OFFSET
0,1
COMMENTS
From Bernard Schott, Oct 23 2019: (Start)
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.
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)
From Chai Wah Wu, Jul 10 2026: (Start)
Proof that a(4*n+5) = 11. First, a(4*n+5) >= 11 follows from the proof in the comment above. Let m = 4*n+5. Since 2^(4n) == 6^n == 6 mod 10, and 2^5 == 2 mod 10, this implies that 2^m = 2^(4*n+5) == 2 mod 10. Therefore 2^(2^m) == 4 mod 11 and 2^(2^m)+1 == 5 mod 11. Similarly 2^m is divisible by 4 and thus 2^(2^m) == 6 mod 10 and thus 2^(2^(2^m)) == 2^6 == 9 mod 11. Since 5*9 == 1 mod 11, this means that a(4*n+5) <= 11 and therefore a(4*n+5) = 11.
Theorem: a(20*n+16) <= 101 for n >= 0.
Proof: Let m = 20*n+16. Since 2^(20*n) == 76^n == 76 mod 100 and 2^16 == 36 mod 100, this implies that 2^m == 76*36 == 36 mod 100. Next, 2^36 == 78 mod 101 implies that 2^(2^m)+1 == 79 mod 101.
Furthermore, 2^(16) == 36 mod 100 and 2^(16*16) = (2^16)^16 == 36 mod 100. This shows that 2^(2^(4n)) == 2^(16^n) == 36 mod 100. Since m is divisible by 4, this means that 2^(2^m) == 36 mod 100 and thus 2^(2^(2^m)) == 78 mod 101. Since 79*78 == 1 mod 101, the proof is complete.
Other terms: a(16) = 101, a(22) = 24606817, a(23) = 243799, a(24) = 30203, a(26) = 6961, a(28) = 16762721, a(31) = 5903, a(32) = 149, a(35) = 1483, a(36) = 101, a(38) = 188527, a(40) = 19577, a(43) = 59, a(44) = 2879.
(End)
REFERENCES
Paulo Ribenboim, The New Book of Prime Number Records, 3rd edition, Springer-Verlag, New York, 1995, p. 89.
FORMULA
From Chai Wah Wu, Oct 21 2019: (Start)
a(n) >= 11 for n >= 0 (conjectured).
a(4n+5) = 11 for n >= 0 (conjectured).
See the proofs of these 2 conjectures in the comments. (End)
From Chai Wah Wu, Jul 10 2026: (Start)
Conjecture 1: a(20*n+16) = 101 for n >= 0.
Conjecture 2: a(220*n+8) = 47 for n >= 0.
Conjecture 3: a(72*n+32) = 149 for n >= 0 and a(72*n+32) is not covered by Conjectures 1 and 2.
Conjecture 4: a(168*n+43) = 59 if n>=0 and n mod 55 <> 26. (End)
PROG
(PARI) a(n) = factor((2^(2^n) + 1)*2^(2^(2^n)) - 1)[1, 1]; \\ Michel Marcus, Oct 22 2013
(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
(Python)
from sympy import primefactors, nextprime
def A227244(n):
if n == 3: return min(primefactors(((m:=1<<(1<<n)) + 1)*(1<<m)-1))
m, p = 1<<n, 11
while True:
if (pow(2, m, p)+1)*pow(2, pow(2, m, p-1), p)%p==1:
return p
p = nextprime(p) # Chai Wah Wu, Jul 04 2026
CROSSREFS
KEYWORD
nonn,hard,more,changed
AUTHOR
Arkadiusz Wesolowski, Oct 20 2013
EXTENSIONS
a(12)-a(14) from Charles R Greathouse IV, Oct 22 2013
STATUS
approved