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%I #53 Apr 03 2023 10:36:10
%S 4,10,82,6562,43046722,1853020188851842,
%T 3433683820292512484657849089282,
%U 11790184577738583171520872861412518665678211592275841109096962
%N Generalized Fermat numbers: 3^(2^n)+1, n >= 0.
%C Generalized Fermat numbers (Ribenboim (1996))
%C 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
%H Vincenzo Librandi, <a href="/A059919/b059919.txt">Table of n, a(n) for n = 0..10</a> (shortened by _N. J. A. Sloane_, Jan 13 2019)
%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 C. K. Caldwell, "Top Twenty" page, <a href="https://t5k.org/top20/page.php?id=28">Generalized Fermat Divisors (base=3)</a>.
%H Wilfrid Keller, <a href="http://www1.uni-hamburg.de/RRZ/W.Keller/GFN03.html">GFN3 factoring status</a>.
%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) = 4; a(n) = (a(n-1)-1)^2 + 1, n >= 1.
%F 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))
%F a(n) = A057727(n)-1. - _R. J. Mathar_, Apr 23 2007
%F 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).
%F 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
%F Sum_{n>=0} 2^n/a(n) = 1/2. - _Amiram Eldar_, Oct 03 2022
%e a(0) = 3^(2^0)+1 = 3^1+1 = 4 = 2*(1)+2 = 2*(empty product)+2;
%e a(1) = 3^(2^1)+1 = 3^2+1 = 10 = 2*(4)+2;
%e a(2) = 3^(2^2)+1 = 3^4+1 = 82 = 2*(4*10)+2;
%e a(3) = 3^(2^3)+1 = 3^8+1 = 6562 = 2*(4*10*82)+2;
%e a(4) = 3^(2^4)+1 = 3^16+1 = 43046722 = 2*(4*10*82*6562)+2;
%e a(5) = 3^(2^5)+1 = 3^32+1 = 1853020188851842 = 2*(4*10*82*6562*43046722)+2;
%p A059919:=n->3^(2^n)+1; seq(A059919(n), n=0..7); # _Wesley Ivan Hurt_, Jan 22 2014
%t Table[3^2^n + 1, {n, 0, 7}] (* _Arkadiusz Wesolowski_, Nov 02 2012 *)
%o (PARI) { for (n=0, 11, write("b059919.txt", n, " ", 3^(2^n) + 1); ) } \\ _Harry J. Smith_, Jun 30 2009
%o (Magma) [3^(2^n) + 1: n in [0..8]]; // _Vincenzo Librandi_, Jun 20 2011
%Y Cf. A000215 (Fermat numbers: 2^(2^n) + 1, n >= 0).
%Y Cf. A059917 ((3^(2^n)+1)/2).
%Y Cf. A199591, A078303, A078304, A152581, A080176, A199592, A152585.
%K easy,nonn
%O 0,1
%A _Henry Bottomley_, Feb 08 2001
%E Edited by _Daniel Forgues_, Jun 19 2011 and Jun 20 2011
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