A100096 An inverse Chebyshev transform of the Jacobsthal numbers.

0, 1, 1, 6, 9, 36, 66, 218, 449, 1332, 2946, 8196, 18954, 50688, 120576, 314586, 761889, 1957092, 4794426, 12194828, 30093854, 76067256, 188595276, 474810276, 1180734234, 2965094536, 7387570516, 18521858088, 46203981924, 115721310552

Offset: 0

Author

Paul Barry, Nov 03 2004

Keywords

• nonn
• easy

Comments

Image of x/(1-x-2*x^2) under the transform g(x)->(1/sqrt(1-4*x^2))*g(x*c(x^2)), where c(x) is the g.f. of the Catalan numbers A000108. This is the inverse of the Chebyshev transform which takes A(x) to ((1-x^2)/(1+x^2))*A(x/(1+x^2)).

Hankel transform is (-1)^n*n. Hankel transform of a(n+1) is A141124. - Paul Barry, Jun 05 2008

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A100095, A100097.

Programs

Mathematica

CoefficientList[Series[x*Sqrt[1-4*x^2]*(3*Sqrt[1-4*x^2]+2*x+1)/(2*(4*x^2-1)*(10*x^2+x-2)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 12 2014 *)

Formula

G.f.: x*sqrt(1-4*x^2)*(3*sqrt(1-4*x^2)+2*x+1)/(2*(4*x^2-1)*(10*x^2+x-2)).

a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*A001045(n-2*k).

Conjecture: 2*(-n+1)*a(n) +(n+3)*a(n-1) +18*(n-2)*a(n-2) +4*(-n-2)*a(n-3) +40*(-n+3)*a(n-4)=0. - R. J. Mathar, Nov 24 2012

a(n) ~ (5/2)^n / 3. - Vaclav Kotesovec, Feb 12 2014