which implies that \(\gcd(a_0,\ldots,a_n)=1\). For the rest, it suffices to show that \(\gcd(s_{n+1}-1,a_{n+1})=1\). Here \(s_{n+1}-1=s_n(s_n-1)\), so it suffices to show that \(\gcd(s_n-1,a_{n+1})=1\) and \(\gcd(s_n,a_{n+1})=1\). The first statement was shown in [6]. For the second, if a prime number \(p\) divides \(s_n\) and \(a_{n+1}\), then \(p>2\) since \(a_{n+1}\) is odd. If \(n\) is even, it follows that \(a_{n+1}=\frac{1}{4}(s_n^2-s_n+2)\equiv \frac{1}{2}\pmod {p}\), not 0, which is a contradiction. If \(n\) is odd, we have \(a_{n+1}=\frac{1}{4}(s_n^2-3s_n+4)\equiv 1\pmod {p}\), not 0, which is a contradiction. That completes the proof that \(Y\) is well-formed.

From the equation for \(X\), the only coordinate linear space of \(Y\) contained in \(X\) is the point \(p \mathrel{\vcenter{:}}= [0,\ldots,0,1]\). Since that has codimension at least 2 in \(X\), \(X\) is well-formed. Also, \(X\) is quasi-smooth outside \(p\), so it has only quotient singularities outside \(p\), and so \(X\) is klt outside \(p\). At the point \(p\), in coordinates \(x_{n+1}=1\), \(X\) is defined by the equation \[0=x_0^2+x_1^3+\cdots+x_n^{s_n}+x_1\cdots x_n\] for \(n\) even, resp. \[0=x_0^2+x_1^3+\cdots+x_n^{s_n}+x_1\cdots x_{n-1}x_n^2\] for \(n\) odd (in \(A^{n+1}/\mu_{a_{n+1}}\)). The same hypersurfaces in \(A^{n+1}\) appeared in our klt variety of conjecturally minimal volume, and we showed that these hypersurfaces in \(A^{n+1}\) have canonical singularities (hence are klt) [6]. The klt, plt, and lc properties are preserved upon dividing by a finite group action that is free in codimension 1 [9]. Therefore, \(X\) is klt.

The divisor \(B=\{x_{n+1}=0\}\cap X\) is quasi-smooth and misses the point \(p\). So the pair \((X,B)\) is étale-locally (near each point of \(B\)) the quotient of \((A^n,A^{n-1})\) by a finite group action, free in codimension 1. Since \(X\) is klt outside \(B\), it follows that the pair \((X,B)\) is plt (hence lc).

Let \(d=s_{n+1}-1\) be the degree of the hypersurface \(X\). Since \(X\) is well-formed and normal, we have \[\begin{align} K_X&=O_X(d-\sum a_j)\\ &=O_X\bigg( (s_{n+1}-1)\bigg(1-\frac{1}{s_0}-\cdots-\frac{1}{s_n}\bigg) -a_{n+1}\bigg)\\ &=O_X(1-a_{n+1}). \end{align}\] The divisor \(B\) on \(X\) is linearly equivalent to \(O_X(a_{n+1})\), and so \(K_X+B\sim O_X(1)\). It follows that \[\begin{align} \mathop{\mathrm{vol}}(K_X+B)&=\mathop{\mathrm{vol}}(O_X(1))\\ &=\frac{d}{a_0\cdots a_{n+1}}\\ &=\frac{(s_{n+1}-1)s_0\cdots s_n}{(s_{n+1}-1)^{n+1}a_{n+1}}\\ &=\frac{1}{(s_{n+1}-1)^{n-1}a_{n+1}}. \end{align}\] Here \(s_{n+1}\) is asymptotic to \(s_n^2\) (with error term on the order of \(s_n\)) as \(n\) goes to infinity, and \(a_{n+1}\sim s_n^2/4\); so the volume of \(K_X+B\) is asymptotic to \(4/s_n^{2n}\). ◻

Proof. (Proposition 6) We prove this by descending induction on \(0\leq k\leq r+1\). For \(k=r+1\), both sides of the equation are equal to \((s_{r+1}-1)-1\). Next, assume that \(0\leq k\leq r\) and the equation holds for \(k+1\), meaning that \[\label{kplus1} (s_{k+1}-1)g_{k+1}t_{k+1}-1 =b_{k+1}\cdots b_{2r-k}w_{k+1}.\tag{1}\] We will prove it for \(k\).

We prove the following lemma at the same time as Proposition 6.

We continue the proof of Proposition 6 for \(k\), using that it holds for \(k+1\). By definition of \(t_k\), we have \[t_k-t_{k+1}=(s_k-1)b_{k+1}\cdots b_{2r+1-k}.\] Using Lemma 7 for the given number \(k\), it follows that \[\begin{gather} (s_k-1)g_kt_k-1=(s_k-1)[s_kg_{k+1}+(s_k-1)b_{k+1}\cdots b_{2r-k}w_{k+1}]\\ \cdot [t_{k+1}+(s_k-1)b_{k+1}\cdots b_{2r+1-k}]-1\\ =(s_{k+1}-1)g_{k+1}t_{k+1}-1+(s_k-1)^2b_{k}\cdots b_{2r+1-k} g_{k+1}\\ +(s_k-1)^2b_{k+1}\cdots b_{2r-k}t_{k+1}w_{k+1} +(s_k-1)^3(b_{k+1}\cdots b_{2r-k})^2b_{2r+1-k}w_{k+1}. \end{gather}\] Since Proposition 6 holds for \(k+1\) (equation 1 ), it follows that \[\begin{gather} (s_k-1)g_kt_k-1=b_{k+1}\cdots b_{2r-k}[(s_k-1)^2b_{k}b_{2r+1-k}g_{k+1} +w_{k+1}+(s_k-1)^2t_{k+1}w_{k+1}\\ +(s_k-1)^3b_{k+1}\cdots b_{2r+1-k}w_{k+1}]. \end{gather}\] By definition, \(b_{2r+1-k}=1+(s_k-1)^2t_{k+1}\). So we can combine the second and third terms in the bracket above into a multiple of \(b_{2r+1-k}\): \[\begin{gather} (s_k-1)g_kt_k-1=b_{k+1}\cdots b_{2r+1-k}[(s_k-1)^2b_{k}g_{k+1}+w_{k+1}\\ +(s_k-1)^3b_{k+1}\cdots b_{2r-k}w_{k+1}]. \end{gather}\] Therefore, to prove Proposition 6 for \(k\) (completing the induction), it suffices to show that: \[\label{last} b_kw_k=(s_k-1)^2b_kg_{k+1}+w_{k+1}+(s_k-1)^3b_{k+1}\cdots b_{2r-k}w_{k+1}.\tag{2}\]

By definition, \[w_k=(s_{k}-1)[B_{r+1,r,\ldots,2r+1-k,k}-B_{r,\ldots,2r+1-k,k} +\cdots-B_{k}+1]-1\] and \[w_{k+1}=s_k(s_{k}-1)[B_{r+1,r,\ldots,2r-k,k+1}-B_{r,\ldots,2r-k,k+1} +\cdots-B_{k+1}+1]-1.\] Therefore, subtracting term by term and using that \(b_k=s_k\), we have \[\begin{gather} b_kw_k-w_{k+1}=s_k(s_k-1)[(b_k-1)b_{2r+1-k}b_{k+1}b_{2r-k}\cdots b_rb_{r+1}\\ -(b_k-1)b_{2r+1-k}b_{k+1}b_{2r-k}\cdots b_r+\cdots +(b_k-1)b_{2r+1-k} -(b_k-1)+1]-(s_k-1) \end{gather}\] \[\begin{align} &=(s_k-1)^2[b_kb_{2r+1-k}b_{k+1}b_{2r-k}\cdots b_rb_{r+1} -b_kb_{2r+1-k}\cdots b_r+\cdots +b_kb_{2r+1-k}-b_k+1]\\ &=(s_k-1)^2B_{k,2r+1-k,\ldots,r,r+1}\\ &=(s_k-1)^2g_k. \end{align}\] So the left side of 2 is \[b_kw_k=(s_k-1)^2g_k+w_{k+1}.\] By Lemma 7 for the given number \(k\), we can expand \(g_k\) here, giving that \[b_kw_k=(s_k-1)^2b_kg_{k+1}+w_{k+1}+(s_k-1)^3b_{k+1}\cdots b_{2r-k}w_{k+1}.\] That proves equation 2 . Thus Proposition 6 holds for \(k\) given that it holds for \(k+1\). The proposition is proved. ◻

Proof. Let \(r\) be a positive integer, and let \(n\) be \(2r\) or \(2r+1\). We first prove the basic properties of Esser’s hypersurface \(V\), namely that it is a well-formed quasi-smooth Calabi–Yau hypersurface. Here \(V\) was defined in section 3 by writing out its equation. That determines the weights \(a_0,\ldots,a_{n+1}\) and degree \(D\) of \(V\), given the requirement that \(\gcd(a_0,\ldots,a_{n+1})=1\). By the form of the equation (a loop for \(n\) odd, \(x_0^2\) plus a loop for \(n\) even), \(V\) is quasi-smooth.

The weighted projective space containing \(V\) is well-formed: if a prime number \(p\) divides all but one weight \(a_i\), then \(p\) divides \(D\) since there is a monomial not involving \(x_i\). Then, there is either a monomial of the form \(x_j^{a}x_i\) in the equation for \(V\) or \(n\) is even and \(i = 0\). In the former case, we get that \(p\) divides \(a_i\) too, contradicting that \(\gcd(a_0,\ldots,a_{n+1})=1\). The same holds in the latter case unless \(p = 2\). If \(p = 2\) divides all weights except \(a_0\) in the \(n\) even case, then \(D \equiv 2 \pmod 4\). The remaining exponents \(b_1,\ldots,b_{2r},v_{2r}\) are odd, so this means the weights \(a_i\) would have to alternate between \(0 \pmod 4\) and \(2 \pmod 4\) moving around the loop part of the equation of \(X\). But the loop has odd length, a contradiction. It follows that \(V\) is well-formed by Proposition 1, together with a look at the equation for \(V\) in dimension 2 (section 3).

By Lemma 9 plus the preceding results, \(V\) is a well-formed quasi-smooth Calabi–Yau hypersurface. More precisely, \(K_V=O_V(D-\sum a_j)=O_V\), and so \(K_V\) is linearly equivalent to zero. As a result, \(K_V\) is Cartier and \(V\) is klt, so \(V\) is canonical.

The inverse \(A^{-1}\) of a loop matrix is written explicitly in Lemma 10. We read off that the charge \(q_{2r+2}\) is the alternating sum \[\begin{align} q_{2r+2}&=\frac{B_{0,2r+1,\ldots,r,r+1}}{b_0\cdots b_{2r+1}v_{2r+1}+1}\\ &= \frac{m_{2r+1}}{b_0\cdots b_{2r+1}v_{2r+1}+1}\\ &= \frac{1}{u_{2r+1}}, \end{align}\] by Proposition 6. The degree \(D\) of \(V\) is the least common multiple of the denominators of the charges \(q_i\); but in this case of a loop matrix, all the charges have the same denominator [5]. Therefore, the degree \(D\) of \(V\) is equal to \(u_{2r+1}\). So the weights of \(V\) are \(a_i=u_{2r+1}q_i\). In particular, the weight \(a_{2r+2}\) is equal to 1, as we want.

By the loop equation of \(V\), the absolute value of the determinant of \(A\) is given by [11]: \[|\det(A)|=b_0\cdots b_{2r+1}v_{2r+1}+1.\] The group \(\mathop{\mathrm{Aut}}_T(V)\) of toric automorphisms of \(V\) is the group of automorphisms of \(V\) that are given by diagonal matrices in the given coordinates. This is related to the degree \(D\) of \(V\) by [11]: \[|\det(A)|=D|\mathop{\mathrm{Aut}}_T(V)|.\] By the computation of the degree \(D\) above, it follows that \(|\mathop{\mathrm{Aut}}_T(V)|=m_{2r+1}\). Moreover, since the equation of \(V\) is a loop, the group \(\mathop{\mathrm{Aut}}_T(V)\) is cyclic [11]. So the group of toric automorphisms of \(V\) is the cyclic group \(\mu_m\) with \(m=m_{2r+1}\). (We will describe the \(\mu_m\)-action explicitly in the proof of Lemma 13.)

Also, because the equation of \(V\) is a loop, the action of \(\mu_m\) on \(V\) is free in codimension 1 [5]. As a result, \(V/\mu_m\) is a klt Calabi–Yau variety (not a pair).

Esser showed that the mld of \(V/\mu_m\) is the smallest charge of the BHK mirror of \(V\) [7]. (The mirror is defined to be the hypersurface whose equation is associated to the transpose of the matrix \(A\).) One mirror charge is \[\begin{align} q_{2r+2}^{\mathsf{T}} & =\frac{B_{r+1,r,r+2,r-1,\ldots,2r+1,0}}{b_0\cdots b_{2r+1}v_{2r+1}+1}\\ &= \frac{u_{2r+1}}{b_0\cdots b_{2r+1}v_{2r+1}+1}\\ &= \frac{1}{m_{2r+1}}. \end{align}\] Since \(A^{\mathsf{T}}\) is a loop matrix, all the mirror charges have the same denominator, and so \(1/m_{2r+1}\) is the smallest mirror charge. Thus \(\mathop{\mathrm{mld}}(V/\mu_m)=1/m\), as we want. (Another way to compute the mld is given in Lemma 13: the variety \(V/\mu_m\) is crepant-birational to the pair \((X,(1-\frac{1}{m})S)\) discussed later.)

It remains to give the analogous argument in even dimensions. Let \(n=2r\). Because the equation of \(V\) is of the form \(x_0^2\) plus a loop, we can write out the inverse matrix \(A^{-1}\) using the formula for the inverse of a loop matrix, Lemma 10. We read off that the charge \(q_{2r+1}\) is the alternating sum \[\begin{align} q_{2r+1}&=\frac{B_{1,2r,\ldots,r,r+1}}{b_1\cdots b_{2r}v_{2r}+1}\\ &=\frac{m_{2r}}{b_1\cdots b_{2r}v_{2r}+1}\\ &=\frac{1}{2u_{2r}}, \end{align}\] by Proposition 6. The degree \(D\) of \(V\) is the least common multiple of the denominators of the charges \(q_i\); so the degree \(D\) is a multiple of \(2u_{2r}\), say \(D=2u_{2r}\lambda\) for a positive integer \(\lambda\). The last weight \(a_{2r+1}\) is then \(Dq_{2r+1}=\lambda\). Using the loop of monomials (starting with \(x_{2r+1}^{v_{2r+1}}x_{r+1}\)) in the equation of \(V\), it follows that all weights \(a_i\) with \(i\neq 0\) are also multiples of \(\lambda\). By the monomial \(x_0^2\) in the equation of \(V\), we have \(a_0=D/2=u_{2r}\lambda\), which is also a multiple of \(\lambda\). Since \(\gcd(a_0,\ldots,a_{2r+1})=1\), it follows that \(\lambda=1\). Thus the degree \(D\) of \(V\) is equal to \(2u_{2r}\), the weights of \(V\) are \(a_i=2u_{2r}q_i\), and the weight \(a_{2r+1}\) is equal to 1, as we want.

Since the equation of \(V\) for \(n=2r\) is \(x_0^2\) plus a loop, the absolute value of the determinant of \(A\) is given by [11]: \[|\det(A)|=2(b_1\cdots b_{2r}v_{2r}+1).\] The group of toric automorphisms is related to the degree \(D\) of \(V\) by [11]: \[|\det(A)|=D|\mathop{\mathrm{Aut}}_T(V)|.\] By the computation of the degree \(D\) above, it follows that \(|\mathop{\mathrm{Aut}}_T(V)|=2m_{2r}\). Moreover, since the equation of \(V\) is \(x_0^2\) plus a loop, the group \(\mathop{\mathrm{Aut}}_T(V)\) is \(\mu_2\) times a cyclic group [11]. So the group of toric automorphisms of \(V\) is the product group \(\mu_2\times\mu_m\) with \(m=m_{2r}\). (We will describe the \(\mu_m\)-action explicitly in Lemma 13. The \(\mu_2\)-action changes the sign of \(x_0\).)

Also, because the equation of \(V\) is \(x_0^2\) plus a loop, the action of \(\mu_m\) on \(V\cap \{x_0\neq 0\}\) is free in codimension 1 [5]. Also, a toric automorphism that fixes the divisor \(V\cap \{x_0=0\}\) can be written as \([x_0,\ldots,x_{n+1}]\mapsto [ex_0,x_1,\ldots,x_{n+1}]\) for some \(e\in \text{\boldsymbol{C}}^*\). We have \(e^2=1\), because the automorphism preserves the equation \(x_0^2+x_1^3x_{2r+1}+\cdots=0\) of \(V\). Since \(m=m_{2r}\) is odd, it follows that no nontrivial element of \(\mu_m\) fixes this divisor. So \(\mu_m\) acts freely in codimension 1 on \(V\), and hence \(V/\mu_m\) is a klt Calabi–Yau variety (not a pair). Given that, Esser showed that \(\mathop{\mathrm{mld}}(V/\mu_m)=1/m\) [7]. (Another way to compute the mld is given in Lemma 13: the variety \(V/\mu_m\) is crepant-birational to the pair \((X,(1-\frac{1}{m})S)\) discussed later.) ◻

Proof. (Theorem 11) We want to construct a klt Calabi–Yau pair \((X,(1-\frac{1}{m})S)\) of dimension \(n\geq 2\) with \(S\) an irreducible divisor (and the number \(m=m_n\) defined above). We define \(X\) as a hypersurface in a weighted projective space \(Y=\text{\boldsymbol{P}}^{n+1}(c_0,\ldots,c_{n+1})\). Namely, if \(n=2r+1\), let \(X\) have the same equation as \(V\) (from section 3) except for the first and last terms: \[0=x_0^2+x_1^{b_1}x_{2r+1}+\cdots+x_r^{b_r}x_{r+2} +x_{r+1}^{b_{r+1}}x_r +\cdots+x_{2r+1}^{b_{2r+1}}x_0 +x_{2r+2}x_{r+1}.\] We also have formulas for the degree \(d\) and the weights \(c_j\) of \(X\). (With these definitions, it is straightforward that each monomial above has degree \(d\).) First, \(d \mathrel{\vcenter{:}}= b_0\cdots b_{2r+1}\). Next, the last weight \(c_{2r+2}\) is \(m_{2r+1}=B_{0,2r+1,1,2r,\ldots,r,r+1}\), the index of Esser’s example. Finally, \[c_{r+1+i} \mathrel{\vcenter{:}}= b_{r+1-i}\cdots b_r b_{r+1}\cdots b_{r+i}B_{0,2r+1,1,2r,\ldots, r-i}\] for \(0\leq i\leq r\), and \[c_{r-i} \mathrel{\vcenter{:}}= b_{r+1-i}\cdots b_rb_{r+1}\cdots b_{r+1+i}B_{0,2r+1,1,2r,\ldots, r-1-i,r+2+i}\] for \(0\leq i\leq r\). (Since a \(B\) with empty subscript is 1, we have \(c_0=b_1\cdots b_{2r+1}\).)

If \(n=2r\), let \(X\) have the same equation as \(V\) except for the second and last terms: \[0=x_0^2+x_1^3+x_2^{b_2}x_{2r}+\cdots+x_r^{b_r}x_{r+2} +x_{r+1}^{b_{r+1}}x_r +\cdots+x_{2r}^{b_{2r}}x_1 +x_{2r+1}^{2}x_{r+1}.\] Again, we have formulas for the degree \(d\) and weights \(c_j\) of \(X\). First, \(d \mathrel{\vcenter{:}}= b_0\cdots b_{2r}\). Next, the last weight \(c_{2r+1}\) is \(m_{2r}=B_{1,2r,2,2r-1,\ldots,r,r+1}\), the index of Esser’s example. Also, \(c_0 \mathrel{\vcenter{:}}= b_1\cdots b_{2r}\). Finally, \[c_{r+1+i} \mathrel{\vcenter{:}}= 2b_{r+1-i}\cdots b_r b_{r+1}\cdots b_{r+i}B_{1,2r,2,2r-1,\ldots, r-i}\] for \(0\leq i\leq r-1\), and \[c_{r-i} \mathrel{\vcenter{:}}= 2b_{r+1-i}\cdots b_rb_{r+1}\cdots b_{r+1+i}B_{1,2r,2,2r-1,\ldots, r-1-i,r+2+i}\] for \(0\leq i\leq r-1\).

Define the divisor \(S\) to be \(X\cap \{x_{n+1}=0\}\). From the form of the equation (a chain as in section 1, or \(x_0^2\) plus a chain), it is immediate that \(X\) and \(S\) are quasi-smooth. (Since \(S\) is a hypersurface of dimension at least 1 in a weighted projective space, it follows that \(S\) is irreducible.) The weight \(c_{n+1}\) (the degree of \(S\)) is equal to the index \(m\) of Esser’s example, Next, we show that \(K_X=O_X(d-\sum c_j)\) is equal to \(O_X(1-m)\), so that \((X,(1-\frac{1}{m})S)\) is a klt Calabi–Yau pair (assuming that \(X\) is well-formed, as we show below). Since \(c_{n+1}=m\), it is equivalent to show that \(d-\sum_{j=0}^n c_j=1\). Assume that \(n=2r+1\). We compute \(d-\sum_{j=0}^n c_j\) by collecting all terms (in the formulas above for \(d\) and the \(c_j\)’s) that are multiples of \(b_{r+1}\cdots b_{2r+1}\), then the remaining terms that are multiples of \(b_{r+2}\cdots b_{2r+1}\), and so on. The result is: \[\begin{gather} d-\sum_{j=0}^{2r+1} c_j=b_{r+1}\cdots b_{2r+1}\bigg[b_0\cdots b_r \bigg( 1-\frac{1}{b_0}-\cdots-\frac{1}{b_r}\bigg)\bigg]\\ -\sum_{i=1}^{r+1}b_{r+1+i}\cdots b_{2r+1}[b_0\cdots b_{r-i}(b_{r+1-i}-1) B_{r+1,r,\ldots,r-1+i,r+2-i}]. \end{gather}\] The first term in brackets is 1, by the properties of the Sylvester numbers \(b_0,\ldots,b_r\). By definition of \(b_{r+i}\), the expression in brackets in the term indexed by \(i\) (for \(1\leq i\leq r+1\)) is \(b_{r+i}-1\). Therefore, the sum telescopes, showing that \(d-\sum_{j=0}^{2r+1}c_j=1\), as we want. For \(n=2r\), the proof is similar: \[\begin{gather} d-\sum_{j=0}^{2r} c_j=b_{r+1}\cdots b_{2r}\bigg[ b_0\cdots b_r \bigg( 1-\frac{1}{b_0}-\cdots-\frac{1}{b_r}\bigg)\bigg]\\ -\sum_{i=1}^{r}b_{r+1+i}\cdots b_{2r}[2b_1\cdots b_{r-i}(b_{r+1-i}-1) B_{r+1,r,\ldots,r-1+i,r+2-i}]. \end{gather}\] Again, the expression in brackets in the term indexed by \(i\) is \(b_{r+i}-1\). So the sum telescopes, showing that \(d-\sum_{j=0}^{2r}c_j=1\), as we want.

The only property that remains to be checked is that \(X\) is well-formed. (We need this in order to justify the formula above for \(K_X\), with \(X\) viewed as a variety rather than a pair.) The main point is to show that the weighted projective space \(Y=\text{\boldsymbol{P}}^{n+1}(c_0,\ldots,c_{n+1})\) is well-formed. Indeed, from there it follows that \(X\) is well-formed when \(n\geq 3\), by Proposition 1. This general result does not apply in dimension 2, but in that case we can check by hand that our example is well-formed. Namely, the example is \(X\subset Y=\text{\boldsymbol{P}}^3(15,10,4,13)\), defined by \(0=x_0^2+x_1^3+x_2^5x_1+x_3^2x_2\). This is well-formed, because \(X\) does not contain the 1-dimensional stratum \(\{ x_2=x_3=0\}\) of \(Y\) with stabilizer \(\mu_5\) or the 1-dimensional stratum \(\{ x_0=x_3=0\}\) with stabilizer \(\mu_2\).

So, for each \(n\geq 2\), we need to show that \(Y\) is well-formed, meaning that \(\gcd(c_0,\ldots,\widehat{c_a},\ldots,c_{n+1})=1\) for each \(0\leq a\leq n+1\). Let \(p\) be a prime number dividing \(c_j\) for each \(j\neq a\); we will derive a contradiction. From looking at the equation of \(X\), we see that it includes a monomial not involving \(x_a\), and so \(p\) divides the degree \(d\) of \(X\). For \(n=2r+1\), the variable \(x_a\) occurs with exponent 1 in some monomial in the equation of \(X\), and so \(p\) divides \(c_a\) as well. For \(n=2r\), we get the same conclusion unless \(a=0\) or \(a=2r+1\). In those cases, the variable \(x_a\) occurs with exponent 2 in some monomial in the equation of \(X\), and so \(p\) divides \(2c_a\). It follows that \(p\) divides \(c_a\) unless \(p=2\). To analyze the case where \(n=2r\) and \(p=2\), use that \(b_1,\ldots,b_{2r}\) are all odd by their definition. So \(c_0=b_1\cdots b_{2r}\) is odd and \(c_{2r+1}=m_{2r}=B_{1,2r,\ldots,r,r+1}\equiv 1-1+1-1+\cdots+1 \equiv 1\pmod{2}\), contradicting that \(2\) divides all but one of the weights \(c_j\). Thus, if \(p\) divides all but one of the weights \(c_j\), then it divides all the weights.

Return to allowing any dimension \(n\geq 2\). We now have a prime number \(p\) that divides the weight \(c_j\) for each \(0\leq j\leq n+1\), and we want to get a contradiction. For \(n=2r+1\), we have \(c_0=b_1\cdots b_{2r+1}\) and \(c_{2r+2}=m_{2r+1}\). By Proposition 6, \(m_{2r+1}u_{2r+1}-1=b_0\cdots b_{2r+1}v_{2r+1}\), and so \(c_0\) and \(c_{2r+2}\) are relatively prime. Next, for \(n=2r\), we have \(c_0=b_1\cdots b_{2r}\) and \(c_{2r+1}=m_{2r}\). By Proposition 6, \(2m_{2r}u_{2r}-1=b_1\cdots b_{2r}v_{2r}\), and so \(c_0\) and \(c_{2r+1}\) are relatively prime. This completes the proof that \(X\) is well-formed. Theorem 11 is proved. ◻

Proof. Let \(n\geq 2\). Let \(W\) be the index-1 cover of the klt Calabi–Yau pair \((X,(1-\frac{1}{m})S)\) of dimension \(n\) from Theorem 11. It follows that \(W\) has canonical singularities and \(K_W\sim 0\). Explicitly, for \(n=2r+1\), \(W\) is the Calabi–Yau hypersurface \[\begin{align} W&=W_{d}\subset \text{\boldsymbol{P}}^{2r+2}(c_0,\ldots,c_{2r+1},1),\\ 0&=x_0^2+x_1^{b_1}x_{2r+1}+\cdots+x_r^{b_r}x_{r+2} +x_{r+1}^{b_{r+1}}x_r +\cdots+x_{2r+1}^{b_{2r+1}}x_0 +x_{2r+2}^mx_{r+1}. \end{align}\] The degree \(d\) of \(W\) is the same as for \(X\), and the exponents \(b_j\) and weights \(c_j\) are the same except for \(j=2r+2\). The last weight is changed from \(c_{2r+2}=m_{2r+1}=m\) to 1. Let the cyclic group \(\mu_m\) act on \(W\) with weights \((0,\ldots,0,-1)\); then the quotient \(W/\mu_m\) is the pair \((X,(1-\frac{1}{m})S)\). (The quotient map \(W\to X\) is given by \([x_0,\ldots,x_{2r+2}]\mapsto [x_0,\ldots,x_{2r+1}, x_{2r+2}^m]\).)

In even dimensions \(n=2r\), the index-1 cover \(W\) of \((X,(1-\frac{1}{m})S)\) has a similar description. Here \(W\) is the Calabi–Yau hypersurface \[\begin{align} W&=W_{d}\subset \text{\boldsymbol{P}}^{2r+1}(c_0,\ldots,c_{2r},1),\\ 0&=x_0^2+x_1^3+x_2^{b_2}x_{2r}+\cdots+x_r^{b_r}x_{r+2} +x_{r+1}^{b_{r+1}}x_r +\cdots+x_{2r}^{b_{2r}}x_1 +x_{2r+1}^{2m}x_{r+1}. \end{align}\] Again, the last weight has been changed from \(c_{2r+1}=m_{2r}=m\) to 1, and the cyclic group \(\mu_m\) acts on \(W\) with weights \((0,\ldots,0,-1)\).

The equations of \(W\) and Esser’s hypersurface \(V\) are the same in the affine charts \(x_{n+1}=1\), and the variable \(x_{n+1}\) has weight 1 in both cases. Therefore, we get a birational map \(\varphi\colon W\dashrightarrow V\) by sending \[[x_0,\ldots,x_n,1]\mapsto [x_0,\ldots,x_n,1].\] Since \(W\) and \(V\) are Calabi–Yau varieties with canonical singularities, \(\varphi\) is automatically crepant.

Moreover, we can define an action of \(\mu_m\) on \(V\) that makes \(\varphi\) \(\mu_m\)-equivariant. Indeed, we can rewrite the action of \(\mu_m\) on \(W\) as \[\begin{align} \zeta([x_0,\ldots,x_n,x_{n+1}])&=[x_0,\ldots,x_n,\zeta^{-1}x_{n+1}]\\ &=[\zeta^{c_0}x_0,\ldots,\zeta^{c_n}x_n,x_{n+1}]. \end{align}\] That suggests defining an action of \(\mu_m\) on \(V\) by the formula: \[\zeta([x_0,\ldots,x_n,x_{n+1}])=[\zeta^{c_0}x_0,\ldots,\zeta^{c_n}x_n, x_{n+1}].\] To show that this action preserves the hypersurface \(V\), it suffices to check this in the affine chart \(x_{n+1}=1\); but there it is clear, because the equation of \(V\) in this chart is the same as the equation of \(W\) in the corresponding chart. Given that, it is clear that the map \(\varphi\) is \(\mu_m\)-equivariant. (It suffices to check this in the affine chart \(x_{n+1}=1\), where the two \(\mu_m\)-actions are given by the same formula.)

Therefore, we have a crepant birational map from \(W/\mu_m\) (viewed as a pair, namely \((X,(1-\frac{1}{m}S))\)) to the klt Calabi–Yau variety \(V/\mu_m\). In particular, this shows more explicitly why \(V/\mu_m\) has mld \(1/m\): because the divisor \(S\) has log discrepancy \(1/m\) with respect to \(V/\mu_m\). ◻

Proof of Theorem 14. Since the equation is a loop or \(x_0^2\) plus a loop, the hypersurface \(V'\) is quasi-smooth.

The weighted projective space containing \(V'\) is well-formed: if a prime number \(p\) divides all but one weight \(a_i'\), then \(p\) divides \(D'\) since there is a monomial not involving \(x_i\). Then, there is either a monomial of the form \(x_j^{a}x_i\) in the equation for \(V'\) or \(n\) is even and \(i = 0\). In the former case, we get that \(p\) divides \(a_i'\) too, contradicting that \(\gcd(a_0',\ldots,a_{n+1}')=1\). The same holds in the latter case unless \(p = 2\). If \(p = 2\) divides all weights except \(a_0'\) in the \(n\) even case, then \(D' \equiv 2 \pmod 4\). The remaining exponents \(b_1,\ldots,b_{2r-1},b'_{2r},v'_{2r}\) are odd, so this means the weights \(a_i'\) would have to alternate between \(0 \pmod 4\) and \(2 \pmod 4\) moving around the loop part of the equation of \(V'\). But the loop has odd length, a contradiction. It follows that \(V'\) is well-formed by Proposition 1, together with a look at the equation for \(V'\) in dimension 2 (above).

By Lemma 15, \(K_{V'} = \mathcal{O}_{V'}(D' - \sum a_j') = \mathcal{O}_{V'}\). Since the hypersurface \(V'\) is klt and has trivial canonical class, it is canonical.

Next, we’ll state and prove expressions for the degree \(D'\) of \(V'\) and the order \(m'\) of a cyclic group action on this hypersurface. Along the way, we’ll also prove that the last two weights \(a_n'\) and \(a_{n+1}'\) equal \(1\). First, we show the following identities involving \(m'_n\) and \(u'_n\): \[\label{indexidentity} \begin{cases} m'_{2r+1} u'_{2r+1} - 1 = b_0\cdots b_{2r}b'_{2r+1}v'_{2r+1} & \text{if }n = 2r+1,\\ 2m'_{2r} u'_{2r} - 1 = b_1 \cdots b_{2r-1}b'_{2r} v'_{2r} &\text{if }n=2r. \end{cases}\tag{7}\] First, when \(n = 2r + 1\), \[\begin{gather} u'_{2r+1} m'_{2r+1} - 1 = u'_{2r+1}(b_0b'_{2r+1}B_{1,2r,\ldots,r,r+1} - 1)-1 \\ = (b_1 \cdots b_{2r} + (s_1-1)B_{r+1,r,\ldots,2r,1})b_0b'_{2r+1}B_{1,2r,\ldots,r,r+1} - u'_{2r+1}-1 \\ = b_0 \cdots b_{2r} b'_{2r+1} B_{1,2r,\ldots,r,r+1} + b_0b'_{2r+1}(s_1 -1)B_{r+1,r,\ldots,2r,1}B_{1,2r,\ldots,r,r+1} - u_{2r+1}' - 1. \end{gather}\] We may now apply Proposition 6 with \(k = 1\) to replace the expression \((s_1 -1)B_{r+1,r,\ldots,2r,1}B_{1,2r,\ldots,r,r+1} = (s_1-1)g_1 t_1\) with \(b_1 \cdots b_{2r}w_1+1\). Hence \[\begin{align} u'_{2r+1} m'_{2r+1} - 1 & = b_0 \cdots b_{2r} b'_{2r+1} (g_1 + w_1) + b_0 b'_{2r+1} - u_{2r+1}' - 1 \\ & = b_0 b_1 \cdots b_{2r} b'_{2r+1} v'_{2r+1} + b_0 b'_{2r+1} - u_{2r+1}' - 1 \\ & = b_0 b_1 \cdots b_{2r} b'_{2r+1} v'_{2r+1}. \end{align}\] Looking at the form \(A^{-1}\) of the loop matrix in Lemma 10, we may read off the last charge for the hypersurface \(V'\) and apply 7 to get: \[\begin{align} q_{2r+2} & = \frac{b_0b'_{2r+1}b_1 b_{2r} \cdots b_r b_{r+1} - b_0b'_{2r+1}b_1 b_{2r} \cdots b_r + \cdots + b_0b'_{2r+1} - b_0 + 1}{b_0 b_1 \cdots b_{2r} b'_{2r+1} v'_{2r+1}+1} \\ & = \frac{m'_{2r+1}}{b_0 b_1 \cdots b_{2r} b'_{2r+1} v'_{2r+1}+1} \\ & = \frac{1}{u'_{2r+1}}. \end{align}\] The degree \(D'\) of \(V'\) is the least common denominator of the charges. All these denominators are the same for a loop potential, so \(D' = u'_{2r+1}\) and the weight \(a'_{2r+2} = 1\). Since \(x_0^2 x_{2r+2}\) is a monomial in the equation for \(V'\), we must have \(a'_0 = (u'_{2r+1}-1)/2\). Then, since \(x_{2r+1}^{b'_{2r+1}}x_0\) is a monomial also, \[a'_{2r+1} = \frac{u'_{2r+1}-(u'_{2r+1}-1)/2}{b'_{2r+1}} = 1.\] Let \(\mathrm{Aut}_T(V')\) be the group of toric automorphisms of \(V'\). The order of this group is related to the degree of the hypersurface and the matrix \(A\) by [11] \[|\det(A)| = D'|\mathrm{Aut}_T(V')|.\] Since \(\det(A) = b_0 \cdots b_{2r} b'_{2r+1} v'_{2r+1} + 1\), the identity 7 implies that \(|\mathrm{Aut}_T(V')| = m_{2r+1}'\). The equation of \(V'\) is a loop, so \(\mathrm{Aut}_T(V')\) is a cyclic group of order \(m' = m'_{2r+1}\). Since the equation of \(V'\) is a loop, the action of \(\mu_{m'}\) is free in codimension \(1\) [5].

Next, if \(n = 2r\), then we prove 7 as follows: \[\begin{gather} 2 u'_{2r+1} m'_{2r+1} - 1 = 2u'_{2r+1}(b_1b'_{2r}B_{2,2r-1,\ldots,r,r+1} - 2)-1 \\ = (2(s_1-1) b_2 \cdots b_{2r-1} + 2s_1 B_{r+1,r,\ldots,2r-1,2})b_1b'_{2r}B_{2,2r-1,\ldots,r,r+1} - 4u'_{2r+1}-1 \\ = (s_1-1)^2 b_1\cdots b_{2r-1} b'_{2r} B_{2,2r-1,\ldots,r,r+1} \\ + b_1b'_{2r}(s_2 -1)B_{r+1,r,\ldots,2r-1,2}B_{2,2r-1,\ldots,r,r+1} - 4u_{2r}' - 1. \end{gather}\] We may now apply Proposition 6 with \(k = 2\) to replace the expression \((s_2 -1)B_{r+1,r,\ldots,2r-1,2}B_{2,2r-1,\ldots,r,r+1} = (s_2-1)g_2 t_2\) with \(b_2 \cdots b_{2r-1} w_2 +1\). Hence \[\begin{align} 2u'_{2r} m'_{2r} - 1 & = b_1 \cdots b_{2r-1} b'_{2r} (4g_2 + w_2) + b_1 b'_{2r} - 4u'_{2r} - 1 \\ & = b_1 \cdots b_{2r-1} b'_{2r} v'_{2r} + b_1 b'_{2r} - 4u'_{2r} - 1 \\ & = b_1 \cdots b_{2r-1} b'_{2r} v'_{2r}. \end{align}\] For \(n = 2r\), the \((n+2) \times (n+2)\) matrix \(A\) expressing the equation of \(V'\) is block diagonal with \(2\) as the top left entry and an \((n+1) \times (n+1)\) loop matrix as the other block. Therefore, we may read off the last charge \(q_{2r+1}\) from \(A^{-1}\) and apply 7 to get: \[\begin{align} q_{2r+1} & = \frac{b_1 b'_{2r} b_2 b_{2r-1} \cdots b_r b_{r+1} - b_1 b'_{2r} b_2 b_{2r-1} \cdots b_r + \cdots + b_1 b'_{2r} - b_1 + 1}{b_1 \cdots b_{2r-1} b'_{2r} v'_{2r} + 1} \\ & = \frac{m'_{2r}}{b_1 \cdots b_{2r-1} b'_{2r} v'_{2r} + 1} \\ & = \frac{1}{2u'_{2r}} \end{align}\] Therefore, the degree \(D'\) is a multiple of \(2u'_{2r}\), say \(D' = 2\lambda u'_{2r}\), so the last weight is \(a'_{n+1} = \lambda\). Following the monomials around the loop, it follows that every weight \(a'_i\) with \(i \neq 0\) is a multiple of \(\lambda\). This contradicts the fact that the weighted projective space containing \(V'\) is well-formed, so in fact we have \(D' = 2u'_{2r}\) and \(a'_{n+1} = 1\). Since \(x_1^3 x_{2r+1}\) is a monomial of \(V'\), we must have \(a'_1 = (D'-1)/3 = (2u'_{2r}-1)/3\). Then, since \(x_{2r}^{b'_{2r}}x_1\) is a monomial also, \[a'_{2r} = \frac{2u'_{2r}-(2u'_{2r}-1)/3}{b'_{2r}} = 1.\] The determinant of the matrix \(A\) in this case is \[|\det(A)| = 2(b_1 \cdots b_{2r-1}b'_{2r}v'_{2r} + 1)\] and \(|\det(A)| = D' |\mathrm{Aut}_T(V')|\) so 7 yields \(|\mathrm{Aut}_T(V')| = 2m'_{2r}\). Because the equation is of the form \(x_0^2\) plus a loop, \(\mathrm{Aut}_T(V') \cong \mu_2 \times \mu_{m'}\) with \(m' = m'_{2r}\). Since the shape of the equation is the same as in the mld example, the same argument from the proof of Theorem 8 shows that the \(\mu_{m'}\)-action is free in codimension \(1\).

This completes the proof of all the properties listed in Theorem 14. ◻

Proof. Suppose that \(n = 2r+1\) is odd. The determinant of the matrix \(A\) encoding the equation of \(V'\) is \(u'_{2r+1} m'_{2r+1}\) by 7 . The degree of the mirror hypersurface of \(V'\) always divides \(u'_{2r+1} m'_{2r+1}/u'_{2r+1} = m'_{2r+1}\). By [5], the action of \(\mu_{m'}\) on \(H^0(V',K_{V'})\) is faithful if and only if the mirror degree actually equals \(m'\). This degree is the least common denominator of the mirror charges of \(V'\), which are the sums of columns of \(A^{-1}\).

We may use Lemma 10 to write the smallest mirror charge as \[\begin{align} q^{\mathsf{T}}_{2r+2} & = \frac{b_{r+1}b_r \cdots b_{2r}b_1 b'_{2r+1} b_0 - b_{r+1}b_r \cdots b_{2r}b_1 b'_{2r+1} + \cdots - b_{r+1} + 1}{u'_{2r+1}m'_{2r+1}} \\ & = \frac{b'_{2r+1} b_1 \cdots b_{2r} + B_{r+1,r,\ldots,2r,1}}{u'_{2r+1}m'_{2r+1}} \\ & = \frac{\frac{1}{2}((2b'_{2r+1}-1)b_1 \cdots b_{2r} + b_1 \cdots b_{2r} + 2B_{r+1,r,\ldots,2r,1})}{u'_{2r+1}m'_{2r+1}} \\ & = \frac{\frac{1}{2}(b_1 \cdots b_{2r} + 1)u'_{2r+1}}{u'_{2r+1}m'_{2r+1}} \\ & = \frac{E_{2r+1}}{m'_{2r+1}}. \end{align}\] If \(\gcd(E_{2r+1},m'_{2r+1}) = 1\), then this proves that the mirror degree is \(m'_{2r+1}\). Conversely, if some prime \(p\) divides \(E_{2r+1}\) and \(m'_{2r+1}\), and the mirror degree were \(m'_{2r+1}\), by following the loop potential, we’d have that \(p\) divides every weight of the mirror, a contradiction.

The same reasoning holds when \(n = 2r\). In that case, \(\mathrm{Aut}_T(V') \cong \mu_2 \times \mu_{m'}\). The \(\mu_2\)-action (which sends \(x_0 \mapsto -x_0\) and leaves the other variables unchanged) is faithful on \(H^0(V',K_{V'})\), so the \(\mu_{m'}\)-action is faithful if and only if the entire group \(\mathrm{Aut}_T(V')\) acts faithfully. The determinant of the matrix \(A\) encoding the equation of \(V'\) is \(4m'_{2r}u'_{2r}\). Using [5] again, \(\mathrm{Aut}_T(V')\) acts faithfully on \(H^0(V',K_{V'})\) if and only if \(2 u_{2r} D^{\mathsf{T}} =4 m'_{2r}u'_{2r}\), where \(D^{\mathsf{T}}\) is the mirror degree. This reduces to \(D^{\mathsf{T}} = 2m'_{2r}\). We compute the smallest mirror charge as \[\begin{align} q^{\mathsf{T}}_{2r+1} & = \frac{b_{r+1}b_r \cdots b_{2r-1}b_2 b'_{2r} b_1 - b_{r+1}b_r \cdots b_{2r-1}b_2 b'_{2r} + \cdots - b_{r+1} + 1}{2u'_{2r}m'_{2r}} \\ & = \frac{(s_1-1)b'_{2r} b_2 \cdots b_{2r-1} + B_{r+1,r,\ldots,2r-1,2}}{2u'_{2r}m'_{2r}} \\ & = \frac{\frac{1}{3}((3b'_{2r}-1)(s_1-1)b_2 \cdots b_{2r-1} + (s_1-1)b_2 \cdots b_{2r-1} + 3B_{r+1,r,\ldots,2r-1,2})}{2u'_{2r}m'_{2r}} \\ & = \frac{\frac{1}{3}(8b_2 \cdots b_{2r-1} + 1)u'_{2r}}{2u'_{2r}m'_{2r}} \\ & = \frac{E_{2r}}{2m'_{2r}}. \end{align}\] The constant \(E_{2r}\) is odd, so if \(\gcd(E_{2r},m'_{2r}) = 1\), then the mirror degree is \(2m'_{2r}\), as required. Conversely, if a prime \(p\) (which must be odd) divides both \(E_{2r}\) and \(m'_{2r}\), and the mirror degree were \(2m'_{2r}\), then the form of the equation of \(V'\) implies that all mirror weights would be divisible by \(p\), a contradiction. ◻

Proof. (Lemma 18) Let \(r\) be a positive integer, and let \(n\) be \(2r\) or \(2r+1\). Let \(b_0,\ldots,b_{n+2}\) be the exponents of Esser’s example, listed in section 3. Then \(b_a\) is equal to the Sylvester number \(s_a\) for \(a\leq r\). Therefore, \[\begin{align} b_{r+1}-1&=(b_r-1)^2\\ &=s_{r+1}\frac{(s_r-1)^2}{s_{r+1}}. \end{align}\] By induction on \(1\leq a\leq r\), it follows that \[b_{r+a}\geq s_{r+a}\frac{(s_{r+1-a}-1)^2}{s_{r+2-a}} \bigg[\frac{(s_{r+2-a}-1)^2}{s_{r+3-a}}\bigg]^{2^{0}} \cdots \bigg[\frac{(s_r-1)^2}{s_{r+1}}\bigg]^{2^{a-2}}.\] As a result, we have \[\begin{align} m_{2r}&=B_{1,2r,\ldots,r,r+1}\\ &=b_1b_{2r}\cdots b_rb_{r+1}-b_1b_{2r}\cdots b_r+\cdots\\ &\geq b_1\cdots b_{r}(b_{r+1}-1)b_{r+2}\cdots b_{2r}\\ &\geq s_1\cdots s_{2r}\bigg[\frac{(s_{1}-1)^2}{s_{2}}\bigg]^{2^{0}} \cdots \bigg[\frac{(s_r-1)^2}{s_{r+1}}\bigg]^{2^{r-1}}\\ &= (s_{2r+1}-1)\frac{1}{2}\bigg[\frac{(s_{1}-1)^2}{s_{2}}\bigg]^{2^{0}} \cdots \bigg[\frac{(s_r-1)^2}{s_{r+1}}\bigg]^{2^{r-1}}\\ &\geq \frac{1}{\alpha}(s_{2r+1}-1). \end{align}\] The proof for \(n=2r+1\) is similar. Here \(m_{2r+1}=B_{0,2r+1,\ldots, r,r+1}\geq b_0\cdots b_r(b_{r+1}-1)b_{r+2}\cdots b_{2r+1}\). The lower bound for \(b_{r+a}\) above holds (by induction) for all \(1\leq a\leq r+1\). We deduce that \[\begin{align} m_{2r+1}&\geq s_0\cdots s_{2r+1}\frac{(s_{0}-1)^2}{s_{1}} \bigg[\frac{(s_{1}-1)^2}{s_{2}}\bigg]^{2^{1}} \cdots \bigg[\frac{(s_r-1)^2}{s_{r+1}}\bigg]^{2^{r}}\\ &=(s_{2r+2}-1)\frac{1}{3} \bigg[\frac{(s_{1}-1)^2}{s_{2}}\bigg]^{2^{1}} \cdots \bigg[\frac{(s_r-1)^2}{s_{r+1}}\bigg]^{2^{r}}\\ &\geq \frac{4}{3\alpha^2}(s_{2r+2}-1). \end{align}\] ◻

Proof. The statement is easy for \(n=2\), and so we can assume that \(n>2\). The index \(m'_n\) is defined in terms of the numbers \(b_0,\ldots,b_{n-1}\) (the same in the small-mld example) together with \(b'_n\) (section 7). For \(n=2r\) with \(r>1\), we have \[\begin{align} b'_{2r}&=\frac{1}{3}[1+8b_2\cdots b_{2r-1}+12B_{r+1,r,\cdots,2r-1,2}]\\ &\geq \frac{1}{3}[8b_2\cdots b_{2r-1}+12(b_2-1)b_3\cdots b_{2r-1}]\\ &=\frac{128}{3}b_3\cdots b_{2r-1}\\ &\geq \frac{16}{9} s_{2r}\frac{(s_1-1)^2}{s_2}\bigg[ \frac{(s_2-1)^2}{s_3}\bigg]^{2^0} \cdots \bigg[\frac{(s_r-1)^2}{s_{r+1}}\bigg]^{2^{r-2}}, \end{align}\] using the formula for \(b_{r+1}-1\) and the lower bounds for \(b_{r+a}\) from the proof of Lemma 18. Therefore, \[\begin{align} m'_{2r}&=B'_{1,2r,\ldots,r,r+1}\\ &\geq b_1\cdots b_r(b_{r+1}-1)b_{r+2}\cdots b_{2r-1}b'_{2r}\\ &\geq \frac{16}{9}s_1\cdots s_{2r}\bigg[\frac{(s_1-1)^2}{s_2}\bigg]^{2^0} \cdots \bigg[\frac{(s_r-1)^2}{s_{r+1}}\bigg]^{2^{r-1}}\\ &\geq \frac{8}{9}(s_{2r}-1)(2s_{2r}-3) \frac{1}{2} \bigg[\frac{(s_{1}-1)^2}{s_{2}}\bigg]^{2^{0}} \cdots \bigg[\frac{(s_r-1)^2}{s_{r+1}}\bigg]^{2^{r-1}}\\ &\geq \frac{8}{9\alpha}(s_{2r}-1)(2s_{2r}-3). \end{align}\] We omit the similar argument for \(n=2r+1\). ◻