A119562 Let 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 + 3.

4, 5, 15, 251, 65523, 4294967267, 18446744073709551555, 340282366920938463463374607431768211331, 115792089237316195423570985008687907853269984665640564039457584007913129639683

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 = 5
		

Programs

PARI

fm2(n) = for(x=0,n,y=2^(2^x)-2^x+3;print1(y","))

Formula

a(n) = A001146(n)-A000079(n)+3 = A119564(n)+2. - R. J. Mathar, May 15 2007

Extensions

Definition corrected by R. J. Mathar, May 15 2007