cp's OEIS Frontend

A119561 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.

%I A119561 #9 Feb 15 2014 12:30:19
%S A119561 4,7,21,265,65553,4294967329,18446744073709551681,
%T A119561 340282366920938463463374607431768211585,
%U A119561 115792089237316195423570985008687907853269984665640564039457584007913129640193
%N A119561 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 A119561 F(1) = 2^(2^1)+1 = 5
%e A119561 M(1) = 2^1-1 = 1
%e A119561 F(1)+M(2)+1 = 7
%t A119561 Table[2^(2^n)+2^n+1,{n,0,10}] (* _Harvey P. Dale_, Jun 24 2011 *)
%o A119561 (PARI) fm(n) = for(x=0,n,y=2^(2^x)+2^x+1;print1(y","))
%K A119561 nonn
%O A119561 0,1
%A A119561 _Cino Hilliard_, May 31 2006