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