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A007909
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Expansion of (1-x)/(1-2*x+x^2-2*x^3).
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11
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1, 1, 1, 3, 7, 13, 25, 51, 103, 205, 409, 819, 1639, 3277, 6553, 13107, 26215, 52429, 104857, 209715, 419431, 838861, 1677721, 3355443, 6710887, 13421773, 26843545, 53687091, 107374183, 214748365, 429496729, 858993459, 1717986919, 3435973837, 6871947673
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OFFSET
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0,4
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COMMENTS
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Equals INVERT transform of (1, 0, 2, 2, 2, ...). - Gary W. Adamson, Apr 28 2009
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REFERENCES
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Kenneth Edwards, Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.
M. E. Larsen, Summa Summarum, A. K. Peters, Wellesley, MA, 2007; see p. 38.
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LINKS
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FORMULA
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G.f.: (1-x)/(1-2*x+x^2-2*x^3).
a(n) = (1/5)*(2^(n+1)+3*cos(n*Pi/2)+sin(n*Pi/2)).
a(n) = Sum_{k=0..floor((n-1)/3)} binomial(n-k-1, 2*k)*2^k. - Paul Barry, Sep 16 2004
a(n) = (1/5)*(2^(n+1) + (-1)^[(n+1)/2] + 2*(-1)^[n/2]). - Ralf Stephan, Jun 09 2005
a(n) = 2*a(n-1)-a(n-2)+2*a(n-3). Sequence is identical to its half second differences from the second term; a(n)+a(n+2)=2^(n+1). - Paul Curtz, Dec 17 2007
a(n+1) = (2^n)*Sum_{k=1..n} (-1)^floor(k/2)/2^k. - Yalcin Aktar, Jul 20 2008
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MAPLE
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U:=n->(1/5)*(2^(n+1)+3*cos(n*Pi/2)+sin(n*Pi/2)); [seq(U(n), n=0..50)];
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MATHEMATICA
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CoefficientList[Series[(1-x)/(1-2*x+x^2-2*x^3), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 17 2012 *)
LinearRecurrence[{2, -1, 2}, {1, 1, 1}, 40] (* Harvey P. Dale, Jul 26 2016 *)
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PROG
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(Magma) I:=[1, 1, 1]; [n le 3 select I[n] else 2*Self(n-1)-Self(n-2)+2*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jun 17 2012
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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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Mogens Esrom Larsen (mel(AT)math.ku.dk)
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EXTENSIONS
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STATUS
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approved
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