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Search: id:A117645
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| A117645 |
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Generalized Fibonacci sequence to simulate a bouncing ball trajectory: f[n]=a+f[n-1]+b*f[n-2]: a=(248/125),b=-1. |
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+0
1
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| 120, 125, 128, 128, 127, 124, 119, 112, 103, 93, 81, 67, 53, 38, 22, 6, 11, 27, 43, 58, 72, 85, 96, 107, 115, 122, 126, 129, 129, 128, 124, 119, 112, 102, 92, 79, 66, 51, 36, 20, 4, 13, 29, 44, 59, 73, 86, 97, 107, 115, 122, 126, 128, 128, 126, 123
(list; graph; listen)
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OFFSET
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0,1
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COMMENT
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This sequence gives a kind of rolling curve: what is odd and interesting is that it gives harmonic behavior with a purely linear approach. This code solves a Binet : Needs["DiscreteMath`RSolve`"]; Clear[f]; f[n_Integer] = Module[{a}, a[n] /. RSolve[{a[n] == (248/125)*a[ n - 1] - a[n - 2], a[0] == 120, a[1] == 125}, a[n], n][[1]] // Simplify] // ToRadicals f[n]=120*Cos[n*ArcTan[Sqrt[249]/124]]+745*Sin[n*ArcTan[Sqrt[249]/124]]/Sqrt[249] The curve as integers is like what a bouncing ball with perfect elasticity would look in an equal time strobe light. This ratio gives tangent curves: b0 = Table[N[f[n + 1]/f[n]], {n, 0, 55}] ListPlot[b0]
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FORMULA
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a(0)=120;a(1)=125 a(n) = (248/125)*a(n - 1) - a(n - 2);a(n)=Abs[Floor[a(n)]]
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MATHEMATICA
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Clear[f] f[0] = 120; f[1] = 125; f[n_] := f[n] = (248/125)*f[n - 1] - f[n - 2] Table[Abs[Floor[f[n]]], {n, 0, 55}] ListPlot[%] M = {{0, 1}, {-1, (248/125)}} v[0] = {120, 125}; v[n_] := v[n] = M.v[n - 1] Table[Abs[Floor[v[n][[1]]]], {n, 0, 55}] ListPlot[%]
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CROSSREFS
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Sequence in context: A056466 A056456 A085218 this_sequence A157425 A030501 A074302
Adjacent sequences: A117642 A117643 A117644 this_sequence A117646 A117647 A117648
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KEYWORD
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nonn,uned,probation
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AUTHOR
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Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Apr 10 2006
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