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 14 results. Next

A000215 Fermat numbers: a(n) = 2^(2^n) + 1.

Original entry on oeis.org

3, 5, 17, 257, 65537, 4294967297, 18446744073709551617, 340282366920938463463374607431768211457, 115792089237316195423570985008687907853269984665640564039457584007913129639937
Offset: 0

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Author

Keywords

Comments

It is conjectured that just the first 5 numbers in this sequence are primes.
An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
For n>0, Fermat numbers F(n) have digital roots 5 or 8 depending on whether n is even or odd (Koshy). - Lekraj Beedassy, Mar 17 2005
This is the special case k=2 of sequences with exact mutual k-residues. In general, a(1)=k+1 and a(n)=min{m | m>a(n-1), mod(m,a(i))=k, i=1,...,n-1}. k=1 gives Sylvester's sequence A000058. - Seppo Mustonen, Sep 04 2005
For n>1 final two digits of a(n) are periodically repeated with period 4: {17, 57, 37, 97}. - Alexander Adamchuk, Apr 07 2007
For 1 < k <= 2^n, a(A007814(k-1)) divides a(n) + 2^k. More generally, for any number k, let r = k mod 2^n and suppose r != 1, then a(A007814(r-1)) divides a(n) + 2^k. - T. D. Noe, Jul 12 2007
From Daniel Forgues, Jun 20 2011: (Start)
The Fermat numbers F_n are F_n(a,b) = a^(2^n) + b^(2^n) with a = 2 and b = 1.
For n >= 2, all factors of F_n = 2^(2^n) + 1 are of the form k*(2^(n+2)) + 1 (k >= 1).
The products of distinct Fermat numbers (in their binary representation, see A080176) give rows of Sierpiński's triangle (A006943). (End)
Let F(n) be a Fermat number. For n > 2, F(n) is prime if and only if 5^((F(n)-1)/4) == sqrt(F(n)-1) (mod F(n)). - Arkadiusz Wesolowski, Jul 16 2011
Conjecture: let the smallest prime factor of Fermat number F(n) be P(F(n)). If F(n) is composite, then P(F(n)) < 3*2^(2^n/2 - n - 2). - Arkadiusz Wesolowski, Aug 10 2012
The Fermat primes are not Brazilian numbers, so they belong to A220627, but the Fermat composites are Brazilian numbers so they belong to A220571. For a proof, see Proposition 3 page 36 on "Les nombres brésiliens" in Links. - Bernard Schott, Dec 29 2012
It appears that this sequence is generated by starting with a(0)=3 and following the rule "Write in binary and read in base 4". For an example of "Write in binary and read in ternary", see A014118. - John W. Layman, Jul 30 2013
Conjecture: the numbers > 5 in this sequence, i.e., 2^2^k + 1 for k>1, are exactly the numbers n such that (n-1)^4-1 divides 2^(n-1)-1. - M. F. Hasler, Jul 24 2015

Examples

			a(0) = 1*2^1 + 1 = 3 = 1*(2*1) + 1.
a(1) = 1*2^2 + 1 = 5 = 1*(2*2) + 1.
a(2) = 1*2^4 + 1 = 17 = 2*(2*4) + 1.
a(3) = 1*2^8 + 1 = 257 = 16*(2*8) + 1.
a(4) = 1*2^16 + 1 = 65537 = 2048*(2*16) + 1.
a(5) = 1*2^32 + 1 = 4294967297 = 641*6700417 = (10*(2*32) + 1)*(104694*(2*32) + 1).
a(6) = 1*2^64 + 1 = 18446744073709551617 = 274177*67280421310721 = (2142*(2*64) + 1)*(525628291490*(2*64) + 1).
		

References

  • M. Aigner and G. M. Ziegler, Proofs from The Book, Springer-Verlag, Berlin, 2nd. ed., 2001; see p. 3.
  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 7.
  • P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 87.
  • James Gleick, Faster, Vintage Books, NY, 2000 (see pp. 259-261).
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.2 Prime Numbers, pp. 78-79.
  • R. K. Guy, Unsolved Problems in Number Theory, A3.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 14.
  • E. Hille, Analytic Function Theory, Vol. I, Chelsea, N.Y., see p. 64.
  • T. Koshy, "The Digital Root Of A Fermat Number", Journal of Recreational Mathematics Vol. 32 No. 2 2002-3 Baywood NY.
  • M. Krizek, F. Luca & L. Somer, 17 Lectures on Fermat Numbers, Springer-Verlag NY 2001.
  • C. S. Ogilvy and J. T. Anderson, Excursions in Number Theory, Oxford University Press, NY, 1966, p. 36.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see pp. 18, 59.
  • C. A. Pickover, The Math Book, Sterling, NY, 2009; see p. 202.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 6-7, 70-75.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 136-137.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 148-149.

Crossrefs

a(n) = A001146(n) + 1 = A051179(n) + 2.
See A004249 for a similar sequence.
Cf. A080176 for binary representation of Fermat numbers.

Programs

  • Haskell
    a000215 = (+ 1) . (2 ^) . (2 ^)  -- Reinhard Zumkeller, Feb 13 2015
    
  • Maple
    A000215 := n->2^(2^n)+1;
  • Mathematica
    Table[2^(2^n) + 1, {n, 0, 8}] (* Alonso del Arte, Jun 07 2011 *)
  • Maxima
    A000215(n):=2^(2^n)+1$ makelist(A000215(n),n,0,10); /* Martin Ettl, Dec 10 2012 */
    
  • PARI
    a(n)=if(n<1,3*(n==0),(a(n-1)-1)^2+1)
    
  • Python
    def a(n): return 2**(2**n) + 1
    print([a(n) for n in range(9)]) # Michael S. Branicky, Apr 19 2021

Formula

a(0) = 3; a(n) = (a(n-1)-1)^2 + 1, n >= 1.
a(n) = a(n-1)*a(n-2)*...*a(1)*a(0) + 2, n >= 0, where for n = 0, we get the empty product, i.e., 1, plus 2, giving 3 = a(0). - Benoit Cloitre, Sep 15 2002 [edited by Daniel Forgues, Jun 20 2011]
The above formula implies that the Fermat numbers (being all odd) are coprime.
Conjecture: F is a Fermat prime if and only if phi(F-2) = (F-1)/2. - Benoit Cloitre, Sep 15 2002
A000120(a(n)) = 2. - Reinhard Zumkeller, Aug 07 2010
If a(n) is composite, then a(n) = A242619(n)^2 + A242620(n)^2 = A257916(n)^2 - A257917(n)^2. - Arkadiusz Wesolowski, May 13 2015
Sum_{n>=0} 1/a(n) = A051158. - Amiram Eldar, Oct 27 2020
From Amiram Eldar, Jan 28 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = A249119.
Product_{n>=0} (1 - 1/a(n)) = 1/2. (End)
a(n) = 2*A077585(n) + 3. - César Aguilera, Jul 26 2023
a(n) = 2*2^A000225(n) + 1. - César Aguilera, Jul 11 2024

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

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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

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

Original entry on oeis.org

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

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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

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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

A100026 Consider all (2n+1)-digit palindromic primes of the form 10...0M0...01 (so that M is a palindrome with <= 2n-1 digits); a(n) = smallest such M.

Original entry on oeis.org

0, 3, 3, 3, 5, 8, 323, 5, 8, 212, 3, 161, 8, 3, 242, 3, 8, 10901, 737, 161, 242, 333, 282, 6, 252, 474, 5, 12921, 8, 131, 18381, 6, 444, 6, 797, 606, 717, 15351, 464, 333, 626, 545, 13031, 161, 747, 191, 323, 636, 32523, 303, 282, 888, 686, 18981, 111, 15951, 12021
Offset: 1

Views

Author

Harvey Dubner (harvey(AT)dubner.com), Nov 20 2004

Keywords

Comments

Is this the same as "Longest palindromic proper substring of A100027(n) or A028989(n+1) that occurs only once in the decimal representation of A100027(n) or A028989(n+1), respectively"? - Felix Fröhlich, Apr 30 2022
A more formal definition may be a(n) = A004151(A028989(n+1) - 10^(2n) - 1) with the convention that A004151(0) = 0. Only in the unlikely situation that A080176 contains undiscovered primes will a(n) = 0 occur for n > 1. - Jeppe Stig Nielsen, Apr 04 2025

Crossrefs

The corresponding palindromic primes are shown in A100027.

Programs

  • Mathematica
    f[n_] := Block[{k = 0, t = Flatten[Join[{1}, Table[0, {n - 1}]]]}, While[s = Drop[t, Min[ -Floor[ Log[10, k]/2], 0]]; k != FromDigits[ Reverse[ IntegerDigits[k]]] || !PrimeQ[ FromDigits[ Join[s, IntegerDigits[k], Reverse[s]]]], k++ ]; k]; Table[ f[n], {n, 56}] (* Robert G. Wilson v, Nov 22 2004 *)

Extensions

More terms from Robert G. Wilson v, Nov 22 2004

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

A268659 Numbers n such that 3*2^n + 1 is a prime factor of a generalized Fermat number 10^(2^m) + 1 for some m.

Original entry on oeis.org

209, 44685, 157169, 303093, 362765, 916773, 2145353
Offset: 1

Views

Author

Arkadiusz Wesolowski, Feb 10 2016

Keywords

References

  • Wilfrid Keller, private communication, 2008.

Crossrefs

Showing 1-10 of 14 results. Next