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

%I A078303 #46 Feb 16 2025 08:32:48
%S A078303 7,37,1297,1679617,2821109907457,7958661109946400884391937,
%T A078303 63340286662973277706162286946811886609896461828097
%N A078303 Generalized Fermat numbers: 6^(2^n) + 1, n >= 0.
%C A078303 The next term is too large to include.
%C A078303 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.
%C A078303 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
%C A078303 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
%C A078303 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
%H A078303 Vincenzo Librandi, <a href="/A078303/b078303.txt">Table of n, a(n) for n = 0..12</a>
%H A078303 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 A078303 C. K. Caldwell, "Top Twenty" page, <a href="https://t5k.org/top20/page.php?id=9">Generalized Fermat Divisors (base=6)</a>.
%H A078303 Wilfrid Keller, <a href="http://www.prothsearch.com/GFN06.html">GFN06 factoring status</a>.
%H A078303 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/GeneralizedFermatNumber.html">Generalized Fermat Number</a>.
%H A078303 OEIS Wiki, <a href="/wiki/Generalized_Fermat_numbers">Generalized Fermat numbers</a>.
%F A078303 a(0) = 7, a(n) = (a(n-1)-1)^2 + 1, n >= 1.
%F 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. - _Daniel Forgues_, Jun 20 2011
%F A078303 Sum_{n>=0} 2^n/a(n) = 1/5. - _Amiram Eldar_, Oct 03 2022
%e A078303 a(0) = 6^1+1 = 7 = 5*(1)+2 = 5*(empty product)+2;
%e A078303 a(1) = 6^2+1 = 37 = 5*(7)+2;
%e A078303 a(2) = 6^4+1 = 1297 = 5*(7*37)+2;
%e A078303 a(3) = 6^8+1 = 1679617 = 5*(7*37*1297)+2;
%e A078303 a(4) = 6^16+1 = 2821109907457 = 5*(7*37*1297*1679617)+2;
%e A078303 a(5) = 6^32+1 = 7958661109946400884391937 = 5*(7*37*1297*1679617*2821109907457)+2;
%t A078303 Table[6^2^n + 1, {n, 0, 6}] (* _Arkadiusz Wesolowski_, Nov 02 2012 *)
%o A078303 (Magma) [6^(2^n) + 1: n in [0..8]]; // _Vincenzo Librandi_, Jun 20 2011
%o A078303 (PARI) a(n)=6^(2^n)+1 \\ _Charles R Greathouse IV_, Jun 21 2011
%Y A078303 Cf. A000215 (Fermat numbers: 2^(2^n) + 1, n >= 0).
%Y A078303 Cf. A019434 (Fermat primes of the form 2^(2^n) + 1).
%Y A078303 Cf. A123669, A123599, A056993, A126032, A178428, A059919, A199591, A078304, A152581, A080176, A199592, A152585.
%K A078303 nonn,easy
%O A078303 0,1
%A A078303 _Eric W. Weisstein_, Nov 21 2002
%E A078303 Edited by _Daniel Forgues_, Jun 22 2011