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

Offset

0,3

Comments

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

Links

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

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-2k)}.

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

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 *)

Crossrefs

Cf. A100095, A100096.

Sequence in context: A024997 A081157 A099177 * A091004 A133467 A005559

Adjacent sequences: A100094 A100095 A100096 * A100098 A100099 A100100

Keyword

nonn,easy

Author

Paul Barry, Nov 03 2004

Status

approved