A144903 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of x/((1-x-x^3)*(1-x)^(k-1)).

0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 2, 1, 1, 0, 1, 3, 3, 2, 1, 0, 1, 4, 6, 5, 3, 1, 0, 1, 5, 10, 11, 8, 4, 2, 0, 1, 6, 15, 21, 19, 12, 6, 3, 0, 1, 7, 21, 36, 40, 31, 18, 9, 4, 0, 1, 8, 28, 57, 76, 71, 49, 27, 13, 6, 0, 1, 9, 36, 85, 133, 147, 120, 76, 40, 19, 9, 0, 1, 10, 45, 121, 218, 280, 267, 196, 116, 59, 28, 13

Offset: 0

Author

Alois P. Heinz, Sep 24 2008

Keywords

• nonn
• tabl

Examples

			Square array (A(n,k)) begins:
  0, 0,  0,  0,  0,   0,   0 ... A000004;
  1, 1,  1,  1,  1,   1,   1 ... A000012;
  0, 1,  2,  3,  4,   5,   6 ... A001477;
  0, 1,  3,  6, 10,  15,  21 ... A000217;
  1, 2,  5, 11, 21,  36,  57 ... A050407;
  1, 3,  8, 19, 40,  76, 133 ... ;
  1, 4, 12, 31, 71, 147, 200 ... A027658;
Antidiagonal triangle (T(n,k)) begins as:
  0;
  0,  1;
  0,  1,  0;
  0,  1,  1,  0;
  0,  1,  2,  1,  1;
  0,  1,  3,  3,  2,  1;
  0,  1,  4,  6,  5,  3,  1;
  0,  1,  5, 10, 11,  8,  4,  2;
  0,  1,  6, 15, 21, 19, 12,  6,  3;
		

Links

• Alois P. Heinz, Antidiagonals n = 0..140, flattened

Crossrefs

Rows 0-4, 6 give: A000004, A000012, A001477, A000217, A050407(n+3), A027658.

Columns 0-9 give: A078012 and A135851(n+2), A078012(n+2) and A135851(n+4), A077868(n-1) for n>0, A050228(n-1) for n>0, A226405, A144898, A144899, A144900, A144901, A144902.

Main diagonal gives: A144904.

Cf. A000930.

Programs

Magma

A000930:= func< n | (&+[Binomial(n-2*j,j): j in [0..Floor(n/3)]]) >;
A144903:= func< n,k | k eq 0 select 0 else (&+[Binomial(n-k+j-2,j)*A000930(k-j-1) : j in [0..k-1]]) >;
[A144903(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Aug 01 2022

Maple

A:= proc(n,k) coeftayl (x/ (1-x-x^3)/ (1-x)^(k-1), x=0, n) end:
seq(seq(A(n, d-n), n=0..d), d=0..13);

Mathematica

(* First program *)
a[n_, k_] := SeriesCoefficient[x/((1-x-x^3)*(1-x)^(k-1)), {x, 0, n}];
Table[a[n-k, k], {n,0,12}, {k,n,0,-1}]//Flatten (* Jean-François Alcover, Jan 15 2014 *)
(* Second Program *)
A000930[n_]:= A000930[n]= Sum[Binomial[n-2*j,j], {j,0,Floor[n/3]}];
T[n_, k_]:= T[n, k]= If[k==0, 0, Sum[Binomial[n-k+j-2,j]*A000930[k-j-1], {j,0,k- 1}]];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Aug 01 2022 *)

SageMath

def A000930(n): return sum(binomial(n-2*j,j) for j in (0..(n//3)))
def A144903(n,k):
    if (k==0): return 0
    else: return sum(binomial(n-k+j-2,j)*A000930(k-j-1) for j in (0..k-1))
flatten([[A144903(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Aug 01 2022

Formula

G.f. of column k: x/((1-x-x^3)*(1-x)^(k-1)).

A(n, n) = A144904(n).

From G. C. Greubel, Aug 01 2022: (Start)

A(n, k) = Sum_{j=0..n-1} binomial(k+j-2, j)*A000930(n-j-1), with A(0, k) = 0.

T(n, k) = Sum_{j=0..k-1} binomial(n-k-j-2, j)*A000930(k-j-1), with T(n, 0) = 0.

T(2*n, n) = A144904(n). (End)