OFFSET
0,8
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.
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.
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).
EXAMPLE
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;
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}]]
CROSSREFS
KEYWORD
AUTHOR
Michael A. Allen, Dec 13 2021
STATUS
approved