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%I #11 Dec 11 2022 06:10:33
%S 6,1154,1773462177794,
%T 9892082352510403757550172975146702122837936996354
%N Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 6.
%C Bisection of A003423.
%F Let alpha = 3 + 2*sqrt(2). Then a(n) = (alpha)^(4^n) + (1/alpha)^(4^n).
%F a(n) = A003423(2*n) = A003499(4^n).
%F Product {n >= 0} ((1 + 2/a(n))/(1 - 2/a(n)^2)) = sqrt(2).
%F From _Peter Bala_, Dec 06 2022: (Start)
%F a(n) = 2*T(4^n,3), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
%F Let b(n) = a(n) - 6. 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)
%Y Cf. A001999, A003423, A003499, A219162, A219163, A219164.
%K nonn,easy
%O 0,1
%A _Peter Bala_, Nov 13 2012
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