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Define F(n) = 2^(2^n)+1 = the 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)+1+2^n-1+1 = 2^(2^n)+2^n+1 = F(n)+2^n.
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%I #9 Feb 15 2014 12:30:19

%S 4,7,21,265,65553,4294967329,18446744073709551681,

%T 340282366920938463463374607431768211585,

%U 115792089237316195423570985008687907853269984665640564039457584007913129640193

%N Define F(n) = 2^(2^n)+1 = the 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)+1+2^n-1+1 = 2^(2^n)+2^n+1 = F(n)+2^n.

%e F(1) = 2^(2^1)+1 = 5

%e M(1) = 2^1-1 = 1

%e F(1)+M(2)+1 = 7

%t Table[2^(2^n)+2^n+1,{n,0,10}] (* _Harvey P. Dale_, Jun 24 2011 *)

%o (PARI) fm(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