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A120582
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Hankel transform of Sum_{k=0..floor(n/2)} binomial(2*k, k).
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2
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1, 2, 0, 0, -16, -32, -64, -128, 0, 0, 1024, 2048, 4096, 8192, 0, 0, -65536, -131072, -262144, -524288, 0, 0, 4194304, 8388608, 16777216, 33554432, 0, 0, -268435456, -536870912, -1073741824, -2147483648, 0, 0, 17179869184, 34359738368, 68719476736, 137438953472, 0, 0, -1099511627776
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OFFSET
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0,2
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COMMENTS
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LINKS
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FORMULA
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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).
G.f.: (1-2*x)*(1+2*x)^2/(1 - 4*x^2 + 16*x^4). - Colin Barker, Jun 27 2013
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
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MATHEMATICA
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LinearRecurrence[{0, 4, 0, -16}, {1, 2, 0, 0}, 51] (* G. C. Greubel, Jun 08 2022 *)
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PROG
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(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
(SageMath)
def C(n): return floor(chebyshev_U(n, 1/2))
def A120582(n): return 2^n*( ((n+1)%2)*(C(n/2) - C((n-2)/2)) + (n%2)*C((n+1)/2) )
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CROSSREFS
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KEYWORD
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easy,sign
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AUTHOR
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STATUS
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approved
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