cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 15 results. Next

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

Views

Author

Henry Bottomley, Feb 08 2001

Keywords

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.
		

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

A080176 Generalized Fermat numbers: 10^(2^n) + 1, n >= 0.

Original entry on oeis.org

11, 101, 10001, 100000001, 10000000000000001, 100000000000000000000000000000001, 10000000000000000000000000000000000000000000000000000000000000001
Offset: 0

Views

Author

Jens Voß, Feb 04 2003

Keywords

Comments

As for standard Fermat numbers 2^(2^n) + 1, a number (2b)^m + 1 (with b > 1) can only be prime if m is a power of 2. On the other hand, out of the first 12 base-10 Fermat numbers, only the first two are primes.
Also, binary representation of Fermat numbers (in decimal, see A000215).

Examples

			a(0) = 10^1 + 1 = 11 = 9*(1) + 2 = 9*(empty product) + 2.
a(1) = 10^2 + 1 = 101 = 9*(11) + 2.
a(2) = 10^4 + 1 = 10001 = 9*(11*101) + 2.
a(3) = 10^8 + 1 = 100000001 = 9*(11*101*10001) + 2.
a(4) = 10^16 + 1 = 10000000000000001 = 9*(11*101*10001*100000001) + 2.
a(5) = 10^32 + 1 = 100000000000000000000000000000001 = 9*(11*101*10001*100000001*10000000000000001) + 2.
		

Crossrefs

Cf. A000215 (Fermat numbers: 2^(2^n) + 1, n >= 0).

Programs

Formula

a(0) = 11; a(n) = (a(n - 1) - 1)^2 + 1.
a(n) = 9*a(n-1)*a(n-2)*...*a(1)*a(0) + 2, n >= 0, where for n = 0, we get 9*(empty product, i.e., 1)+ 2 = 11 = a(0). - Daniel Forgues, Jun 20 2011
Sum_{n>=0} 2^n/a(n) = 1/9. - Amiram Eldar, Oct 03 2022

Extensions

Edited by Daniel Forgues, Jun 19 2011

A078303 Generalized Fermat numbers: 6^(2^n) + 1, n >= 0.

Original entry on oeis.org

7, 37, 1297, 1679617, 2821109907457, 7958661109946400884391937, 63340286662973277706162286946811886609896461828097
Offset: 0

Views

Author

Eric W. Weisstein, Nov 21 2002

Keywords

Comments

The next term is too large to include.
As for standard Fermat numbers 2^(2^n) + 1, a number (2b)^m + 1 (with b > 1) can only be prime if m is a power of 2. On the other hand, out of the first 13 base-6 Fermat numbers, only the first three are primes.
Either the sequence of (standard) Fermat numbers contains infinitely many composite numbers or the sequence of base-6 Fermat numbers contains infinitely many composite numbers (cf. https://mathoverflow.net/a/404235/1593). - José Hernández, Nov 09 2021
Since all powers of 6 are congruent to 6 (mod 10), all terms of this sequence are congruent to 7 (mod 10). - Daniel Forgues, Jun 22 2011
There are only 5 known Fermat primes of the form 2^(2^n) + 1: {3, 5, 17, 257, 65537}. There are only 2 known base-10 generalized Fermat primes of the form 10^(2^n) + 1: {11, 101}. - Alexander Adamchuk, Mar 17 2007

Examples

			a(0) = 6^1+1 = 7 = 5*(1)+2 = 5*(empty product)+2;
a(1) = 6^2+1 = 37 = 5*(7)+2;
a(2) = 6^4+1 = 1297 = 5*(7*37)+2;
a(3) = 6^8+1 = 1679617 = 5*(7*37*1297)+2;
a(4) = 6^16+1 = 2821109907457 = 5*(7*37*1297*1679617)+2;
a(5) = 6^32+1 = 7958661109946400884391937 = 5*(7*37*1297*1679617*2821109907457)+2;
		

Crossrefs

Cf. A000215 (Fermat numbers: 2^(2^n) + 1, n >= 0).
Cf. A019434 (Fermat primes of the form 2^(2^n) + 1).

Programs

Formula

a(0) = 7, a(n) = (a(n-1)-1)^2 + 1, n >= 1.
a(n) = 5*a(n-1)*a(n-2)*...*a(1)*a(0) + 2, n >= 0, where for n = 0, we get 5*(empty product, i.e., 1)+ 2 = 7 = a(0). This implies that the terms are pairwise coprime. - Daniel Forgues, Jun 20 2011
Sum_{n>=0} 2^n/a(n) = 1/5. - Amiram Eldar, Oct 03 2022

Extensions

Edited by Daniel Forgues, Jun 22 2011

A078304 Generalized Fermat numbers: 7^(2^n)+1, n >= 0.

Original entry on oeis.org

8, 50, 2402, 5764802, 33232930569602, 1104427674243920646305299202, 1219760487635835700138573862562971820755615294131238402
Offset: 0

Views

Author

Eric W. Weisstein, Nov 21 2002

Keywords

Comments

From Daniel Forgues, Jun 19 2011: (Start)
Generalized Fermat numbers 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 the current sequence) (Ribenboim (1996)).
All factors of generalized Fermat numbers F_n(a,b) := a^(2^n)+b^(2^n), a >= 2, n >= 0, are of the form k*2^m+1, k >= 1, m >=0 (Riesel (1994, 1998)). (This only expresses that the factors are odd, which means that it only applies to odd generalized Fermat numbers.) (End)

Examples

			a(0) = 7^1+1 = 8 = 6*(1)+2 = 6*(empty product)+2.
a(1) = 7^2+1 = 50 = 6*(8)+2.
a(2) = 7^4+1 = 2402 = 6*(8*50)+2.
a(3) = 7^8+1 = 5764802 = 6*(8*50*2402)+2.
a(4) = 7^16+1 = 33232930569602 = 6*(8*50*2402*5764802)+2.
a(5) = 7^32+1 = 1104427674243920646305299202 = 6*(8*50*2402*5764802*33232930569602)+2.
		

Crossrefs

Cf. A000215 (Fermat numbers: 2^(2^n)+1, n >= 0).

Programs

Formula

a(0) = 8, a(n)=(a(n-1)-1)^2+1, n >= 1.
a(n) = 6*a(n-1)*a(n-2)*...*a(1)*a(0) + 2, n >= 0, where for n = 0, we get 6*(empty product, i.e., 1)+ 2 = 8 = a(0). This means that the GCD of any pair of terms is 2. - Daniel Forgues, Jun 20 2011
Sum_{n>=0} 2^n/a(n) = 1/6. - Amiram Eldar, Oct 03 2022

Extensions

Edited by Daniel Forgues, Jun 19 2011

A152585 Generalized Fermat numbers: 12^(2^n) + 1, n >= 0.

Original entry on oeis.org

13, 145, 20737, 429981697, 184884258895036417, 34182189187166852111368841966125057, 1168422057627266461843148138873451659428421700563161428957815831003137
Offset: 0

Views

Author

Cino Hilliard, Dec 08 2008

Keywords

Comments

There appears to be no divisibility rule for this sequence.
13 is the only prime up to 12^(2^15)+1.

Examples

			a(0) = 12^1+1 = 13 = 11(1)+2 = 11(empty product)+2.
a(1) = 12^2+1 = 145 = 11(13)+2.
a(2) = 12^4+1 = 20737 = 11(13*145)+2.
a(3) = 12^8+1 = 429981697 = 11(13*145*20737)+2.
a(4) = 12^16+1 = 184884258895036417 = 11(13*145*20737*429981697)+2.
a(5) = 12^32+1 = 34182189187166852111368841966125057 = 11(13*145*20737*429981697*184884258895036417)+2.
		

Crossrefs

Cf. A000215 (Fermat numbers: 2^(2^n)+1, n >= 0).

Programs

  • Magma
    [12^(2^n) + 1: n in [0..8]]; // Vincenzo Librandi, Jun 20 2011
    
  • Mathematica
    Table[12^2^n + 1, {n, 0, 6}] (* Arkadiusz Wesolowski, Nov 02 2012 *)
  • PARI
    g(a,n) = if(a%2,b=2,b=1);for(x=0,n,y=a^(2^x)+b;print1(y","))
    
  • Python
    def A152585(n): return (1<<2*(m:=1<Chai Wah Wu, Jul 19 2022

Formula

a(0) = 13; a(n)=(a(n-1)-1)^2 + 1, n >= 1.
a(n) = 11*a(n-1)*a(n-2)*...*a(1)*a(0) + 2, n >= 0, where for n = 0, we get 11*(empty product, i.e., 1)+ 2 = 13 = a(0). This implies that the terms, all odd, are pairwise coprime. - Daniel Forgues, Jun 20 2011
Sum_{n>=0} 2^n/a(n) = 1/11. - Amiram Eldar, Oct 03 2022

Extensions

Edited by Daniel Forgues, Jun 19 2011

A199591 Generalized Fermat numbers: 5^(2^n) + 1, n >= 0.

Original entry on oeis.org

6, 26, 626, 390626, 152587890626, 23283064365386962890626, 542101086242752217003726400434970855712890626
Offset: 0

Views

Author

Arkadiusz Wesolowski, Nov 08 2011

Keywords

Examples

			a(0) = 5^(2^0) + 1 = 5^1 + 1 = 6 = 4*(2^0) + 2;
a(1) = 5^(2^1) + 1 = 5^2 + 1 = 26 = 4*(2^1*3) + 2;
a(2) = 5^(2^2) + 1 = 5^4 + 1 = 626 = 4*(2^2*3*13) + 2;
a(3) = 5^(2^3) + 1 = 5^8 + 1 = 390626 = 4*(2^3*3*13*313) + 2;
a(4) = 5^(2^4) + 1 = 5^16 + 1 = 152587890626 = 4*(2^4*3*13*313*195313) + 2;
a(5) = 5^(2^5) + 1 = 5^32 + 1 = 23283064365386962890626 = 4*(2^5*3*13*313*195313*76293945313) + 2;
		

Crossrefs

Programs

  • Magma
    [5^2^n+1 : n in [0..6]];
    
  • Mathematica
    Table[5^2^n + 1, {n, 0, 6}]
  • PARI
    for(n=0, 6, print1(5^2^n+1, ", "))

Formula

a(0) = 6; a(n) = (a(n-1)-1)^2 + 1, n >= 1.
a(0) = 6, a(1) = 26; a(n) = a(n-1) + 4*5^(2^(n-1))*Product_{i=0..n-2} a(i), n >= 2.
a(0) = 6, a(1) = 26; a(n) = a(n-1)^2 - 2*(a(n-2)-1)^2, n >= 2.
a(0) = 6; a(n) = 4*(Product_{i=0..n-1} a(i)) + 2, n >= 1.
a(n) = A152578(n) - 1.
Sum_{n>=0} 2^n/a(n) = 1/4. - Amiram Eldar, Oct 03 2022

A199592 Generalized Fermat numbers: 11^(2^n) + 1, n >= 0.

Original entry on oeis.org

12, 122, 14642, 214358882, 45949729863572162, 2111377674535255285545615254209922, 4457915684525902395869512133369841539490161434991526715513934826242
Offset: 0

Views

Author

Arkadiusz Wesolowski, Nov 08 2011

Keywords

Examples

			a(0) = 11^(2^0) + 1 = 11^1 + 1 = 12 = 10*(2^0) + 2;
a(1) = 11^(2^1) + 1 = 11^2 + 1 = 122 = 10*(2^1*6) + 2;
a(2) = 11^(2^2) + 1 = 11^4 + 1 = 14642 = 10*(2^2*6*61) + 2;
a(3) = 11^(2^3) + 1 = 11^8 + 1 = 214358882 = 10*(2^3*6*61*7321) + 2;
a(4) = 11^(2^4) + 1 = 11^16 + 1 = 45949729863572162 = 10*(2^4*6*61*7321*107179441) + 2;
a(5) = 11^(2^5) + 1 = 11^32 + 1 = 2111377674535255285545615254209922 = 10*(2^5*6*61*7321*107179441*22974864931786081) + 2;
		

Crossrefs

Programs

  • Magma
    [11^2^n+1 : n in [0..6]]
    
  • Mathematica
    Table[11^2^n + 1, {n, 0, 6}]
  • PARI
    for(n=0, 6, print1(11^2^n+1, ", "))

Formula

a(0) = 12; a(n) = (a(n-1)-1)^2 + 1, n >= 1.
a(0) = 12, a(1) = 122; a(n) = a(n-1) + 10*11^(2^(n-1))*Product_{i=0..n-2} a(i), n >= 2.
a(0) = 12, a(1) = 122; a(n) = a(n-1)^2 - 2*(a(n-2)-1)^2, n >= 2.
a(0) = 12; a(n) = 10*(Product_{i=0..n-1} a(i)) + 2, n >= 1.
a(n) = A152583(n) - 1.
Sum_{n>=0} 2^n/a(n) = 1/10. - Amiram Eldar, Oct 03 2022

A268657 Numbers k such that 3*2^k + 1 is a prime factor of a generalized Fermat number 3^(2^m) + 1 for some m.

Original entry on oeis.org

6, 12, 18, 30, 36, 41, 66, 189, 201, 209, 276, 408, 438, 534, 2208, 3168, 3189, 3912, 34350, 42294, 44685, 48150, 54792, 55182, 59973, 80190, 157169, 213321, 303093, 382449, 709968, 801978, 916773, 1832496, 2145353, 2291610, 2478785, 5082306, 7033641, 10829346
Offset: 1

Views

Author

Arkadiusz Wesolowski, Feb 10 2016

Keywords

References

  • Wilfrid Keller, private communication, 2008.

Crossrefs

Programs

  • PARI
    for(k=1,+oo,p=3*2^k+1;if(ispseudoprime(p),t=znorder(Mod(3,p));bitand(t,t-1)==0&&print1(k,", "))) \\ Jeppe Stig Nielsen, Oct 30 2020

A268661 Numbers n such that 5*2^n + 1 is a prime factor of a generalized Fermat number 3^(2^m) + 1 for some m.

Original entry on oeis.org

3, 55, 127, 13165, 240937, 819739, 1282755
Offset: 1

Views

Author

Arkadiusz Wesolowski, Feb 10 2016

Keywords

References

  • Wilfrid Keller, private communication, 2008.

Crossrefs

A152581 Generalized Fermat numbers: a(n) = 8^(2^n) + 1, n >= 0.

Original entry on oeis.org

9, 65, 4097, 16777217, 281474976710657, 79228162514264337593543950337, 6277101735386680763835789423207666416102355444464034512897
Offset: 0

Views

Author

Cino Hilliard, Dec 08 2008

Keywords

Comments

These numbers are all composite. We rewrite 8^(2^n) + 1 = (2^(2^n))^3 + 1.
Then by the identity a^n + b^n = (a+b)*(a^(n-1) - a^(n-2)*b + ... + b^(n-1)) for odd n, 2^(2^n) + 1 divides 8^(2^n) + 1. All factors of generalized Fermat numbers F_n(a,b) := a^(2^n)+b^(2^n), a >= 2, n >= 0, are of the form k*2^m+1, k >= 1, m >=0 (Riesel (1994)). - Daniel Forgues, Jun 19 2011

Examples

			For n = 3, 8^(2^3) + 1 = 16777217. Similarly, (2^8)^3 + 1 = 16777217. Then 2^8 + 1 = 257 and 16777217/257 = 65281.
		

Crossrefs

Cf. A000215 (Fermat numbers: 2^(2^n) + 1, n >= 0).

Programs

Formula

a(0)=9, a(n) = (a(n-1) - 1)^2 + 1, n >= 1.
Sum_{n>=0} 2^n/a(n) = 1/7. - Amiram Eldar, Oct 03 2022

Extensions

Edited by Daniel Forgues, Jun 19 2011
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