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%I #18 Dec 11 2022 06:09:39
%S 4,194,1416317954,4023861667741036022825635656102100994
%N Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 4.
%C Bisection of A003010.
%C a(4) has 147 digits and a(5) has 586 digits. - _Harvey P. Dale_, Mar 03 2020
%F Let alpha = 2 + sqrt(3). Then a(n) = (alpha)^(4^n) + (1/alpha)^(4^n).
%F a(n) = A003010(2*n) = A003500(4^n).
%F Product_{n >= 0} ((1 + 2/a(n))/(1 - 2/a(n)^2)) = sqrt(3).
%F From _Peter Bala_, Dec 06 2022: (Start)
%F a(n) = 2*T(4^n,2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
%F Let b(n) = a(n) - 4. 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&,4,5] (* _Harvey P. Dale_, Mar 03 2020 *)
%o (PARI) a(n)={if(n==0,4,a(n-1)^4-4*a(n-1)^2+2)} \\ _Edward Jiang_, Sep 11 2014
%Y Cf. A001999, A003010, A003500, A219162, A219164, A219165.
%K nonn,easy
%O 0,1
%A _Peter Bala_, Nov 13 2012
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