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

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.
