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A020797
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Decimal expansion of 1/sqrt(40).
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7
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1, 5, 8, 1, 1, 3, 8, 8, 3, 0, 0, 8, 4, 1, 8, 9, 6, 6, 5, 9, 9, 9, 4, 4, 6, 7, 7, 2, 2, 1, 6, 3, 5, 9, 2, 6, 6, 8, 5, 9, 7, 7, 7, 5, 6, 9, 6, 6, 2, 6, 0, 8, 4, 1, 3, 4, 2, 8, 7, 5, 2, 4, 2, 6, 3, 9, 6, 2, 9, 7, 2, 1, 9, 3, 1, 9, 6, 1, 9, 1, 1, 0, 6, 7, 2, 1, 2, 4, 0, 5, 4, 1, 8, 9, 6, 5, 0, 1, 4
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OFFSET
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0,2
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COMMENTS
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With offset 1, decimal expansion of sqrt(5/2). - Eric Desbiaux, May 01 2008
sqrt(5/2) appears as a coordinate in a degree-5 integration formula on 13 points in the unit sphere [Stroud & Secrest]. - R. J. Mathar, Oct 12 2011
With offset 2, decimal expansion of sqrt(250). - Michel Marcus, Nov 04 2013
The regular continued fraction of 1/sqrt(40) = 1/(2*sqrt(10)) is [0; 6, 3, repeat(12, 3)], and the convergents are given by A(n-1)/B(n-1), n >= 0, with A(-1) = 0, A(n-1) = A041067(n) and B(-1) = 1, B(n-1) = A041066(n).
The regular continued fraction of sqrt(5/2) = sqrt(10)/2 is [1; repeat(1, 1, 2)], and the convergents are given in A295333/A295334.
sqrt(10)/2 is one of the catheti of the rectangular triangle with hypotenuse sqrt(13)/2 = A295330 and the other cathetus sqrt(3)/2 = A010527. This can be constructed from a regular hexagon inscribed in a circle with a radius of 1 unit. If the vertex V_0 has coordinates (x, y) = (1, 0) and the midpoint M_4 = (0, -sqrt(3)/2) then the point L = (sqrt(10)/2, 0) is obtained as intersection of the x-axis and a circle around M_4 with radius taken from the distance between M_4 and V_1 = (1/2, sqrt(3)/2) which is sqrt(13)/2. (End)
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LINKS
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FORMULA
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EXAMPLE
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1/sqrt(40) = 0.15811388300841896659994467722163592668597775696626084134287...
sqrt(5/2) = 1.5811388300841896659994467722163592668597775696626084134287...
sqrt(250) = 15.811388300841896659994467722163592668597775696626084134287...
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MATHEMATICA
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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