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Find a (presumably, generalized hypergeometric-based function $f(n,a,k)$), yielding for $n=1, a=\frac{1}{2}$,the rational function (ratio of fifth-degree polynomials) \begin{equation} f(1,\frac{1}{2},k)=\frac{64 k^5 +128 k^4 -340 k^3 -1032 k^2 -1099 k -384}{k (2 k-1) (2 k+5) (4 k-1) (4 k+1)} \end{equation} and for $n=1, a=1$, \begin{equation} f(1,1,k)=\frac{8 k^5+36 k^4 -82 k^3 -681 k^2 -1366 k -885}{128 (k+2) (k+3) (k+4) (4 k+5) (4 k+7)} \end{equation} These are, respectively, eqs. (25) and (8) in arXiv:1207.1297v2, "Bures and Hilbert-Schmidt $2 \times 2$ Determinantal Moments". Presumably, for general $n$, we have ratios of $5 n$-degree polynomials in $k$. (These ratios pertain to the Bures case. In the Hilbert-Schmidt counterpart, general formulas--incorporating a $_5F_4$ hypergeometric function--yielding ratios of $3 n$-degree polynomials have previously been found, though not yet rigorously demonstrated arXiv:1109.2560, sec. D.6.)

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Are there any other conditions this function must satisfy? Otherwise, there is a trivial solution: $$\frac{(2-2a) \left(64 k^5+128 k^4-340 k^3-1032 k^2-1099 k-384\right)}{k (2 k-1) (2 k+5) (4 k-1) (4 k+1)}+\frac{(2 a-1) \left(8 k^5+36 k^4-82 k^3-681 k^2-1366 k-885\right)}{128 (k+2) (k+3) (k+4) (4 k+5) (4 k+7)}$$

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Yes, there are other conditions that must be satisfied. In general, $f(n,a,k)$ should yield ratios of $5 n$-degree polynomials in $k$ (as indicated, but not prominently enough perhaps, in the original posting). Our objective is to develop a "Bures"--degree $5 n$--counterpart to the detailed "Hilbert-Schmidt" (HS)--degree $3 n$--results featured in the two arXiv preprints indicated in the original posting. Computational limitations prevent us from directly finding the $n \geq 2$ counterparts, but the (HS) line of reasoning employed in sec. D.4 of arXiv:1109.2560 may be transferable. –  Paul Slater May 13 '13 at 21:00
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