A113187 Inverse of twin-prime related triangle A111125.

1, -3, 1, 10, -5, 1, -35, 21, -7, 1, 126, -84, 36, -9, 1, -462, 330, -165, 55, -11, 1, 1716, -1287, 715, -286, 78, -13, 1, -6435, 5005, -3003, 1365, -455, 105, -15, 1, 24310, -19448, 12376, -6188, 2380, -680, 136, -17, 1, -92378, 75582, -50388, 27132, -11628, 3876, -969, 171, -19, 1, 352716, -293930, 203490

Offset: 0

Author

Paul Barry, Oct 17 2005

Keywords

• easy
• sign
• tabl

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

Examples

			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)
		

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