Confluent hypergeometric function for positive integers Do any of you know a simple form for the confluent hypergeometric function with positive integers that involve simple functions? 
What I actually need to compute is $_1F_1(n,n + m,z)$.
I know for $m=0$, $_1F_1(n,n,z)=e^z$ and for $m=1$, the hypergeometric function is related to the incomplete gamma.
Are there some results for $m\ge 2$?
Many thanks in advance for my first question.
P
 A: many thanks for help. Having a look at some other properties Gradshteyn and Ryzhik, see 3.383. Case $1^{11}$, pag 347, I think I found the result I was looking for. Here it is:
$_1F_1(n,n+m,z)=\frac{z^{1-n-m}}{B(m,n)}\int_0^zt^{n-1}(z-t)^{m-1}e^tdt$.
Developing the binomial the integral becomes:
$\int_0^zt^{n-1}(z-t)^{m-1}e^tdt=\sum_{\ell=0}^{m-1}z^\ell(-1)^{m-1-\ell}\beta(m-1,\ell)\int_0^zt^{n+m-\ell-2}e^tdt$
where $\beta(m-1,\ell)$ denotes the binomial coefficient.
The last integral can be solved by parts and is known. One can also write it in terms of incomplete gamma $\gamma(a,x)$ and finally would get a finite sum:
$_1F_1(n,n+m,z)=\frac{(-1)^nz^{1-n-m}}{B(m,n)}\sum_{\ell=0}^nz^{\ell}\beta(m-1,\ell)\gamma(n+m-\ell-1, -z)$.
Please note that $\gamma(n+m-\ell-1, -z)$ is real because $n+m-\ell-1$ is an integer.
I checked numerically and it works. Just a remark, I tried using the results in Wikipedia https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions
but I do not get the correct answer. Something seems wrong in the results that are mentioned.
Thanks once more.
P
A: For $m = 2$, it involves both $e^z$ and the incomplete gamma function. Let, 
$$_1F_1(n,n+2,z) =\,_1F_1(v-1,v+1,z)$$
for symmetry. Then,
$$_1F_1(v-1,v+1,z) = v\,e^z - n(-v)^{-v}(v + z - 1)\big(\Gamma(v) - \Gamma(v, -z)\big)$$
For $m>2$, you can find more identities using WA.
A: As an example of using the contiguous relations from http://dlmf.nist.gov/13.3#i
We convert:
$_{1}F_{1}(n,n,z)=e^{z}$
$_{1}F_{1}(n,n+1,z)=n\cdot z^{-n}\cdot\gamma(n,-z)$ 
To
$_{1}F_{1}(3,2,z)$
--
We can take http://dlmf.nist.gov/13.3#E2 
$_{1}F_{1}(3,2,z)=\frac{-b\cdot\left(1-b-z\right)\cdot_{1}F_{1}\left(3,3,z\right)-z\cdot\left(b-a\right)\cdot_{1}F_{1}(3,4,z)}{b\cdot\left(b-1\right)}$
$=\frac{-3\cdot\left(-2-z\right)\cdot_{1}F_{1}\left(3,3,z\right)-z\cdot0\cdot_{1}F_{1}(3,4,z)}{2\cdot3}$
$_{1}F_{1}(3,2,z)=\frac{\left(2+z\right)}{2}e^{z}$ 
Sort of a chess game with an infinite board and new rules.
We can always start from
 $_{1}F_{1}(n,n,z)=e^{z}$
and $_{1}F_{1}(n,n+1,z)=n\cdot z^{-n}\cdot\gamma(n,-z)$ 
Lets consider: n=1,m=4
Then using 13.3.2
$2\cdot _{1}F_{1}(1,1,z)+2\cdot (-1-z)\cdot _{1}F_{1}(1,2,z)=-z\cdot (1)\cdot _{1}F_{1}(1,3,z)$
$6\cdot _{1}F_{1}(1,2,z)+3\cdot (-2-z)\cdot _{1}F_{1}(1,3,z)=-z\cdot (2)\cdot_{1}F_{1}(1,4,z)$
Notice that in the straight forward application we have singularities that occur when b<2 or =a .  These can result in the calculation being 0 times the answer we want.  The cases b<=0 is easy to toss out since it's not allowed in the series expression.  b=1 has to be handled by something like 13.3.4.  b=a is also of this nature.
