%I #33 Jan 14 2024 08:59:45
%S 4,12,144,33024,2147549184,9223372041149743104,
%T 170141183460469231750134047789593657344,
%U 57896044618658097711785492504343953926975274699741220483192166611388333031424
%N a(n) = 2^(2^n) + 2^(2^(n+1)-1).
%C a(n) is the long leg of the Pythagorean triangle whose short leg is the n-th Fermat number, A000215(n), and whose hypotenuse is a(n) + 1.
%C [A000215(n), a(n), a(n) + 1] is a primitive Pythagorean triple of the form [2*k + 1, 2*k^2 + 2*k, 2*k^2 + 2*k + 1] where k = A058891(n).
%F a(n) = A000215(n) + A058891(n+1) - 1.
%F a(n) = sqrt(Integral_{x=1..A000215(n)} (x^3-x) dx).
%F a(n) = 2*A058891(n)^2 + 2*A058891(n).
%F sqrt(a(n) + (a(n)+1)) = sqrt((a(n)+1)^2 - a(n)^2) = A000215(n).
%t Table[2^(2^n) + 2^(2^(n + 1) - 1), {n, 0, 7}] (* _Paul F. Marrero Romero_, Jan 13 2024 *)
%o (Python)
%o for n in range(0,8):
%o print(2**(2**n)+2**(2**(n+1)-1))
%Y Cf. A000215, A058891.
%K nonn,easy
%O 0,1
%A _César Aguilera_, Aug 31 2023