0,2

Triangle A029653 less first column. In general, the product (1/(1-x),x/(1-x))*(1+m*x,x) yields the Riordan array ((1+(m-1)x)/(1-x)^2,x/(1-x)) with general term T(n,k)=(m*n-(m-1)*k+1)*C(n+1,k+1)/(n+1). This is the reversal of the (1,m)-Pascal triangle, less its first column. - Paul Barry (pbarry(AT)wit.ie), Mar 01 2006

The column sequences give, for k=0..10: A005408 (odd numbers), A000290 (squares), A005330, A002415, A005585, A040977, A050486, A053347, A054333, A054334, A057788.

Linked to Chebyshev polynomials by the fact that this triangle with interpolated zeros in the rows and columns is a scaled version of A053120.

Row sums are A033484. Diagonal sums are A001911(n+1) or F(n+4)-2. Factors as (1/(1-x),x/(1-x))*(1+2x,x). Inverse is A110814 or (-1)^(n-k)*A104709.

This triangle is a subtriangle of the [2,1] Pascal triangle A029653 (omit there the first column).

Number triangle T(n, k)=C(n, k)*(2n-k+1)/(k+1)=2*C(n+1, k+1)-C(n, k); Riordan array ((1+x)/(1-x)^2, x/(1-x)); As a number square read by anti-diagonals, T(n, k)=C(n+k, k)(2n+k+1)/(k+1).

Equals A007318 * an infinite bidiagonal matrix with 1's in the main diagonal and 2's in the subdiagonal. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 01 2007

Binomial transform of an infinite lower triangular matrix with all 1's in the main diagonal, all 2's in the subdiagonal and the rest zeros. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 12 2007

As a number triangle, rows begin

1;

3,1;

5,4,1;

7,9,5,1;

9,16,14,6,1;

11,25,30,20,7,1;

As a number square, rows begin

1,_1,_1,__1,__1,__1,...

3,_4,_5,__6,__7,__8,...

5,_9,14,_20,_27,_35,...

7,16,30,_50,_77,112,...

9,25,55,105,182,294,...

Cf. A029655.

Sequence in context: A117853 A104734 A029655 this_sequence A124883 A131809 A016574

Adjacent sequences: A110810 A110811 A110812 this_sequence A110814 A110815 A110816

easy,nonn,tabl

Paul Barry (pbarry(AT)wit.ie), Aug 05 2005