A026377 a(n) = number of integer strings s(0),...,s(n) counted by array T in A026374 that have s(n)=4; also a(n) = T(2n,n-2).

1, 9, 58, 330, 1770, 9198, 46928, 236736, 1185645, 5909805, 29362806, 145570230, 720606705, 3563543025, 17610412600, 86989143480, 429579843435, 2121099312195, 10472653252550, 51708363376950, 255326054688320, 1260886172311524, 6227515552731528, 30762417293645400

Offset: 2

Author

Clark Kimberling

Keywords

• nonn
• changed

Comments

Number of lattice paths from (0,0) to (2n,4), using steps U=(1,1), D=(1,-1) and at even levels(zero, positive and negative) also H=(2,0). Example: a(3)=9 because we have UUUUUD, UUUUDU, UUUDUU, UUDUUU, UDUUUU, DUUUUU, HUUUU, UUHUU and UUUUH. - Emeric Deutsch, Jan 30 2004

Links

• G. C. Greubel, Table of n, a(n) for n = 2..1000

Crossrefs

Cf. A026374.

Programs

Maple

series( (3*x-1+(1-6*x+7*x^2)/sqrt(5*x^2-6*x+1))/(2*x^2), x=0, 30); # Mark van Hoeij, Apr 18 2013

Mathematica

CoefficientList[Series[(3*x-1+(1-6*x+7*x^2)/Sqrt[5*x^2-6*x+1])/(2*x^2), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 07 2013 *)

PARI

x='x+O('x^66); Vec((3*x-1+(1-6*x+7*x^2)/sqrt(5*x^2-6*x+1))/(2*x^2)) /* Joerg Arndt, Apr 19 2013 */

Formula

From Emeric Deutsch, Jan 30 2004: (Start)

a(n) = [t^(n+2)](1+3t+t^2)^n.

a(n) = Sum_{j=ceiling((n+2)/2)..n} (3^(2j-n-2)*binomial(n, j)*binomial(j, n+2-j)). (End)

From Paul Barry, Sep 20 2004: (Start)

E.g.f.: exp(3x) * BesselI(2, 2x).

a(n) = Sum_{k=0..n} binomial(n, k)*binomial(2k, k+2). (End)

Conjecture: n*(n+4)*a(n) - 3*(n+2)*(2*n+3)*a(n-1) + 5*(n+2)*(n+1)*a(n-2) = 0. - R. J. Mathar, Nov 24 2012

G.f.: (3*x - 1 + (1-6*x+7*x^2)/sqrt(5*x^2-6*x+1))/(2*x^2). - Mark van Hoeij, Apr 18 2013

a(n) ~ 5^(n+1/2)/(2*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 07 2013

Assuming offset 0: a(n) = C(2*n+4,n)*hypergeom([-n,-n-4],[-3/2-n],-1/4). - Peter Luschny, May 09 2016