login
Hankel transform of Sum_{k=0..floor(n/2)} binomial(2*k, k).
2

%I #16 Jun 09 2022 02:22:57

%S 1,2,0,0,-16,-32,-64,-128,0,0,1024,2048,4096,8192,0,0,-65536,-131072,

%T -262144,-524288,0,0,4194304,8388608,16777216,33554432,0,0,-268435456,

%U -536870912,-1073741824,-2147483648,0,0,17179869184,34359738368,68719476736,137438953472,0,0,-1099511627776

%N Hankel transform of Sum_{k=0..floor(n/2)} binomial(2*k, k).

%C Hankel transform of A100066(n+1).

%H G. C. Greubel, <a href="/A120582/b120582.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,4,0,-16).

%F a(n) = 2^n*(-sqrt(3)*cos(5*Pi*n/6 + Pi/3)/6 + (sqrt(3)/3 - 1/2)*sin(5*Pi*n/6 + Pi/3) + (sqrt(3)/3 + 1/2)*cos(Pi*n/6 + Pi/6) + sqrt(3)*sin(Pi*n/6 + Pi/6)/6).

%F G.f.: (1-2*x)*(1+2*x)^2/(1 - 4*x^2 + 16*x^4). - _Colin Barker_, Jun 27 2013

%F a(n) = 2^(n-1)*( (1+(-1)^n)*(ChebyshevU(n/2, 1/2) - ChebyshevU((n-2)/2, 1/2)) + (1 -(-1)^n)*ChebyshevU((n+1)/2, 1/2)). - _G. C. Greubel_, Jun 08 2022

%t LinearRecurrence[{0,4,0,-16}, {1,2,0,0}, 51] (* _G. C. Greubel_, Jun 08 2022 *)

%o (Magma) I:=[1,2,0,0]; [n le 4 select I[n] else 4*Self(n-2) -16*Self(n-4): n in [1..51]]; // _G. C. Greubel_, Jun 08 2022

%o (SageMath)

%o def C(n): return floor(chebyshev_U(n, 1/2))

%o def A120582(n): return 2^n*( ((n+1)%2)*(C(n/2) - C((n-2)/2)) + (n%2)*C((n+1)/2) )

%o [A120582(n) for n in (0..50)] # _G. C. Greubel_, Jun 08 2022

%Y Cf. A100066, A120580.

%K easy,sign

%O 0,2

%A _Paul Barry_, Jun 15 2006