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.

4, 7, 21, 265, 65553, 4294967329, 18446744073709551681, 340282366920938463463374607431768211585, 115792089237316195423570985008687907853269984665640564039457584007913129640193

Offset: 0

Author

Cino Hilliard, May 31 2006

Keywords

• nonn

Examples

			F(1) = 2^(2^1)+1 = 5
M(1) = 2^1-1 = 1
F(1)+M(2)+1 = 7
		

Programs

Mathematica

Table[2^(2^n)+2^n+1,{n,0,10}] (* Harvey P. Dale, Jun 24 2011 *)

PARI

fm(n) = for(x=0,n,y=2^(2^x)+2^x+1;print1(y","))