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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)

The rows of this triangle are symmetric up to sign. Row sums = 2 after row 0. Unsigned row sums = A116466. Row squared sums = A116467. Central terms of odd rows: T(2*n+1,n+1) = |A064310(n)|.

Both triangles A112555 and A114700 have the property that the m-th matrix power of the triangles satisfy T^m = I + m*(T - I). So it is curious that the row squared sums of A112555 is a bisection of the unsigned row sums of A114700.

a(n)/a(n-1) tends to -4.

a(n)/a(n-1) tends to -5.

a(n)/a(n-1) tends to -6.

a(n)/a(n-1) tends to -7.

a(n)/a(n-1) tends to -8.

a(n)/a(n-1) tends to -9.

First term < 0: a(27) = -60053864762402471338497.