cp's OEIS Frontend

This is the m=3, t=3 member of a two-parameter family of triangles such that T(n,k) is the number of tilings of an (n+(t-1)*k) X 1 board using k (1,m-1;t)-combs and n-k unit square tiles. A (1,g;t)-comb is composed of a line of t unit square tiles separated from each other by gaps of width g.

T(3*j+r-2*k,k) is the coefficient of x^k in (f(j,x))^(3-r)*(f(j+1,x))^r for r=0,1, where f(n,x) is a Narayana's cows polynomial defined by f(n,x)=f(n-1,x)+x*f(n-3,x)+delta(n,0) where f(n<0,x)=0.

T(n+6-2*k,k) is the number of subsets of {1,2,...,n} of size k such that no two elements in a subset differ by 3 or 6.

Values generate solutions to the recurrence a(n) = a(n-1) + k(k+1)* a(n-2), a(0)=1, a(1) = k(k+1)+1. Values and sequences associated with this table are included in A072024.

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.

The fourth diagonal is 1, 2, 5, 11, 21, ..., which is 1 + A000292. The fifth diagonal is 0, 2, 7, 18, 39, 75, 132, 217, 338, 504, 725, 1012, ..., which is A051743.

The array A007318 is generated by placing A000012 on both edges with the same Pascal-like recurrence, and the array A059259 uses edges defined by A000012 and A059841. - R. J. Mathar, Jan 21 2008

From Michael A. Allen, Nov 30 2021: (Start)

T(n,n-k) is the (n,k)-th entry of the (1/(1-x^3), x/(1-x)) Riordan array.

Sums of rows give A077947.

Sums of antidiagonals give A079962. (End)

If the triangle is viewed as a square array S(m, k) = T(m+k, k), 0 <= m, 0 <= k, its first row is (1,0,1,0,1,...) with e.g.f. cosh(x), g.f. 1/(1-x^2) and subsequent rows have g.f. 1/((1+x)^n*(1-x^2)) (substitute x for -x in g.f. for A059259).

By column, S(m, k) is the coefficient of [x^m] in the generating function Sum_{i=0..k} (-1)^i/(1-x)^(i+1).

This is a rational generating function down column k with a power of (1-x) in the denominator; therefore column k is a polynomial in m respectively n. - Mathew Englander, May 14 2014

Column k multiplied by k! seems to correspond to row k of A054651, considered as a polynomial and then evaluated on the negative integers. For example, row 5 of A054651 represents the polynomial x^5 - 5*x^4 + 25*x^3 + 5*x^2 + 94*x + 120. Evaluating that for x = -1, x = -2, x = -3, ... gives (0, -360, -1440, -4080, -9600, -19920, -37680, ...) which is 5! times column 5 of this triangle. - Mathew Englander, May 23 2014

This triangle provides a solution to a question in the mathematics of gambling. For 0 < p < 1 and positive integers N and G with N < G, suppose you begin with N dollars and make repeated wagers, each time winning 1 dollar with probability p and losing 1 dollar with probability 1-p. You continue betting 1 dollar at a time until you have either G dollars (your Goal) or 0 (bankrupt). What is the probability of reaching your Goal before going bankrupt, as a function of p, N, and G? (This is a type of one-dimensional random walk.) Answer: Let Q_m_(x) be the polynomial whose coefficients are given by row m-1 of the triangle (e.g., Q_6_(x) = 1 - 4x + 7x^2 - 6x^3 + 3x^4). Then, the probability of reaching G dollars before going bankrupt is p^(G-N)*Q_N_(p)/Q_G_(p). - Mathew Englander, May 23 2014

From Paul Curtz, Mar 17 2017: (Start)

Consider the triangle Ja(n+1,k) (here, but generally Ja(n,k)) composed of the triangle a(n) prepended with a column of 0's, i.e.,

0;

0, 1;

0, 1, 0;

0, 1, -1, 1;

0, 1, -2, 2, 0;

0, 1, -3, 4, -2, 1;

0, 1, -4, 7, -6, 3, 0;

0, 1, -5, 11, -13, 9, -3, 1;

... .

The row sums are 0, 1, 1, ... = A057427(n), the most elementary autosequence of the first kind (a sequence of the first kind has 0's as main diagonal of its array of successive differences).

The row sums of the absolute values are A001045(n).

Ja applied to a sequence written in its reluctant form yields an autosequence of the first kind. Example: the reluctant form of A001045(n) is 0, 0, 1, 0, 1, 1, 0, 1, 1, 3, 0, 1, 1, 3, 5, ... = Jl.

Jl multiplied by Ja gives the triangle Jal:

0;

0, 1;

0, 1, 0;

0, 1, -1, 3;

0, 1, -2, 6, 0;

0, 1, -3, 12, -10, 11;

0, 1, -4, 21, -30, 33, 0;

0, 1, -5, 33, -65, 99, -63, 43;

... .

The row sums are A001045(n). (End)