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Fermat numbers of order 7 or F(n,7) = 2^(2^n)+7.
6

%I #19 Sep 08 2022 08:45:30

%S 9,11,23,263,65543,4294967303,18446744073709551623,

%T 340282366920938463463374607431768211463,

%U 115792089237316195423570985008687907853269984665640564039457584007913129639943

%N Fermat numbers of order 7 or F(n,7) = 2^(2^n)+7.

%C This sequence is equivalent to F(n)+ 6 or 2^(2^n)+ 1 + 6. This sequence does not appear to have any special divisibility properties. Fermat numbers of order 5 which are found in A063486, have the divisibility property if n is even, then 7 divides F(n,5). After the first 2 terms the ending digit is the same for all F(n,m) and is (6+m) mod 10.

%H Vincenzo Librandi, <a href="/A130730/b130730.txt">Table of n, a(n) for n = 0..11</a>

%H Tigran Hakobyan, <a href="http://arxiv.org/abs/1601.04946">On the unboundedness of common divisors of distinct terms of the sequence a(n)=2^2^n+d for d>1</a>, arXiv:1601.04946 [math.NT], 2016.

%F F(n,m): The n-th Fermat number of order m = 2^(2^n)+ m. The traditional Fermat numbers are F(n,1) or Fermat numbers of order 1 in this nomenclature.

%t Table[(2^(2^n) + 7), {n, 0, 15}] (* _Vincenzo Librandi_, Jan 09 2013 *)

%o (PARI) fplusm(n,m)= { local(x,y); for(x=0,n, y=2^(2^x)+m; print1(y",") ) }

%o (Magma) [2^(2^n)+7: n in [0..11]]; // _Vincenzo Librandi_, Jan 09 2013

%Y Cf. A063486, A130729.

%K nonn

%O 0,1

%A _Cino Hilliard_, Jul 05 2007