A251634 Numerators of inverse Riordan triangle of Riordan triangle A029635. Riordan (1/(1-x), x/(1+2*x)). Triangle read by rows for 0 <= m <= n.

1, 1, 1, 1, -1, 1, 1, 3, -3, 1, 1, -5, 9, -5, 1, 1, 11, -23, 19, -7, 1, 1, -21, 57, -61, 33, -9, 1, 1, 43, -135, 179, -127, 51, -11, 1, 1, -85, 313, -493, 433, -229, 73, -13, 1, 1, 171, -711, 1299, -1359, 891, -375, 99, -15, 1, 1, -341, 1593, -3309, 4017, -3141, 1641, -573, 129, -17, 1

Offset: 0

Author

Wolfdieter Lang, Jan 09 2015

Keywords

• sign
• tabl
• easy

Comments

The denominators are given by 2*A130321(n,m).

The rational lower triangular matrix with entries R(n,m) = T(n,m)/(2*A130321(n,m)) = T(n,m)/2^(n-m+1) for n >= m >= 0 and 0 otherwise is the inverse of the Riordan matrix A029635.

R is the rational Riordan triangle (1/(2-x), x/(1+x)).

The numerator triangle T is the Riordan array (1/(1-x), x/(1+2*x)). From the o.g.f. of the column sequences of R and T(n,m) = 2^(n-m+1)*R(n,m).

Row sums of the rational triangle R are [1/2, seq(3/2^(n+1), for n >= 1)].

Row sums of the present triangle T give [repeat(1,2,)].

Alternating row sums of the rational triangle R give (-1)^n*A102900(n)/2^(n+1), n >= 0: 1/2, -1/4, 7/8, -25/16, 103/32, -409/64, 1639/128, -6553/256, 26215/512, ... .

Alternating row sums of the present triangle T give A084567.

The inverse of the T Riordan matrix is ((1-3*x)/(1-2*x), x/(1-2*x)) = A251636.

Equals A248810 when the first column (m = 0) of ones is removed. - Georg Fischer, Jul 26 2023

Examples

			The triangle T(n,m) begins:
  n\m  0    1    2     3     4     5    6    7    8   9 ...
  0:   1
  1:   1    1
  2:   1   -1    1
  3:   1    3   -3     1
  4:   1   -5    9    -5     1
  5:   1   11  -23    19    -7     1
  6:   1  -21   57   -61    33    -9    1
  7:   1   43 -135   179  -127    51  -11    1
  8:   1  -85  313  -493   433  -229   73  -13    1
  9:   1  171 -711  1299 -1359   891 -375   99  -15   1
  ...
The rational Riordan triangle R(n,m) begins:
  n\m  0      1      2      3     4    5  ...
  0:  1/2
  1:  1/4    1/2
  2:  1/8   -1/4    1/2
  3:  1/16   3/8   -3/4    1/2
  4:  1/32  -5/16   9/8   -5/4   1/2
  5:  1/64  11/3  -23/1   19/8  -7/4  1/2
  ...
For more rows see the link.
		

Links

• Wolfdieter Lang, Eleven rows of the triangle, and rational triangle.

Crossrefs

Cf. A029635, A084567, A102900, A130321, A251636, A248810, A331333.

Programs

Maple

A251634 := proc(n, k) local S; S := proc(n, k) option remember; `if`(k = 0, 1,
`if`(k > n, 0, S(n-1, k-1)/k - 2*S(n-1, k))) end: k!*S(n, k) end:
seq(seq(A251634(n, k), k=0..n)), n=0..9); # Peter Luschny, Jan 19 2020

Formula

O.g.f. of the row polynomials P(n,x) = Sum_{m=0..n} R(n,m)*x^m of the rational triangle R: G(z,x) = Sum_{n>=0} P(n,x)*z^n = (1+z)/((2-z)*(1+(1-x)*z)).

O.g.f. column m of the rational triangle R: (1/(2-x))*(x/(1+x))^m, m >= 0 (Riordan property of R).

O.g.f. column m of the numerator triangle T: (1/(1-x))*(x/(1+2*x))^m, m >= 0. (Riordan property of T).

T(n, k) = k!*S(n, k) where S(n, k) is recursively defined by:

if k = 0 then 1 else if k > n then 0 else S(n-1, k-1)/k - 2*S(n-1, k). - Peter Luschny, Jan 19 2020