%I #18 Dec 11 2022 06:11:17
%S 5,527,77132286527,35395236908668169265765137996816180039862527
%N Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 5.
%C Bisection of A003487.
%C The next term -- a(4) -- has 175 digits. - _Harvey P. Dale_, Jun 09 2017
%F Let alpha = 1/2*(5 + sqrt(21)). Then a(n) = (alpha)^(4^n) + (1/alpha)^(4^n).
%F a(n) = A003487(2*n) = A003501(4^n).
%F Product_{n>=0} ((1 + 2/a(n))/(1 - 2/a(n)^2)) = sqrt(7/3).
%F From _Peter Bala_, Dec 06 2022: (Start)
%F a(n) = 2*T(4^n,5/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
%F Let b(n) = a(n) - 5. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. (End)
%t NestList[#^4-4#^2+2&,5,5] (* _Harvey P. Dale_, Jun 09 2017 *)
%Y Cf. A001999, A003487, A003501, A219162, A219163, A219165.
%K nonn,easy
%O 0,1
%A _Peter Bala_, Nov 13 2012