A144898 Expansion of x/((1-x-x^3)*(1-x)^4).

0, 1, 5, 15, 36, 76, 147, 267, 463, 775, 1262, 2011, 3150, 4867, 7438, 11268, 16951, 25358, 37766, 56047, 82945, 122482, 180553, 265798, 390880, 574358, 843432, 1237966, 1816384, 2664311, 3907237, 5729077, 8399372, 12313154, 18049371, 26456513, 38778103

Offset

0,3

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (5,-10,11,-9,7,-4,1).

Formula

G.f.: x/((1-x-x^3)*(1-x)^4).

From G. C. Greubel, Jul 27 2022: (Start)

a(n) = Sum_{j=0..floor((n+3)/3)} binomial(n-2*j+3, j+4).

a(n) = A099567(n+3, 4). (End)

Maple

a:= n-> (Matrix(7, (i, j)-> if i=j-1 then 1 elif j=1 then [5, -10, 11, -9, 7, -4, 1][i] else 0 fi)^n)[1, 2]: seq(a(n), n=0..40);

Mathematica

CoefficientList[Series[ x/((1-x-x^3)(1-x)^4), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 06 2013 *)

Prog

(Magma)
A144898:= func< n | n eq 0 select 0 else (&+[Binomial(n-2*j+3, j+4): j in [0..Floor((n+3)/3)]]) >;
[A144898(n): n in [0..40]]; // G. C. Greubel, Jul 27 2022
(SageMath)
def A144898(n): return sum(binomial(n-2*j+3, j+4) for j in (0..((n+3)//3)))
[A144898(n) for n in (0..40)] # G. C. Greubel, Jul 27 2022

Crossrefs

5th column of A144903.

Cf. A000930, A050228, A077868, A144899, A144900, A144901, A144902, A144903, A144904, A226405.

Cf. A078012, A099567, A135851.

Sequence in context: A086716 A046776 A360486 * A163250 A053808 A111926

Adjacent sequences: A144895 A144896 A144897 * A144899 A144900 A144901

Keyword

nonn,easy

Author

Alois P. Heinz, Sep 24 2008

Status

approved