A227244 - OEIS

%I #43 Oct 23 2019 15:53:53

%S 11,79,1114111,

%T 29758566933990262223857743147232792318290386059069624958140599090033674317463551,

%U 16267,11,563,139,47,11,131107211,47,163430017,11,563

%N Smallest prime factor of (2^(2^n) + 1)*2^(2^(2^n)) - 1.

%C From _Bernard Schott_, Oct 23 2019: (Start)

%C G_n = (2^(2^n) + 1) * 2^(2^(2^n)) - 1 = F_n * 2^(F_n -1) - 1 with F_n is the n-th Fermat number.

%C 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]

%C Proof that the 1st conjecture of Chai Wah Wu is right.

%C G_n = F_n * 2^(F_n -1) - 1.

%C 1) a(n) <> 2

%C G_n is clearly odd, so 2 does not divide G_n and a(n) > 2.

%C 2) a(n) <> 3

%C 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.

%C 3) a(n) <> 5

%C 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}.

%C Fn - 1 == 0 (mod 4) but when q == 0 (mod 4) , 2^q == 6 (mod 10).

%C G_n == 7 * 6 - 1 = 41 = 1 (mod 10) and G_n is not divisible by 5 ==> a(n) > 5.

%C 4) a(n) <> 7

%C 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).

%C 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,

%C 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).

%C G_n is not divisible by 7 ==> a(n) > 7.

%C 5) a(0) = a(5) = 11 and the conjecture is proved.

%C (End)

%D Paulo Ribenboim, The New Book of Prime Number Records, 3rd edition, Springer-Verlag, New York, 1995, p. 89.

%F From _Chai Wah Wu_, Oct 21 2019: (Start)

%F a(n) >= 11 for n >= 0 (conjectured).

%F a(4n+5) = 11 for n >= 0 (conjectured).

%F (End)

%o (PARI) a(n) = factor((2^(2^n) + 1)*2^(2^(2^n)) - 1)[1,1]; \\ _Michel Marcus_, Oct 22 2013

%o (PARI) a(n)=my(N=2^n);forprime(p=3,,if((Mod(2,p)^N+1)*Mod(2,p)^lift(Mod(2,p-1)^N)==1,return(p))) \\ _Charles R Greathouse IV_, Oct 22 2013

%Y Cf. A000215.

%K nonn,hard

%O 0,1

%A _Arkadiusz Wesolowski_, Oct 20 2013

%E a(12)-a(14) from _Charles R Greathouse IV_, Oct 22 2013