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
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Programs
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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
Comments