Is $u_n$ where $\left\{\begin{matrix}u_{1}=5 \\ u_{n}=\frac{2^{u_{n-1}}+1}{3} \end{matrix}\right.$ always prime?

$\left\{\begin{matrix}u_{1}=5 \\ u_{n}=\frac{2^{u_{n-1}}+1}{3} \end{matrix}\right.$

I conjecture that $u_{n}$ is prime number. But I can not prove it. So I want to know my conjecture is right or wrong.

• Why do you think this works? – Alfred Yerger Jul 1 '15 at 16:33
• Reminds me of a famous conjecture by Fermat, disproved by Euler... – lhf Jul 1 '15 at 17:08

No. WolframAlpha says that $u_4$ is not prime and is equal to $$1676083\times \\26955961001\times\\ 29608434354586376051669975373338765263536609888911641073166192\\ 42535637290590853367799328108998193136129252550026666912268005\\ 07277398580985624625950496168983999760414855301693388419156899841$$
• I put "Is (2^683+1)/3 prime?" (where I determined earlier that $u_1=5, u_2=11, u_3=683$ are prime). – n55 Jul 1 '15 at 16:21