A129818 Riordan array (1/(1+x), x/(1+x)^2), inverse array is A039599.

1, -1, 1, 1, -3, 1, -1, 6, -5, 1, 1, -10, 15, -7, 1, -1, 15, -35, 28, -9, 1, 1, -21, 70, -84, 45, -11, 1, -1, 28, -126, 210, -165, 66, -13, 1, 1, -36, 210, -462, 495, -286, 91, -15, 1, -1, 45, -330, 924, -1287, 1001, -455, 120, -17, 1, 1, -55, 495, -1716, 3003, -3003, 1820, -680, 153, -19, 1

Offset: 0

Author

Philippe Deléham, Jun 09 2007

Keywords

• sign
• tabl

Comments

This sequence is up to sign the same as A129818. - T. D. Noe, Sep 30 2011

Row sums: A057078. - Philippe Deléham, Jun 11 2007

Subtriangle of the triangle given by (0, -1, 0, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 19 2012

This triangle provides the coefficients of powers of x^2 for the even-indexed Chebyshev S polynomials (see A049310): S(2*n,x) = Sum_{k=0..n} T(n,k)*x^(2*k), n >= 0. - Wolfdieter Lang, Dec 17 2012

If L(x^n) := C(n) = A000108(n) (Catalan numbers), then the polynomials P_n(x) := Sum_{k=0..n} T(n,k)*x^k are orthogonal with respect to the inner product given by (f(x),g(x)) := L(f(x)*g(x)). - Michael Somos, Jan 03 2019

Examples

			Triangle T(n,k) begins:
  n\k  0   1    2     3     4     5    6    7    8   9 10 ...
   0:  1
   1: -1   1
   2:  1  -3    1
   3: -1   6   -5     1
   4:  1 -10   15    -7     1
   5: -1  15  -35    28    -9     1
   6:  1 -21   70   -84    45   -11    1
   7: -1  28 -126   210  -165    66  -13    1
   8:  1 -36  210  -462   495  -286   91  -15    1
   9: -1  45 -330   924 -1287  1001 -455  120  -17   1
  10:  1 -55  495 -1716  3003 -3003 1820 -680  153 -19  1
  ... Reformatted by _Wolfdieter Lang_, Dec 17 2012
Recurrence from the A-sequence A115141:
15 = T(4,2) = 1*6 + (-2)*(-5) + (-1)*1.
(0, -1, 0, -1, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, ...) begins:
  1
  0,  1
  0, -1,   1
  0,  1,  -3,   1
  0, -1,   6,  -5,  1
  0,  1, -10,  15, -7,  1
  0, -1,  15, -35, 28, -9, 1. - _Philippe Deléham_, Mar 19 2012
Row polynomial for n=3 in terms of x^2: S(6,x) = -1 + 6*x^2 -5*x^4 + 1*x^6, with Chebyshev's S polynomial. See a comment above. - _Wolfdieter Lang_, Dec 17 2012
Boas-Buck type recurrence: -35 = T(5,2) = (5/3)*(-1*1 +1*(-5) - 1*15) = -3*7 = -35. - _Wolfdieter Lang_, Jun 03 2020
		

Links

• Vincenzo Librandi, Rows n = 1..101, flattened
• Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
• Paul Barry and Aoife Hennessy, The Euler-Seidel Matrix, Hankel Matrices and Moment Sequences, J. Int. Seq. 13 (2010) # 10.8.2, example 15.
• Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.

Crossrefs

Cf. A039599, A085478, A123970, A321620.

Programs

Maple

# The function RiordanSquare is defined in A321620.
RiordanSquare((1 - sqrt(1 - 4*x))/(2*x), 10):
LinearAlgebra[MatrixInverse](%); # Peter Luschny, Jan 04 2019

Mathematica

max = 10; Flatten[ CoefficientList[#, y] & /@ CoefficientList[ Series[ (1 + x)/(1 + (2 - y)*x + x^2), {x, 0, max}], x]] (* Jean-François Alcover, Sep 29 2011, after Wolfdieter Lang *)

Sage

@CachedFunction
def A129818(n,k):
    if n< 0: return 0
    if n==0: return 1 if k == 0 else 0
    h = A129818(n-1,k) if n==1 else 2*A129818(n-1,k)
    return A129818(n-1,k-1) - A129818(n-2,k) - h
for n in (0..9): [A129818(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

Formula

T(n,k) = (-1)^(n-k)*A085478(n,k) = (-1)^(n-k)*binomial(n+k,2*k).

Sum_{k=0..n} T(n,k)*A000531(k) = n^2, with A000531(0)=0. - Philippe Deléham, Jun 11 2007

Sum_{k=0..n} T(n,k)*x^k = A033999(n), A057078(n), A057077(n), A057079(n), A005408(n), A002878(n), A001834(n), A030221(n), A002315(n), A033890(n), A057080(n), A057081(n), A054320(n), A097783(n), A077416(n), A126866(n), A028230(n+1) for x = 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16, respectively. - Philippe Deléham, Nov 19 2009

O.g.f.: (1+x)/(1+(2-y)*x+x^2). - Wolfdieter Lang, Dec 15 2010

O.g.f. column k with leading zeros (Riordan array, see NAME): (1/(1+x))*(x/(1+x)^2)^k, k >= 0. - Wolfdieter Lang, Dec 15 2010

From Wolfdieter Lang, Dec 20 2010: (Start)

Recurrences from the Z- and A-sequences for Riordan arrays. See the W. Lang link under A006232 for details and references.

T(n,0) = -1*T(n-1,0), n >= 1, from the o.g.f. -1 for the Z-sequence (trivial result).

T(n,k) = Sum_{j=0..n-k} A(j)*T(n-1,k-1+j), n >= k >= 1, with A(j):= A115141(j) = [1,-2,-1,-2,-5,-14,...], j >= 0 (o.g.f. 1/c(x)^2 with the A000108 (Catalan) o.g.f. c(x)). (End)

T(n,k) = (-1)^n*A123970(n,k). - Philippe Deléham, Feb 18 2012

T(n,k) = -2*T(n-1,k) + T(n-1,k-1) - T(n-2,k), T(0,0) = T(1,1) = 1, T(1,0) = -1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Mar 19 2012

A039599(m,n) = Sum_{k=0..n} T(n,k) * C(k+m) where C(n) are the Catalan numbers. - Michael Somos, Jan 03 2019

Equals the matrix inverse of the Riordan square (cf. A321620) of the Catalan numbers. - Peter Luschny, Jan 04 2019

Boas-Buck type recurrence for column k >= 0 (see Aug 10 2017 comment in A046521 with references): T(n,k) = ((1 + 2*k)/(n - k))*Sum_{j = k..n-1} (-1)^(n-j)*T(j,k), with input T(n,n) = 1, and T(n,k) = 0 for n < k. - Wolfdieter Lang, Jun 03 2020