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Can we get a closed form for the following series:

$$\sum\limits_{x=1}^c \dfrac{(x+c-1)!}{x!} {}_2F_1(x+c,x,x+1,z)$$

where $c$ is a positive integer and $z$ is a real number less than -1?

Any suggestions or hints are appreciated.

1 Answers1

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$$\sum\limits_{x=1}^c \dfrac{(x+c-1)!}{x!} {}_2F_1(x+c,x,x+1,z)=\frac{(c-1)!}{(z-1)^{2c-1}}P_c(z),$$

where $P_c(z)$ is a polynomial in $z$ of degree $2c-2$, the first few are

$$P_1(z)=1$$ $$P_2(z)=z^2-4 z+5$$ $$P_3(z)=z^4-6 z^3+16 z^2-24 z+19$$ $$P_4(z)=z^6-8 z^5+29 z^4-64 z^3+97 z^2-104 z+69$$ $$P_5(z)=z^8-10 z^7+46 z^6-130 z^5+256 z^4-380 z^3+446 z^2-410 z+251$$ $$P_6(z)=z^{10}-12 z^9+67 z^8-232 z^7+562 z^6-1024 z^5+1484 z^4-1792 z^3+1847 z^2-1572 z+923$$ $$P_7(z)=z^{12}-14 z^{11}+92 z^{10}-378 z^9+1093 z^8-2380 z^7+4096 z^6-5810 z^5+7071 z^4-7630 z^3+7344 z^2-5992 z+3431$$ these polynomials may well have appeared before...

Carlo Beenakker
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    In a way of suggestion -- maybe $Q_c(u):=\frac12\left(P_c(1-u)+\frac{u^{2c-1}-2^{2c-1}}{u-2}\right)$ are more recognizable. They go like$$ \begin{array}{rl} Q_1(u)&=0\ Q_2(u)&=1\ Q_3(u)&=u+5\ Q_4(u)&=u^2+6 u+22\ Q_5(u)&=u^3+7 u^2+29 u+93\ Q_6(u)&=u^4+8 u^3+37 u^2+130 u+386\ Q_7(u)&=u^5+9 u^4+46 u^3+176 u^2+562 u+1586\ Q_8(u)&=u^6+10 u^5+56 u^4+232 u^3+794 u^2+2380 u+6476\ Q_9(u)&=u^7+11 u^6+67 u^5+299 u^4+1093 u^3+3473 u^2+9949 u+26333\ Q_{10}(u)&=u^8+12 u^7+79 u^6+378 u^5+1471 u^4+4944 u^3+14893 u^2+41226 u+106762 \end{array} $$ – მამუკა ჯიბლაძე Feb 05 '17 at 19:17
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    In fact their generating function seems to be$$\sum_{c=1}^\infty Q_c(u)t^c=\frac{2t^2}{(1-4t) \left(1+\sqrt{1-4 t}-2ut\right)}$$ – მამუკა ჯიბლაძე Feb 05 '17 at 21:10
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    And for the $P_c$ we then get$$\sum_{c\geqslant1}P_c(z)t^c=\frac{t \left(1+\sqrt{1-4 t}+2 t (1-z)\right)}{\sqrt{1-4 t} \left(1+\sqrt{1-4 t}-2 t (1-z)\right) \left(1-t (1-z)^2\right)}$$(sorry, did not try to simplify this). – მამუკა ჯიბლაძე Feb 05 '17 at 21:33
  • @მამუკა ჯიბლაძე --- brilliant! – Carlo Beenakker Feb 05 '17 at 21:54