A349841 Triangle T(n,k) built by placing all ones on the left edge, [1,0,0,0,0] repeated on the right edge, and filling the body using the Pascal recurrence T(n,k) = T(n-1,k) + T(n-1,k-1).

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 4, 1, 1, 1, 5, 10, 10, 5, 2, 0, 1, 6, 15, 20, 15, 7, 2, 0, 1, 7, 21, 35, 35, 22, 9, 2, 0, 1, 8, 28, 56, 70, 57, 31, 11, 2, 0, 1, 9, 36, 84, 126, 127, 88, 42, 13, 2, 1

Offset: 0

Author

Michael A. Allen, Dec 13 2021

Keywords

• easy
• nonn
• tabl

Comments

This is the m=5 member in the sequence of triangles A007318, A059259, A118923, A349839, A349841 which have all ones on the left side, ones separated by m-1 zeros on the other side, and whose interiors obey Pascal's recurrence.

T(n,k) is the (n,n-k)-th entry of the (1/(1-x^5),x/(1-x)) Riordan array.

For n>0, T(n,n-1) = A002266(n+4).

For n>1, T(n,n-2) = A008732(n-2).

For n>2, T(n,n-3) = A122047(n-1).

Sums of rows give A349842.

Sums of antidiagonals give A349843.

Examples

			Triangle begins:
  1;
  1,   0;
  1,   1,   0;
  1,   2,   1,   0;
  1,   3,   3,   1,   0;
  1,   4,   6,   4,   1,   1;
  1,   5,  10,  10,   5,   2,   0;
  1,   6,  15,  20,  15,   7,   2,   0;
  1,   7,  21,  35,  35,  22,   9,   2,   0;
  1,   8,  28,  56,  70,  57,  31,  11,   2,   0;
  1,   9,  36,  84, 126, 127,  88,  42,  13,   2,   1;
		

Links

• Michael A. Allen and Kenneth Edwards, On Two Families of Generalizations of Pascal's Triangle, J. Int. Seq. 25 (2022) Article 22.7.1.

Crossrefs

Other members of sequence of triangles: A007318, A059259, A118923, A349839.

Columns: A000012, A001477, A000217, A000292, A000332, A323228.

Diagonals: A079998, A002266, A008732, A122047.

Programs

Mathematica

Flatten[Table[CoefficientList[Series[(1 - x*y)/((1 - (x*y)^5)(1 - x - x*y)), {x, 0, 20}, {y, 0, 10}], {x, y}][[n+1,k+1]],{n,0,10},{k,0,n}]]

Formula

G.f.: (1-x*y)/((1-(x*y)^5)(1-x-x*y)) in the sense that T(n,k) is the coefficient of x^n*y^k in the series expansion of the g.f.

T(n,0) = 1.

T(n,n) = delta(n mod 5,0).

T(n,1) = n-1 for n>0.

T(n,2) = (n-1)*(n-2)/2 for n>1.

T(n,3) = (n-1)*(n-2)*(n-3)/6 for n>2.

T(n,4) = (n-1)*(n-2)*(n-3)*(n-4)/24 for n>3.

T(n,5) = C(n-1,5) + 1 for n>4.

T(n,6) = C(n-1,6) + n - 6 for n>5.

For 0 <= k < n, T(n,k) = (n-k)*Sum_{j=0..floor(k/5)} binomial(n-5*j,n-k)/(n-5*j).

The g.f. of the n-th subdiagonal is 1/((1-x^5)(1-x)^n).