A099578 a(n) = binomial(floor((3n+2)/2), floor(n/2)).

1, 1, 4, 5, 21, 28, 120, 165, 715, 1001, 4368, 6188, 27132, 38760, 170544, 245157, 1081575, 1562275, 6906900, 10015005, 44352165, 64512240, 286097760, 417225900, 1852482996, 2707475148, 12033222880, 17620076360, 78378960360, 114955808528

Offset: 0

Author

Paul Barry, Oct 23 2004

Keywords

• easy
• nonn

Comments

Main diagonal of triangle A099575.

With offset 2, this is the number of compositions of n-1 into floor(n/2) parts. - T. D. Noe, Jan 05 2007

From Petros Hadjicostas, Jul 19 2018: (Start)

We clarify the above comment by T. D. Noe. The number of compositions of N into K positive parts is C(N-1, K-1). This was proved by MacMahon in 1893 (and probably by others before him). The number of compositions of N into K nonnegative parts is C(N+K-1, K-1) because for every composition b_1 + ... + b_K = N with b_i >= 0 for all i, we may create another composition c_1 + ... + c_K = N+K with c_i = b_i + 1 >= 1.

The statement of T. D. Noe above means that, for n>=2, a(n-2) is the number of compositions of N = n-1 into K = floor(n/2) nonnegative parts. Thus, a(n-2) = C(N+K-1, K-1) = C(n-1+floor(n/2)-1, floor(n/2)-1) = C(floor((3(n-2)+2)/2), floor((n-2)/2)).

This interpretation is important for T. D. Noe's comments for sequence A030077, whose unknown general formula remains an unsolved problem (as of July 2018).

It should be noted, however, that for most authors "composition" means "composition into positive parts". The phrase "weak composition" is sometimes used for a "composition into nonnegative parts".

(End)

Examples

			From _Petros Hadjicostas_, Jul 19 2018: (Start)
With n=2 there are a(2-2) = a(0) = 1 compositions of 2-1 = 1 into floor(2/2) = 1 nonnegative parts, namely 1 (only).
With n=3 there are a(3-2) = a(1) = 1 compositions of 3-1 = 2 into floor(3/2) = 1 nonnegative parts, namely 2 (only).
With n=4 there are a(4-2) = a(2) = 4 compositions of 4-1 = 3 into floor(4/2) = 2 nonnegative parts, namely 0+3, 3+0, 1+2, and 2+1.
With n=5 there are a(5-2) = a(3) = 5 compositions of 5-1 = 4 into floor(5/2) = 2 nonnegative parts, namely 0+4, 4+0, 1+3, 3+1, and 2+2.
With n=6 there are a(6-2) = a(4) = 21 compositions of 6-1 = 5 into floor(6/2) = 3 nonnegative parts, namely the 3 permutations of 1+1+3, the 3 permutations of 1+2+2, the 3 permutations of 0+0+5, the 6 permutations of 0+1+4, and the 6 permutations of 0+2+3.
(End)
		

Links

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

Crossrefs

Cf. A025174 (bisection), A030077, A045721 (bisection), A099575, A127040.

Programs

Magma

[(&+[Binomial(n+j, j): j in [0..Floor(n/2)]]): n in [0..40]]; // G. C. Greubel, Jul 24 2022

Maple

A099578:=n->binomial(floor((3*n+2)/2), floor(n/2)); seq(A099578(k), k=0..50); # Wesley Ivan Hurt, Nov 01 2013

Mathematica

Table[Binomial[Floor[(3n+2)/2], Floor[n/2]], {n, 0, 50}] (* Wesley Ivan Hurt, Nov 01 2013 *)
CoefficientList[Series[-((Sqrt[4 -27 x^2] -2(Cos[1/3 ArcSin[(3 Sqrt[3] x)/2]] + Sqrt[3] Sin[2/3 ArcSin[(3 Sqrt[3] x)/2]]))/(3 x Sqrt[4 -27 x^2])), {x, 0, 20}], x] (* Benedict W. J. Irwin, Aug 15 2016 *)
a[n_] := Binomial[2*n+1, n]*Hypergeometric2F1[-n, n+1, -2*n-1, -1]; Flatten[Table[a[n], {n, 0, 29}]] (* Detlef Meya, Dec 25 2023 *)

PARI

a(n) = binomial((3*n+2)\2, n\2); \\ Michel Marcus, Nov 02 2013

SageMath

[binomial((3*n+2)//2, n//2) for n in (0..40)] # G. C. Greubel, Jul 24 2022

Formula

a(n) = Sum_{k=0..floor(n/2)} binomial(n+k, k).

40*n*(n+1)*a(n) +36*n*(n-2)*a(n-1) -6*(45*n^2-23)*a(n-2) -27*(3*n-4)*(3*n-5)*a(n-3) = 0. - R. J. Mathar, Oct 30 2014

From Benedict W. J. Irwin, Aug 15 2016: (Start)

G.f.: -( (sqrt(4-27*x^2) - 2*(cos(arcsin(3*sqrt(3)*x/2)/3) + sqrt(3)*sin(2*arcsin(3*sqrt(3)*x/2)/3)) )/( 3*x*sqrt(4-27*x^2)) ).

E.g.f.: Hypergeometric2F3(2/3,4/3;1/2,1,3/2;27*x^2/16) + x*Hypergeometric2F3(4/3,5/3;3/2,3/2,2;27*x^2/16).

(End)

Recurrence: 4*n*(n+1)*(6*n-1)*a(n) = 18*n*a(n-1) + 3*(3*n-2)*(3*n-1)*(6*n+5)*a(n-2). - Vaclav Kotesovec, Aug 15 2016

a(n) = binomial(2*n+1, n)*hypergeom([-n, n+1], [-2*n-1], -1). - Detlef Meya, Dec 25 2023