A059917 -id:A059917 - cp's OEIS frontend

A059919 Generalized Fermat numbers: 3^(2^n)+1, n >= 0.

4, 10, 82, 6562, 43046722, 1853020188851842, 3433683820292512484657849089282, 11790184577738583171520872861412518665678211592275841109096962

Offset: 0

Henry Bottomley, Feb 08 2001

Generalized Fermat numbers (Ribenboim (1996))

F_n(a) := F_n(a,1) = a^(2^n) + 1, a >= 2, n >= 0, can't be prime if a is odd (as is the case for this sequence). - Daniel Forgues, Jun 19-20 2011

			a(0) = 3^(2^0)+1 = 3^1+1 = 4 = 2*(1)+2 = 2*(empty product)+2;
a(1) = 3^(2^1)+1 = 3^2+1 = 10 = 2*(4)+2;
a(2) = 3^(2^2)+1 = 3^4+1 = 82 = 2*(4*10)+2;
a(3) = 3^(2^3)+1 = 3^8+1 = 6562 = 2*(4*10*82)+2;
a(4) = 3^(2^4)+1 = 3^16+1 = 43046722 = 2*(4*10*82*6562)+2;
a(5) = 3^(2^5)+1 = 3^32+1 = 1853020188851842 = 2*(4*10*82*6562*43046722)+2;

Vincenzo Librandi, Table of n, a(n) for n = 0..10 (shortened by _N. J. A. Sloane_, Jan 13 2019)
Anders Björn and Hans Riesel, Factors of Generalized Fermat Numbers, Mathematics of Computation, Vol. 67, No. 221, Jan., 1998, pp. 441-446.
C. K. Caldwell, "Top Twenty" page, Generalized Fermat Divisors (base=3).
Wilfrid Keller, GFN3 factoring status.
Eric Weisstein's World of Mathematics, Generalized Fermat Number.
OEIS Wiki, Generalized Fermat numbers.

Cf. A000215 (Fermat numbers: 2^(2^n) + 1, n >= 0).
Cf. A059917 ((3^(2^n)+1)/2).
Cf. A199591, A078303, A078304, A152581, A080176, A199592, A152585.

[3^(2^n) + 1: n in [0..8]]; // Vincenzo Librandi, Jun 20 2011

A059919:=n->3^(2^n)+1; seq(A059919(n), n=0..7); # Wesley Ivan Hurt, Jan 22 2014

Table[3^2^n + 1, {n, 0, 7}] (* Arkadiusz Wesolowski, Nov 02 2012 *)

a(n) = { 3^(2^n) + 1 } \\ Harry J. Smith, Jun 30 2009

a(0) = 4; a(n) = (a(n-1)-1)^2 + 1, n >= 1.
a(n) = A011764(n)+1 = A059918(n+1)/A059918(n) = (A059917(n+1)-1)/(A059917(n)-1) = (A059723(n)/A059723(n+1))*(A059723(n+2)-A059723(n+1))/(A059723(n+1)-A059723(n))
a(n) = A057727(n)-1. - R. J. Mathar, Apr 23 2007
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).
The above formula implies the GCD of any pair of terms is 2, which means that the terms of (3^(2^n)+1)/2 (A059917) are pairwise coprime. - Daniel Forgues, Jun 20 & 22 2011
Sum_{n>=0} 2^n/a(n) = 1/2. - Amiram Eldar, Oct 03 2022

Edited by Daniel Forgues, Jun 19 2011 and Jun 20 2011

A093625 Primes of the form (3^m + 1)/2.

Original entry on oeis.org

2, 5, 41, 21523361, 926510094425921, 1716841910146256242328924544641

Offset: 1

Hugo Pfoertner, Apr 19 2004

Primes can only occur for m=2^k. The next term (if it exists) has k >= 21 (i.e., more than 10^6 decimal digits). - Don Reble, Apr 28 2004. [For the latest results about the search, see A171381. - N. J. A. Sloane, Jun 14 2010]

a(7) >= (3^(2^24) + 1)/2, if it exists. - Tyler Busby, Mar 22 2023

			a(3)=41 because it can be written as (3^4 + 1)/2 = 82/2.

A171381 is the main entry for this sequence. Cf. also A059917, A138083.

Erroneous terms deleted by N. J. A. Sloane, Jun 14 2010

A059918 a(n) = (3^(2^n)-1)/2.

Original entry on oeis.org

1, 4, 40, 3280, 21523360, 926510094425920, 1716841910146256242328924544640, 5895092288869291585760436430706259332839105796137920554548480

Offset: 0

Henry Bottomley, Feb 08 2001

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

Harry J. Smith, Table of n, a(n) for n = 0..11

Cf. A059917 (numerators).
Cf. A059723, A059917, A059919, A011764.
Cf. A007089, A136308.

Array[(3^(2^#) - 1)/2 &, 8, 0] (* Michael De Vlieger, Feb 05 2022 *)

{ for (n=0, 11, write("b059918.txt", n, " ", (3^(2^n) - 1)/2); ) } \\ Harry J. Smith, Jun 30 2009

a(n) = a(n-1)*(3^(2^(n-1))+1) with a(0) = 1.
a(n) = (3^(2^n)-1)/2 = (A059723(n+1)-A059723(n))/A059723(n) = A059917(n)-1 = a(n-1)*A059919(n-1) = a(n-1)*(A011764(n-1)+1)
1 = Sum_{n>=0} 3^(2^n)/a(n+1). 1 = 3/4 + 9/40 + 81/3280 + 6561/21523360 + ...; with partial sums: 3/4, 39/40, 3279/3280, 21523359/21523360, ..., (a(n)-1)/a(n), ... . - Gary W. Adamson, Jun 22 2003
A136308(n) = A007089(a(n)). - Jason Kimberley, Dec 19 2012

A145008 Reduced numerators of the convergents to 2 = sqrt(4) using the recursion x -> (4/x + x)/2.

Original entry on oeis.org

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

Offset: 1

Cino Hilliard, Sep 28 2008

The recursion x -> (n/x + x)/2 converges to a square root of n.
These are the numerators of the first order Newton method to solve x^2-4=f(x)=0, starting at x=1 as the initial estimate: x -> x-f(x)/f'(x), where f'(x)=2x is the first derivative. - R. J. Mathar, Oct 07 2008
The equivalent sequence for n=9 starting from x=1 is 5, 17, 257,.., apparently A000215. - R. J. Mathar, Oct 14 2008

			(4/1+1)/2 = 5/2 = 2.5
(4/5/2+5/2)/2 = 41/20 = 2.05
(4/(41/20)+41/20)/2 = 3281/1640 = 2.000609...

Wikipedia, Newton's method.

Cf. A059917.

g(n,p) = x=1;for(j=1,p,x=(n/x+x)/2;print1(numerator(x)","))
g(4,8)

Divided the right hand side of formula in the first comment by 2. - R. J. Mathar, Oct 14 2008

cp's OEIS Frontend

A059919 Generalized Fermat numbers: 3^(2^n)+1, n >= 0.

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A093625 Primes of the form (3^m + 1)/2.

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A059918 a(n) = (3^(2^n)-1)/2.

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A145008 Reduced numerators of the convergents to 2 = sqrt(4) using the recursion x -> (4/x + x)/2.

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