Asymptotics of sequence depending on Tricomi's function I'm dealing with the following sequence
$$ p_n = U(a, a - n, 1)$$
where $a > 0$ and $U$ is Tricomi's function.
I suspect that asymptotically when $n \to \infty$ its behaviour is a power law (evidence)
$$ p_n \sim 1/n^a $$
but I cannot arrive to anything with known properties (I've already checked the NIST handbook).
Any ideas are welcome, thanks in advance.
EDIT: Found stronger evidence: $\beta=a$
EDIT: Specified which asymptote
 A: There are some details in the following derivation which will be omitted to keep this answer at a reasonable length.  What follows is a set of clotheslines on which a rigorous proof can hopefully be hung.
We will use the integral representation
$$
\begin{align}
U(a,a-n,1) &= \frac{1}{\Gamma(a)} \int_0^\infty e^{-t} t^{a-1} (1+t)^{-n-1}\,dt \\
&= \frac{1}{\Gamma(a)} \int_0^\infty \exp f_{a,n}(t)\,dt.
\end{align}
$$
(see equation 13.2.5 in Abramowitz and Stegun).  The function $f_{a,n}(t)$ critical points at $t \approx -n$ and $t \approx \frac{a-1}{n}$, so in the domain of integration the function will be maximized at $t=0$ if $a>1$ for $n$ large enough.  One can check explicitly that this is also true when $a=1$.  If $a>1$ then $f_{a,n}(t)$ is maximized at the point $t \approx \frac{a-1}{n}$.
Let's consider the case $0 < a \leq 1$.  We have
$$
\begin{align}
f_{a,n}(t) &= -t + (a-1)\log t - (n+1)\log(1+t) \\
&= -t + (a-1)\log t - (n+1)t + O(nt^2) \\
&\approx (a-1)\log t - (n+2)t,
\end{align}
$$
near $t=0$, so that as $n \to \infty$ the Laplace method yields
$$
\begin{align}
\int_0^\infty \exp f_{a,n}(t)\,dt &\sim \int_0^\infty \exp\Bigl[(a-1)\log t - (n+2)t\Bigr]\,dt \\
&= \frac{\Gamma(a)}{(n+2)^a} \\
&\sim \frac{\Gamma(a)}{n^a}.
\end{align}
$$
Now let's consider $a>1$.  In this case the function $f_{a,n}(t)$ has a maximum inside the range of integration at $t = t^* \approx \frac{a-1}{n}$.  Explicitly this is
$$
\begin{align}
t^* &= \frac{-n+a-3}{2}+\frac{1}{2}\sqrt{n^2+(-2a+6)n+a^2-2a+5} \\
&= \frac{-n+a-3}{2}+\frac{n}{2}\sqrt{1+\frac{-2a+6}{n}+\frac{a^2-2a+5}{n^2}} \\
&= \frac{-n+a-3}{2}+\frac{n}{2}\left[1+\frac{1}{2}\left(\frac{-2a+6}{n}+\frac{a^2-2a+5}{n^2}\right) \right. \\
&\qquad\qquad \left.- \frac{1}{8}\left(\frac{-2a+6}{n}+\frac{a^2-2a+5}{n^2}\right)^2 + O\left(\frac{1} {n^3}\right)\right] \\
&= \frac{-n+a-3}{2}+\frac{n}{2}\left[1 + \frac{-a+3}{n} + \frac{2a-2}{n^2} + O\left(\frac{1} {n^3}\right)\right] \\
&= \frac{-n+a-3}{2} + \frac{n}{2} + \frac{-a+3}{2} + \frac{a-1}{n} + O\left(\frac{1} {n^2}\right) \\
&= \frac{a-1}{n} + O\left(\frac{1} {n^2}\right),
\end{align}
$$
where we used the fact that $\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + O(x^3)$ to get the third equality.  We then have
$$
\begin{align}
f_{a,n}(t) &= -t + (a-1)\log t - (n+1)\log(1+t) \\
&= \alpha(a,n) + \beta(a,n)(t-t^*)^2 + \text{higher order terms},
\end{align}
$$
and we calculate
$$
\alpha(a,n) = (a-1)\Bigl[\log(a-1)-\log(n)-1\Bigr] + o(1)
$$
and
$$
\beta(a,n) = -\frac{n^2}{2 (a-1)} + O(n).
$$
By the Laplace method we conclude that
$$
\begin{align}
\int_0^\infty \exp f_{a,n}(t)\,dt &\sim e^{\alpha(a,n)} \int_{-\infty}^{\infty} \exp\Bigl[ \beta(a,n) t^2 \Bigr]\,dt \\
&\sim \exp\Bigl\{(a-1)\Bigl[\log(a-1)-\log(n)-1\Bigr]\Bigr\} \int_{-\infty}^{\infty} \exp\Bigl[ -\frac{n^2}{2 (a-1)} t^2 \Bigr]\,dt \\
&= e^{1-a} \sqrt{\frac{2\pi}{a-1}} \left(\frac{a-1}{n}\right)^a.
\end{align}
$$
In summary,

If $0 < a \leq 1$ then
  $$
U(a,a-n,1) \sim \frac{1}{n^a}
$$
  and if $a > 1$ then
  $$
U(a,a-n,1) \sim \frac{e^{1-a}}{\Gamma(a)} \sqrt{\frac{2\pi}{a-1}} \left(\frac{a-1}{n}\right)^a.
$$

