A059917 a(n) = (3^(2^n) + 1)/2 = A059919(n)/2, n >= 0.

2, 5, 41, 3281, 21523361, 926510094425921, 1716841910146256242328924544641, 5895092288869291585760436430706259332839105796137920554548481

Offset: 0

Author

Henry Bottomley, Feb 08 2001

Keywords

• nonn
• frac

Comments

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

Numerator of b(n) where b(n) = (1/2)*(b(n-1) + 1/b(n-1)), b(0)=2. - Vladeta Jovovic, Aug 15 2002

From Daniel Forgues, Jun 22 2011: (Start)

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.

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

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

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

Examples

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

Links

• Harry J. Smith, Table of n, a(n) for n = 0..11
• A. Granville, Using Dynamical Systems to Construct Infinitely Many Primes, arXiv:1708.06953 [math.NT], 2017.
• A. Granville, Using Dynamical Systems to Construct Infinitely Many Primes, The American Mathematical Monthly 125, no. 6 (2018), 483-496. DOI: 10.1080/00029890.2018.1447732
• N. J. A. Sloane, A Nasty Surprise in a Sequence and Other OEIS Stories, Experimental Mathematics Seminar, Rutgers University, Oct 10 2024, Youtube video; Slides [Mentions this sequence]

Crossrefs

Cf. A059918, A059919. Primes are in A093625.

Programs

GAP

List([0..10],n->(3^(2^n)+1)/2); # Muniru A Asiru, Aug 07 2018

Magma

[(3^(2^n)+1)/2: n in [0..10]]; // Vincenzo Librandi, May 16 2015

Maple

seq((3^(2^n)+1)/2,n=0..11); # Muniru A Asiru, Aug 07 2018

Mathematica

Table[(3^(2^n) + 1)/2, {n, 0, 10}] (* Vincenzo Librandi, May 16 2015 *)

PARI

{ for (n=0, 11, write("b059917.txt", n, " ", (3^(2^n) + 1)/2); ) } \\ Harry J. Smith, Jun 30 2009

Formula

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).

a(0) = 2; a(n) = ((2*a(n-1) - 1)^2 + 1)/2, n >= 1. - Daniel Forgues, Jun 22 2011