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Generalized Fermat numbers: 12^(2^n) + 1, n >= 0.
13

%I #40 Apr 03 2023 10:36:11

%S 13,145,20737,429981697,184884258895036417,

%T 34182189187166852111368841966125057,

%U 1168422057627266461843148138873451659428421700563161428957815831003137

%N Generalized Fermat numbers: 12^(2^n) + 1, n >= 0.

%C There appears to be no divisibility rule for this sequence.

%C 13 is the only prime up to 12^(2^15)+1.

%H Vincenzo Librandi, <a href="/A152585/b152585.txt">Table of n, a(n) for n = 0..12</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 C. K. Caldwell, "Top Twenty" page, <a href="https://t5k.org/top20/page.php?id=11">Generalized Fermat Divisors (base=12)</a>.

%H Wilfrid Keller, <a href="http://www.prothsearch.com/GFN12.html">GFN12 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) = 13; a(n)=(a(n-1)-1)^2 + 1, n >= 1.

%F 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

%F Sum_{n>=0} 2^n/a(n) = 1/11. - _Amiram Eldar_, Oct 03 2022

%e a(0) = 12^1+1 = 13 = 11(1)+2 = 11(empty product)+2.

%e a(1) = 12^2+1 = 145 = 11(13)+2.

%e a(2) = 12^4+1 = 20737 = 11(13*145)+2.

%e a(3) = 12^8+1 = 429981697 = 11(13*145*20737)+2.

%e a(4) = 12^16+1 = 184884258895036417 = 11(13*145*20737*429981697)+2.

%e a(5) = 12^32+1 = 34182189187166852111368841966125057 = 11(13*145*20737*429981697*184884258895036417)+2.

%t Table[12^2^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) [12^(2^n) + 1: n in [0..8]]; // _Vincenzo Librandi_, Jun 20 2011

%o (Python)

%o def A152585(n): return (1<<2*(m:=1<<n))*3**m+1 # _Chai Wah Wu_, Jul 19 2022

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

%Y Cf. A059919, A199591, A078303, A078304, A152581, A080176, A199592.

%K easy,nonn

%O 0,1

%A _Cino Hilliard_, Dec 08 2008

%E Edited by _Daniel Forgues_, Jun 19 2011