2000 Ray Ballinger's idea
n^n+1 is prime only if n=2^k,
in that case n^n+1 = (2^k)^(2^k)+1 = 2^(k*2^k)+1 is prime
only if k*2^k is a power of 2, say k=2^m,
in that case n^n+1 = 2^(2^m*2^2^m)+1 = 2^2^(m+2^m)+1
It is the Fermat number F(m+2^m) = (2^2^m)^(2^2^m)+1

0 F(1) = 2^2+1 = 5 is prime
1 F(3) = 4^4+1 = 257 is prime
2 F(6) 1071*2^8+1 (1855 Clausen)
3 F(11) 39*2^13+1 (1899 Cunningham)
4 F(20) composite (1987 Buell & Young) no factor found yet
5 F(37) 1275438465*2^39+1 (1991 Gostin)
6 F(70) .
7 F(135) .
8 F(264) .
9 F(521) .
10 F(1034) .
11 F(2059) 591909*2^2063+1 (2000 Ballinger)
12 F(4108) .
13 F(8205) .
14 F(16398) .
15 F(32783) .
16 F(65552) .
17 F(131089) .
18 F(262162) .
19 F(524307) .
20 F(1048596) .
21 F(2097173) .
22 F(4194326) .
23 F(8388631) .
24 F(16777240) .
   F(18233954) largest Fermat number, with a known factor, so far
25 F(33554457)  .

[worth of some extra out of limit search]



extending his idea to n^n^n+1

n^(n^n)+1 is prime only if n^n is a power of 2
and it is possible only if n=2^k
n^n = (2^k)^(2^k) = 2^(k*2^k)
in that case n^n^n+1 = (2^k)^(2^(k*2^k))+1 = 2^(k*2^(k+2^k))+1
and it can be a prime only if k=2^m
in that case
n^n^n+1 = 2^((2^m)*2^((2^m)*(2^2^m)))+1 = 2^2^(m+2^(m+2^m))+1
It is the Fermat number F(m+2^(m+2^m)) = (2^2^m)^(2^2^m)^(2^2^m)+1


n^n^n+1 can be a prime only if n=2^2^m, 
where in that case n^n^n+1 = F(m+2^(m+2^m))

0 F(2) = 2^2^2+1 = 17 is prime
1 F(9) 37*2^16+1 (1903 Western)
2 F(66) 7551*2^69+1 (1977 Shippee)
3 F(2051) .
4 F(1048580) .
5 F(137438953477)



[n^n^n^n+1]


J. Ray Ballinger (1953-2014)
[Where is his grave? Tampa, FL?]