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Can

$$\begin{align}\sum_{m = 0}^n \frac{Y_{m, n, 4}}{m + s} & = - (s^2 + (n + 1) s + n (n - 1)) \frac{\prod^{n - 2}_{^{} k = 1} (s - k)}{\prod^n_{k = 0} (s + k)}\\ & = - (s^2 + (n + 1) s + n (n - 1)) \left( \frac{(- 1)^n \Gamma (n - s) \Gamma (s)}{\Gamma (1 - s) \Gamma (n + s + 1)} \right)\end{align}$$

be written as a hypergeometric function? We have the related sum \begin{align} \sum_{m = 0}^n \frac{Y_{m, n, 3}}{m + s} & = \sum_{m = 0}^n (- 1)^{m + n - 1} \frac{\Gamma (m + n - 1)}{\Gamma (m + 1)^2 \Gamma (n - m - 1)(m + s)}\\ & = - \frac{(- 1)^n}{s} \;_3 F_ 2 \left( \begin{array}{lll} s & (2 - n) & (n - 1)\\ 1 & s + 1 & \end{array} \middle| 1 \right) \end{align}

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Yes, it can be shown that

$\begin{array}{ll} Y_{m, n, 4} & = {Res} (\hat{y}_{n, 4} (s), s = - m)\\ & = \lim_{s = - m} (s + m) \hat{y}_{n, 4} (s)\\ & = \left\{ \begin{array}{ll} 0 & n \leqslant 2\\ \frac{(- 1)^{m + n + 1}}{(m!^{})^2} \left( \prod_{k = 0}^{m - 1} (n - k) \right) \left( \prod_{k = 1}^{m - 2} (n + k) \right) (n^2 - (m + 1) n + m (m - 1)) & n \geqslant 3 \end{array} \right. \end{array}$

so that

$\begin{array}{ll} \begin{array}{l} \hat{y}_{n, 4} (s) \end{array} & = \sum_{m = 0}^{\infty} \frac{Y_{m, n, 4}}{m + s}\\ & = \sum_{m = 0}^{\infty} \frac{(- 1)^{m + n + 1} (- 1)^m \Gamma (- n + m) \Gamma (n + m - 1) (n^2 - (m + 1) n + m (m - 1))}{\Gamma (- n) \Gamma (n + 1) (m!)^2 (m + s)}\\ & = \end{array}$

$ - \frac{(- 1)^n}{s} 5 F 4 \left( \begin{array}{l} \begin{array}{lllll} {}[s] & [- n] & [n - 1] & \left[ - \frac{n}{2} + \frac{1}{2} - \frac{1}{2} \hspace{0.17em} \sqrt{- 3 \hspace{0.17em} n^2 + 6 \hspace{0.17em} n + 1} \right] & \left[ - \frac{n}{2} + \frac{1}{2} + \frac{1}{2} \hspace{0.17em} \sqrt{- 3 \hspace{0.17em} n^2 + 6 \hspace{0.17em} n + 1} \right] \end{array}\\ \begin{array}{l} \begin{array}{llll} {}[1] & [s + 1] & \left[ - \frac{n}{2} - \frac{1}{2} - \frac{1}{2} \hspace{0.17em} \sqrt{- 3 \hspace{0.17em} n^2 + 6 \hspace{0.17em} n + 1} \right] & \left[ - \frac{n}{2} - \frac{1}{2} + \frac{1}{2} \hspace{0.17em} \sqrt{- 3 \hspace{0.17em} n^2 + 6 \hspace{0.17em} n + 1} \right] \end{array} \end{array} \end{array} \right) $

well, whatever, tired of fighting latex, you get the point,

$\sum _{m=0}^{n}{\frac {{\it Y4} \left( m,n \right) }{m+s}}=-{\frac { \left( -1 \right) ^{n} {\mbox{$_5$F$_4$}(s,-n,n-1,-n/2+1/2-1/2\,\sqrt {-3\,{n}^{2}+6\,n+1},-n/2+1/2+1/2\,\sqrt {-3\,{n}^{2}+6\,n+1};\,1,s+1,-n/2-1/2-1/2\,\sqrt {-3\,{n}^{2}+6\,n+1},-n/2-1/2+1/2\,\sqrt {-3\,{n}^{2}+6\,n+1};\,1)} }{s}}+{\frac {n \left( n-1 \right) \Gamma \left( 2\,n \right) {\mbox{$_6$F$_5$}(1,1,2\,n,s+1+n,n/2+3/2-1/2\,\sqrt {-3\,{n}^{2}+6\,n+1},n/2+3/2+1/2\,\sqrt {-3\,{n}^{2}+6\,n+1};\,n+2,n+2,2+n+s,n/2+1/2+1/2\,\sqrt {-3\,{n}^{2}+6\,n+1},n/2+1/2-1/2\,\sqrt {-3\,{n}^{2}+6\,n+1};\,1)} \left( -1 \right) ^{3\,n}\sin \left( \pi \, \left( n+1 \right) \right) }{ \left( \left( n+1 \right) ! \right) ^{2} \left( s+1+n \right) \pi }} $

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