A127361 a(n) = Sum_{k=0..n} binomial(n, floor(k/2))*(-2)^(n-k).

1, -1, 4, -7, 22, -46, 130, -295, 790, -1870, 4864, -11782, 30148, -73984, 187534, -463687, 1168870, -2902870, 7293640, -18161170, 45541492, -113576596, 284470564, -710118262, 1777323772, -4439253196, 11105933440, -27749232700, 69403169200

Offset: 0

Author

Paul Barry, Jan 11 2007

Keywords

• easy
• sign

Comments

Hankel transform is 3^n. In general, for r >= 0, the sequence given by Sum_{k=0..n} binomial(n, floor(k/2))*(-r)^(n-k) has Hankel transform (r+1)^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.

Second binomial transform is A026641. - Philippe Deléham, Mar 14 2007

Signed version of A100098. - Philippe Deléham, Nov 25 2007

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200

Crossrefs

Cf. A000108, A026641, A061554.

Programs

Magma

m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (1+2*x-Sqrt(1-4*x^2))/(2*Sqrt(1-4*x^2)*(1+x-Sqrt(1-4*x^2))) )); // G. C. Greubel, Feb 17 2019

Maple

a:=n->add(binomial(n,floor(k/2))*(-2)^(n-k),k=0..n): seq(a(n),n=0..30); # Muniru A Asiru, Feb 18 2019

Mathematica

CoefficientList[Series[(1/Sqrt[1-4*x^2])*(1+x*(1-Sqrt[1-4*x^2]) / (2*x^2)) /(1+2*x*(1-Sqrt[1-4*x^2])/(2*x^2)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 13 2014 *)

PARI

my(x='x+O('x^30)); Vec( (1+2*x-sqrt(1-4*x^2))/(2*sqrt(1-4*x^2)*(1+x-sqrt(1-4*x^2))) ) \\ G. C. Greubel, Feb 17 2019

Sage

((1+2*x-sqrt(1-4*x^2))/(2*sqrt(1-4*x^2)*(1+x-sqrt(1-4*x^2))) ).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 17 2019

Formula

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

a(n) = Sum_{k=0..n} A061554(n,k)*(-2)^k. - Philippe Deléham, Nov 25 2007

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

Conjecture: 2*n*a(n) + (5*n-4)*a(n-1) - 2*(4*n-3)*a(n-2) - 20*(n-2)*a(n-3) = 0. - R. J. Mathar, Nov 30 2012

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

Extensions

More terms from Vincenzo Librandi, Feb 15 2014