OFFSET
0,2
COMMENTS
Row sums are (-1)^n*A000984. Diagonal sums are (-1)^n*A014301(n+1). An interesting factorization is (1/sqrt(1+4x)),(sqrt(1+4x)-1)/2)(1/(1+x),x/(1+x)).
The Z-sequence for this Riordan array is [-3,1], and the A-sequence is [1,-2,1]. For the Z- and A-sequence of Riordan arrays see the W. Lang link, with references, under A006232. - Wolfdieter Lang, Oct 18 2012
This triangle appears in the formula (x-1/x)^(2*n+1) = sum(T(n,k)*(x^(2*k+1) - 1/x^(2*k+1)),k=0..n), n >= 0. Proof from the inversion of the formula given in an Oct 18 2012 comment on A111125, due to the Riordan property. - Wolfdieter Lang, Nov 14 2012
LINKS
Indranil Ghosh, Rows 0..125 of triangle, flattened
FORMULA
Riordan array ((sqrt(1+4x)-1)/(2x*sqrt(1+4x)), (1+2x-sqrt(1+4x))/(2x)).
T(n, k)=(-1)^(n-k)*C(2n+1, n+k+1); T(n, k)=sum{j=0..n, (-1)^(n-k)*C(2n-j, n-j)C(j, k)}.
O.g.f. column k: ((2-c(-x))/(1+4*x))*(1-c(-x))^k, with the o.g.f. c(x) of A000108 (Catalan), k>=0. From the Riordan property given above. - Wolfdieter Lang, Oct 17 2012
O.g.f. of the row polynomials R(n,x) = sum(T(n,k)*x^k,k=0..n): ((2-c(-z))/(1+4*z))/(1-x*(1-c(-z))) = 1/((1+4*z)*(x-(1-x)^2*z))*(x+2*x*z-2*z + (1+x)*z*c(-z)), with the o.g.f. c(x) of A000108. - Wolfdieter Lang, Oct 18 2012
EXAMPLE
Triangle T(n,k) begins:
n\k 0 1 2 3 4 5 6 7 8 9 ...
0: 1
1: -3 1
2: 10 -5 1
3: -35 21 -7 1
4: 126 -84 36 -9 1
5: -462 330 -165 55 -11 1
6: 1716 -1287 715 -286 78 -13 1
7: -6435 5005 -3003 1365 -455 105 -15 1
8: 24310 -19448 12376 -6188 2380 -680 136 -17 1
9: -92378 75582 -50388 27132 -11628 3876 -969 171 -19 1
... Reformatted by Wolfdieter Lang, Oct 17 2012
From Wolfdieter Lang, Oct 18 2012: (Start)
Recurrence from the Z-sequence [-3,1] (see a comment above): T(3,0) = -3*T(2,0) + 1*T(2,1) = -3*10 + (-5) = -35.
Recurrence from the A-sequence [1,-2,1]: T(5,1) = 1*T(4,0) -2*T(4,1) + 1*T(4,2) = 126 -2*(-84) +36 = 330. (End)
CROSSREFS
KEYWORD
AUTHOR
Paul Barry, Oct 17 2005
STATUS
approved