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A099938
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Consider the sequence of circles C0, C1, C2, C3 ..., where C0 is a half-circle of radius 1. C1 is the largest circle that fits into C0 and has radius 1/2. C(n+1) is the largest circle that fits inside C0 but outside C(n), etc. Sequence gives the curvatures (reciprocals of the radii) of the circles.
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2
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2, 4, 18, 100, 578, 3364, 19602, 114244, 665858, 3880900, 22619538, 131836324, 768398402, 4478554084, 26102926098, 152139002500, 886731088898, 5168247530884, 30122754096402, 175568277047524, 1023286908188738, 5964153172084900, 34761632124320658
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OFFSET
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1,1
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COMMENTS
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The numbers a(2), a(4), a(6) etc. are squares and a(1), a(3), a(5) ... are twice squares. Furthermore, a(1) - 2, a(3) - 2, a(5) - 2 etc. are squares and a(2) - 2, a(4) - 2, a(6) - 2 etc. are twice square.
C(n) is centered at (x(n), y(n)), where x(n) = sqrt(1 - 2/a(n)) and y(n) = 1/a(n). - David Wasserman, Feb 28 2008
C(n) is tangent to C0 because sqrt(x(n)^2 + y(n)^2) + y(n) = 1 and C(n) is tangent to C(n+1) because sqrt[(x(n+1) - x(n))^2 + (y(n+1) - y(n))^2] = y(n) + y(n+1). - David Wasserman, Feb 28 2008
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LINKS
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FORMULA
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a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3). G.f.: -2*x*(2*x^2-5*x+1) / ((x-1)*(x^2-6*x+1)). - Colin Barker, Jan 07 2013
a(n) = 1/2*(2 + (3 - 2*sqrt(2))^n*(3 + 2*sqrt(2)) + (3 - 2*sqrt(2))*(3 + 2*sqrt(2))^n). - Colin Barker, Jun 05 2016
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MATHEMATICA
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PROG
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(PARI) Vec(-2*x*(2*x^2-5*x+1)/((x-1)*(x^2-6*x+1)) + O(x^30)) \\ Colin Barker, Jun 05 2016
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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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Hartmut Neubauer (hartmut.f.neubauer(AT)t-online.de), Nov 12 2004
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EXTENSIONS
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STATUS
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approved
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