A127040 a(n) = binomial(floor((3n+4)/2),floor(n/2)).

1, 1, 5, 6, 28, 36, 165, 220, 1001, 1365, 6188, 8568, 38760, 54264, 245157, 346104, 1562275, 2220075, 10015005, 14307150, 64512240, 92561040, 417225900, 600805296, 2707475148, 3910797436, 17620076360, 25518731280, 114955808528

Offset

0,3

Comments

With offset 2, the number of compositions of n into floor(n/2) parts, which is an upper bound for A007874.

Links

T. D. Noe, Table of n, a(n) for n=0..200

Formula

From Benedict W. J. Irwin, Aug 16 2016: (Start)

G.f.: (-1 + (2*cos(arcsin(3*sqrt(3)*x/2)/3))/sqrt(4-27*x^2) + 3*x^3*2F1(4/3,5/3;5/2;27*x^2/4))/(3*x^2).

E.g.f.: 2F3(4/3,5/3;1/2,3/2,2;27*x^2/16) + x*2F3(4/3,5/3;1,3/2,5/2;27*x^2/16).

(End)

D-finite with recurrence 8*(n+2)*(n+1)*a(n) -84*(n-1)*(n+1)*a(n-1) +6*(-33*n^2+54*n-8)*a(n-2) +9*(63*n^2-63*n-16)*a(n-3) +108*(3*n-5)*(3*n-7)*a(n-4)=0. - R. J. Mathar, Feb 08 2021

Maple

seq(sum(binomial(n+k, k-1), k=0..ceil((n+1)/2)), n=0..28); # Zerinvary Lajos, Apr 11 2007

Mathematica

CoefficientList[Series[(-1 + (2 Cos[1/3 ArcSin[(3 Sqrt[3] x)/2]])/Sqrt[4 - 27 x^2] + 3 x^3 Hypergeometric2F1[4/3, 5/3, 5/2, (27 x^2)/4])/(3 x^2), {x, 0, 20}], x] (* Benedict W. J. Irwin, Aug 16 2016 *)
Table[Binomial[Floor[(3 n + 4)/2], Floor[n/2]], {n, 0, 28}] (* Michael De Vlieger, Aug 18 2016 *)

Prog

(PARI) a(n) = binomial((3*n+4)\2, n\2); \\ Michel Marcus, Sep 09 2016

Crossrefs

Cf. A004319 (bisection), A025174 (bisection), A099578.

Sequence in context: A115761 A320047 A249221 * A041011 A152118 A041056

Adjacent sequences: A127037 A127038 A127039 * A127041 A127042 A127043

Keyword

nonn

Author

T. D. Noe, Jan 03 2007

Status

approved