cp's OEIS Frontend

Row sums = A134508

Row sums are Fibonacci(n+2). Diagonal sums are A016116. - Paul Barry, Jul 07 2004

Riordan array (1/(1-x), x/(1-x^2)). Matrix inverse is A106180. - Paul Barry, Apr 24 2005

As an infinite lower triangular matrix * [1,2,3,...] = A055244. - Gary W. Adamson, Dec 23 2008

From Emeric Deutsch, Jun 18 2010: (Start)

T(n,k) is the number of alternating parity increasing subsequences of {1,2,...,n} of size k, starting with an odd number (Terquem's problem, see the Riordan reference, p. 17). Example: T(8,5)=6 because we have 12345, 12347, 12367, 12567, 14567, and 34567.

T(n,k) is the number of alternating parity increasing subsequences of {1,2,...,n,n+1} of size k, starting with an even number. Example: T(7,4)=5 because we have 2345, 2347, 2367, 2567, and 4567. (End)

From L. Edson Jeffery, Mar 01 2011: (Start)

This triangle can be constructed as follows. Interlace two copies of the table of binomial coefficients to get the preliminary table

1

1

1 1

1 1

1 2 1

1 2 1

1 3 3 1

1 3 3 1

...,

then shift each entire r-th column up r rows, r=0,1,2,.... Also, a signed version of this sequence (A187660 in tabular form) begins with

1;

1, -1;

1, -1, -1;

1, -2, -1, 1;

1, -2, -3, 1, 1;

...

(compare with A066170, A130777). Let T(N,k) denote the k-th entry in row N of the signed table. Then, for N>1, row N gives the coefficients of the characteristic function p_N(x) = Sum_{k=0..N} T(N,k)*x^(N-k) = 0 of the N X N matrix U_N=[(0 ... 0 1);(0 ... 0 1 1);...;(0 1 ... 1);(1 ... 1)]. Now let Q_r(t) be a polynomial with recurrence relation Q_r(t)=t*Q_(r-1)(t)-Q_(r-2)(t) (r>1), with Q_0(t)=1 and Q_1(t)=t. Then p_N(x)=0 has solutions Q_(N-1)(phi_j), where phi_j=2*(-1)^(j-1)*cos(j*Pi/(2*N+1)), j=1,2,...,N.

For example, row N=3 is {1,-2,-1,1}, giving the coefficients of the characteristic function p_3(x) = x^3-2*x^2-x+1 = 0 for the 3 X 3 matrix U_3=[(0 0 1);(0 1 1);(1 1 1)], with eigenvalues Q_2(phi_j)=[2*(-1)^(j-1)*cos(j*Pi/7)]^2-1, j=1,2,3. (End)

Given the signed polynomials (+--++--,...) of the triangle, the largest root of the n-th row polynomial is the longest (2n+1) regular polygon diagonal length, with edge = 1. Example: the largest root to x^3 - 2x^2 - x + 1 = 0 is 2.24697...; the longest heptagon diagonal, sin(3*Pi/7)/sin(Pi/7). - Gary W. Adamson, Sep 06 2011

Given the signed polynomials from Gary W. Adamson's comment, the largest root of the n-th polynomial also equals the length from the center to a corner (vertex) of a regular 2*(2*n+1)-sided polygon with side (edge) length = 1. - L. Edson Jeffery, Jan 01 2012

Put f(x,0) = 1 and f(x,n) = x + 1/f(x,n-1). Then f(x,n) = u(x,n)/v(x,n), where u(x,n) and v(x,n) are polynomials. The flattened triangles of coefficients of u and v are both essentially A046854, as indicated by the Mathematica program headed "Polynomials". - Clark Kimberling, Oct 12 2014

From Jeremy Dover, Jun 07 2016: (Start)

T(n,k) is the number of binary strings of length n+1 starting with 0 that have exactly k pairs of consecutive 0's and no pairs of consecutive 1's.

T(n,k) is the number of binary strings of length n+2 starting with 1 that have exactly k pairs of consecutive 0's and no pairs of consecutive 1's. (End)

The original name of this sequence was: Triangle giving coefficients of characteristic function of n X n matrix in which the left upper half and the antidiagonal are filled with 1's and the right lower half is filled with 0's. As was pointed out by L. Edson Jeffery this is only correct if we multiply each triangle row by (-1)^n. For the straightforward version of the coefficients of the characteristic polynomials see A187660. - Johannes W. Meijer, Aug 08 2011