Let $(E):\ \sum_{k=0}^n P_k(z)g(z+k)=0$ with the $P_k\in\mathbb C[z]$ a finite differences equation in $\mathbb C$.
Is it true that every any entire solution $g$ of (E) of exponential type $<\pi$ is an exponential polynomial?
Thanks in advance
Let $(E):\ \sum_{k=0}^n P_k(z)g(z+k)=0$ with the $P_k\in\mathbb C[z]$ a finite differences equation in $\mathbb C$. Is it true that every any entire solution $g$ of (E) of exponential type $<\pi$ is an exponential polynomial? Thanks in advance 


The answer is no. Function $g(z)=(e^z1)/z$ is entire, of exponential type $1$, is not an exponential polynomial, but it satisfies the equation $$(z+2)g(z+2)(1+e)(z+1)g(z+1)+ezg(z)=0.$$ Ref. JP Bezivin et F. Gramain, Solutiuons entieres d'un systeme d'equations aux differences, Ann. Inst. Fourier, Grenoble, 43 (1993) 792814. Notice that you can modify this example to achieve arbitrarily small exponential type. Edit. There exists also such equations with entire solutions of smaller growth, of order less than one. This paper has explicit examples with order $1/3$ and $1/5$: K. Ishizaki and N. Yanagihara, WimanValiron method for difference equations, Nagoya Math. J., 175 (2004) 75102. 

