cp's OEIS Frontend

A219162 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 3.

%I A219162 #18 Dec 11 2022 06:10:07
%S A219162 3,47,4870847,562882766124611619513723647
%N A219162 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 3.
%C A219162 Bisection of A001566. Compare the following remarks with A001999.
%C A219162 The present sequence is the case x = 3 of the following general remarks. For other cases see A219163 (x = 4), A219164 (x = 5) and A219165 (x = 6).
%C A219162 Let x > 2 and let alpha := {x + sqrt(x^2 - 4)}/2. Define a sequence a(n) (which depends on x) by setting a(n) = alpha^(4^n) + (1/alpha)^(4^n). Then it is easy to verify that the sequence a(n) satisfies the recurrence equation a(n+1) = a(n)^4 + 4*a(n)^2 - 2 with the initial condition a(0) = x.
%C A219162 We have the product expansion sqrt((x + 2)/(x - 2)) = Product_{n >= 0} ((1 + 2/a(n))/(1 - 2/a(n)^2)).
%F A219162 Let alpha = 1/2*(3 + sqrt(5)) then a(n) = (alpha)^(4^n) + (1/alpha)^(4^n).
%F A219162 a(n) = A001566(2*n) = A000032(2*4^n).
%F A219162 Product {n >= 0} ((1 + 2/a(n))/(1 - 2/a(n)^2)) = sqrt(5).
%F A219162 From _Peter Bala_, Dec 06 2022: (Start)
%F A219162 a(n) = 2*T(4^n,3/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
%F A219162 Let b(n) = a(n) - 3. 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)
%o A219162 (PARI) a(n)={if(n==0,3,a(n-1)^4-4*a(n-1)^2+2)} \\ _Edward Jiang_, Sep 11 2014
%Y A219162 Cf. A000032, A001566, A001999, A002814, A145502, A219163, A219164, A219165.
%K A219162 nonn,easy
%O A219162 0,1
%A A219162 _Peter Bala_, Nov 13 2012