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 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 Table of n, a(n) for n=0..14. 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

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Last modified June 2 10:58 EDT 2023. Contains 363097 sequences. (Running on oeis4.)