A127358 a(n) = Sum_{k=0..n} binomial(n, floor(k/2))*2^(n-k).

1, 3, 8, 21, 54, 138, 350, 885, 2230, 5610, 14088, 35346, 88596, 221952, 555738, 1391061, 3480870, 8708610, 21783680, 54483510, 136254964, 340729788, 852000828, 2130354786, 5326563004, 13317759588, 33296999120, 83247698100, 208129274400, 520343244300

Offset: 0

Author

Paul Barry, Jan 11 2007

Keywords

• easy
• nonn

Comments

Hankel transform is (-1)^n. In general, given r >= 0, the sequence given by Sum_{k=0..n} binomial(n, floor(k/2))*r^(n-k) has Hankel transform (1-r)^n. The sequence is the image of the sequence with g.f. (1+x)/(1-2*x) under the Chebyshev mapping 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.

Examples

			a(3) = 21 = (12 + 6 + 2 + 1), where the top row of M^3 = (12, 6, 2, 1).
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000
• Isaac DeJager, Madeleine Naquin, Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.

Crossrefs

Cf. A107430. - Philippe Deléham, Sep 16 2009

Programs

Mathematica

Table[Sum[Binomial[n,Floor[k/2]]2^(n-k),{k,0,n}],{n,0,30}] (* Harvey P. Dale, Jun 03 2012 *)
CoefficientList[Series[(1 + 2*x - Sqrt[1 - 4*x^2])/(2*Sqrt[1 - 4*x^2]*(x - 1 + Sqrt[1 - 4*x^2])), {x, 0, 50}], x] (* G. C. Greubel, May 22 2017 *)

PARI

my(x='x+O('x^50)); Vec((1 + 2*x - sqrt(1 - 4*x^2))/(2*sqrt(1 - 4*x^2)*(x - 1 + sqrt(1 - 4*x^2)))) \\ G. C. Greubel, May 22 2017

Formula

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

a(n) = 2*a(n-1) + A054341(n-1). a(n) = Sum_{k=0..n} A126075(n,k). - Philippe Deléham, Mar 03 2007

a(n) = Sum_{k=0..n} A061554(n,k)*2^k. - Philippe Deléham, Dec 04 2009

From Gary W. Adamson, Sep 07 2011: (Start)

a(n) is the sum of top row terms of M^n, M is an infinite square production matrix as follows:

2, 1, 0, 0, 0, ...

1, 0, 1, 0, 0, ...

0, 1, 0, 1, 0, ...

0, 0, 1, 0, 1, ...

0, 0, 0, 1, 0, ...

... (End)

D-finite with recurrence 2*n*a(n) + (-5*n-4)*a(n-1) + 2*(-4*n+13)*a(n-2) + 20*(n-2)*a(n-3) = 0. - R. J. Mathar, Nov 30 2012

a(n) ~ 3 * 5^n / 2^(n+1). - Vaclav Kotesovec, Feb 13 2014