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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.

%I A251634 #33 Apr 11 2025 07:59:09
%S A251634 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,
%T A251634 -9,1,1,43,-135,179,-127,51,-11,1,1,-85,313,-493,433,-229,73,-13,1,1,
%U A251634 171,-711,1299,-1359,891,-375,99,-15,1,1,-341,1593,-3309,4017,-3141,1641,-573,129,-17,1
%N 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.
%C A251634 The denominators are given by 2*A130321(n,m).
%C A251634 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.
%C A251634 R is the rational Riordan triangle (1/(2-x), x/(1+x)).
%C A251634 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).
%C A251634 Row sums of the rational triangle R are [1/2, seq(3/2^(n+1), for n >= 1)].
%C A251634 Row sums of the present triangle T give [repeat(1,2,)].
%C A251634 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, ... .
%C A251634 Alternating row sums of the present triangle T give A084567.
%C A251634 The inverse of the T Riordan matrix is ((1-3*x)/(1-2*x), x/(1-2*x)) = A251636.
%C A251634 Equals A248810 when the first column (m = 0) of ones is removed. - _Georg Fischer_, Jul 26 2023
%H A251634 Wolfdieter Lang, <a href="/A251634/a251634.pdf">Eleven rows of the triangle, and rational triangle.</a>
%F A251634 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)).
%F A251634 O.g.f. column m of the rational triangle R: (1/(2-x))*(x/(1+x))^m, m >= 0 (Riordan property of R).
%F A251634 O.g.f. column m of the numerator triangle T: (1/(1-x))*(x/(1+2*x))^m, m >= 0. (Riordan property of T).
%F A251634 T(n, k) = k!*S(n, k) where S(n, k) is recursively defined by:
%F A251634 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
%e A251634 The triangle T(n,m) begins:
%e A251634   n\m  0    1    2     3     4     5    6    7    8   9 ...
%e A251634   0:   1
%e A251634   1:   1    1
%e A251634   2:   1   -1    1
%e A251634   3:   1    3   -3     1
%e A251634   4:   1   -5    9    -5     1
%e A251634   5:   1   11  -23    19    -7     1
%e A251634   6:   1  -21   57   -61    33    -9    1
%e A251634   7:   1   43 -135   179  -127    51  -11    1
%e A251634   8:   1  -85  313  -493   433  -229   73  -13    1
%e A251634   9:   1  171 -711  1299 -1359   891 -375   99  -15   1
%e A251634   ...
%e A251634 The rational Riordan triangle R(n,m) begins:
%e A251634   n\m  0      1      2      3     4    5  ...
%e A251634   0:  1/2
%e A251634   1:  1/4    1/2
%e A251634   2:  1/8   -1/4    1/2
%e A251634   3:  1/16   3/8   -3/4    1/2
%e A251634   4:  1/32  -5/16   9/8   -5/4   1/2
%e A251634   5:  1/64  11/3  -23/1   19/8  -7/4  1/2
%e A251634   ...
%e A251634 For more rows see the link.
%p A251634 A251634 := proc(n, k) local S; S := proc(n, k) option remember; `if`(k = 0, 1,
%p A251634 `if`(k > n, 0, S(n-1, k-1)/k - 2*S(n-1, k))) end: k!*S(n, k) end:
%p A251634 seq(seq(A251634(n, k), k=0..n)), n=0..9); # _Peter Luschny_, Jan 19 2020
%Y A251634 Cf. A029635, A084567, A102900, A130321, A251636, A248810, A331333.
%K A251634 sign,tabl,easy
%O A251634 0,8
%A A251634 _Wolfdieter Lang_, Jan 09 2015