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A145180
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Continued cotangent recurrence a(n+1) = a(n)^3 + 3*a(n) and a(1) = 6.
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12
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OFFSET
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1,1
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COMMENTS
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General formula for continued cotangent recurrences type:
a(n+1) = a(n)3 + 3*a(n) and a(1)=k is following:
a(n) = Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))].
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LINKS
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FORMULA
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a(n+1)=a(n)^3 + 3*a(n) and a(1)=6
a(n)=Floor[((6+Sqrt[6^2+4])/2)^(3^(n-1))]
a(n) divides a(n+1) and b(n) = a(n+1)/a(n) satisfies the recurrence b(n+1) = b(n)^3 - 3*b(n-1)^2 + 3. See A002813. - Peter Bala, Nov 23 2012
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MATHEMATICA
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a = {}; k = 6; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
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Table[Floor[((6 + Sqrt[40])/2)^(3^(n - 1))], {n, 1, 5}] (* Artur Jasinski *)
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CROSSREFS
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Cf. A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189 (k = 1 to 15 with k=4 being A006267(n+1)).
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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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