Find a (presumably, generalized hypergeometricbased function $f(n,a,k)$), yielding for $n=1, a=\frac{1}{2}$,the rational function (ratio of fifthdegree 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 k1) (2 k+5) (4 k1) (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 HilbertSchmidt $2 \times 2$ 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 HilbertSchmidt counterpart, general formulasincorporating a $_5F_4$ hypergeometric functionyielding ratios of $3 n$degree polynomials have previously been found, though not yet rigorously demonstrated arXiv:1109.2560, sec. D.6.)
Find a generalized hypergeometricbased function yielding certain ratios of fifthdegree polynomials
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Are there any other conditions this function must satisfy? Otherwise, there is a trivial solution: $$\frac{(22a) \left(64 k^5+128 k^4340 k^31032 k^21099 k384\right)}{k (2 k1) (2 k+5) (4 k1) (4 k+1)}+\frac{(2 a1) \left(8 k^5+36 k^482 k^3681 k^21366 k885\right)}{128 (k+2) (k+3) (k+4) (4 k+5) (4 k+7)}$$

$\begingroup$ 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 "HilbertSchmidt" (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. $\endgroup$ – Paul Slater May 13 '13 at 21:00