This site is supported by donations to The OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A321198 Triangular Riordan matrix T = R^(-1) for triangular Riordan matrix R = (1/(1 + x^2 - x^3), x/(1 + x^2 - x^3) given in A321196. 3
 1, 0, 1, 1, 0, 1, -1, 2, 0, 1, 2, -2, 3, 0, 1, -5, 5, -3, 4, 0, 1, 8, -12, 9, -4, 5, 0, 1, -21, 21, -21, 14, -5, 6, 0, 1, 42, -56, 40, -32, 20, -6, 7, 0, 1, -96, 114, -108, 66, -45, 27, -7, 8, 0, 1, 222, -270, 225, -180, 100, -60, 35, -8, 9, 0, 1 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,8 COMMENTS Riordan triangle T = (f(t), t*f(t)), with f(t) = F^{[-1]}(t)/t, where F^{[-1]}(t) is the compositional inverse of t = F(x) = x/(1 + x^2 - x^3). The expansion of f(t) is given by {(-1)^n*A001005(n)}_{n >= 0}, the sequence of column k = 0. This gives the inverse matrix (with upper diagonals filled with zeros) of the Riordan matrix from A321196 for any finite dimension. The inverse of the Riordan matrix ((1/(1 + x^2 + x^3), x/(1 + x^2 + x^3)) is obtained from the triangle t(n, k) = (-1)^(n-k)*T(n, k), with vanishing upper diagonals. The row sums give A321199. The alternating row sums give A321200. The finite A- and Z-sequences of this inverse Riordan triangle of (F(x)/x, F(x)) are A = [1, 0, +1, -1] generated by 1/(F(x)/x), and Z = [0, +1, -1] generated by 1/F(x) - 1/x. See the W. Lang link for A- and Z- sequences in A006232 with references. For the Boas-Buck column recurrences of Riordan triangles see the Aug 10 2017 remark in A046521, also for two references. For this Bell-type Riordan triangle the Boas-Buck sequence b is generated by B(t) = (log(f(t))' = (1/(1/f(t) - t^2*f(t) + 2*t^3*f(t)^2) - 1)/t, and b(n) = (-1)^(n)*A176806(n+1), for n >= 0, because the parity of e_3 in the rows n of A321201 coincides with the one of n. See A321203 for the multinomials with negative signs for odd row numbers. LINKS FORMULA Recurrence (from A-and Z-sequences): T(n, k) = 0 for n < k, T(0, 0) = 1. Z: T(n, 0) = T(n-1, 1) - T(n-1, 2), n >= 1; A: T(n, k) = T(n-1, k-1) + T(n-1, k+1) - T(n-1, k+2), n >= k >= 0. Recurrence for column k (Boas-Buck type): T(n, n) = 1; T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} b(n-1-j)*T(j, k), n >= m+1 >= 1. For b see the Boas-Buck comment above. G.f of row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k: G(x, z) = f(z)/(1 - x*z*f(z)), with f(z) = F^{[-1]}(z)/z, where F^{[-1]}(z) is the compositional inverse of z = F(y) = y/(1 + y^2 - y^3). G.f of column k: Gcol(k, x) = x^k*f(x)^{k+1}. EXAMPLE The triangle begins: n\k     0     1     2     3    4    5   6   7  8  9  10 ... ----------------------------------------------------------- 0:      1 1:      0     1 2:      1     0     1 3:     -1     2     0     1 4:      2    -2     3     0    1 5:     -5     5    -3     4    0    1 6:      8   -12     9    -4    5    0   1 7:    -21    21   -21    14   -5    6   0   1 8:     42   -56    40   -32   20   -6   7   0  1 9:    -96   114  -108    66  -45   27  -7   8  0  1 10:   222  -270   225  -180  100  -60  35  -8  9  0   1 ... ------------------------------------------------------------ Recurrence (from A- and Z-sequence): Z: T(5, 0) = T(4, 1) - T(4, 2) = -2 - 3 = -5. A: T(5, 2) = T(4, 1) + T(4, 3) - T(4, 4) = - 2 + 0 - 1 = -3. Recurrence column k = 2 (Boas-Buck type sequence b = (-1)^(n+1)* = {0, 2, -3, 6, ...}): T(5, 2) = (3/3)*(b(2)*T(2, 2) + b(1)*T(3, 2) + b(0)*T(4, 2)) = -3*1 + 2*0 + 0*4 = -3. CROSSREFS Cf. A001005, A176806, A321196, A321199, A321200, A321201, A321203. Sequence in context: A166291 A162986 A319203 * A128584 A080099 A268040 Adjacent sequences:  A321195 A321196 A321197 * A321199 A321200 A321201 KEYWORD sign,tabl AUTHOR Wolfdieter Lang, Nov 12 2018 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified March 19 15:02 EDT 2019. Contains 321330 sequences. (Running on oeis4.)