A059723 -id:A059723 - 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

A059722 a(n) = n(2n^2 - 2*n + 1).

Original entry on oeis.org

0, 1, 10, 39, 100, 205, 366, 595, 904, 1305, 1810, 2431, 3180, 4069, 5110, 6315, 7696, 9265, 11034, 13015, 15220, 17661, 20350, 23299, 26520, 30025, 33826, 37935, 42364, 47125, 52230, 57691, 63520, 69729, 76330, 83335, 90756, 98605, 106894, 115635, 124840

Offset: 0

Henry Bottomley, Feb 07 2001

Mean of the first four nonnegative powers of 2n+1, i.e., ((2n+1)^0 + (2n+1)^1 + (2n+1)^2 + (2n+1)^3)/4. E.g., a(2) = (1 + 3 + 9 + 27)/4 = 10.
Equatorial structured meta-diamond numbers, the n-th number from an equatorial structured n-gonal diamond number sequence. There are no 1- or 2-gonal diamonds, so 1 and (n+2) are used as the first and second terms since all the sequences begin as such. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Starting with offset 1 = row sums of triangle A143803. - Gary W. Adamson, Sep 01 2008
Form an array from the antidiagonals containing the terms in A002061 to give antidiagonals 1; 3,3; 7,4,7; 13,8,8,13; 21,14,9,14,21; and so on. The difference between the sum of the terms in n+1 X n+1 matrices and those in n X n matrices is a(n) for n>0. - J. M. Bergot, Jul 08 2013
Sum of the numbers from (n-1)^2 to n^2. - Wesley Ivan Hurt, Sep 08 2014

Harry J. Smith, Table of n, a(n) for n = 0..1000
Bruno Berselli, Table of sequences with the formula m*(m+1)^2 - (k+2)*m^2 (table includes this sequence for k = 2-m, m >= 0).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000330, A005900, A081436, A100178, A100179, A059722: "equatorial" structured diamonds; A000447: "polar" structured meta-diamond; A006484 for other structured meta numbers; and A100145 for more on structured numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004

Cf. A000217, A000290, A001844, A002061, A002414, A053698, A059721, A059723, A136392, A143803.

[n*(2*n^2 - 2*n + 1) : n in [0..50]]; // Wesley Ivan Hurt, Sep 08 2014

A059722:=n->n*(2*n^2 - 2*n + 1): seq(A059722(n), n=0..50); # Wesley Ivan Hurt, Sep 08 2014

Table[n (2 n^2 - 2 n + 1), {n, 0, 50}] (* Wesley Ivan Hurt, Sep 08 2014 *)

a(n) = { n*(2*n^2 - 2*n + 1) } \\ Harry J. Smith, Jun 28 2009

a(n) = A053698(2*n-1)/4.
a(n) = Sum_{j=1..n} ((n+j-1)^2-j^2+1). - Zerinvary Lajos, Sep 13 2006
From R. J. Mathar, Sep 02 2008: (Start)
G.f.: x*(1 + x)*(1 + 5*x)/(1 - x)^4.
a(n) = A002414(n-1) + A002414(n).
a(n+1) - a(n) = A136392(n+1). (End)
a(n) = (A000290(n) + A000290(n+1)) * (A000217(n+1) - A000217(n)). - J. M. Bergot, Nov 02 2012
a(n) = n * A001844(n-1). - Doug Bell, Aug 18 2015
a(n) = A000217(n^2) - A000217(n^2-2*n). - Bruno Berselli, Jun 26 2018
E.g.f.: exp(x)*x*(1 + 4*x + 2*x^2). - Stefano Spezia, Jun 20 2021

Edited with new definition by N. J. A. Sloane, Aug 29 2008

A059721 Mean of first six positive powers of n, i.e., (n + n^2 + n^3 + n^4 + n^5 + n^6)/6.

Original entry on oeis.org

0, 1, 21, 182, 910, 3255, 9331, 22876, 49932, 99645, 185185, 324786, 542906, 871507, 1351455, 2034040, 2982616, 4274361, 6002157, 8276590, 11228070, 15009071, 19796491, 25794132, 33235300, 42385525, 53545401, 67053546, 83289682, 102677835, 125689655, 152847856

Offset: 0

Henry Bottomley, Feb 07 2001

			a(2) = (2 + 4 + 8 + 16 + 32 + 64)/6 = 126/6 = 21.

Harry J. Smith, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A059722, A059723.

Table[Total[n^Range[6]]/6,{n,0,30}]  (* Harvey P. Dale, Jan 29 2011 *)

a(n) = { if(n==1, 1, (n^7 - n)/(6*n - 6)) } \\ Harry J. Smith, Jun 28 2009

a(n) = (n^7 - n)/(6n - 6) = A053700(n)*n/6.

G.f.: x*(1+14*x+56*x^2+42*x^3+7*x^4)/(1-x)^7. [Colin Barker, May 08 2012]

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

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

A261066 a(n) = (7^(2^n) - 1) / 2^(n+2).

Original entry on oeis.org

6, 150, 180150, 519264540150, 8628341205030630049260150, 4764689404827483203666304150636608674826622242700150

Offset: 1

Marco Ripà, Aug 08 2015

nonn

(m^(2^n)-1)/2^(n+2) is an integer for any odd value of m and n>0.

The next term, a(7), has 106 decimal digits.

			a(3) = (7^8 - 1)/2^5 = 180150.

Cf. A059723, A068531, A097421. Twice A359500.

[(7^(2^n)-1)/2^(n+2): n in [1..10]]; // Vincenzo Librandi, Aug 09 2015

Table[(7^(2^n) - 1)/2^(n + 2), {n, 1, 7}] (* Vincenzo Librandi, Aug 09 2015 *)

first(m)=vector(m,i, (7^(2^i)-1)/2^(i+2)); /* Anders Hellström, Aug 08 2015 */

A261066(n)=7^2^n>>(n+2) \\ M. F. Hasler, Aug 11 2015

a(n) == 150 (mod 10^4) for all n > 1. - M. F. Hasler, Aug 11 2015

a(5) corrected by Vincenzo Librandi, Aug 09 2015

A261067 a(n) = (11^(2^n) - 1)/2^(n + 2).

Original entry on oeis.org

15, 915, 6698715, 717964529118315, 16495138082306681918325119173515, 17413733142679306233865281770975943513633443105435651232476307915

Offset: 1

Marco Ripà, Aug 08 2015

(m^(2^n) - 1)/2^(n + 2) is an integer for any odd value of m and n > 0.

In particular, for m = 11, a(n) is a multiple of 15.

			a(3) = (11^8 - 1)/2^5 = 6698715.

Cf. A059723, A068531, A068533, A097421.

[(11^(2^n)-1)/2^(n+2): n in [1..6]]; // Vincenzo Librandi, Aug 14 2015

A261067:=n->(11^(2^n) - 1)/2^(n + 2): seq(A261067(n), n=1..8); # Wesley Ivan Hurt, Jan 27 2017

Table[(11^(2^n) - 1)/2^(n + 2), {n, 10}] (* Vincenzo Librandi, Aug 14 2015 *)

first(m)=vector(m,i, (11^(2^i)-1)/2^(i+2)); /* Anders Hellström, Aug 08 2015 */

a(n) = 15*A068533(n). - Michel Marcus, Aug 14 2015

cp's OEIS Frontend

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

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A059722 a(n) = n*(2*n^2 - 2*n + 1).

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A059721 Mean of first six positive powers of n, i.e., (n + n^2 + n^3 + n^4 + n^5 + n^6)/6.

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

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

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A261066 a(n) = (7^(2^n) - 1) / 2^(n+2).

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A059722 a(n) = n(2n^2 - 2*n + 1).