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A219163
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Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 4.
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3
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OFFSET
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0,1
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COMMENTS
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a(4) has 147 digits and a(5) has 586 digits. - Harvey P. Dale, Mar 03 2020
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LINKS
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FORMULA
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Let alpha = 2 + sqrt(3). Then a(n) = (alpha)^(4^n) + (1/alpha)^(4^n).
Product_{n >= 0} ((1 + 2/a(n))/(1 - 2/a(n)^2)) = sqrt(3).
a(n) = 2*T(4^n,2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
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)
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MATHEMATICA
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PROG
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(PARI) a(n)={if(n==0, 4, a(n-1)^4-4*a(n-1)^2+2)} \\ Edward Jiang, Sep 11 2014
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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