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A100067
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a(n) = Sum_{k=0..floor(n/2)} binomial(n,k)*2^(n-2*k).
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5
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1, 2, 6, 14, 38, 92, 240, 590, 1510, 3740, 9476, 23564, 59372, 147968, 371636, 927374, 2324870, 5805740, 14538660, 36322340, 90898228, 227153192, 568235696, 1420236524, 3551943388, 8878506392, 22201466280, 55498465400, 138766221800, 346895496200, 867316299260, 2168213189390
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OFFSET
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0,2
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COMMENTS
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An inverse Chebyshev transform of x/(1-2*x), where the Chebyshev transform of g(x) is ((1-x^2)/(1+x^2))*g(x/(1+x^2)) and the inverse transform maps a g.f. A(x) to (1/sqrt(1-4*x^2))*A(x*c(x^2)) where c(x) is the g.f. of the Catalan numbers A000108. In general, Sum_{k=0..floor(n/2)} binomial(n,k) * r^(n-2*k) has g.f. 2*x/(sqrt(1-4*x^2)*(r*sqrt(1-4*x^2) + 2*x - r)). - corrected by Vaclav Kotesovec, Dec 06 2012
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LINKS
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FORMULA
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G.f.: x/(sqrt(1-4*x^2)*(sqrt(1-4*x^2)+x-1)).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*2^(n-2*k).
a(n) = Sum_{k=0..n} binomial(n, (n-k)/2)*(1 + (-1)^(n-k)*2^k/2.
Recurrence: 2*n*(3*n-7)*a(n) = (15*n^2 - 35*n + 8)*a(n-1) + 4*(6*n^2 - 20*n + 11)*a(n-2) - 20*(n-2)*(3*n-4)*a(n-3). - Vaclav Kotesovec, Dec 06 2012
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MATHEMATICA
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CoefficientList[Series[x/(Sqrt[1-4*x^2]*(Sqrt[1-4*x^2]+x-1)), {x, 0, 20}], x] (* Vaclav Kotesovec, Dec 06 2012 *)
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PROG
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(PARI) my(x='x+O('x^66)); Vec(x/(sqrt(1-4*x^2)*(sqrt(1-4*x^2)+x-1))) \\ Joerg Arndt, May 12 2013
(Magma) m:=2; [(&+[Binomial(n, k)*m^(n-2*k): k in [0..Floor(n/2)]]): n in [0..40]]; // G. C. Greubel, Jun 08 2022
(SageMath) m=2; [sum(binomial(n, k)*m^(n-2*k) for k in (0..n//2)) for n in (0..40)] # G. C. Greubel, Jun 08 2022
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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