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A145508
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a(n+1)=a(n)^2+2*a(n)-2 and a(1)=8
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2
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OFFSET
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1,1
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COMMENTS
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General formula for a(n+1)=a(n)^2+2*a(n)-2 and a(1)=k+1 is a(n)=Floor[((k + Sqrt[k^2 + 4])/2)^(2^((n+1) - 1))
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LINKS
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FORMULA
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From Peter Bala, Nov 12 2012: (Start)
a(n) = alpha^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1, where alpha := 1/2*(9 + sqrt(77)). a(n) = 1 (mod 7).
Recurrence: a(n) = 10*{product {k = 1..n-1} a(k)} - 2 with a(1) = 8.
Product {n = 1..inf} (1 + 1/a(n)) = 10/sqrt(77).
Product {n = 1..inf} (1 + 2/(a(n) + 1)) = sqrt(11/7).
(End)
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MATHEMATICA
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aa = {}; k = 8; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
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k =7; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}] (*Artur Jasinski*)
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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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