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A019863
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Decimal expansion of sin(3*Pi/10) (sine of 54 degrees).
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25
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8, 0, 9, 0, 1, 6, 9, 9, 4, 3, 7, 4, 9, 4, 7, 4, 2, 4, 1, 0, 2, 2, 9, 3, 4, 1, 7, 1, 8, 2, 8, 1, 9, 0, 5, 8, 8, 6, 0, 1, 5, 4, 5, 8, 9, 9, 0, 2, 8, 8, 1, 4, 3, 1, 0, 6, 7, 7, 2, 4, 3, 1, 1, 3, 5, 2, 6, 3, 0, 2, 3, 1, 4, 0, 9, 4, 5, 1, 2, 2, 4, 8, 5, 3, 6, 0, 3, 6, 0, 2, 0, 9, 4, 6, 9, 5, 5, 6, 8
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OFFSET
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0,1
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COMMENTS
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Midsphere radius of regular icosahedron with unit edges.
Also half of the golden ratio (A001622). - Stanislav Sykora, Jan 30 2014
Andris Ambainis (see Aaronson link) observes that combining the results of Barak-Hardt-Haviv-Rao with Dinur-Steurer yields the maximal probability of winning n parallel repetitions of a classical CHSH game (see A201488) asymptotic to this constant to the power of n, an improvement on the naive probability of (3/4)^n. (All the random bits are received upfront but the players cannot communicate or share an entangled state.) - Charles R Greathouse IV, May 15 2014
This is the height h of the isosceles triangle in a regular pentagon, in length units of the circumscribing radius, formed by a side as base and two adjacent radii. h = sin(3*Pi/10) = cos(Pi/5) (radius 1 unit). - Wolfdieter Lang, Jan 08 2018
Also the limiting value(L) of "r" which is abscissa of the vertex of the parabola F(n)*x^2 - F(n+1)*x + F(n + 2)(where F(n)=A000045(n) are the Fibonacci numbers and n>0). - Burak Muslu, Feb 24 2021
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LINKS
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Stanislav Sykora, Table of n, a(n) for n = 0..2000
Scott Aaronson, The NEW Ten Most Annoying Questions in Quantum Computing (2014)
Boaz Barak, Moritz Hardt, Ishay Haviv, and Anup Rao, Rounding Parallel Repetitions of Unique Games (2008)
Irit Dinur and David Steurer, Analytical approach to parallel repetition, arXiv:1305.1979 [cs.CC], 2013-2014.
Wikipedia, Exact trigonometric constants
Wikipedia, Platonic solid
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FORMULA
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Equals (1+sqrt(5))/4 = cos(Pi/5) = sin(3*Pi/10). - R. J. Mathar, Jun 18 2006
Equals 2F1(4/5,1/5;1/2;3/4) / 2 = A019827 + 1/2. - R. J. Mathar, Oct 27 2008
Equals A001622 / 2. - Stanislav Sykora, Jan 30 2014
phi / 2 = (i^(2/5) + i^(-2/5)) / 2 = i^(2/5) - (sin(Pi/5))*i = i^(-2/5) + (sin(Pi/5))*i = i^(2/5) - (cos(3*Pi/10))*i = i^(-2/5) + (cos(3*Pi/10))*i. - Jaroslav Krizek, Feb 03 2014
Equals 1/A134972. - R. J. Mathar, Jan 17 2021
Equals 2*A019836*A019872. - R. J. Mathar, Jan 17 2021
Equals (A094214 + 1)/2 or 1/(2*A094214). - Burak Muslu, Feb 24 2021
Equals hypergeom([-2/5, -3/5], [6/5], -1) = hypergeom([-1/5, 3/5], [6/5], 1) = hypergeom([1/5, -3/5], [4/5], 1). - Peter Bala, Mar 04 2022
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EXAMPLE
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0.80901699437494742410229341718281905886015458990288143106772431135263...
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MAPLE
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Digits:=100; evalf((1+sqrt(5))/4); # Wesley Ivan Hurt, Mar 27 2014
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MATHEMATICA
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RealDigits[(1 + Sqrt[5])/4, 10, 111] (* Robert G. Wilson v *)
RealDigits[Sin[54 Degree], 10, 120][[1]] (* Harvey P. Dale, Apr 21 2018 *)
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PROG
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(PARI) (1+sqrt(5))/4 \\ Charles R Greathouse IV, Jan 16 2012
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CROSSREFS
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Cf. A001622, A019827.
Platonic solids midradii: A020765 (tetrahedron), A020761 (octahedron), A010503 (cube), A239798 (dodecahedron).
Sequence in context: A197617 A005076 A296182 * A350747 A243456 A246772
Adjacent sequences: A019860 A019861 A019862 * A019864 A019865 A019866
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KEYWORD
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nonn,cons,easy
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AUTHOR
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N. J. A. Sloane
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STATUS
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approved
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