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Define F(n) = 2^(2^n)+1 = n-th Fermat number, M(n) = 2^n-1 = the n-th Mersenne number. Then a(n) = F(n)+M(n)-1 = 2^(2^n) + 2^n - 1.
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%I #8 Oct 01 2013 17:58:25

%S 2,5,19,263,65551,4294967327,18446744073709551679,

%T 340282366920938463463374607431768211583,

%U 115792089237316195423570985008687907853269984665640564039457584007913129640191

%N Define F(n) = 2^(2^n)+1 = n-th Fermat number, M(n) = 2^n-1 = the n-th Mersenne number. Then a(n) = F(n)+M(n)-1 = 2^(2^n) + 2^n - 1.

%C The first 5 entries are primes. Are there infinitely many primes in this sequence?

%F a(n) = A119561(n)-2=A000215(n)+A000225(n)-1. - _R. J. Mathar_, Apr 22 2007

%e F(2) = 2^(2^2)+1 = 17, M(2) = 2^2-1 = 3, F(2)+ M(2) - 1 = 19

%o (PARI) fm3(n) = for(x=0,n,y=2^(2^x)+2^x-1;print1(y","))

%K nonn

%O 0,1

%A _Cino Hilliard_, May 31 2006

%E Edited by _N. J. A. Sloane_, Jun 03 2006