I have a combinatoric problem where I ran into the term
$$ (a-1)_{[k]}\ {}_2F_1(1, a, a-k, -1) $$
where the $(a-1)_{[k]}$ indicates the falling factorial
After playing around with it and building up a relation to the generating function of A119258 I find that
$$ (a-1)_{[k]}\ {}_2F_1(1, a, a-k, -1) = \frac{k!}{2^{(k+1)}} \frac{d^a}{d x^a} \left(\frac{x^k}{(1-x)(1-2x)^k}\right)_{x=0} $$
and from this I can potentially extract a recursion to build up with or something similar, but it feels like this should have a cleaner expression (probably as some kind ratio of falling factorials in $k$) given the very specific form of the hypergeometric function provided there
Is there a nicer way to express this?
It turns out this is pretty simple, by looking at first differences we can find that
$$ \frac{d^a}{d x^a} \left(\frac{x^k}{(1-x)(1-2x)^k}\right)_{x=0} = \sum_{n=0}^{a} {n \choose k} 2^{(n-k)} $$
and so we get
$$ (a-1)_{[k]}\ {}_2F_1(1, a, a-k, -1) = \sum_{n=0}^{a} (n)_{[k]} 2^{(n-2k-1)} $$
although I'd still be much happier to have this expressed as some kind of series in $a$ instead