A127359 a(n) = Sum_{k=0..n} binomial(n, floor(k/2))*3^(n-k).

1, 4, 14, 48, 162, 544, 1820, 6080, 20290, 67680, 225684, 752448, 2508468, 8362176, 27875064, 92919168, 309734850, 1032458080, 3441543140, 11471842880, 38239537852, 127465249344, 424884399624, 1416281802368, 4720940242612, 15736469278144, 52454901060680

Offset: 0

Author

Paul Barry, Jan 11 2007

Keywords

• easy
• nonn

Comments

Hankel transform is (-2)^n. In general, given r>=0, the sequence given by Sum_{k=0..n} C(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)/(1-3x) under the Chebyshev mapping g(x)->(1/sqrt(1-4x^2))*g(xc(x^2)), where c(x) is the g.f. of the Catalan numbers A000108.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..300
• Isaac DeJager, Madeleine Naquin, and Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.

Crossrefs

Cf. A107430. - Philippe Deléham, Sep 16 2009

Cf. A000108 (Catalan numbers).

Programs

GAP

a:=[1, 4, 14];; for n in [4..30] do a[n]:=(2*(5*n-2)*a[n-1] +4*(3*n-14)*a[n-2] -40*(n-3)*a[n-3])/(3*(n-1)); od; a; # G. C. Greubel, Dec 15 2019

Magma

I:=[1, 4, 14]; [n le 3 select I[n] else (2*(5*n-2)*Self(n-1) + 4*(3*n - 14)*Self(n-2) -40*(n-3)*Self(n-3))/(3*(n-1)): n in [1..30]]; // G. C. Greubel, Dec 15 2019

Maple

A127359:=n->sum(binomial(n,floor(k/2))*3^(n-k), k=0..n): seq(A127359(n), n=0..30); # Wesley Ivan Hurt, Mar 14 2015

Mathematica

Table[Sum[Binomial[n, Floor[k/2]]*3^(n-k), {k,0,n}], {n,0,30}] (* Vaclav Kotesovec, Oct 19 2012 *)

PARI

a(n) = sum(j=0, n, binomial(n, j\2)*3^(n-j));
vector(31, n, a(n-1)) \\ G. C. Greubel, Dec 15 2019

Sage

[sum(binomial(n, floor(j/2))*3^(n-j) for j in (0..n)) for n in (0..30)] # G. C. Greubel, Dec 15 2019

Formula

G.f.: (1/sqrt(1-4*x^2))*(1+x*c(x^2))/(1-3*x*c(x^2)), where c(x) = (1 - sqrt(1 - 4*x))/(2*x).

a(n) = Sum_{k=0..n} A061554(n,k)*3^k. - Philippe Deléham, Dec 04 2009

Recurrence: 3*n*a(n) = 2*(5*n + 3)*a(n-1) + 4*(3*n - 11)*a(n-2) - 40*(n-2)*a(n-3). - Vaclav Kotesovec, Oct 19 2012

a(n) ~ 4*10^n/3^(n+1). - Vaclav Kotesovec, Oct 19 2012