simpler expression for terminating 3F2 series with negative unit argument

Question

I am hoping to evaluate a simpler expression for the following: $${}_3F_2(-n,1,1; a, (a+3)/2; -1)$$ Here $n,a \in \mathbb{N}$ and $a$ is odd. I am also interested in the asymptotics in $n \in \mathbb{N}$ for fixed $a$. There are numerous transformations for hypergeometric series with unit argument, but I was not able to find any with negative unity as the argument. Considering that two of the parameters in the numerator are unity, I was hoping that a nice expression would be nearby.

I would appreciate if someone can help with references or ideas.

Answer 1

With $f_n$ denoting the hypergeometric function, it can be checked to satisfy the inhomogeneous rank 2 recurrence equation: $$ (n+a+1)\left(n+\frac{a+5}{2}\right)f_{n+2} - 3 (n+2)\left(n+\frac{a+3}{2}\right) f_{n+1} + 2 (n+1)(n+2) f_n = \frac{a^2-1}{2} $$

In[52]:= Table[-(1/2) (a^2 - 1) + (n + a + 1) (n + (a + 3)/2 + 1) y[
      n + 2] - 3 (n + 2) (n + (a + 3)/2) y[n + 1] + 
    2 (n + 1) (n + 2) y[n] /. 
   y -> Function[k, 
     HypergeometricPFQ[{-k, 1, 1}, {a, (a + 3)/2}, -1]], {n, 0, 
   6}] // Together

Out[52]= {0, 0, 0, 0, 0, 0, 0}

From this recurrence equation, using techniques of Birkhoff and Trjitzinsky, "Analytic theory of singular difference equations", Acta Math., 60 (1932), pp. 1–89, one gets: $$ f_n = 2^n n^{(1-3a)/2} \left(c_1 + \mathcal{O}\left(n^{-1}\right) \right) + \frac{1}{n} \left(c_2 + \mathcal{O}\left(n^{-1}\right) \right) + \frac{a^2-1}{2} \frac{\log(n)}{n} \left(1 + \mathcal{O}\left(n^{-1}\right) \right) $$

Also note, that for special case of $a=1$, closed-form is easy to find: $$ f_n = {}_3F_2\left(-n,1,1; 1,2; -1\right) = {}_2F_1\left(-n,1; 2; -1\right) = \frac{2^{n+1}-1}{n+1} $$

One could also try use the following integral representation for $f_n$, valid for $a>1$: $$ f_n = \frac{a+1}{2} \int_0^1 \left(1-t\right)^{(a-1)/2} \cdot {}_2F_1\left(-n, 1; a; -t\right) \mathrm{d} t $$

In[79]:= Table[
 Simplify[HypergeometricPFQ[{-n, 1, 1}, {a, (a + 3)/2}, -1] == 
   Assuming[
    a > 1, ((a + 1)/2)*
     Integrate[(1 - t)^((a - 1)/2)*
       HypergeometricPFQ[{-n, 1}, {a}, -t], {t, 0, 1}]]], {n, 1, 5}]

Out[79]= {True, True, True, True, True}