A110162 Riordan array ((1-x)/(1+x), x/(1+x)^2).

1, -2, 1, 2, -4, 1, -2, 9, -6, 1, 2, -16, 20, -8, 1, -2, 25, -50, 35, -10, 1, 2, -36, 105, -112, 54, -12, 1, -2, 49, -196, 294, -210, 77, -14, 1, 2, -64, 336, -672, 660, -352, 104, -16, 1, -2, 81, -540, 1386, -1782, 1287, -546, 135, -18, 1, 2, -100, 825, -2640, 4290, -4004, 2275, -800, 170, -20, 1

Offset: 0

Author

Paul Barry, Jul 14 2005

Keywords

• easy
• sign
• tabl

Comments

Inverse of Riordan array A094527. Rows sums are A099837. Diagonal sums are A110164. Product of Riordan array A102587 and inverse binomial transform (1/(1+x), x/(1+x)).

Coefficients of polynomials related to Cartan matrices of types C_n and B_n: p(x, n) = (-2 + x)*p(x, n - 1) - p(x, n - 2), with p(x,0) = 1; p(x,1) = 2-x; p(x,2) = x^2-4*x-2. - Roger L. Bagula, Apr 12 2008

From Wolfdieter Lang, Nov 16 2012: (Start)

The alternating row sums are given in A219233.

For n >= 1 the row polynomials in the variable x^2 are R(2*n,x):=2*T(2*n,x/2) with Chebyshev's T-polynomials. See A127672 and also the triangle A127677.

(End)

From Peter Bala, Jun 29 2015: (Start)

Riordan array has the form ( x*h'(x)/h(x), h(x) ) with h(x) = x/(1 + x)^2 and so belongs to the hitting time subgroup H of the Riordan group (see Peart and Woan).

T(n,k) = [x^(n-k)] f(x)^n with f(x) = (1 - 2*x + sqrt(1 - 4*x))/2. In general the (n,k)th entry of the hitting time array ( x*h'(x)/h(x), h(x) ) has the form [x^(n-k)] f(x)^n, where f(x) = x/( series reversion of h(x) ). (End)

Examples

			Triangle T(n,k) begins:
m\k  0    1    2     3     4     5     6    7    8   9 10 ...
0:   1
1:  -2    1
2:   2   -4    1
3:  -2    9   -6     1
4:   2  -16   20    -8     1
5:  -2   25  -50    35   -10     1
6:   2  -36  105  -112    54   -12     1
7:  -2   49 -196   294  -210    77   -14    1
8:   2  -64  336  -672   660  -352   104  -16    1
9:  -2   81 -540  1386 -1782  1287  -546  135  -18   1
10:  2 -100  825 -2640  4290 -4004  2275 -800  170 -20  1
... Reformatted and extended by _Wolfdieter Lang_, Nov 16 2012
Row polynomial n=2: P(2,x) = 2 - 4*x + x^2. R(4,x):= 2*T(4,x/2) = 2 - 4*x^2 + x^4. For P and R see a comment above. - _Wolfdieter Lang_, Nov 16 2012.
		

Links

• G. C. Greubel, Rows n=0..100 of triangle, flattened
• Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
• Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group and Chebyshev polynomials, arXiv:2502.13673 [math.CO], 2025.
• P. Peart and W.-J. Woan, A divisibility property for a subgroup of Riordan matrices, Discrete Applied Mathematics, Vol. 98, Issue 3, Jan 2000, 255-263.
• T. M. Richardson, The Reciprocal Pascal Matrix, arXiv:1405.6315 [math.CO], 2014.

Crossrefs

Cf. A128411. See A127677 for an almost identical triangle.

Cf. A136674, A053122.

Programs

Magma

/* As triangle */ [[(-1)^(n-k)*(Binomial(n+k,n-k) + Binomial(n+k-1,n-k-1)): k in [0..n]]: n in [0.. 12]]; // Vincenzo Librandi, Jun 30 2015

Mathematica

Table[If[n==0 && k==0, 1, (-1)^(n-k)*(Binomial[n+k, n-k] + Binomial[n+k-1, n-k-1])], {n, 0, 15}, {k, 0, n}]//Flatten (* G. C. Greubel, Dec 16 2018 *)

PARI

{T(n,k) = (-1)^(n-k)*(binomial(n+k,n-k) + binomial(n+k-1,n-k-1))};
for(n=0, 12, for(k=0, n, print1(T(n,k), ", "))) \\ G. C. Greubel, Dec 16 2018

Sage

[[(-1)^(n-k)*(binomial(n+k,n-k) + binomial(n+k-1,n-k-1)) for k in range(n+1)] for n in range(12)] # G. C. Greubel, Dec 16 2018

Formula

T(n,k) = (-1)^(n-k)*(C(n+k,n-k) + C(n+k-1,n-k-1)), with T(0,0) = 1. - Paul Barry, Mar 22 2007

From Wolfdieter Lang, Nov 16 2012: (Start)

O.g.f. row polynomials P(n,x) := Sum(T(n,k)*x^k, k=0..n): (1-z^2)/(1+(x-2)*z+z^2) (from the Riordan property).

O.g.f. column No. k: ((1-x)/(1+x))*(x/(1+x)^2)^k, k >= 0.

T(0,0) = 1, T(n,k) = (-1)^(n-k)*(2*n/(n+k))*binomial(n+k,n-k), n>=1, and T(n,k) = 0 if n < k. (From the Chebyshev T-polynomial formula due to Waring's formula.)

(End)

T(n,k) = -2*T(n-1,k) + T(n-1,k-1) - T(n-2,k), T(0,0)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 29 2013