A059918 -id:A059918 - 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

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

Original entry on oeis.org

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

Offset: 0

Henry Bottomley, Feb 08 2001

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

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

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]

Cf. A059918, A059919. Primes are in A093625.

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

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

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

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

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

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

A136308 a(n) = (10^2^n - 1)/9.

Original entry on oeis.org

1, 11, 1111, 11111111, 1111111111111111, 11111111111111111111111111111111, 1111111111111111111111111111111111111111111111111111111111111111

Offset: 0

Ctibor O. Zizka, Mar 22 2008

More generally, reading in base B >= 2: a(n) = (B^2^n - 1)/(B-1).
Recurrence: a(n) = a(n-1)*(B^K + 1) and a(0)=1 where K = floor(log_B a(n-1)) + 1.
B = 2 gives A051179; B = 3 gives A059918.

Cf. A000042 (repunits).
In other bases: A051179, A059918.
Cf. A000079, A007088, A007089, A007953.

A136308 := func; [A136308(n):n in[0..7]];

(10^2^Range[0, 10] - 1)/9 (* G. C. Greubel, Apr 19 2021 *)

[(10^2^n -1)/9 for n in (0..10)] # G. C. Greubel, Apr 19 2021

a(n) = a(n-1)*(10^K + 1) and a(0)=1 where K=floor(log_10 a(n-1)) + 1 = 2^n + 1.
a(n) = A000042(A000079(n)) = A007088(A051179(n)) = A007089(A059918(n)).
A007953(a(n)) = 2^n. - Stefano Spezia, Mar 27 2025

Edited by Jason Kimberley, Dec 18 2012

cp's OEIS Frontend

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

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

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A136308 a(n) = (10^2^n - 1)/9.

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