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Timothy Chow
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Is the following result about Hankel determinants known or a simple consequence of some known results? Let

$f(x) = \frac{1}{{1 - \frac{{a{x^{m + 2}}}}{{1 - \frac{{b{x^{m + 2}}}}{{1 - {x^2}g(x)}}}}}} = \sum\limits_{n \ge 0} {{f_n}{x^n}} $$f(x) = \frac{\displaystyle 1}{{\displaystyle 1 - \frac{{a x^{m + 2}}}{\displaystyle {1 - \frac{{b{x^{m + 2}}}}{\displaystyle {1 - {x^2}g(x)}}}}}} = \sum\limits_{n \ge 0} {{f_n}{x^n}} $

and

$g(x) = \sum\limits_{n \ge 0} {{g_n}{x^n}} $

be formal power series. Then the Hankel determinants

${d_f}(n) = \det \left( {{f_{i + j}}} \right)_{i,j = 0}^n$ and

${d_g}(n) = \det \left( {{g_{i + j}}} \right)_{i,j = 0}^n$

are closely related. To simplify notation let

${d_g}(n) = 1$ for $n < 0.$

Then for $n \ge 0$

$${d_f}(n + m + 1) = (-1)^{m+1\choose 2} {a^{m+n+1}}{b^n}{d_g}(n - 2)$$

Remark: For $m = 0$ this result is known and follows from the fact that the moment generating function of monic orthogonal polynomials can be represented as a continued fraction.

Is the following result about Hankel determinants known or a simple consequence of some known results? Let

$f(x) = \frac{1}{{1 - \frac{{a{x^{m + 2}}}}{{1 - \frac{{b{x^{m + 2}}}}{{1 - {x^2}g(x)}}}}}} = \sum\limits_{n \ge 0} {{f_n}{x^n}} $

and

$g(x) = \sum\limits_{n \ge 0} {{g_n}{x^n}} $

be formal power series. Then the Hankel determinants

${d_f}(n) = \det \left( {{f_{i + j}}} \right)_{i,j = 0}^n$ and

${d_g}(n) = \det \left( {{g_{i + j}}} \right)_{i,j = 0}^n$

are closely related. To simplify notation let

${d_g}(n) = 1$ for $n < 0.$

Then for $n \ge 0$

$${d_f}(n + m + 1) = (-1)^{m+1\choose 2} {a^{m+n+1}}{b^n}{d_g}(n - 2)$$

Remark: For $m = 0$ this result is known and follows from the fact that the moment generating function of monic orthogonal polynomials can be represented as a continued fraction.

Is the following result about Hankel determinants known or a simple consequence of some known results? Let

$f(x) = \frac{\displaystyle 1}{{\displaystyle 1 - \frac{{a x^{m + 2}}}{\displaystyle {1 - \frac{{b{x^{m + 2}}}}{\displaystyle {1 - {x^2}g(x)}}}}}} = \sum\limits_{n \ge 0} {{f_n}{x^n}} $

and

$g(x) = \sum\limits_{n \ge 0} {{g_n}{x^n}} $

be formal power series. Then the Hankel determinants

${d_f}(n) = \det \left( {{f_{i + j}}} \right)_{i,j = 0}^n$ and

${d_g}(n) = \det \left( {{g_{i + j}}} \right)_{i,j = 0}^n$

are closely related. To simplify notation let

${d_g}(n) = 1$ for $n < 0.$

Then for $n \ge 0$

$${d_f}(n + m + 1) = (-1)^{m+1\choose 2} {a^{m+n+1}}{b^n}{d_g}(n - 2)$$

Remark: For $m = 0$ this result is known and follows from the fact that the moment generating function of monic orthogonal polynomials can be represented as a continued fraction.

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Johann Cigler
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An identity for Hankel determinants

Is the following result about Hankel determinants known or a simple consequence of some known results? Let

$f(x) = \frac{1}{{1 - \frac{{a{x^{m + 2}}}}{{1 - \frac{{b{x^{m + 2}}}}{{1 - {x^2}g(x)}}}}}} = \sum\limits_{n \ge 0} {{f_n}{x^n}} $

and

$g(x) = \sum\limits_{n \ge 0} {{g_n}{x^n}} $

be formal power series. Then the Hankel determinants

${d_f}(n) = \det \left( {{f_{i + j}}} \right)_{i,j = 0}^n$ and

${d_g}(n) = \det \left( {{g_{i + j}}} \right)_{i,j = 0}^n$

are closely related. To simplify notation let

${d_g}(n) = 1$ for $n < 0.$

Then for $n \ge 0$

$${d_f}(n + m + 1) = (-1)^{m+1\choose 2} {a^{m+n+1}}{b^n}{d_g}(n - 2)$$

Remark: For $m = 0$ this result is known and follows from the fact that the moment generating function of monic orthogonal polynomials can be represented as a continued fraction.