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A223067
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A sequence related to the period T of a simple gravity pendulum for arbitrary amplitudes.
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5
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1, 1, 11, 173, 22931, 1319183, 233526463, 2673857519, 39959591850371, 8797116290975003, 4872532317019728133, 1657631603843299234219, 247098748783812523360613, 77729277912104164732573547, 1503342018433974345747514544039
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OFFSET
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0,3
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COMMENTS
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For small angles the period T of a simple gravity pendulum obeys Christiaan Huygens’s law, i.e. T = 2*Pi*sqrt(L/g) with L the length of the pendulum and g the acceleration due to gravity. For arbitrary amplitudes the period T is given below, see Wikipedia. The Taylor series expansion of T as a function of the angular displacement phi leads for the numerators of the even powers of phi to the sequence given above and for the denominators to A223068.
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REFERENCES
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C. D. Andriesse and Sally Miedema, Huygens: The Man Behind the Principle, Ch. 8, 2005.
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LINKS
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FORMULA
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T = 2*Pi*sqrt(L/g)*(2/Pi)*K(sin(phi/2)) with K(k) the complete elliptic integral of the first kind.
T = 2*Pi*sqrt(L/g)/M(1,cos(phi/2)) where M(x,y) = (Pi/4)*((x+y)/(K((x-y)/(x+y)) is the arithmetic-geometric mean of x and y. - Johannes W. Meijer, Dec 28 2016
Let S = Sum_{n>=0} (-1)^n*euler(2*n)*x^n/(2*n) then a(n) = numerator(1/(2*n)! * [x^n] exp(S)). - Peter Luschny, Jan 05 2017
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EXAMPLE
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T = 2*Pi*sqrt(L/g) * (1 + (1/16)*phi^2 + (11/3072)*phi^4 + (173/737280)*phi^6 + … ).
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MAPLE
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nmax:=14: f := series((2/Pi)*EllipticK(sin(phi/2)), phi, 2*nmax+1): for n from 0 to nmax do a(n):= numer(coeff(f, phi, 2*n)) od: seq(a(n), n=0..nmax); # End first program.
nmax:=14: f := series(1/((Pi/4)*(1+cos(phi/2))/EllipticK((1-cos(phi/2))/(1+cos(phi/2)))), phi, 2*nmax+1): for n from 0 to nmax do a(n):= numer(coeff(f, phi, 2*n)) od: seq(a(n), n=0..nmax); # End second program. - Johannes W. Meijer, Dec 28 2016
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MATHEMATICA
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s = Series[EllipticK[Sin[t/2]^2 ], {t, 0, 60}]; CoefficientList[s/Pi, t^2] // Numerator (* Jean-François Alcover, Oct 07 2014 *)
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PROG
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(Sage)
P.<x> = PowerSeriesRing(QQ, default_prec=2*prec)
g = lambda x: exp(sum((-1)^k*euler_number(2*k)*x^k/(2*k) for k in (1..prec+1)))
R = P(g(x)).coefficients()
return [numerator(R[n]/factorial(2*n)) for n in (0..prec)]
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CROSSREFS
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KEYWORD
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nonn,easy,frac
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AUTHOR
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STATUS
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approved
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