I tried up to some level - but not getting a general term. Let me know once you are done.

Definitely

I have done it up to n = 100, getting all prime numbers.

Hi! I add my answer as a file attachment. Please, answer me. Your sincerely: Anna Tomova

Hi! I add my answer as a file attachment. I generalized the problem and write 2^n +3^n as a product of more natural numbers, whose number depends on multipliers of n. Please, answer me. Your sincerely: Anna Tomova

Hi. You didn't answer my answers, but I still think about Your question. I think that the more interesting questions could be: Can we write 2^n+3^n as a product of two prime  numbers?Can we write x^n+y^n as a product of two (or more) prime numbers? My examples are made by http://www.wolframalpha.com/ and they are shown in - below: Please, answer me. Your sincerely: Anna Tomova.

Dear Anna

Thank you very much for your inputs. 

Hi! I generalized Your question in the folloving file attachment. It is written on Bulgarian , but I hope that You can understand it, because there are a lot of mathematical symbols. Please, answer me. Your sincerely: Anna Tomova.

Today I looked again your question. The serie (or the infinite sequence) with the general term 2 ^n +3^n is divergent, why should get   any formula for its general term in the form of a product of two integers? However, for odd n = 2k + 1, we have: 2 ^ (2k + 1) + 3 ^ (2k + 1) = 5 (2 ^ (2k) - 2 ^ (2k - 1) 3 + ... 3^(2k)). Please, answer me. Your sincerely: Anna Tomova.