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%I #27 Jun 09 2022 02:25:21
%S 1,2,6,14,38,92,240,590,1510,3740,9476,23564,59372,147968,371636,
%T 927374,2324870,5805740,14538660,36322340,90898228,227153192,
%U 568235696,1420236524,3551943388,8878506392,22201466280,55498465400,138766221800,346895496200,867316299260,2168213189390
%N a(n) = Sum_{k=0..floor(n/2)} binomial(n,k)*2^(n-2*k).
%C 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
%C Generally (for r>1), a(n) ~ (r + 1/r)^n. - _Vaclav Kotesovec_, Dec 06 2012
%C Hankel transform is A088138(n+1). - _Paul Barry_, Jun 16 2009
%H Vincenzo Librandi, <a href="/A100067/b100067.txt">Table of n, a(n) for n = 0..1000</a>
%F G.f.: x/(sqrt(1-4*x^2)*(sqrt(1-4*x^2)+x-1)).
%F a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*2^(n-2*k).
%F a(n) = Sum_{k=0..n} binomial(n, (n-k)/2)*(1 + (-1)^(n-k)*2^k/2.
%F 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
%F a(n) ~ 5^n/2^n. - _Vaclav Kotesovec_, Dec 06 2012
%t CoefficientList[Series[x/(Sqrt[1-4*x^2]*(Sqrt[1-4*x^2]+x-1)), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Dec 06 2012 *)
%o (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
%o (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
%o (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
%Y Cf. A027306, A100068, A100069.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Nov 02 2004
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