Is there a function with the property$ f(n)=f^{(n)}(a)$ Is there a not identically zero, real-analytic function $f\colon\mathbb R\to\mathbb R$, which satisfies
$$f(n)=f^{(n)}(a),n\in\mathbb N \text{ or }\mathbb N^+?$$
and $a\in \mathbb R$
I saw a special case when $a=0$
I try to solve it by :
$$f(x)=e^{cx}$$
$$f(n)=e^{nc}$$
$$f^{(n)}(x)=c^ne^{cx}$$
$$f^{(n)}(a)=c^ne^{ca}$$
so $$e^{nc}=c^ne^{ca}$$
so $$c=\frac{nW(\frac{a-n}{n})}{a-n}$$
the problem is we always see  n with c 
but the special case when a=0 give 
$$c=\frac{nW(\frac{0-n}{n})}{0-n}$$
$$c=\frac{W(\frac{-1}{1})}{-1}=-W(-1)$$
I think there is no solution when $a\neq 0$
may be there is another function can solve it
Is there any solution in general?
thanks for all
 A: I can't explicitly find an example, so perhaps turning to an existence proof, that can also be used to construct an example. To do so, consider the operator $\mathcal L_a: C^\infty \times \mathbb N \to \mathbb R$, where $\mathcal L_a\{f,n\} = f(n) - f^{(n)}(a)$. Then inspect Banach fixed-point theorem, if you can choose a norm in $C^\infty \times \mathbb N$ where $\mathcal L_a$ is a contraction, then jack-pot. Or, if it is not a contraction, for many reasonable norms, then you can say there is probably no such function in Banach space.
Hope this helps.
A: So, an easy answer to a seemingly related problem:
$f(x)=\sin(x)$ has the property that for each $n\in\mathbb{N}$, $f^{(n)}(a)=f\left(\frac{\pi}{2}n\right)$, where $a=2\pi k$ for any $k\in\mathbb{Z}$.
What one would like to do is just replace $f$ with $g(x)=\sin(\frac{\pi}{2}x)$, but of course this messes up the derivative.  I do not immediately see how to fix this, but I also do not see why solving the equation $f^{(n)}(a)=f(\frac{\pi}{2}n)$ should be fundamentally different than solving $f^{(n)}(a)=f(n)$.  Perhaps this can shed light on the question though.
A: I don't know about a general solution. But if we assume the function is invertible, then:
$$ f(n) = f^n(a)$$ 
$$f^{-1}f(n) = n= f^{n-1}(a)$$  
Then, $$f^0(a) = a = 1$$ 
$$f(1) = f(a)= 2$$
 $$f(2) = f^2(1) = 3$$ and soon. In this case $f$ is an increment function. 
