OFFSET
0,2
COMMENTS
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, Mar 01 2006
The column sequences give, for k=0..10: A005408 (odd numbers), A000290 (squares), A000330, 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).
This is the iterated partial sums triangle of A005408 (odd numbers). Such iterated partial sums of arithmetic progression sequences have been considered by Narayana Pandit (see the Mar 20 2015 comment on A000580 where the MacTutor History of Mathematics archive link and the Gottwald et al. reference, p. 338, are given). - Wolfdieter Lang, Mar 23 2015
LINKS
G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
FORMULA
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 antidiagonals, 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, 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, Dec 12 2007
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0)=T(1,1)=1, T(1,0)=3, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 30 2013
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(7 + 9*x + 5*x^2/2! + x^3/3!) = 7 + 16*x + 30*x^2/2! + 50*x^3/3! + 77*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 21 2014
T(n, k) = ps(1, 2; k, n-k) with ps(a, d; k, n) = sum(ps(a, d; k-1, j), j=0..n) and input ps(a, d; 0, j) = a + d*j. See the iterated partial sums comment from Mar 23 2015 above. - Wolfdieter Lang, Mar 23 2015
From Franck Maminirina Ramaharo, May 21 2018: (Start)
T(n,k) = coefficients in the expansion of ((x + 2)*(x + 1)^n - 2)/x.
T(n,k) = A097207(n,n-k).
G.f.: (y + 1)/((y - 1)*(x*y + y - 1)).
E.g.f.: ((x + 2)*exp(x*y + y) - 2*exp(y))/x.
(End)
EXAMPLE
The number triangle T(n, k) begins
n\k 0 1 2 3 4 5 6 7 8 9 10 11
0: 1
1: 3 1
2: 5 4 1
3: 7 9 5 1
4: 9 16 14 6 1
5: 11 25 30 20 7 1
6: 13 36 55 50 27 8 1
7: 15 49 91 105 77 35 9 1
8: 17 64 140 196 182 112 44 10 1
9: 19 81 204 336 378 294 156 54 11 1
10: 21 100 285 540 714 672 450 210 65 12 1
11: 23 121 385 825 1254 1386 1122 660 275 77 13 1
... reformatted by Wolfdieter Lang, Mar 23 2015
As a number square S(n, k) = T(n+k, k), 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, ...
MATHEMATICA
Table[2*Binomial[n + 1, k + 1] - Binomial[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Oct 19 2017 *)
PROG
(PARI) for(n=0, 10, for(k=0, n, print1(2*binomial(n+1, k+1) - binomial(n, k), ", "))) \\ G. C. Greubel, Oct 19 2017
CROSSREFS
KEYWORD
AUTHOR
Paul Barry, Aug 05 2005
STATUS
approved