Evaluation of hypergeometric type continued fraction Is there a (possibly hypergeometric-type) explicit evaluation of the
continued fraction
$$a-\dfrac{1.(c+d)}{2a-\dfrac{2.(2c+d)}{3a-\dfrac{3.(3c+d)}{4a-\ddots}}}$$
Even the special case $d=0$, $a=1$ would be interesting. Note that
$$\tanh^{-1}(z)=\dfrac{z}{1-\dfrac{1^2z^2}{3-\dfrac{2^2z^2}{5-\ddots}}}$$
but that does not seem to help.
 A: This is found in [1] $\S 82$, Satz 5.  It covers the case where the numerator $a_n$ is polynomial of degree $2$ in $n$  and the denominator $b_n$ is degree $1$.  
If I plugged in correctly, we get for your continued fraction:  Let
$a, c, d$ be complex numbers satisfying: $c \ne 0, a \ne 0, a^2 \ne 4c$, and
$(a^2-4c)/a^2$ is not a negative real.  Then the value is
$$
{\frac { \left( d+c \right) \sqrt {{a}^{2}-4\,c}+a \left( c-d
 \right) }{2c} 
\;{\mbox{$_2$F$_1$}\left(1,{\frac {d+c}{c}};{\frac { \left( 3\,c+d \right) \sqrt {{a}^{2}-4\,c}+a \left( c-d \right) }{2c\sqrt {{a}^{2}-4\,c}}};{\frac {\sqrt {{a}^{2}-4\,c}-a}{2\sqrt {{a}^{2}-4\,c}}}\right)}^{-1}}
$$
The sign of the square-root is chosen so that $\displaystyle\frac{a}{\sqrt{a^2-4c}}$ has positive real part.
(The numerator ${{}_2F_1}$ has a zero in there, so it turns out to be constant.)
[1] Oskar Perron, Die lehre von den Kettenbrüchen, 2 Auflage 1929
A: Thanks to Nemo's suggestion, the answer for $d=0$ is
$$\dfrac{a-\delta}{2c}\sum_{n\ge0}\dfrac{n!}{((3+a/\delta)/2)_n}\left(\dfrac{1-a/\delta}{2}\right)^n=\dfrac{1}{a-\dfrac{1^2c}{2a-\dfrac{2^2c}{3a-\ddots}}}\;,$$
with $\delta=\sqrt{a^2-4c}$.
