# S(n) = n^n + 1

Can the number S(n) = n^n+1 be prime? It is easy to show that a necessary condition is that n is of the form n=2^(2^k). The number S(n) therefore becomes a Fermat number F(m)=2^(2^m)+1 where m=k+2^k (Sloane's A006127).

Here is the status of such numbers. It seems extremely unlikely that any more of the numbers S(n) would be prime.

kmnstatus of S(n)
012prime, 2^2 + 1 = 5
134prime, 4^4 + 1 = 257
2616composite, 16^16 + 1 has factor 1071*2^8 + 1
311256composite, 256^256 + 1 has factor 39*2^13 + 1
42065536composite, but no factor known
537composite, S(n) has factor 1275438465*2^39 + 1
670status unknown
7135status unknown
8264status unknown
9521status unknown
101034status unknown
112059composite, S(n) has factor 591909*2^2063 + 1
124108status unknown
138205status unknown
1416398status unknown
1532783status unknown
1665552status unknown
17131089status unknown
18262162status unknown
19524307status unknown
201048596status unknown
212097173status unknown
224194326status unknown
238388631status unknown
2416777240status unknown
2533554457status unknown
2667108890status unknown