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I need to make a table of ${}_3 F_2\left(\frac{a}2+\frac14, \frac{a}2+\frac34, \frac{a+b}2;\; a+1,\frac{a+b}2+1;\;1\right)$ for integer $a, b,$ $0\le a\le N_1$, $0\le b\le N_2$, with precision of 50 decimal places in mantissa.

Currently I'm trying to use Wolfram Mathematica for this task, but it appears to evaluate a single value for over a minute when $a$ and $b$ are $\sim50$. And as $N_1\approx50$ and $N_2\approx 300$, this appears to be too long a computation.

So I'm looking for ways to speed up this calculation. Namely, is there a way to reduce this function to something simpler for these particular arguments? Are there some relations which would allow to easily compute it for given $a_i, b_i$, having computed for several other values (like the relation $\Gamma(x+1)=x\Gamma(x)$, which lets one easily find $\Gamma(x+1)$ knowing $\Gamma(x)$ without computing $\Gamma$ once more)?

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  • $\begingroup$ I think that the only interesting transformation formula you can apply is the Thomae’s transformation formula. However, your ${}_3 F_2$ is not balanced, so chances that the ${}_3 F_2$ reduces to lower hypergeometric series is rather slim. Just a small remark, but if you remove a 1 in the denominator the problem becomes easy. $\endgroup$ – Noud Nov 25 '13 at 13:11
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See "Computing hypergeometric functions rigorously" by Fredrik Johansson. He addresses your question in Section 7.5 where he states:

The functions $_3F_2$, $_4F_3$, etc. have a $1/z$ transformation analogous to (10), but no other such formulas for generic parameters [21, 16.8]. We could cover $|z| \gg 1$ by evaluating $_pF_q$-series directly, but other methods are needed on the annulus surrounding the unit circle. Convergence accelerations schemes such as [69, 7, 56] can be effective, but the $D$-finite analytic continuation approach (with the singular expansion at $z = 1$) is likely better since effective error bounds are known and since complexity-reduction techniques apply. A study remains to be done. We note that an important special function related to $_{n+1}F_n$, the polylogarithm $\text{Li}_s(z)$, is implemented for general complex $s$, $z$ in Arb, using direct series for $|z| \ll 1$ and the Hurwitz zeta function otherwise [35]. Also $s$-derivatives are supported.

I don't know if this will directly answer your question, but you may find a solution in the references cited in this part of the paper.

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Hint:

$_3F_2\left(\dfrac{a}{2}+\dfrac{1}{4},\dfrac{a}{2}+\dfrac{3}{4},\dfrac{a+b}{2};a+1,\dfrac{a+b}{2}+1;1\right)$

$=\sum\limits_{n=0}^\infty\dfrac{\left(\dfrac{a}{2}+\dfrac{1}{4}\right)^{(n)}\left(\dfrac{a}{2}+\dfrac{3}{4}\right)^{(n)}\left(\dfrac{a+b}{2}\right)^{(n)}}{(a+1)^{(n)}\left(\dfrac{a+b}{2}+1\right)^{(n)}n!}$

$=\sum\limits_{n=0}^\infty\dfrac{\left(\dfrac{a}{2}+\dfrac{1}{4}\right)^{(n)}\left(\dfrac{a}{2}+\dfrac{3}{4}\right)^{(n)}\dfrac{a+b}{2}}{(a+1)^{(n)}\left(n+\dfrac{a+b}{2}\right)n!}$

$=\dfrac{a+b}{2}\int_0^1\sum\limits_{n=0}^\infty\dfrac{\left(\dfrac{a}{2}+\dfrac{1}{4}\right)^{(n)}\left(\dfrac{a}{2}+\dfrac{3}{4}\right)^{(n)}x^{n+\frac{a+b}{2}-1}}{(a+1)^{(n)}n!}~dx$

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  • $\begingroup$ I don't quite see how an integral could help speed up calculation... Could you elaborate? $\endgroup$ – Ruslan Sep 6 '16 at 16:04

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