|
%I #26 Jul 26 2023 14:31:19
%S 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 -9,1,1,43,-135,179,-127,51,-11,1,1,-85,313,-493,433,-229,73,-13,1,1,
%U 171,-711,1299,-1359,891,-375,99,-15,1,1,-341,1593,-3309,4017,-3141,1641,-573,129,-17,1
%N 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 The denominators are given by 2*A130321(n,m).
%C 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 R is the rational Riordan triangle (1/(2-x), x/(1+x)).
%C 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 Row sums of the rational triangle R are [1/2, seq(3/2^(n+1), for n >= 1)].
%C Row sums of the present triangle T give [repeat(1,2,)].
%C 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 Alternating row sums of the present triangle T give A084567.
%C The inverse of the T Riordan matrix is ((1-3*x)/(1-2*x), x/(1-2*x) = A251636.
%C Equals A248810 when the first column (m = 0) of ones is removed. - _Georg Fischer_, Jul 26 2023
%H Wolfdieter Lang, <a href="/A251634/a251634.pdf">Eleven rows of the triangle, and rational triangle.</a>
%F 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 O.g.f. column m of the rational triangle R: (1/(2-x))*(x/(1+x))^m, m >= 0 (Riordan property of R).
%F 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 T(n, k) = k!*S(n, k) where S(n, k) is recursively defined by:
%F 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 The triangle T(n,m) begins:
%e n\m 0 1 2 3 4 5 6 7 8 9 10 ...
%e 0: 1
%e 1: 1 1
%e 2: 1 -1 1
%e 3: 1 3 -3 1
%e 4: 1 -5 9 -5 1
%e 5: 1 11 -23 19 -7 1
%e 6: 1 -21 57 -61 33 -9 1
%e 7: 1 43 -135 179 -127 51 -11 1
%e 8: 1 -85 313 -493 433 -229 73 -13 1
%e 9: 1 171 -711 1299 -1359 891 -375 99 -15 1
%e ...
%e The rational Riordan triangle R(n,m) begins:
%e n\m 0 1 2 3 4 5 ...
%e 0: 1/2
%e 1: 1/4 1/2
%e 2: 1/8 -1/4 1/2
%e 3: 1/16 3/8 -3/4 1/2
%e 4: 1/32 -5/16 9/8 -5/4 1/2
%e 5: 1/64 11/3 -23/1 19/8 -7/4 1/2
%e ...
%e For more rows see the link.
%p A251634 := proc(n, k) local S; S := proc(n, k) option remember; `if`(k = 0, 1,
%p `if`(k > n, 0, S(n-1, k-1)/k - 2*S(n-1, k))) end: k!*S(n, k) end:
%p seq(seq(A251634(n, k), k=0..n)), n=0..9); # _Peter Luschny_, Jan 19 2020
%Y Cf. A029635, A084567, A102900, A130321, A251636, A248810, A331333.
%K sign,tabl,easy
%O 0,8
%A _Wolfdieter Lang_, Jan 09 2015
|
