Orthogonal polynomials and Hankel Determinants for certain Bernoulli and Euler Polynomials

Theorem 1.1. If $b_k = B_{2k+1}(\frac{x+1}{2})$ , then for $n \geq 0$ we have (1.5) $$H_n(b_k) = (-1)^{\binom{n+1}{2}} \left(\frac{x}{2}\right)^{n+1} \prod_{\ell=1}^n \left(\frac{\ell^4(x^2 - \ell^2)}{4(2\ell+1)(2\ell-1)}\right)^{n+1-\ell}.$$ In this paper we will prove this result and similar identities for certain sequences of Euler polynomials. In the process we establish some mutual connections between Hankel determinants, orthogonal polynomials, and certain continued fractions. We begin by recalling some basic but important identities in Section 2, which is followed, in Section 3, by some necessary background on orthogonal polynomials and continued fractions. Our main results on Bernoulli and Euler polynomials are then stated and proved in Sections 4 and 5, respectively. Finally, in Section 6 we consider the relationship between $H_n(c_k)$ and $H_n(c_{k+1})$ , with some consequences for earlier results.

Lemma 2.1. Let x be a variable or a complex number. Then (2.5) $$H_n(x^k c_k) = x^{n(n+1)} H_n(c_k).$$ Proof. We consider the determinant of the matrix in (1.1), with $x^k c_k$ in place of $c_k$ . We divide the second row by x, the third row by $x^2$ , etc., and finally the (n+1)th row by $x^n$ . Then, similarly, we divide the 2nd column by x, etc., up to the (n+1)th column which we divide by $x^n$ . What remains is the determinant of $(c_{i+j})$ , while the total power of x taken out is $2(1+2+\cdots+n)=n(n+1)$ , which completes the proof. The next lemma can be found, with proof, in [11]; it is also mentioned and used in various other publications, for instance in [12, Lemma 15].

Lemma 2.2. Let $(c_0, c_1, ...)$ be a sequence and x a number or a variable. If <span id="page-3-0"></span> $$c_k(x) = \sum_{j=0}^k \binom{k}{j} c_j x^{k-j},$$ then for all $n \geq 0$ we have $$(2.6) H_n(c_k(x)) = H_n(c_k).$$ By the identity (2.1), this lemma shows that $H_n(B_k(x)) = H_n(B_k)$ , as already mentioned in the Introduction. Similarly, applying both (2.6) and (2.5) to (2.2), we see that (2.7) $$H_n(E_k(x)) = 2^{-n(n+1)} H_n(E_k).$$

Lemma 3.1. If $\mu$ is the measure in (3.3), there exists a unique sequence of monic polynomials $P_n(y)$ of degree n, n = 0, 1, ..., and a sequence of positive numbers $(\zeta_n)_{n>1}$ , with $\zeta_0=1$ , such that <span id="page-4-4"></span>(3.4) $$\int_{\mathbb{R}} P_m(y) P_n(y) d\mu(y) = \zeta_n \delta_{m,n},$$ where $\delta_{m,n}$ is the Kronecker delta function. Furthermore, for all $n \geq 1$ we have $\zeta_n = H_n(\mathbf{c})/H_{n-1}(\mathbf{c}), \text{ and for } n \geq 0,$ <span id="page-4-2"></span>(3.5) $$P_n(y) = \frac{1}{H_{n-1}(\mathbf{c})} \det \begin{pmatrix} c_0 & c_1 & \cdots & c_n \\ c_1 & c_2 & \cdots & c_{n+1} \\ \vdots & \vdots & \ddots & \vdots \\ c_{n-1} & c_n & \cdots & c_{2n-1} \\ 1 & y & \cdots & y^n \end{pmatrix},$$ where the polynomials $P_n(y)$ satisfy the 3-term recurrence relation $P_0(y) = 1$ , $P_1(y) = y + s_0$ , and <span id="page-4-3"></span>(3.6) $$P_{n+1}(y) = (y+s_n)P_n(y) - t_n P_{n-1}(y) \qquad (n \ge 1),$$ for some sequences $(s_n)_{n\geq 0}$ and $(t_n)_{n\geq 1}$ . We now multiply both sides of (3.5) by $y^r$ and replace $y^j$ by $c_j$ , which includes replacing the constant term 1 by $c_0$ for r=0. (Similar evaluations apply in the rest of this paper, at $y^k = a_k$ including k = 0, for some sequence $(a_n)$ . Then for $0 \le r \le n-1$ the last row of the matrix in (3.5) is identical with one of the previous rows, and thus the determinant is 0. When r=n, the determinant is $H_n(\mathbf{c})$ . We therefore have the following result. <span id="page-5-4"></span>Corollary 3.2. With the sequence $(c_k)$ and the polynomials $P_n(y)$ as above, we have (3.7) $$y^r P_n(y) \Big|_{y^k = c_k} = \begin{cases} 0 & \text{when } 0 \le r \le n - 1, \\ H_n(\mathbf{c}) / H_{n-1}(\mathbf{c}) & \text{when } r = n. \end{cases}$$ This corollary, by the way, is consistent with (3.2). Another important consequence of Lemma 3.1 will be an essential ingredient in most of our results. For the sake of completeness we give a proof of this well-known result. <span id="page-5-1"></span>Corollary 3.3. With the sequence $(t_n)$ as in (3.6), we have <span id="page-5-0"></span>(3.8) $$H_n(\mathbf{c}) = t_1^n t_2^{n-1} \cdots t_{n-1}^2 t_n \qquad (n \ge 0).$$

Lemma 3.4. Let $\mathbf{c} = (c_k)_{k \geq 0}$ be a sequence of numbers with $c_0 \neq 0$ , and suppose that its generating function is written in the form (3.9) $$\sum_{k=0}^{\infty} c_k t^k = \frac{c_0}{1 + s_0 t - \frac{t_1 t^2}{1 + s_1 t - \frac{t_2 t^2}{1 + s_2 t - \ddots}}},$$ where both sides are considered as formal power series. Then the sequences $(s_n)$ and $(t_n)$ are the same as in (3.6), and <span id="page-5-2"></span>(3.10) $$H_n(\mathbf{c}) = c_0^{n+1} t_1^n t_2^{n-1} \cdots t_{n-1}^2 t_n.$$ With the exception of the factor $c_0^{n+1}$ , the identities (3.10) and (3.8) are the same. The difference comes from the assumption $c_0 = 1$ in (3.3) and in Lemma 3.1, which can be suitably relaxed. We finish this section with some definitions and results which will also be needed later. Following the usage in books such as [5] or [14], we write <span id="page-6-0"></span>(3.11) $$b_0 + \mathop{\mathbf{K}}_{m=1}^{\infty} \left( a_m / b_m \right) = b_0 + \mathop{\mathbf{K}} \left( a_m / b_m \right) = b_0 + \frac{a_1}{b_1 + \frac{a_2}{b_0 + \cdots}}$$ for an infinite continued fraction. The nth approximant is expressed by <span id="page-6-1"></span>(3.12) $$b_0 + \mathbf{K}_{m=1} (a_m/b_m) = b_0 + \frac{a_1}{b_1 + \cdot \cdot + \frac{a_n}{b_n}} = \frac{A_n}{B_n},$$ and $A_n$ , $B_n$ are called the nth numerator and denominator, respectively. The continued fraction (3.11) is said to converge if the sequence of approximants in (3.12) converges. In this case, the limit is called the value of the continued fraction (3.11). Two continued fractions are said to be equivalent if and only if they have the same sequences of approximants. In other words, we have <span id="page-6-2"></span> $$b_0 + \mathbf{K}_{m=1}^n (a_m/b_m) = d_0 + \mathbf{K}_{m=1}^n (c_m/d_m)$$ if and only if there exists a sequence of nonzero complex numbers $(r_m)_{m\geq 0}$ with $r_0=1$ , such that $$(3.13) d_m = r_m b_m, c_{m+1} = r_{m+1} r_m a_{m+1} (m \ge 0);$$ see [5, Eq. (1.4.2)]. We also require the following special case of the more general concept of a contraction; see, e.g., [5, p. 16].

Lemma 3.6. An even canonical contraction of $b_0 + \mathbf{K}(a_m/b_m)$ exists if and only if $b_{2k} \neq 0$ for $k \geq 1$ , and we have <span id="page-6-3"></span> $$(3.14) \quad b_0 + \mathop{\mathbf{K}}_{m=1}^{\infty} \left( a_m / b_m \right) = b_0$$ $$+ \frac{a_1 b_2}{a_2 + b_1 b_2 - \frac{a_2 a_3 b_4 / b_2}{a_4 + b_3 b_4 + a_3 b_4 / b_2 - \frac{a_4 a_5 b_6 / b_4}{a_6 + b_5 b_6 + a_5 b_6 / b_4 - \frac{a_6 a_7 b_8 / b_6}{\cdot}}.$$ In particular, with $b_0 = 0$ , $b_k = 1$ for $k \ge 1$ , $a_1 = 1$ , and $a_k = \alpha_{k-1}t$ $(k \ge 1)$ , for some variable t, we have <span id="page-7-0"></span>(3.15) $$\frac{1}{1 - \frac{\alpha_1 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 - \frac{\alpha_2 t}{1 -$$ Similarly, an odd canonical contraction gives <span id="page-7-4"></span>(3.16) $$1 + \frac{\alpha_1 t}{1 - (\alpha_1 + \alpha_2)t - \frac{\alpha_2 \alpha_3 t^2}{1 - (\alpha_3 + \alpha_4)t - \frac{\alpha_4 \alpha_5 t^2}{1 - (\alpha_5 + \alpha_6)t - \frac{\alpha_6 \alpha_7 t^2}{\cdot \cdot \cdot}}}$$ for the continued fraction on the left-hand side of (3.15).

Theorem 4.1. Let $b_k = B_{2k+1}(\frac{x+1}{2})$ , and let $(W_n(y;x))_{n\geq 0}$ be the sequence of polynomials orthogonal in y with respect to the sequence $(b_k)$ , that is, (4.1) $$y^r W_n(y;x) \Big|_{y^k = b_h} = 0 \qquad (0 \le r \le n-1).$$ Then we have $W_0(y;x)=1$ , $W_1(y;x)=y+\sigma_0$ , and for $n\geq 1$ , $$(4.2) W_{n+1}(y;x) = (y+\sigma_n)W_n(y;x) + \tau_n W_{n-1}(y;x),$$ where (4.3) $$\sigma_n = \binom{n+1}{2} - \frac{x^2 - 1}{4}$$ and $\tau_n = \frac{n^4(x^2 - n^2)}{4(2n+1)(2n-1)}$ . Since $b_0 = B_1(\frac{x+1}{2}) = \frac{x}{2}$ (see Table 2), Lemma 3.4 with $c_0 = x/2$ , $s_j = \sigma_j$ , and $t_j = -\tau_j$ immediately gives Theorem 1.1 as a corollary. By Lemma 3.4, for the proof of Theorem 4.1 it suffices to prove the following lemma. On the left-hand side we will have a formal power series which could also be interpreted as an asymptotic expansion. <span id="page-7-2"></span>Lemma 4.2. We have the continued fraction expansion <span id="page-7-3"></span>(4.4) $$\sum_{k=0}^{\infty} B_{2k+1}(\frac{x+1}{2})z^{2k} = \frac{\frac{x}{2}}{1 + \sigma_0 z^2 + \frac{\tau_1 z^4}{1 + \sigma_1 z^2 + \frac{\tau_2 z^4}{1 + \sigma_2 z^2 + \cdots}}}.$$ An important tool for proving Lemma 4.2, as well as the results in the next section, is the polygamma function. For an integer $n \ge 0$ it is defined by $$\psi^{(n)}(z) := \frac{\mathrm{d}^{n+1}}{\mathrm{d}z^{n+1}} (\log \Gamma(z)),$$ where $\Gamma(z)$ is the gamma function. For n=0 we have $\psi^{(0)}(z)=\psi(z)=\Gamma'(z)/\Gamma(z)$ , the well-known and important digamma function, and $\psi^{(1)}(z)=\psi'(z)$ is sometimes called the trigamma function.

Lemma 4.3. We have the formal power series <span id="page-8-0"></span>(4.5) $$\sum_{k=0}^{\infty} B_{2k+1}(\frac{x+1}{2})z^{2k} = \frac{1}{2z^2} \left( \psi'\left(\frac{1}{z} + \frac{1-x}{2}\right) - \psi'\left(\frac{1}{z} + \frac{1+x}{2}\right) \right).$$

Theorem 5.1. For $\nu = 0, 1, 2$ , let $c_k^{(\nu)} := E_{2k+\nu}(\frac{x+1}{2})$ , and let $(q_n^{(\nu)}(y;x))_{n\geq 0}$ be the sequence of monic polynomials orthogonal in y with respect to the sequence $(c_k^{(\nu)})$ , that is, (5.3) $$y^r q_n^{(\nu)}(y;x) \bigg|_{y^k = c_k^{(\nu)}} = 0 \qquad (0 \le r \le n-1).$$ Then we have $q_0^{(\nu)} = 1$ , $q_1^{(\nu)}(y; x) = y + \sigma_0^{(\nu)}$ , and for $n \ge 1$ , (5.4) $$q_{n+1}^{(\nu)}(y;x) = (y + \sigma_n^{(\nu)})q_n^{(\nu)}(y;x) + \tau_n^{(\nu)}q_{n-1}^{(\nu)}(y;x),$$ where for $\nu = 0, 1, 2$ , <span id="page-11-10"></span>(5.5) $$\sigma_n^{(\nu)} = (2n+1)(n+\frac{\nu}{2}) - \frac{x^2-1}{4}, \qquad \tau_n^{(\nu)} = \frac{n^2}{4} \left(x^2 - (2n+\nu-1)^2\right).$$ Now Lemma 3.4 with $c_0 = c_0^{(\nu)} = E_{\nu}(\frac{x+1}{2})$ , with $s_j = \sigma_j^{(\nu)}$ , and $t_j = -\tau_j^{(\nu)}$ immediately gives the following Hankel determinants. <span id="page-11-9"></span>Corollary 5.2. Let $c_k^{(\nu)} = E_{2k+\nu}(\frac{x+1}{2})$ for $\nu = 0, 1, 2$ . Then for all $n \ge 0$ we have <span id="page-11-2"></span>(5.6) $$H_n(c_k^{(\nu)}) = (-1)^{\binom{n+1}{2}} E_{\nu}(\frac{x+1}{2})^{n+1} \prod_{\ell=1}^n \left(\tau_{\ell}^{(\nu)}\right)^{n+1-\ell},$$ or more explicitly, <span id="page-11-0"></span>(5.7) $$H_n(c_k^{(0)}) = (-1)^{\binom{n+1}{2}} \prod_{\ell=1}^n \left( \frac{\ell^2}{4} (x^2 - (2\ell - 1)^2) \right)^{n+1-\ell},$$ <span id="page-11-11"></span>(5.8) $$H_n(c_k^{(1)}) = (-1)^{\binom{n+1}{2}} \left(\frac{x}{2}\right)^{n+1} \prod_{\ell=1}^n \left(\frac{\ell^2}{4} (x^2 - (2\ell)^2)\right)^{n+1-\ell},$$ <span id="page-11-1"></span>(5.9) $$H_n(c_k^{(2)}) = (-1)^{\binom{n+1}{2}} \left(\frac{x^2 - 1}{4}\right)^{n+1} \prod_{\ell=1}^n \left(\frac{\ell^2}{4} (x^2 - (2\ell + 1)^2)\right)^{n+1-\ell}.$$ To obtain (5.7)–(5.9) from (5.6), we only need to notice that <span id="page-11-3"></span> $$E_0(\frac{x+1}{2}) = 1,$$ $E_1(\frac{x+1}{2}) = \frac{x}{2},$ $E_2(\frac{x+1}{2}) = \frac{x^2 - 1}{4};$ see Table 2. The proof of Theorem 5.1 will be similar in nature to that of Theorem 4.1. In particular, by Lemma 3.4 it suffices to prove the following lemma.

Lemma 5.3. For $\nu = 0, 1, 2$ we define the formal power series (5.10) $$F^{(\nu)}(z) := \sum_{k=0}^{\infty} E_{2k+\nu}(\frac{x+1}{2})z^{2k},$$ where the dependence on x is implied. Then we have <span id="page-11-8"></span>(5.11) $$F^{(\nu)}(z) = \frac{E_{\nu}(\frac{x+1}{2})}{1 + \sigma_0^{(\nu)} z^2 + \frac{\tau_1^{(\nu)} z^4}{1 + \sigma_1^{(\nu)} z^2 + \frac{\tau_2^{(\nu)} z^4}{\cdot \cdot \cdot}}.$$ The proof of this, in turn, relies on the following important connection with the digamma function.

Lemma 5.4. With $F^{(\nu)}(z)$ , $\nu = 0, 1, 2$ , as defined in (5.10), we have <span id="page-11-5"></span> $$(5.12) F^{(0)}(z) = \frac{\psi(\frac{1}{2z} + \frac{3+x}{4}) - \psi(\frac{1}{2z} + \frac{1+x}{4}) + \psi(\frac{1}{2z} + \frac{3-x}{4}) - \psi(\frac{1}{2z} + \frac{1-x}{4})}{2z},$$ <span id="page-11-4"></span> $$(5.13) F^{(1)}(z) = \frac{-\psi(\frac{1}{2z} + \frac{3+x}{4}) + \psi(\frac{1}{2z} + \frac{1+x}{4}) + \psi(\frac{1}{2z} + \frac{3-x}{4}) - \psi(\frac{1}{2z} + \frac{1-x}{4})}{2z^2},$$ <span id="page-11-6"></span>(5.14) $$F^{(2)}(z) = \frac{F^{(0)}(z) - 1}{z^2}$$ .

Proposition 6.1. ([15, Prop. 1.2]). With notation as above, for a given sequence a we have $$(6.4) H_n(a_{k+1}) = H_n(a_k) \cdot d_n,$$ and <span id="page-15-6"></span>(6.5) $$H_n(a_{k+2}) = H_n(a_k) \cdot \left(\prod_{\ell=1}^{n+1} t_\ell\right) \cdot \sum_{\ell=-1}^n \frac{d_\ell^2}{\prod_{j=1}^{\ell+1} t_j}.$$

Proposition 6.2. ([8, Eq. (2.4)]). For a given sequence $\mathbf{a}$ and $(s_n)$ as defined above, we have <span id="page-15-7"></span> $$(6.6) s_n = -\frac{1}{H_{n-1}(a_{k+1})} \left( \frac{H_{n-1}(a_k)H_n(a_{k+1})}{H_n(a_k)} + \frac{H_n(a_k)H_{n-2}(a_{k+1})}{H_{n-1}(a_k)} \right).$$ Set $\mathbf{c}^{(\nu)} = (c_0^{(\nu)}, c_1^{(\nu)}, \dots)$ and also recall (6.1) and (6.2). We now use Propositions 6.1 and 6.2 with $\mathbf{a} = \mathbf{c}^{(0)}$ . With (6.2) and (6.4) we get $H_n(c_k^{(2)})$ , provided we know $d_n$ (see below). With (3.10) we then get $\tau_n^{(2)}$ , and (6.5) and (6.6) together give $\sigma_n^{(2)}$ . So altogether we would have everything we need to know for the case $\nu = 2$ in Section 5, confirming this part of Theorem 5.1 and Corollary 5.2. The identity (6.5) would also give us $H_n(c_k^{(4)})$ . It remains to determine the factor $d_n$ in (6.4). For this purpose we use the following lemma.

Lemma 6.3. For $\nu = 0, 1, 2$ , let the sequences $\left(\sigma_n^{(\nu)}\right)_{n \geq 0}$ and $\left(\tau_n^{(\nu)}\right)_{n \geq 1}$ be defined as in (5.5). Furthermore, let <span id="page-16-0"></span> $$J^{(\nu)} := \begin{pmatrix} -\sigma_0^{(\nu)} & 1 & 0 & 0 & \cdots \\ -\tau_1^{(\nu)} & -\sigma_1^{(\nu)} & 1 & 0 & \cdots \\ 0 & -\tau_2^{(\nu)} & -\sigma_2^{(\nu)} & 1 & \cdots \\ \vdots & \vdots & \ddots & \ddots & \ddots \end{pmatrix}$$ and let $d_n^{(\nu)} := \det J_n^{(\nu)}$ be the determinant of the (n+1)th leading principal submatrix of $J^{(\nu)}$ . Then (6.7) $$d_n^{(0)} = \prod_{\ell=0}^n \frac{x^2 - (2\ell+1)^2}{4}.$$

Definition 3.5. Let $A_n, B_n$ be the nth numerator and denominator, respectively, of a continued fraction $\operatorname{cf}_1 := b_0 + \mathbf{K}(a_m/b_m)$ , and let $C_n, D_n$ be the corresponding quantities of $\operatorname{cf}_2 := d_0 + \mathbf{K}(c_m/d_m)$ . Then Then $\operatorname{cf}_2$ is called an even canonical contraction of $\operatorname{cf}_1$ if $$C_n = A_{2n}, \qquad D_n = B_{2n} \qquad (n \ge 0),$$ and is called an odd canonical contraction of $\operatorname{cf}_1$ if $$C_0 = \frac{A_1}{B_1}$$ , $D_0 = 1$ , $C_n = A_{2n+1}$ , $D_n = B_{2n+1}$ $(n \ge 1)$ . We will now state three identities that will be used in later sections; see [5, pp. 16-18], [14, pp. 83-85], or [19, pp. 21-22] for proofs and further details.