This site is supported by donations to The OEIS Foundation.

 Please make a donation to keep the OEIS running. We are now in our 55th year. In the past year we added 12000 new sequences and reached 8000 citations (which often say "discovered thanks to the OEIS"). We need to raise money to hire someone to manage submissions, which would reduce the load on our editors and speed up editing. Other ways to donate

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A227244 Smallest prime factor of (2^(2^n) + 1)*2^(2^(2^n)) - 1. 0
 11, 79, 1114111, 29758566933990262223857743147232792318290386059069624958140599090033674317463551, 16267, 11, 563, 139, 47, 11, 131107211, 47, 163430017, 11, 563 (list; graph; refs; listen; history; text; internal format)
 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) REFERENCES Paulo Ribenboim, The New Book of Prime Number Records, 3rd edition, Springer-Verlag, New York, 1995, p. 89. LINKS FORMULA From Chai Wah Wu, Oct 21 2019: (Start) a(n) >= 11 for n >= 0 (conjectured). a(4n+5) = 11 for n >= 0 (conjectured). (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 CROSSREFS Cf. A000215. Sequence in context: A155619 A126506 A159663 * A026897 A021024 A127021 Adjacent sequences:  A227241 A227242 A227243 * A227245 A227246 A227247 KEYWORD nonn,hard AUTHOR Arkadiusz Wesolowski, Oct 20 2013 EXTENSIONS a(12)-a(14) from Charles R Greathouse IV, Oct 22 2013 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified December 12 07:31 EST 2019. Contains 329948 sequences. (Running on oeis4.)