[Background:] Looking at the powerseries for the gamma-function

$ \Gamma(1+x) = 1 + a_1 x + a_2 x^2 - a_3 * x^3 + ... $

then we can arrive at a decomposition

$ \Gamma(1+x) = r(x) + g(x) $

where g(x) is constructed by the sum of the (taylor-expansions of) geometric series

```
1 1 1 1 1 1
g(x) = --- - --- + ---*--- - ---*--- + ...
1+x 2+x 3+x 2! 4+x 3!
[edit:corrected a sign-error]
```

and from this the powerseries for r(x)

$ r(x) = \Gamma(1+x) - g(x) $

[end background]

That function r(x) begins with

$ r(x) = 1/e + 0.21938 * x + 0.09784 *x^2 + \ldots $

The function has then some nice properties. By heuristics and inspection of its powerseries it seems for instance, that

apparently it is entire, has no zero except that $ lim_{x-> \infty} r(-x) = 0 $

$ r(0) = 1/e $ where $ e = \exp(1) $

$ r(k) = r(k-1)*k + 1/e $

Just today I found, that in fact this is the incomplete gamma- function as defined/implemented in mathematica as "gamma(1+x,1)". But this may not be of concern here, because I want to understand how to think the other way round:

Question 1:

Assume we had only the functional relation and the initial value

$ r(x) = r(x-1)*x +1/e $

$ r(0) = 1/e $what else would we need to make r(x) unique and arrive at the solution

$ r(x) = \Gamma(1+x)-g(x) $

?

Question 2:

Is there any way to generalize that construction scheme to get some function f(x) where the functional equation depends on a constant parameter c =/= 0

$ f(x) = f(x-1)*x + c $

For instance let $ c=1/2 $ . What would a -for instance convex - function $ f(x)$ look like?

[update]: Question 2 seems to be easy - at the integer x f(x) is simply a scaling of r(x) by c and e: $ f(x)= r(x)*c*e $ so I'd assume the same can be assumed for fractional x.