A321201 -id:A321201 - cp's OEIS frontend

A321203 Irregular triangle T giving the coefficients of x^n = x^{2e2 + 3e3} of (1 + x^2 + x^3)^n, with the pair of nonnegative numbers [e2, e3] listed in row n of A321201, for n >= 2.

2, 3, 6, 20, 15, 20, 105, 168, 70, 84, 504, 1260, 252, 1320, 2310, 495, 7920, 924, 12870, 10296, 10010, 45045, 3432, 3003, 100100, 45045, 120120, 240240, 12870, 74256, 680680, 194480, 18564, 1113840, 1225224, 48620, 1058148, 4232592, 831402, 542640, 8817900, 6046560, 184756

Offset: 2

Wolfdieter Lang, Nov 05 2018

The row length is r(n), with r(n) = A008615(n+2) for n >= 2:  [1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 3, 2, 3, 3, 3, 3, 4, 3, 4, ...].
The row sums give A176806(n).
For n = 0 with the trivial [e2, e3] = [0, 0] solution the multinomial is 1 with the row sum A176806(0) = 1. For n = 1 there is no solution (with row sum set to A176806(1) = 0).
This multinomial array for pairs [e2, e3] with 2*2e2 + 3*e3 = n, with nonnegative numbers e2 and e3, is obtained from the multinomial array n!/(e1!*e2!*e3!) with n = e1 + e2 + e3, giving the coefficient x_1^{e1}* x_2^{e2}*x_3^{e3} of (x_1 + x_2 + x_3)^n. Here, in order to find the coefficients of (1 + x^2 + x^3)^n, one sets x_1 = 1, x_2 = x^2 and x_3 = x^3. Hence n = e1 + e2 + e3, and the power of x^n becomes n = 2*e2 + 3*e3. Therefore, e1 = n - (e2 + e3), and the array gives n!/((n-(e2+e3))!*e2!*e3!).

			The triangle T(n, m), and the row sums begin:
n\m        0        1        2       3  ...  Row sums A176806(n)
2:         2                                         2
3:         3                                         3
4:         6                                         6
5:        20                                        20
6:        15       20                               35
7:       105                                       105
8:       168       70                              238
9:        84      504                              588
10:     1260      252                             1512
11:     1320     2310                             3630
12:      495     7920      924                    9339
13:    12870    10296                            23166
14:    10010    45045     3432                   58487
15:     3003   100100    45045                  148148
16:   120120   240240    12870                  373230
17:    74256   680680   194480                  949416
18:    18564  1113840  1225224   48620         2406248
19:  1058148  4232592   831402                 6122142
20:   542640  8817900  6046560  184756        15591856
...
------------------------------------------------------------------------------
n = 8: (1 + x^2 + x^3)^8 has coefficients 238 of x^n arising from the two [e2, e3] pairs [1, 2] and [4, 0], given in row n = 8 of A321201. The multinomial values are 8!/((8-3)!*1!*2!) = 168 and 8!/((8-4)!*4!*0!) = 70, summing to 238.

Cf. A008615, A130561, A176806, A321201, A319204.

T(n, m) is obtained from the pair(s) [e2, e3] given in row n of A321201 by n!/((n - (e2 +e3))!*e2!*e3!), for n >= 2 and m = 1, 2, ..., A008615(n+2).

A321202 Row sums of the irregular triangle A321201.

Original entry on oeis.org

1, 1, 2, 2, 5, 3, 7, 7, 9, 9, 15, 11, 18, 18, 21, 21, 30, 24, 34, 34, 38, 38, 50, 42, 55, 55, 60, 60, 75, 65, 81, 81, 87, 87, 105, 93, 112, 112, 119, 119, 140, 126, 148, 148, 156, 156, 180, 164, 189, 189, 198, 198, 225, 207, 235, 235, 245, 245, 275, 255, 286, 286, 297

Offset: 2

Wolfdieter Lang, Nov 05 2018

nonn

Total number of parts in the partitions of n into parts of size 2 and 3. - Andrew Howroyd, Nov 10 2018

Andrew Howroyd, Table of n, a(n) for n = 2..1000
Index entries for linear recurrences with constant coefficients, signature (-1,1,3,2,-2,-3,-1,1,1).

Cf. A008615, A321201.

row[n_] := Reap[Do[If[2 e2 + 3 e3 == n, Sow[{e2, e3}]], {e2, 0, n/2}, {e3, 0, n/3}]][[2, 1]];
a[n_] := row[n] // Flatten // Total;
Table[a[n], {n, 2, 100}] (* Jean-François Alcover, Nov 23 2018 *)

Vec((1 + 2*x + 2*x^2)/((1 + x + x^2)^2*(1 + x)^2*(1 - x)^3) + O(x^60)) \\ Andrew Howroyd, Nov 10 2018

a(n) = Sum_{k=1..2*A008615(n+2)} A321201(n, k),  n >= 2.
From Andrew Howroyd, Nov 10 2018: (Start)
G.f.: x^2*(1 + 2*x + 2*x^2)/((1 + x + x^2)^2*(1 + x)^2*(1 - x)^3).
a(n) = Sum_{k=0..floor(n/6)} 2*k + (n-6*k)/2 for even n.
a(n) = Sum_{k=0..floor((n-3)/6)} 2*k + 1 + (n-3-6*k)/2 for odd n.
(End)

Terms a(27) and beyond from Andrew Howroyd, Nov 10 2018

A001005 Number of ways of partitioning n points on a circle into subsets only of sizes 2 and 3.

Original entry on oeis.org

1, 0, 1, 1, 2, 5, 8, 21, 42, 96, 222, 495, 1177, 2717, 6435, 15288, 36374, 87516, 210494, 509694, 1237736, 3014882, 7370860, 18059899, 44379535, 109298070, 269766655, 667224480, 1653266565, 4103910930, 10203669285, 25408828065, 63364046190, 158229645720, 395632288590, 990419552730

Offset: 0

N. J. A. Sloane

a(n) is also the number of rooted trees on n nodes such that each node has 0, 2, or 3 children. - Patrick Devlin, Mar 04 2012

a(n) is the number of Motzkin paths that have no flatsteps (F) at ground level and avoid a factor of the form FMF with M a Motzkin path (possibly empty). For example, a(5) = 5 counts UDUFD, UFDUD, UFUDD, UUDFD, UUFDD but not UFFFD. Proof: Such a path can have at most one flatstep at height 1 before the first return to ground level or else the first component will contain an FMF. Hence, with a dot denoting concatenation, such a path is either empty or has the form U.P1.D.P2 or the form U.P1.F.P2.D.P3 where P1, P2, P3 are all paths of the type being counted. Hence the gf F(x) = 1 + x^2 + x^3 + 2*x^4 + ... satisfies F = 1 + x^2*F^2 + x^3*F^3. - David Callan, Nov 21 2021

			a(7)=21: 7 rotations of [12][34][567], 7 rotations of [12][45][367], 7 rotations of [12][37][456]. - _Len Smiley_, Jun 18 2005
From _Wolfdieter Lang_, Nov 05 2018: (Start)
a(7) = b(8)/8, where b(8) = (d^7/dx^7)((1 + x^2 + x^3)^8)/7! evaluated for x = 0, which is 168, and 168/8 = 21.
a(7) =(1/8)*8!/((8-(2+1))!*2!*1!) =(1/8)*8!/(5!*2!)= 168/8 = 21, from the only solution [e2, e3] = [2, 1] of 2*e2 + 3*e3 = 7. (End)

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Paul Barry, Riordan arrays, generalized Narayana triangles, and series reversion, Linear Algebra and its Applications, 491 (2016) 343-385.
F. R. Bernhart & N. J. A. Sloane, Emails, April-May 1994
Isaac DeJager, Madeleine Naquin, and Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 396
T. S. Motzkin, Relations between hypersurface cross ratios and a combinatorial formula for partitions of a polygon, for permanent preponderance and for non-associative products, Bull. Amer. Math. Soc., 54 (1948), 352-360.
Len Smiley, a(7) and a(8)
Index entries for sequences related to rooted trees

Cf. A321201.

a:=proc(n::nonnegint) local k,j; a(n):=0; for k from 0 to floor(n/2) do for j from 0 to floor(n/3) do if (2*k+3*j=n) then a(n):=a(n)+(n)!/(k!*j!*(n-k-j+1)!) fi od od; print(a(n)) end proc; seq(a(i),i=0..30); # Len Smiley, Jun 18 2005
A001005 := n -> ifelse(n=0, 1, add(binomial(n, k-1)*binomial(k, n-2*k)/k, k = 1 + iquo(n-1,3)..iquo(n,2))): seq(A001005(n), n=0..35); # Peter Luschny, Oct 18 2022

Table[Sum[(n)!/(k!*j!*(n - k - j + 1)!) * KroneckerDelta[2*k + 3*j - n], {k, 0, Floor[n/2]}, {j, 0, Floor[n/3]}], {n, 0, 20}] (* Ricardo Bittencourt, Jun 09 2013 *)
CoefficientList[ InverseSeries[x/(1+x^2+x^3) + O[x]^66]/x, x] (* Jean-François Alcover, Feb 15 2016, after Joerg Arndt*)

Vec(serreverse(x/(1+x^2+x^3)+O(x^66))/x) /* Joerg Arndt, Aug 19 2012 */

G.f. for a(n-1), with a(-1) = 0, satisfies A(x)=x*(1+A(x)^2+A(x)^3). - Christian G. Bower, Dec 15 1999
a(n) = Sum_{k=0..floor(n/2)} Sum_{j=0..floor(n/3)} n!/(k!*j!*(n-k-j+1)!)*[2*k+3*j=n]. - Len Smiley, Jun 18 2005
Recurrence: 2*(n+1)*(2*n+3)*(26*n+1)*a(n) = -(n-1)*(26*n^2 + 53*n + 18)*a(n-1) + 6*(n-1)*(78*n^2 + 42*n - 25)*a(n-2) + 31*(n-2)*(n-1)*(26*n+27)*a(n-3). - Vaclav Kotesovec, Aug 14 2013
a(n) ~ c*d^n/n^(3/2), where d = ((6371-624*sqrt(78))^(1/3)+(6371+624*sqrt(78))^(1/3)-1)/12 = 2.610718613276039349818649... is the root of the equation 4d^3 + d^2 - 18d - 31 = 0 and c = d^2 / (2*sqrt(Pi)*sqrt(1 + 3*d + sqrt(1 + 3*d))) = 0.559628309722556021604897336422272... - Vaclav Kotesovec, Aug 14 2013, updated Jun 27 2018
a(n) = Sum_{k=1..floor(n/2)} C(n,k-1)*C(k,n-2k)/k, n > 0. - Michael D. Weiner, Mar 02 2015
From Wolfdieter Lang, Nov 05 2018: (Start)
The o.g.f of a(n) is G(x) = F^[-1](x)/x, where F^[-1](x) is the compositional inverse of F(y) = y/(1 + y^2 + y^3), that is F(F^[-1](x)) = x, identically. (Compare this with the g.f. given above, and see the Pari and Mathematica programs below.)
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} (n+1)!/(n+1 - (e2 + e3))!*e2!*e3!) (from the multinomial formula for (x1 + x2 + x3)^(n+1)). For the solutions of 2*e2 + 3*e3 = n see the array A321201.
(End)

More terms from Christian G. Bower, Dec 15 1999

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

Original entry on oeis.org

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.

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)

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}.

A319204 Sequence used for the Boas-Buck type recurrence for Riordan triangle A319203.

Original entry on oeis.org

0, -2, -3, 6, 20, -5, -105, -98, 420, 1008, -990, -6501, -2574, 31603, 52052, -107250, -411944, 81328, 2343042, 2413456, -9883800, -25327722, 23371634, 168185131, 77113020, -835281800, -1452148815, 2847865635, 11561517870, -1613666430, -66318892875, -72637680690, 280330495200, 750725215020

Offset: 0

Wolfdieter Lang, Oct 29 2018

sign

See A319203 for the Boas-Buck type recurrence.

			a(5) = (1/6!)*[d^6/dx^6 (1 - x^2 - x^3)^6] for x = 0, which is -5.
a(5) = +15 - 20 = -5; from the sum of the signed row n=6 in A321203, with parity of e2 + e3 from A321201 even and odd.

Cf. A319203, A321201, A321203.

O.g.f.: (log(f(x)))' = (1/(1/f(x) + x^2*f(x) + 2*x^3*f(x)^2) - 1)/x, with the expansion of f given in A319201. f(x) = F^{[-1]}(x)/x, where F(t) = t/(1 - t^2 - t^3).
a(n) = (1/(n+1)!)*[d^(n+1)/dx^(n+1) (1 - x^2 - x^3)^(n+1)] evaluated at x = 0, for n >= 0. (Cf. Joerg Arndt's conjecture for A176806, which is proved there.)
a(n-1) = Sum_{2*e + 3*e3 = n} (-1)^(e2+e3)*n!/((n - (e2+e3))!*e2!*e3!), n >= 2, with a(0) = 0. The pairs (e2, e3) are given in A321201; see also the multinomial coefficient table A321203 and add the sign factors.

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A321203 Irregular triangle T giving the coefficients of x^n = x^{2*e2 + 3*e3} of (1 + x^2 + x^3)^n, with the pair of nonnegative numbers [e2, e3] listed in row n of A321201, for n >= 2.

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A321202 Row sums of the irregular triangle A321201.

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A001005 Number of ways of partitioning n points on a circle into subsets only of sizes 2 and 3.

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

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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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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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A319204 Sequence used for the Boas-Buck type recurrence for Riordan triangle A319203.

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A321203 Irregular triangle T giving the coefficients of x^n = x^{2e2 + 3e3} of (1 + x^2 + x^3)^n, with the pair of nonnegative numbers [e2, e3] listed in row n of A321201, for n >= 2.