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Generalized Fermat numbers: a(n) = 8^(2^n) + 1, n >= 0.
9

%I #30 Oct 03 2022 04:19:24

%S 9,65,4097,16777217,281474976710657,79228162514264337593543950337,

%T 6277101735386680763835789423207666416102355444464034512897

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

%C These numbers are all composite. We rewrite 8^(2^n) + 1 = (2^(2^n))^3 + 1.

%C 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

%H Charles R Greathouse IV, <a href="/A152581/b152581.txt">Table of n, a(n) for n = 0..10</a>

%H Anders Björn and Hans Riesel, <a href="http://www.jstor.org/stable/2584996">Factors of Generalized Fermat Numbers</a>, Mathematics of Computation, Vol. 67, No. 221, Jan., 1998, pp. 441-446.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/GeneralizedFermatNumber.html">Generalized Fermat Number</a>.

%H OEIS Wiki, <a href="/wiki/Generalized_Fermat_numbers">Generalized Fermat numbers</a>.

%F a(0)=9, a(n) = (a(n-1) - 1)^2 + 1, n >= 1.

%F Sum_{n>=0} 2^n/a(n) = 1/7. - _Amiram Eldar_, Oct 03 2022

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

%t Table[8^2^n + 1, {n, 0, 6}] (* _Arkadiusz Wesolowski_, Nov 02 2012 *)

%o (PARI) g(a,n) = if(a%2,b=2,b=1);for(x=0,n,y=a^(2^x)+b;print1(y","))

%o (Magma) [8^(2^n) + 1: n in [0..8]]; // _Vincenzo Librandi_, Jun 20 2011

%o (PARI) a(n)=1<<(3*2^n)+1 \\ _Charles R Greathouse IV_, Jul 29 2011

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

%Y Cf. A059919, A199591, A078303, A078304, A080176, A199592, A152585.

%K nonn,easy

%O 0,1

%A _Cino Hilliard_, Dec 08 2008

%E Edited by _Daniel Forgues_, Jun 19 2011