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A100096
An inverse Chebyshev transform of the Jacobsthal numbers.
4
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
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
Sequence in context: A046634 A166624 A025628 * A192172 A057397 A038263
KEYWORD
nonn,easy
AUTHOR
Paul Barry, Nov 03 2004
STATUS
approved