Let $${p_n}(x) = \sum\limits_{j = 0}^{n } {{{( - 1)}^{n - j}}p(n,j){x^j}} $$ be orthogonal polynomials satisfying $${p_n}(x) = (x - {s_{n - 1}}){p_{n - 1}}(x) - {t_{n - 2}}{p_{n - 2}}(x)$$ with moments ${M_n}.$ Then it is easy to prove that $$\det \left( {p(i + 1,j)} \right)_{i,j = 0}^{n - 1} = {M_n}.$$ It seems that moreover $$\det \left( {p(i + k,j)} \right)_{i,j = 0}^{n - 1} = \frac{{\det \left( {{M_{n + i + j}}} \right)_{i,j = 0}^{k - 1}}}{{\prod\limits_{\ell = 0}^{k - 1} {t_\ell ^{k - 1 - \ell }} }}.$$ Is this a known result?
I can only prove it for some special cases. For example if $s_0=1$, $s_n=2$ for $n>0$ and $t_n=1$ then $p(n,j)=\binom{n+j}{n-j}$ and $\det \left( {\left( {\begin{array}{*{20}{c}} {i + j + k} \\ {i - j + k} \end{array}} \right)} \right)_{i,j = 0}^{n - 1} = \det \left( {{C_{n + i + j}}} \right)_{i,j = 0}^{k - 1},$ where ${C_n}=\frac{1}{n+1}\binom{2n}{n}$ is a Catalan number.
