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 12 results. Next

A178428 5 followed by the generalized Fermat numbers 6^(2^n)+1 (A078303).

Original entry on oeis.org

5, 7, 37, 1297, 1679617, 2821109907457, 7958661109946400884391937, 63340286662973277706162286946811886609896461828097
Offset: 0

Views

Author

Roger L. Bagula, May 27 2010

Keywords

Comments

If a(0) = 3, the recursion formula gives the Fermat numbers (A000215).
With a(0) = 3 instead of 5; a(n) = 2 + product_{i=0..n-1} a(i), n >= 1.
The recurrence equation for generalized Fermat numbers F_n(a) = a^(2^n)+1,
a >= 2, n >= 0, is F_{n}(a) = (F_{n-1}(a)-1)^2 + 1. - Daniel Forgues, Jun 22 2011

Crossrefs

Programs

  • Mathematica
    a[0] := 5;
    a[n_] := a[n] = Product[a[i], {i, 0, n - 1}] + 2;
    Table[a[n], {n, 0, 10}]

Formula

a(0) = 5; a(n) = 2 + product_{i=0..n-1} a(i), n >= 1.
From Daniel Forgues, Jun 22 2011: (Start)
The motivation for this sequence comes from the recurrence for generalized Fermat numbers 6^(2^n)+1 (A078303)
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. (End)

Extensions

Definition simplified by the Assoc. Eds. of the OEIS - May 28 2010
Edited by Daniel Forgues, Jun 22 2011

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

Original entry on oeis.org

4, 10, 82, 6562, 43046722, 1853020188851842, 3433683820292512484657849089282, 11790184577738583171520872861412518665678211592275841109096962
Offset: 0

Views

Author

Henry Bottomley, Feb 08 2001

Keywords

Comments

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

Examples

			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;
		

Crossrefs

Cf. A000215 (Fermat numbers: 2^(2^n) + 1, n >= 0).
Cf. A059917 ((3^(2^n)+1)/2).

Programs

Formula

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

Extensions

Edited by Daniel Forgues, Jun 19 2011 and Jun 20 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

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

A268663 Numbers n such that 5*2^n + 1 is a prime factor of a generalized Fermat number 6^(2^m) + 1 for some m.

Original entry on oeis.org

127, 4687, 1777515
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

A273947 Prime factors of generalized Fermat numbers of the form 6^(2^m) + 1 with m >= 0.

Original entry on oeis.org

7, 17, 37, 257, 353, 1297, 1697, 2753, 18433, 65537, 80897, 98801, 145601, 763649, 3360769, 4709377, 13631489, 50307329, 376037377, 2483027969, 3191106049, 4926056449, 51808043009, 152605556737, 916326983681, 1268357529601, 6597069766657, 40711978221569
Offset: 1

Views

Author

Arkadiusz Wesolowski, Jun 05 2016

Keywords

Comments

Primes p other than 5 such that the multiplicative order of 6 (mod p) is a power of 2.

References

  • Hans Riesel, Common prime factors of the numbers A_n=a^(2^n)+1, BIT 9 (1969), pp. 264-269.

Crossrefs

Cf. A023394, A072982, A078303, A268663, A273945 (base 3), A273946 (base 5), A273948 (base 7), A273949 (base 11), A273950 (base 12).

Programs

  • Mathematica
    Select[Prime@Range[4, 10^5], IntegerQ@Log[2, MultiplicativeOrder[6, #]] &]
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