G_n = (2^(2^n) + 1) * 2^(2^(2^n))  1 = F_n * 2^(F_n 1)  1 with F_n is the nth Fermat number.
For the first four terms, a(n) = F_n * 2^(F_n 1)  1 because the numbers G_n are then primes. It is in 1968 that Williams and Zarnke showed that G_3 = a(3) with 80 digits was a prime number. [Ribenboim]
Proof that the 1st conjecture of Chai Wah Wu is right.
G_n = F_n * 2^(F_n 1)  1.
1) a(n) <> 2
G_n is clearly odd, so 2 does not divide G_n and a(n) > 2.
2) a(n) <> 3
For n >= 2, F_n == 5 (mod 6) [Proof by induction with F_(n+1) = (F_n  1)^2 + 1], so F_n  1 == 4 (mod 6). Yet, if q even, 2^q == 4 (mod 6), so G_n == 5 * 4 1 == 19 == 1 (mod 6) ==> G_n == 1 (mod 3). G_n is not divisible by 3 ==> a(n) > 3.
3) a(n) <> 5
For n >= 2, the final two digits of F_n are periodically repeated with period 4: {17, 57, 37, 97}, hence F_n = 7 (mod 10) and F_n  1 ends with {16, 56, 36, 96}.
Fn  1 == 0 (mod 4) but when q == 0 (mod 4) , 2^q == 6 (mod 10).
G_n == 7 * 6  1 = 41 = 1 (mod 10) and G_n is not divisible by 5 ==> a(n) > 5.
4) a(n) <> 7
If n is even, F_n == 3 (mod 7) and if n is odd, F_n = 5 (mod 7). We have 2^q == 1, 2 or 4 (mod 7).
If n even, G_n == 3*1  1 == 2 (mod 7) or G_n == 3*2  1 === 5 (mod 7) or G(n) = 3*4  1 == 6 (mod 7), and,
if n odd, G_n == 5*1  1 == 4 (mod 7) or G_n == 5*2  1 === 9 == 2 (mod 7) or G(n) = 5*4  1 == 19 == 5 (mod 7).
G_n is not divisible by 7 ==> a(n) > 7.
5) a(0) = a(5) = 11 and the conjecture is proved.
(End)
