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A102879
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A Chebyshev transform of the first kind of the central binomial numbers.
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2
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1, 2, 4, 14, 48, 162, 556, 1934, 6784, 23954, 85044, 303294, 1085712, 3898962, 14040156, 50678814, 183309312, 664263714, 2411050084, 8764098158, 31899231088, 116244082178, 424064770188, 1548543412398, 5659898710912
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OFFSET
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0,2
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COMMENTS
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Image of 1/sqrt(1-4x) under the mapping g(x)->((1-x^2)/(1+x^2))*g(x/(1+x^2)).
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LINKS
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FORMULA
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G.f.: (1-x^2)/sqrt(1-4*x+2*x^2-4*x^3+x^4).
a(n) = n*Sum_{k=0..floor(n/2)} C(n-k, k)*(-1)^k*C(2(n-2k), n-2k)/(n-k).
Conjecture: n*(n-3)*a(n) - 2*(2*n-1)*(n-3)*a(n-1) + 2*(2-4*n+n^2)*a(n-2) - 2*(n-1)*(2*n-7)*a(n-3) + (n-1)*(n-4)*a(n-4) = 0. - R. J. Mathar, Nov 09 2012
Conjecture verified using the differential equation x*(x^2+1)*(x^2-4*x+1)*g'' + (4*x^4-10*x^3+2*x^2+2*x-2)*g' + 4*(x^2-x+1)*g = 0 satisfied by the g.f. - Robert Israel, Aug 28 2018
a(n) ~ 3^(1/4) * (2 + sqrt(3))^n / sqrt(2*Pi*n). - Vaclav Kotesovec, Nov 02 2023
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MAPLE
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f:= gfun:-rectoproc({n*(n-3)*a(n) -2*(2*n-1)*(n-3)*a(n-1) +2*(2-4*n+n^2)*a(n-2) -2*(n-1)*(2*n-7)*a(n-3) +(n-1)*(n-4)*a(n-4), a(0)=1, a(1)=2, a(2)=4, a(3)=14}, a(n), remember):
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MATHEMATICA
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CoefficientList[Series[(1-x^2)/Sqrt[1-4*x+2*x^2-4*x^3+x^4], {x, 0, 30}], x] (* G. C. Greubel, Mar 31 2019 *)
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PROG
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(PARI) my(x='x+O('x^30)); Vec((1-x^2)/sqrt(1-4*x+2*x^2-4*x^3+x^4)) \\ G. C. Greubel, Mar 31 2019
(Magma) R<x>:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( (1-x^2)/Sqrt(1-4*x+2*x^2-4*x^3+x^4) )); // G. C. Greubel, Mar 31 2019
(Sage) ((1-x^2)/sqrt(1-4*x+2*x^2-4*x^3+x^4)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Mar 31 2019
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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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