A100097 An inverse Chebyshev transform of the Pell numbers.

0, 1, 2, 8, 20, 64, 172, 512, 1416, 4096, 11468, 32768, 92248, 262144, 739832, 2097152, 5925520, 16777216, 47429900, 134217728, 379536440, 1073741824, 3036661032, 8589934592, 24294699120, 68719476736, 194363001272, 549755813888, 1554924811376, 4398046511104

Offset: 0

Author

Paul Barry, Nov 03 2004

Keywords

• nonn
• easy

Comments

Image of x/(1-2*x-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)).

Links

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

Crossrefs

Cf. A100095, A100096.

Programs

Mathematica

CoefficientList[Series[x*Sqrt[1-4*x^2]*(Sqrt[1-4*x^2]+2*x)/((1-4*x^2)*(1-8*x^2)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 12 2014 *)

Formula

G.f.: x*sqrt(1-4*x^2)*(sqrt(1-4*x^2)+2*x)/((1-4*x^2)*(1-8*x^2)).

a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*A000129(n-2*k).

Conjecture: (-n+2)*a(n) +(-n+3)*a(n-1) +4*(3*n-7)*a(n-2) +4*(3*n-10)*a(n-3) +32*(-n+3)*a(n-4) +32*(-n+4)*a(n-5)=0. - R. J. Mathar, Nov 24 2012

Recurrence: (n-2)*a(n) = 4*(3*n-7)*a(n-2) - 32*(n-3)*a(n-4). - Vaclav Kotesovec, Feb 12 2014

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

a(2*n) = 8^n/(2*sqrt(2)) - 2^n * (2*n-1)!! * hypergeom([1, n+1/2], [n+1], 1/2)/(4*n!), a(2*n+1) = 8^n. - Vladimir Reshetnikov, Oct 13 2016