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

0,4

Comments

Image of x/(1-x-2x^2) under the transform g(x)->(1/sqrt(1-4x^2)g(xc(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

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-2k) ).

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

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 *)

Crossrefs

Cf. A100095, A100097.

Sequence in context: A046634 A166624 A025628 * A192172 A057397 A038263

Adjacent sequences: A100093 A100094 A100095 * A100097 A100098 A100099

Keyword

nonn,easy

Author

Paul Barry, Nov 03 2004

Status

approved