%I #59 Oct 20 2024 23:47:21

%S 2,5,41,3281,21523361,926510094425921,1716841910146256242328924544641,

%T 5895092288869291585760436430706259332839105796137920554548481

%N a(n) = (3^(2^n) + 1)/2 = A059919(n)/2, n >= 0.

%C Average of first 2^(n+1) powers of 3 divided by average of first 2^n powers of 3.

%C Numerator of b(n) where b(n) = (1/2)*(b(n-1) + 1/b(n-1)), b(0)=2. - _Vladeta Jovovic_, Aug 15 2002

%C From _Daniel Forgues_, Jun 22 2011: (Start)

%C Since for the generalized Fermat numbers 3^(2^n)+1 (A059919), we have a(n) = 2*a(n-1)*a(n-2)*...*a(1)*a(0) + 2, n >= 0, where for n = 0, we get 2*(empty product, i.e., 1) + 2 = 4 = a(0). This formula implies that the GCD of any pair of terms of A059919 is 2, which means that the terms of (3^(2^n)+1)/2 (A059917) are pairwise coprime.

%C 2, 5, 41, 21523361, 926510094425921 are prime. 3281 = 17*193. (End)

%C a(0), a(1), a(2), a(4), a(5), and a(6) are prime. Conjecture: a(n) is composite for all n > 6. - _Thomas Ordowski_, Dec 26 2012

%C This may be a primality test for Mersenne numbers. a(2) = 41 == -1 mod 7 (=M3), a(4) = 21523361 == 30 == -1 mod 31 (=M5). However, a(10) is not == -1 mod M11. - _Nobuyuki Fujita_, May 16 2015

%H Harry J. Smith, <a href="/A059917/b059917.txt">Table of n, a(n) for n = 0..11</a>

%H A. Granville, <a href="https://arxiv.org/abs/1708.06953">Using Dynamical Systems to Construct Infinitely Many Primes</a>, arXiv:1708.06953 [math.NT], 2017.

%H A. Granville, <a href="https://doi.org/10.1080/00029890.2018.1447732">Using Dynamical Systems to Construct Infinitely Many Primes</a>, The American Mathematical Monthly 125, no. 6 (2018), 483-496. DOI: 10.1080/00029890.2018.1447732

%H N. J. A. Sloane, <a href="https://www.youtube.com/watch?v=3RAYoaKMckM">A Nasty Surprise in a Sequence and Other OEIS Stories</a>, Experimental Mathematics Seminar, Rutgers University, Oct 10 2024, Youtube video; <a href="https://sites.math.rutgers.edu/~zeilberg/expmath/sloane85BD.pdf">Slides</a> [Mentions this sequence]

%F a(n) = a(n-1)*(3^(2^(n-1)) + 1) - 3^(2^(n-1)) = A059723(n+1)/A059723(n) = A059918(n) + 1 = a(n-1)*A059919(n-1) - A011764(n-1).

%F a(0) = 2; a(n) = ((2*a(n-1) - 1)^2 + 1)/2, n >= 1. - _Daniel Forgues_, Jun 22 2011

%e a(2) = Average(1,3,9,27,81,243,729,2187)/Average(1,3,9,27) = 410/10 = 41.

%p seq((3^(2^n)+1)/2,n=0..11); # _Muniru A Asiru_, Aug 07 2018

%t Table[(3^(2^n) + 1)/2, {n, 0, 10}] (* _Vincenzo Librandi_, May 16 2015 *)

%o (PARI) { for (n=0, 11, write("b059917.txt", n, " ", (3^(2^n) + 1)/2); ) } \\ _Harry J. Smith_, Jun 30 2009

%o (Magma) [(3^(2^n)+1)/2: n in [0..10]]; // _Vincenzo Librandi_, May 16 2015

%o (GAP) List([0..10],n->(3^(2^n)+1)/2); # _Muniru A Asiru_, Aug 07 2018

%Y Cf. A059918, A059919. Primes are in A093625.

%K nonn,frac

%O 0,1

%A _Henry Bottomley_, Feb 08 2001