A118650 Center column a(2n,n) of A106597.

1, 3, 14, 72, 388, 2150, 12134, 69370, 400398, 2328162, 13616892, 80022612, 472133638, 2794893246, 16592160944, 98743919468, 588913687966, 3518968100982, 21062444053040, 126256612255260, 757853797478188

Offset

0,2

Comments

Number of lattice paths from (0,0) to (n,n) using steps (1,0), (0,1), and (s,s) for s>=1. [Joerg Arndt, Jul 01 2011]

Links

Table of n, a(n) for n=0..20.

Formula

G.f.: (1-x)/sqrt(1-8*x+12*x^2-4*x^3). - Mark van Hoeij, Apr 16 2013

Conjecture: n*a(n) +3*(-3*n+2)*a(n-1) +4*(5*n-8)*a(n-2) +2*(-8*n+21)*a(n-3) +2*(2*n-7)*a(n-4)=0. - R. J. Mathar, Nov 10 2013

From Emanuele Munarini, Feb 06 2017: (Start)

a(n) = Sum_{k=0..n} binomial(2*k,k)*(-1)^(n-k) * Sum_{j=0..k} binomial(k,j)*binomial(k+1,n-k-j)*2^(k-j).

Proof of Mathar's recurrence.

Let A(t) be the g.f. of the coefficients a(n). Then we have the identity (1 - 9*t + 20*t^2 - 16*t^3 + 4*t^4)*A'(t) = (3 - 8*t + 6*t^2 - 2*t^3)*A(t).

Let R be the incremental ratio, i.e. the operator defined by RA(t) = (A(t)-A(0))/t, giving the g.f. of the shifted sequence a(n+1), then we have

R^3A'(t) - 9*R^2A'(t) + 20*RA'(t) - 16*A'(t) + 4*t*A'(t) = 3*R^3A(t) - 8*R^2A(t) + 6*RA(t) - 2*A(t), from which we obtain the recurrence:

(n+4)*a(n+4)-3*(3*n+10)*a(n+3)+4*(5*n+12)*a(n+2)-2*(8*n+11)*a(n+1)+2*(2*n+1)*a(n)=0.

This proves the above conjecture.

(End)

Mathematica

Table[Sum[Binomial[2k, k](-1)^(n-k)Sum[Binomial[k, j]Binomial[k+1, n-k-j]2^(k-j), {j, 0, k}], {k, 0, n}], {n, 0, 40}] (* Emanuele Munarini, Feb 06 2017 *)

Prog

(Maxima) makelist(sum(binomial(2*k, k)*(-1)^(n-k)*sum(binomial(k, j)*binomial(k+1, n-k-j)*2^(k-j), j, 0, k), k, 0, n), n, 0, 12); /* Emanuele Munarini, Feb 06 2017 */

Crossrefs

Cf. A106597.

Sequence in context: A009637 A098648 A026295 * A180187 A295104 A080238

Adjacent sequences: A118647 A118648 A118649 * A118651 A118652 A118653

Keyword

nonn

Author

Joshua Zucker, May 10 2006

Status

approved