A001005 -id:A001005 - cp's OEIS frontend

A319201 Expansion of f(t) = F^{[-1]}(t)/t, where F(x) = x/(1 - x^2 - x^3).

1, 0, -1, -1, 2, 5, -2, -21, -14, 72, 138, -165, -803, -143, 3575, 4732, -11674, -36244, 15130, 195738, 152456, -799102, -1700272, 2042975, 11038183, 2582670, -53547795, -76684530, 185864265, 618689190, -231325605, -3506922585, -2974386450, 14866619160, 33459332610, -38401746930, -223156727472

Offset: 0

Wolfdieter Lang, Nov 06 2018

sign

The compositional inverse F^{[-1]} of F(x) = x/(1 - x^2 - x^3) (appearing in the Riordan matrix of the Bell type R = (F(x)/x, F(x)) from A104578) is needed for the inverse matrix of R, the Riordan matrix R^(-1) = (f(t), t*f(t)).
This function f(t) is also needed for the A- and Z-sequences of this Riordan matrix R, namely A(n) = [t^n](1/f(t)) = A319202(n) and Z(n) = [t^n]((1/t)*(1/f(t) - 1)) = A319202(n+1).
For the expansion of the compositional inverse of  x/(1 + x^2 + x^3) see A001005.

			a(8) = (1/9)*[x^8] (1- x^2 - x^3)^9 = (1/9)*(-126) = -14.
a(8) = (1/9)*(- 9!/(6!*1!*2!) + 9!/(5!*0!*4!)) = -14, from the two solutions for [e2, e3], namely [1, 2] (parity odd) and [0, 4] (parity even).

Cf. A001005, A104578, A319202.

f := series(x/(1 - x^2 -x^3), x, 40):
r := gfun:-seriestoseries(f, 'revogf'):
gf := convert(r, polynom) / x:
seq(coeff(gf,x,n), n=0..36); # Peter Luschny, Nov 09 2018

a(n) = [t^n] G(t), where G(t) =  F^{[-1]}(t)/t, and F(x) = x/(1 - x^2 - x^3).
O.g.f.: G(t) can be computed by Lagrange inversion of F.
a(n) = b(n+1)/(n+1), for n >= 0, where b(n+1) is the coefficient of x^n of (1 - x^2 - x^3)^(n+1). This follows from the Lagrange inversion series for G(x) = F^{[-1]}(x)/x.
a(n) = (1/(n+1))*(Sum_{2*e2 + 3*e3 = n} (-1)^{e2 + e3}*(n+1)!/(n+1 - (e2 + e3))!*e2!*e3!) (from the multinomial formula for (x1 + x2 + x3)^(n+1)) with x1 = 1, x2 = -x^2 and x3 = -x^3. For the solutions of 2*e2 + 3*e3 = n >= 2, and the parity of e2 + e3, see the array A321201.

A365730 G.f. satisfies A(x) = 1 + x^4A(x)^4(1 + x*A(x)).

Original entry on oeis.org

1, 0, 0, 0, 1, 1, 0, 0, 4, 9, 5, 0, 22, 78, 91, 35, 140, 680, 1224, 969, 1254, 5985, 14630, 17710, 17710, 55660, 164450, 269100, 299520, 593775, 1805076, 3681405, 4951692, 7594752, 20173560, 47303520, 76404460, 110676324, 239784864, 589602585, 1106339923

Offset: 0

Seiichi Manyama, Sep 17 2023

nonn

Cf. A365727, A365728, A365729, A365731.
Cf. A001005, A001006, A365724.
Cf. A215341.

a(n) = sum(k=0, n\4, binomial(k, n-4*k)*binomial(n+1, k))/(n+1);

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/4)} binomial(k,n-4*k) * binomial(n+1,k).

A365724 G.f. satisfies A(x) = 1 + x^3A(x)^3(1 + x*A(x)).

Original entry on oeis.org

1, 0, 0, 1, 1, 0, 3, 7, 4, 12, 45, 55, 77, 286, 546, 728, 1960, 4760, 7548, 15504, 39729, 75582, 140448, 336490, 723327, 1366200, 2992990, 6758895, 13522275, 28094040, 63183315, 133231800, 273896532, 600805296, 1305229332, 2720740792, 5843241088, 12797739672

Offset: 0

Seiichi Manyama, Sep 17 2023

nonn

Cf. A308616, A365723, A365725, A365726.
Cf. A001005, A001006, A365730.
Cf. A114997.

a(n) = sum(k=0, n\3, binomial(k, n-3*k)*binomial(n+1, k))/(n+1);

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/3)} binomial(k,n-3*k) * binomial(n+1,k).

A176806 Consider asymmetric 1-D random walk with set of possible jumps {-1,+1,+2}. Sequence gives number of paths of length n ending at origin.

Original entry on oeis.org

1, 0, 2, 3, 6, 20, 35, 105, 238, 588, 1512, 3630, 9339, 23166, 58487, 148148, 373230, 949416, 2406248, 6122142, 15591856, 39729000, 101432982, 259049230, 662421643, 1695149220, 4341026900, 11125755615, 28530984915, 73213888650, 187980163110, 482906682675

Offset: 0

Sergey Perepechko, Apr 26 2010

It appears that a(n) is the coefficient of x^n in the expansion (1+x^2+x^3)^n. - Joerg Arndt, Jul 01 2011 [For the proof see the formula section. - Wolfdieter Lang, Nov 05 2018]

			a(3) = 3: (+2-1-1) or (-1+2-1) or (-1-1+2).
From _Wolfdieter Lang_, Nov 05 2018: (Start)
a(8) = (1/8!)*(d/dt)^8 (1 + t^2 + t^3)^8 becomes for t = 0: 238. (See the comment with the conjecture by _Joerg Arndt_, now proved.)
a(8) = 168 + 70 = 238, the row sum of row n = 8 of A321203, arising from the two [e2, e3] pairs [1, 2] and [4, 0], given in row n = 8 of A321201.
(End)

Cf. A001005, A321201, A321203.

a:=n->add(binomial(n,k)*binomial(k,3*k-n),k=floor((n+2)/3)..floor(n/2));

Table[Sum[Binomial[n, k]*Binomial[k, 3*k-n], {k, Floor[(n+2)/3], Floor[n/2]}], {n, 0, 30}] (* Vaclav Kotesovec, Mar 01 2016 *)

a(n) = Sum_{k=floor((n+2)/3)..floor(n/2)} binomial(n,k)*binomial(k,3*k-n).
G.f. g(x) satisfies (31*x^3 + 18*x^2 - x - 4)*g(x)^3 + (x+3)*g(x) + 1 = 0.
Recurrence: 2*n*(2*n-1)*(52*n-79)*a(n) + (n-1)*(52*n^2-79*n+36)*a(n-1) - 6*(n-1)*(156*n^2-315*n+106)*a(n-2) - 31*(n-1)*(n-2)*(52*n-27)*a(n-3) = 0.
a(n) ~ c * d^n / sqrt(Pi*n), where d = 2.61071861327603934981864900838405862... is the root of the equation -31 - 18*d + d^2 + 4*d^3 = 0 and c = 0.57803237802255683003114674597591616... is the root of the equation -31 + 324*c^2 - 1248*c^4 + 1664*c^6 = 0. - Vaclav Kotesovec, Mar 01 2016
From Wolfdieter Lang, Nov 05 2018: (Start)
G.f. G(x) = x*(d/dx)log(F^{[-1]}) = f(x)/(1 - (x*f(x))^2 - 2*(x*f(x))^3) = f(x)/(3 - 2*f(x) + (x*f(x))^2), where f(x) = F^{[-1]}(x)/x, and F^{[-1]})(x) is the compositional inverse of F(y) = x/(1 + x^2 + x^3); that is, F(F^{[-1]}(x)) = x, identically. This G can be proved to solve the equation given above for the g.f. g, if one applies the identity for f (as done above in the last formula for G): (x*f(x))^3 + (x*f(x))*2 - f(x) + 1 = 0 (following from the equation for F^{[-1]}). The expansion of f is given in A001005.
The g.f. G(x) can be computed from the general Lagrange series for the function h(t) with derivative h(t)" = 1/phi(t), where phi(t) = (1 + t^2 + t^3), and the inversion of x = y/phi(y) = F(y). Then one finds G(x) = (d/dx)h(F^{[-1]}(x)) = (1/phi(F^{[-1]}(x)))*(d/dx)F^{[-1]}(x), which becomes with the above mentioned identity for f(x) = F^{[-1]}(x)/x the result G(x) = f(x)/(3 - 2*f(x) + (x*f(x))^2).
From this special Lagrange series derivation and the proof that the g.f. g from above coincides with G, the conjecture, given by Joerg Arndt as a comment above, has been proved. This uses [t^n]phi(t)^n = (1/n!)*(d/dt)^n phi(t)^n, evaluated at t = 0, which appears in the considered Lagrange series.
a(n) = Sum_{2*e + 3*e3 = n} n!/((n - (e2+e3))!*e2!*e3!), n >= 2, with a(0) = 1 and a(1) = 0. This is the row sum of the irregular table A321203 of these multinomial numbers for the solutions for the pairs (e2, e3). The pairs of solutions are given in A321201.
(End)

A321196 Riordan triangle T = (1/(1 + x^2 - x^3), x/(1 + x^2 - x^3)).

Original entry on oeis.org

1, 0, 1, -1, 0, 1, 1, -2, 0, 1, 1, 2, -3, 0, 1, -2, 3, 3, -4, 0, 1, 0, -6, 6, 4, -5, 0, 1, 3, -1, -12, 10, 5, -6, 0, 1, -2, 12, -4, -20, 15, 6, -7, 0, 1, -3, -7, 30, -10, -30, 21, 7, -8, 0, 1, 5, -16, -15, 60, -20, -42, 28, 8, -9, 0, 1

Offset: 0

Wolfdieter Lang, Nov 09 2018

This is the (ordinary) convolution triangle based on A077961 (the column k = 0 of T).
The row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k, with R(-1, x) = 0, appear in the Cayley-Hamilton formula for nonnegative powers of a 3 X 3 matrix with Det M = sigma(3; 3) = x1*x2*x3 = +1, sigma(3; 2) := x1*x2 + x1*x*3 + x2*x^3 = +1 and Tr M = sigma(3; 1) = x1 + x2 = x, where x1, x2, and x3 are the eigenvalues of M, and sigma the elementary symmetric functions, as M^n = R(n-2, x)*M^2 + (-R(n-3, x) + R(n-4, x))*M + R(n-3, x)*1_3, for n >= 3, where M^0 = 1_3 is the 3 X 3 unit matrix.
For the Cayley-Hamilton formula for 3 X 3 matrices with Det M = +1, sigma(3,2) = -1 and Tr(M) = x see A104578.
The row sums give A133872 (repeat(1, 1, 0, 0)). The alternating row sums give A057597(n+2), for n >= 0.
The Riordan triangle (1/(1 + x^2 + x^3), x/(1 + x^2 + x^3)) has entries t(n, m) = (-1)^(n-m)*T(n, m) (from the g.f. G(-x, -z), where the g.f. G of T is given below).
The inverse of Riordan T is T^{-1}, given in A321198.

			The triangle T(n, k) 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:   1   2  -3   0   1
5:  -2   3   3  -4   0   1
6:   0  -6   6   4  -5   0  1
7:   3  -1 -12  10   5  -6  0  1
8:  -2  12  -4 -20  15   6 -7  0  1
9:  -3  -7  30 -10 -30  21  7 -8  0  1
10:  5 -16 -15  60 -20 -42 28  8 -9  0  1
...
Cayley-Hamilton formula for the matrix TS(x) =[[x,-1,1], [1,0,0], [0,1,0]] with Det(TS(x)) = +1, sigma(3, 2) = +1, and Tr(TS(x)) = x. For n = 3: TS(x)^3 = R(1, x)*TS(x)^2 + (-R(0, x) + R(-1, x))*TS(x) + R(0, x)*1_3 = x*TS(x)^2 - TS(x) + 1_3. Compare this for x = -1 with r^3 = R(3)*r^2 + (-R(2) + R(1))*r + R(2)*1 = r^2 - r + 1, where r = 1/t = A192918, with the tribonacci constant t = A058265, and R(n) = A057597(n) = R(n-2, -1).
Recurrence: T(5, 2) = T(4, 1) - T(3, 2) + T(2, 2) = 1 -(-1) + 1 = 3.
Boas-Buck type recurrence with B = {0, -2, 3, ...}:
  T(5, 2) = ((2+1)/(5-2))*(3*1 + (-2)*0 + 0*(-3)) = 1*3 = 3.
Z- and A-recurrence with A(n) = {1, 0, -1, 1, -1, ...} and Z(n) = A(n+1):
  T(4, 0) = 0*T(3, 0) - 1*T(3, 1) + 1*T(3, 2) - 1*T(3, 3) = 0 + 2 + 0 - 1 = 1.
  T(5, 2) = 1*T(4, 1) + 0*T(4, 2) - 1*T(4, 3) + 1*T(4, 4) = 2 + 0 + 0 + 1 = 3.

Cf. A057597, A058265, A077961, A001005, A104578, A112455, A133872, A192918, A321197, A321198.

T[n_, k_] /; 0 <= k <= n := T[n, k] = T[n - 1, k - 1] - T[n - 2, k] + T[n - 3, k]; T[0, 0] = 1; T[, ] = 0;
Table[T[n, k], {n, 0, 10}, {k, 0, n}] (* Jean-François Alcover, Jul 06 2019 *)

# uses[riordan_array from A256893]
riordan_array(1/(1 + x^2 - x^3), x/(1 + x^2 - x^3), 11) # Peter Luschny, Nov 13 2018

T(n, k) = T(n-1, k-1) - T(n-2, k) + T(n-3, k), T(0, 0) = 1, T(n,k) = 0 if n < k or if k < 0. (Cf. A104578.)
The Riordan property T = (G(x), x*G(x)) with G(x) = 1/(1 + x^2 - x^3) implies the following.
G.f. of row polynomials R(n, x) is G(x, z) = 1/(1 - x*z + z^2 - z^3).
G.f. of column sequence k: x^k/(1 + x^2 - x^3)^(k+1), k >= 0.
Boas-Buck recurrence (see the Aug 10 2017 remark in A046521, also for two references):
T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} B(n-1-j)*T(j, k), for n >= 1, k = 0,1, ..., n-1, and input T(n, n) = 1, for n >= 0. Here B(n) = [x^n]*(d/dx)log(G(x)) = x*(-2 + 3*x)/(1 + x^2 - x^3) = (-1)^n*A112455(n+1), for n >= 0.
Recurrences from the A- and Z- sequences (see the W. Lang link under A006232 with references), which are A(n) = A321197(n) and Z(n) = A(n+1).
  T(0, 0) = 1, T(n, k) = 0 for n < k, and
  T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1, and
  T(n, k) = Sum_{j=0..n-k} A(j)*T(n-1, k-1+j), for n >= m >= 1.

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.

Original entry on oeis.org

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

Offset: 0

Wolfdieter Lang, Nov 12 2018

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.

			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.

Cf. A001005, A176806, A321196, A321199, A321200, A321201, A321203.

T[n_, k_] := T[n - 1, k - 1] + T[n - 1, k + 1] - T[n - 1, k + 2] /; n >= k >= 0;Table[T[n, k], {n, 0, 10}, {k, 0, n}] (* James C. McMahon, Dec 10 2023 *)

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}.

A321199 Row sums of Riordan triangle A321198.

Original entry on oeis.org

1, 1, 2, 2, 4, 2, 7, -5, 16, -40, 74, -196, 431, -999, 2433, -5575, 13600, -32264, 77794, -188630, 456078, -1113702, 2716432, -6656382, 16351231, -40236773, 99308252, -245471638, 608083547, -1508944093, 3750434282, -9336906118, 23277132752, -58112501728, 145266946742, -363574425058

Offset: 0

Wolfdieter Lang, Nov 12 2018

The alternating row sums of triangle A321198 are given in A321200.

Cf. A001005, A077961, A321198, A321200.

a(n) = Sum_{k=0..n} A321198(n, k), n >= 0.
G.f.: f(x)/(1 - x*f(x)), with f(x) = F^{[-1]}(x)/x = Sum_{n >= 0}
(-1)^(n+1)*A001005(n)*x^n, where F^{[-1]}(x) is the compositional inverse of F(y) = y/(1 + y^2 - y^3) (see A077961 for F(y)/y).

A321200 Alternating row sums of Riordan triangle A321198.

Original entry on oeis.org

1, -1, 2, -4, 8, -18, 39, -89, 204, -472, 1110, -2616, 6231, -14909, 35861, -86705, 210364, -512480, 1252350, -3069638, 7544818, -18589202, 45907708, -113608590, 281698359, -699748003, 1741102844, -4338995332, 10828981851, -27063384783, 67722954114, -169674183372, 425590855116, -1068654838488

Offset: 0

Wolfdieter Lang, Nov 12 2018

The row sums of triangle A321198 are given in A321199.

Cf. A001005, A077961, A321198, A321199.

a(n) = Sum_{k=0..n} A321198(n, k), n >= 0.

G.f.: f(x)/(1 + x*f(x)), with f(x) = F^{[-1]}(x)/x = Sum_{n >= 0} (-1)^(n+1)*A001005(n)*x^n, where F^{[-1]}(x) is the compositional inverse of F(y) = y/(1 + y^2 - y^3) (see A077961 for F(y)/y).

A321197 a(n) gives the A-sequence for the Riordan matrix (1/(1 + x^2 - x^3), x/(1 + x^2 - x^3)) from A321196.

Original entry on oeis.org

1, 0, -1, 1, -1, 3, -4, 10, -20, 42, -98, 210, -492, 1122, -2607, 6149, -14443, 34463, -82238, 197574, -476918, 1154402, -2807516, 6845016, -16743674, 41067512, -100967539, 248843095, -614546545, 1520779665

Offset: 0

Wolfdieter Lang, Oct 30 2018

sign

See the recurrence formula for A321196 from the A- and Z-sequences.

Cf. A001005, A321196.

a(n) = [t^n] (1/f(t)), where f(t) = F^{[-1]}(t)/t, with the compositional inverse F^{[-1]}(t) of F(x) = 1/(1 + x^2 - x^3). The expansion of f is given by (-1)^n*A001005(n), for n >= 0.

A365609 G.f. satisfies A(x) = 1 + x^2A(x)^4(1 + x*A(x)).

Original entry on oeis.org

1, 0, 1, 1, 4, 9, 27, 78, 231, 715, 2193, 6954, 21999, 70840, 228896, 746650, 2447757, 8072208, 26745627, 89002364, 297344960, 996865397, 3352918429, 11310307593, 38256171642, 129718262583, 440855654827, 1501451066767, 5123671576890, 17516503865294

Offset: 0

Seiichi Manyama, Sep 17 2023

nonn

Cf. A001005, A025250, A055113, A217358.

Cf. A365690.

a(n) = sum(k=0, n\2, binomial(k, n-2*k)*binomial(n+2*k+1, k)/(n+2*k+1));

a(n) = Sum_{k=0..floor(n/2)} binomial(k,n-2*k) * binomial(n+2*k+1,k) / (n+2*k+1).

cp's OEIS Frontend

A319201 Expansion of f(t) = F^{[-1]}(t)/t, where F(x) = x/(1 - x^2 - x^3).

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Maple

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A365730 G.f. satisfies A(x) = 1 + x^4*A(x)^4*(1 + x*A(x)).

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PARI

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A365724 G.f. satisfies A(x) = 1 + x^3*A(x)^3*(1 + x*A(x)).

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PARI

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A176806 Consider asymmetric 1-D random walk with set of possible jumps {-1,+1,+2}. Sequence gives number of paths of length n ending at origin.

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Maple

Mathematica

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A321196 Riordan triangle T = (1/(1 + x^2 - x^3), x/(1 + x^2 - x^3)).

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Mathematica

Sage

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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.

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Mathematica

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A321199 Row sums of Riordan triangle A321198.

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A321200 Alternating row sums of Riordan triangle A321198.

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A321197 a(n) gives the A-sequence for the Riordan matrix (1/(1 + x^2 - x^3), x/(1 + x^2 - x^3)) from A321196.

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A365730 G.f. satisfies A(x) = 1 + x^4A(x)^4(1 + x*A(x)).

A365724 G.f. satisfies A(x) = 1 + x^3A(x)^3(1 + x*A(x)).

A365609 G.f. satisfies A(x) = 1 + x^2A(x)^4(1 + x*A(x)).