A181997 -id:A181997 - cp's OEIS frontend

A001002 Number of dissections of a convex (n+2)-gon into triangles and quadrilaterals by nonintersecting diagonals.

1, 1, 3, 10, 38, 154, 654, 2871, 12925, 59345, 276835, 1308320, 6250832, 30142360, 146510216, 717061938, 3530808798, 17478955570, 86941210950, 434299921440, 2177832612120, 10959042823020, 55322023332420, 280080119609550, 1421744205767418, 7234759677699954

Offset: 0

N. J. A. Sloane

a(n+1) is number of (2,3)-rooted trees on n nodes.
This sequence appears to be a transform of the Fibonacci numbers A000045. This sequence is to the Fibonacci numbers as the Catalan numbers A000108 is to the all ones sequence. See link to Mathematica program. - Mats Granvik, Dec 30 2017
a(n) is the number of parking functions of size n avoiding the patterns 231, 312, and 321. - Lara Pudwell, Apr 10 2023

			a(3)=10 because a convex pentagon can be dissected in 5 ways into triangles (draw 2 diagonals from any of the 5 vertices) and in 5 ways into a triangle and a quadrilateral (draw any of the 5 diagonals).

F. Bergeron, G. Labelle and P. Leroux, Combinatorial Species and Tree-Like Structures, Cambridge, 1998, p. 211 (3.2.73-74)
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).

Alois P. Heinz, Table of n, a(n) for n = 0..1372 (first 101 terms from T. D. Noe)
Ayomikun Adeniran and Lara Pudwell, Pattern avoidance in parking functions, Enumer. Comb. Appl. 3:3 (2023), Article S2R17.
Paul Barry, On the Central Coefficients of Bell Matrices, J. Int. Seq. 14 (2011) # 11.4.3, example 7.
D. Birmajer, J. B. Gil, and M. D. Weiner, Colored partitions of a convex polygon by noncrossing diagonals, arXiv preprint arXiv:1503.05242 [math.CO], 2015.
I. M. H. Etherington, Non-associate powers and a functional equation, Math. Gaz. 21 (1937), 36-39; addendum 21 (1937), 153.
I. M. H. Etherington, On non-associative combinations, Proc. Royal Soc. Edinburgh, 59 (Part 2, 1938-39), 153-162.
I. M. H. Etherington, Some problems of non-associative combinations, Edinburgh Math. Notes, 32 (1940), 1-6.
I. M. H. Etherington, Some problems of non-associative combinations (I), Edinburgh Math. Notes, 32 (1940), pp. i-vi. [Annotated scanned copy].
A. Erdelyi and I. M. H. Etherington, Some problems of non-associative combinations (II), Edinburgh Math. Notes, 32 (1940), pp. vii-xiv.
Mats Granvik, Mathematica program to compute the sequence as a transform of the Fibonacci numbers.
Nancy S. S. Gu, Nelson Y. Li, and Toufik Mansour, 2-Binary trees: bijections and related issues, Discr. Math., 308 (2008), 1209-1221.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 395
Elżbieta Liszewska and Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.
T. Motzkin, The hypersurface cross ratio, Bull. Amer. Math. Soc., 51 (1945), 976-984.
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.
Thomas M. Richardson, The three 'R's and Dual Riordan Arrays, arXiv:1609.01193 [math.CO], 2016.
A. Schuetz and G. Whieldon, Polygonal Dissections and Reversions of Series, arXiv preprint arXiv:1401.7194 [math.CO], 2014.
Index entries for reversions of series

n*a(n) = A038112(n-1), n > 0.

Cf. A104978, A181997, A181998, A209441, A209442, A213684 (log).

List([0..25], n->Sum([0..Int(n/2)],k->Binomial(2*n-k,n+k)*Binomial(n+k,k)/(n+1))); # Muniru A Asiru, Mar 30 2018

a:= proc(n) option remember; `if`(n<2, 1, (n*(22*n-11)*
      a(n-1) + (9*n-6)*(3*n-4)*a(n-2))/(5*n*(n+1)))
    end:
seq(a(n), n=0..25);  # Alois P. Heinz, Jan 21 2021

Rest[CoefficientList[InverseSeries[Series[y - y^2 - y^3, {y, 0, 30}], x], x]]
a[n_] := CatalanNumber[n]*Hypergeometric2F1[1/2-n/2, -n/2, -2n, -4]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jan 20 2015, after Peter Luschny *)
a[n_] := a[n] = If[n == 0, 1, Sum[a[i] a[n - 1 - i], {i, 0, n - 1}] + Sum[a[i] a[j] a[n - 2 - i - j], {i, 0, n - 2}, {j, 0, n - 2 - i}]];
Table[a[n], {n, 0, 30}] (* Li Han, Jan 02 2021 *)

T(n,k):=if n<0 or k<0 then 0 else if nVladimir Kruchinin, Oct 03 2014 */

a(n)=if(n<0,0,polcoeff(serreverse(x-x^2-x^3+x^2*O(x^n)),n+1))

a(n)=if(n<0,0,sum(k=0,n\2,(2*n-k)!/k!/(n-2*k)!)/(n+1)!)

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

{Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
{a(n)=local(A=1); A=1+(1/x)*sum(m=1, n+1, Dx(m-1, (x^2+x^3 +x^2*O(x^n))^m/m!)); polcoeff(A, n)}  \\ Paul D. Hanna, Jun 22 2012

{Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
{a(n)=local(A=1); A=exp(sum(m=1, n+1, Dx(m-1, (x^2+x^3 +x^2*O(x^n))^m/x/m!)+x*O(x^n))); polcoeff(A, n)}  \\ Paul D. Hanna, Jun 22 2012

A001002 = lambda n: catalan_number(n)*hypergeometric([1/2-n/2, -n/2], [-2*n], -4) if n>0 else 1
[A001002(n).n(100).round() for n in range(24)] # Peter Luschny, Oct 03 2014

G.f. (offset 1) is series reversion of x - x^2 - x^3.
a(n) = (1/(n+1))*Sum_{k=ceiling(n/2)..n} binomial(n+k, k)*binomial(k, n-k). - Len Smiley
D-finite with recurrence 5*n*(n+1) * a(n) = 11*n*(2*n-1) * a(n-1) + 3*(3*n-2)*(3*n-4) * a(n-2). - Len Smiley
G.f.: (4*sin(asin((27*x+11)/16)/3)-1)/(3*x). - Paul Barry, Feb 02 2005
G.f. satisfies: A(x) = 1 + x*A(x)^2 + x^2*A(x)^3. - Paul D. Hanna, Jun 22 2012
Antidiagonal sums of triangle A104978 which has g.f. F(x,y) that satisfies: F = 1 + x*F^2 + x*y*F^3. - Paul D. Hanna, Mar 30 2005
a(n) = Sum_{k=0..floor(n/2)} C(2*n-k, n+k)*C(n+k, k)/(n+1). - Paul D. Hanna, Mar 30 2005
G.f. satisfies: x = Sum_{n>=1} 1/(1+x*A(x))^(2*n) * Product_{k=1..n} (1 - 1/(1+x*A(x))^k). - Paul D. Hanna, Apr 05 2012
G.f.: 1 + (1/x)*Sum_{n>=1} d^(n-1)/dx^(n-1) (x^2+x^3)^n/n!. - Paul D. Hanna, Jun 22 2012
G.f.: exp( Sum_{n>=1} d^(n-1)/dx^(n-1) ((x^2+x^3)^n/x)/n! ). - Paul D. Hanna, Jun 22 2012
Logarithmic derivative yields A213684. - Paul D. Hanna, Jun 22 2012
a(n) ~ 3^(3*n+3/2) / (2 * sqrt(2*Pi) * 5^(n+1/2) * n^(3/2)). - Vaclav Kotesovec, Mar 09 2014
a(n) = Catalan(n)*hypergeom([1/2-n/2, -n/2], [-2*n], -4) for n>0. - Peter Luschny, Oct 03 2014
a(n) = [x^n] 1/(1 - x - x^2)^(n+1)/(n + 1). - Ilya Gutkovskiy, Mar 29 2018
a(n) = -Sum_{i=1..n} A217596(i) * a(n-i) for n>0. - Muhammed Sefa Saydam, Jan 27 2025
a(n) = -Sum_{i=1..n+2} A217596(i) * A217596(n-i+2) for n >= 0. - Muhammed Sefa Saydam, Jul 24 2025

Revised by Emeric Deutsch and Len Smiley, Jun 05 2005

A214693 G.f. A(x) satisfies: x = Sum_{n>=1} 1/A(x)^(6n) Product_{k=1..n} (1 - 1/A(x)^(2*k-1)).

Original entry on oeis.org

1, 1, 4, 34, 338, 3691, 42623, 510949, 6289912, 78972928, 1006665781, 12985611054, 169115724583, 2219614920740, 29318819296959, 389331204757856, 5192978617937181, 69522908878900079, 933674035184058960, 12571898958515379108, 169651868248129552194

Offset: 0

Paul D. Hanna, Jul 26 2012

nonn

Compare the g.f. to the identity:
G(x) = Sum_{n>=0} 1/G(x)^(2*n) * Product_{k=1..n} (1 - 1/G(x)^(2*k-1))
which holds for all power series G(x) such that G(0)=1.

			G.f.: A(x) = 1 + x + 4*x^2 + 34*x^3 + 338*x^4 + 3691*x^5 + 42623*x^6 +...
The g.f. satisfies:
x = (A(x)-1)/A(x)^7 + (A(x)-1)*(A(x)^3-1)/A(x)^16 + (A(x)-1)*(A(x)^3-1)*(A(x)^5-1)/A(x)^27 + (A(x)-1)*(A(x)^3-1)*(A(x)^5-1)*(A(x)^7-1)/A(x)^40 +
(A(x)-1)*(A(x)^3-1)*(A(x)^5-1)*(A(x)^7-1)*(A(x)^9-1)/A(x)^55 +...

Cf. A214690, A214692, A214694, A214695, A181997 (variant).

{a(n)=if(n<0, 0, polcoeff(1 + serreverse(x - 4*x^2 - 2*x^3 + 22*x^4 + 49*x^5 + 49*x^6 + 27*x^7 + 8*x^8 + x^9 +x^2*O(x^n)), n))}

{a(n)=local(A=[1, 1]); for(i=1, n, A=concat(A, 0); A[#A]=-polcoeff(sum(m=1, #A, 1/Ser(A)^(6*m)*prod(k=1, m, 1-1/Ser(A)^(2*k-1))), #A-1)); A[n+1]}
for(n=0, 25, print1(a(n), ", "))

G.f. satisfies: 1+x = A(y) where y = x - 4*x^2 - 2*x^3 + 22*x^4 + 49*x^5 + 49*x^6 + 27*x^7 + 8*x^8 + x^9, which is the g.f. of row 3 in triangle A214690.

G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(n*(n+6)) * Product_{k=1..n} (A(x)^(2*k-1) - 1).

A209441 G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(5n) Product_{k=1..n} (1 - 1/A(x)^k).

Original entry on oeis.org

1, 1, 4, 30, 260, 2463, 24656, 256493, 2745149, 30031677, 334334789, 3775539592, 43145236171, 498018527632, 5798165437701, 68009060597311, 802908842472516, 9533509909631074, 113774810189434083, 1363985826416978416, 16418865502303963429, 198369001060550654651

Offset: 0

Paul D. Hanna, Apr 08 2012

nonn

Compare the g.f. to the identity:
G(x) = Sum_{n>=0} 1/G(x)^n * Product_{k=1..n} (1 - 1/G(x)^k)
which holds for all power series G(x) such that G(0)=1.

			G.f.: A(x) = 1 + x + 4*x^2 + 30*x^3 + 260*x^4 + 2463*x^5 + 24656*x^6 +...
The g.f. satisfies:
x = (A(x)-1)/A(x)^6 + (A(x)-1)*(A(x)^2-1)/A(x)^13 + (A(x)-1)*(A(x)^2-1)*(A(x)^3-1)/A(x)^21 + (A(x)-1)*(A(x)^2-1)*(A(x)^3-1)*(A(x)^4-1)/A(x)^30 + (A(x)-1)*(A(x)^2-1)*(A(x)^3-1)*(A(x)^4-1)*(A(x)^5-1)/A(x)^40 +...

G. C. Greubel, Table of n, a(n) for n = 0..900

Cf. A001002, A181997, A181998, A209442, A214695 (variant).

nmax = 20; aa = ConstantArray[0,nmax]; aa[[1]] = 1; Do[AGF = 1+Sum[aa[[n]]*x^n,{n,1,j-1}]+koef*x^j; sol=Solve[SeriesCoefficient[Sum[Product[(1-1/AGF^m)/AGF^5,{m,1,k}],{k,1,j}],{x,0,j}]==0,koef][[1]]; aa[[j]]=koef/.sol[[1]],{j,2,nmax}]; Flatten[{1,aa}] (* Vaclav Kotesovec, Dec 01 2014 *)
CoefficientList[1+InverseSeries[Series[x - 4*x^2 + 2*x^3 + 20*x^4 - 19*x^5 - 100*x^6 + 3*x^7 + 403*x^8 + 808*x^9 + 861*x^10 + 584*x^11 + 262*x^12 + 76*x^13 + 13*x^14 + x^15, {x, 0, 20}], x],x] (* Vaclav Kotesovec, Dec 01 2014 *)

{a(n)=if(n<0, 0, polcoeff(1 + serreverse(x - 4*x^2 + 2*x^3 + 20*x^4 - 19*x^5 - 100*x^6 + 3*x^7 + 403*x^8 + 808*x^9 + 861*x^10 + 584*x^11 + 262*x^12 + 76*x^13 + 13*x^14 + x^15 +x^2*O(x^n)), n))}

{a(n)=local(A=[1, 1]); for(i=1, n, A=concat(A, 0); A[#A]=-polcoeff(sum(m=1, #A, 1/Ser(A)^(5*m)*prod(k=1, m, 1-1/Ser(A)^k)), #A-1)); A[n+1]}
for(n=0, 25, print1(a(n), ", "))

G.f. satisfies: 1+x = A(y) where y = x - 4*x^2 + 2*x^3 + 20*x^4 - 19*x^5 - 100*x^6 + 3*x^7 + 403*x^8 + 808*x^9 + 861*x^10 + 584*x^11 + 262*x^12 + 76*x^13 + 13*x^14 + x^15.

G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(n*(n+11)/2) * Product_{k=1..n} (A(x)^k - 1).

A181998 G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(4n) Product_{k=1..n} (1 - 1/A(x)^k).

Original entry on oeis.org

1, 1, 3, 18, 124, 935, 7443, 61510, 522467, 4532452, 39985628, 357641094, 3235846003, 29565353095, 272429349163, 2528938553028, 23629834081955, 222080711420655, 2098112946860819, 19915641133236764, 189853287434733709, 1816924035668823659, 17450483777418686431

Offset: 0

Paul D. Hanna, Apr 05 2012

nonn

Compare the g.f. to the identity:
G(x) = Sum_{n>=0} 1/G(x)^n * Product_{k=1..n} (1 - 1/G(x)^k)
which holds for all power series G(x) such that G(0)=1.

			G.f.: A(x) = 1 + x + 3*x^2 + 18*x^3 + 124*x^4 + 935*x^5 + 7443*x^6 +...
The g.f. satisfies:
x = (A(x)-1)/A(x)^5 + (A(x)-1)*(A(x)^2-1)/A(x)^11 + (A(x)-1)*(A(x)^2-1)*(A(x)^3-1)/A(x)^18 + (A(x)-1)*(A(x)^2-1)*(A(x)^3-1)*(A(x)^4-1)/A(x)^26 + (A(x)-1)*(A(x)^2-1)*(A(x)^3-1)*(A(x)^4-1)*(A(x)^5-1)/A(x)^35 +...

Cf. A001002, A181997, A209441, A209442, A214694 (variant).

nmax = 20; aa = ConstantArray[0,nmax]; aa[[1]] = 1; Do[AGF = 1+Sum[aa[[n]]*x^n,{n,1,j-1}]+koef*x^j; sol=Solve[SeriesCoefficient[Sum[Product[(1-1/AGF^m)/AGF^4,{m,1,k}],{k,1,j}],{x,0,j}]==0,koef][[1]]; aa[[j]]=koef/.sol[[1]],{j,2,nmax}]; Flatten[{1,aa}] (* Vaclav Kotesovec, Dec 01 2014 *)
CoefficientList[1+InverseSeries[Series[x - 3*x^2 + 11*x^4 + x^5 - 30*x^6 - 42*x^7 - 26*x^8 - 8*x^9 - x^10, {x, 0, 20}], x],x] (* Vaclav Kotesovec, Dec 01 2014 *)

{a(n)=if(n<0,0,polcoeff(1 + serreverse(x - 3*x^2 + 11*x^4 + x^5 - 30*x^6 - 42*x^7 - 26*x^8 - 8*x^9 - x^10 +x^2*O(x^n)),n))}

{a(n)=local(A=[1,1]);for(i=1,n,A=concat(A,0);A[#A]=-polcoeff(sum(m=1,#A,1/Ser(A)^(4*m)*prod(k=1,m,1-1/Ser(A)^k)),#A-1));A[n+1]}
for(n=0,25,print1(a(n),", "))

G.f. satisfies: 1+x = A(y) where y = x - 3*x^2 + 11*x^4 + x^5 - 30*x^6 - 42*x^7 - 26*x^8 - 8*x^9 - x^10.

G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(n*(n+9)/2) * Product_{k=1..n} (A(x)^k - 1).

A209442 G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(6n) Product_{k=1..n} (1 - 1/A(x)^k).

Original entry on oeis.org

1, 1, 5, 45, 470, 5365, 64766, 813012, 10505163, 138800397, 1866712401, 25470265992, 351717013269, 4906153922941, 69030042202001, 978531875343171, 13961726654230994, 200351151383453293, 2889692388200640136, 41867983817065377259, 609091785100828769195

Offset: 0

Paul D. Hanna, Apr 08 2012

nonn

Compare the g.f. to the identity:
G(x) = Sum_{n>=0} 1/G(x)^n * Product_{k=1..n} (1 - 1/G(x)^k)
which holds for all power series G(x) such that G(0)=1.

			G.f.: A(x) = 1 + x + 5*x^2 + 45*x^3 + 470*x^4 + 5365*x^5 + 64766*x^6 +...
The g.f. satisfies:
x = (A(x)-1)/A(x)^7 + (A(x)-1)*(A(x)^2-1)/A(x)^15 + (A(x)-1)*(A(x)^2-1)*(A(x)^3-1)/A(x)^24 + (A(x)-1)*(A(x)^2-1)*(A(x)^3-1)*(A(x)^4-1)/A(x)^34 + (A(x)-1)*(A(x)^2-1)*(A(x)^3-1)*(A(x)^4-1)*(A(x)^5-1)/A(x)^45 +...

G. C. Greubel, Table of n, a(n) for n = 0..835

Cf. A001002, A181997, A181998, A209441.

nmax = 20; aa = ConstantArray[0,nmax]; aa[[1]] = 1; Do[AGF = 1+Sum[aa[[n]]*x^n,{n,1,j-1}]+koef*x^j; sol=Solve[SeriesCoefficient[Sum[Product[(1-1/AGF^m)/AGF^6,{m,1,k}],{k,1,j}],{x,0,j}]==0,koef][[1]]; aa[[j]]=koef/.sol[[1]],{j,2,nmax}]; Flatten[{1,aa}] (* Vaclav Kotesovec, Dec 01 2014 *)
CoefficientList[1+InverseSeries[Series[x - 5*x^2 + 5*x^3 + 30*x^4 - 65*x^5 - 191*x^6 + 378*x^7 + 1557*x^8 + 103*x^9 - 8551*x^10 - 23911*x^11 - 37958*x^12 - 41831*x^13 - 34156*x^14 - 21179*x^15 - 10015*x^16 - 3571*x^17 - 933*x^18 - 169*x^19 - 19*x^20 - x^21, {x, 0, 20}], x],x] (* Vaclav Kotesovec, Dec 01 2014 *)

{a(n)=if(n<0, 0, polcoeff(1 + serreverse(x - 5*x^2 + 5*x^3 + 30*x^4 - 65*x^5 - 191*x^6 + 378*x^7 + 1557*x^8 + 103*x^9 - 8551*x^10 - 23911*x^11 - 37958*x^12 - 41831*x^13 - 34156*x^14 - 21179*x^15 - 10015*x^16 - 3571*x^17 - 933*x^18 - 169*x^19 - 19*x^20 - x^21 +x^2*O(x^n)), n))}

{a(n)=local(A=[1, 1]); for(i=1, n, A=concat(A, 0); A[#A]=-polcoeff(sum(m=1, #A, 1/Ser(A)^(6*m)*prod(k=1, m, 1-1/Ser(A)^k)), #A-1)); A[n+1]}
for(n=0, 25, print1(a(n), ", "))

G.f. satisfies: 1+x = A(y) where y = x - 5*x^2 + 5*x^3 + 30*x^4 - 65*x^5 - 191*x^6 + 378*x^7 + 1557*x^8 + 103*x^9 - 8551*x^10 - 23911*x^11 - 37958*x^12 - 41831*x^13 - 34156*x^14 - 21179*x^15 - 10015*x^16 - 3571*x^17 - 933*x^18 - 169*x^19 - 19*x^20 - x^21.

G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(n*(n+13)/2) * Product_{k=1..n} (A(x)^k - 1).

A214670 Triangle, read by rows of n(n+1)/2 terms, where row n equals the coefficients in the series reversion of the function G(x,n)-1 such that: x = Sum_{m>=1} 1/G(x,n)^(nm) * Product_{k=1..m} (1 - 1/G(x,n)^k), for n>=1.

Original entry on oeis.org

1, 1, -1, -1, 1, -2, -1, 4, 4, 1, 1, -3, 0, 11, 1, -30, -42, -26, -8, -1, 1, -4, 2, 20, -19, -100, 3, 403, 808, 861, 584, 262, 76, 13, 1, 1, -5, 5, 30, -65, -191, 378, 1557, 103, -8551, -23911, -37958, -41831, -34156, -21179, -10015, -3571, -933, -169, -19, -1

Offset: 1

Paul D. Hanna, Jul 25 2012

The row sums are a signed version of A005014. [From _Olivier Gérard_, Jun 26 2012, in an email to the seqfan list, which suggested that the g.f. A(x,y) is a generalization of the g.f. for A005014.]

			Consider the family of power series G(x,n) that satisfy:
x = Sum_{m>=1} 1/G(x,n)^(n*m) * Product_{k=1..m} (1 - 1/G(x,n)^k).
Examples of sequences with g.f. G(x,n) are:
n=2: A001002 = [1, 1, 1, 3, 10, 38, 154, 654, 2871, 12925, ...];
n=3: A181997 = [1, 1, 2, 9, 46, 259, 1539, 9484, 59961, ...];
n=4: A181998 = [1, 1, 3, 18, 124, 935, 7443, 61510, 522467, ...];
n=5: A209441 = [1, 1, 4, 30, 260, 2463, 24656, 256493, 2745149, ...];
n=6: A209442 = [1, 1, 5, 45, 470, 5365, 64766, 813012, 10505163, ...]; ...
Observe that Series_Reversion( G(x,n) - 1 ) is given by the polynomials:
n=1: x;
n=2: x - x^2 - x^3;
n=3: x - 2*x^2 - x^3 + 4*x^4 + 4*x^5 + x^6;
n=4: x - 3*x^2 + 11*x^4 + x^5 - 30*x^6 - 42*x^7 - 26*x^8 - 8*x^9 - x^10;
n=5: x - 4*x^2 + 2*x^3 + 20*x^4 - 19*x^5 - 100*x^6 + 3*x^7 + 403*x^8 + 808*x^9 + 861*x^10 + 584*x^11 + 262*x^12 + 76*x^13 + 13*x^14 + x^15; ...
This triangle of coefficients in the above polynomials begins:
[1];
[1, -1, -1];
[1, -2, -1, 4, 4, 1];
[1, -3, 0, 11, 1, -30, -42, -26, -8, -1];
[1, -4, 2, 20, -19, -100, 3, 403, 808, 861, 584, 262, 76, 13, 1];
[1, -5, 5, 30, -65, -191, 378, 1557, 103, -8551, -23911, -37958, -41831, -34156, -21179, -10015, -3571, -933, -169, -19, -1];
[1, -6, 9, 40, -145, -261, 1384, 2897, -8980, -38710, -14146, 258401, 990407, 2170834, 3426095, 4198850, 4137440, 3336534, 2220430, 1221799, 554027, 205250, 61206, 14351, 2550, 323, 26, 1];
[1, -7, 14, 49, -266, -245, 3325, 2596, -36710, -70556, 281645, 1413916, 1184890, -10255248, -54012830, -156371880, -329973512, -552895722, -765517470, -895408431, -896614676, -774834055, -580511469, -377792286, -213512611, -104550572, -44163315, -15985147, -4910774, -1263620, -267378, -45321, -5918, -559, -34, -1]; ...

Paul D. Hanna, Rows 1..12, flattened.

Cf. A001002, A181997, A181998, A209441, A209442, A005014 (row sums), A214669.

Cf. A214690 (variant).

{T(n,k)=local(Axy=x*y);Axy=sum(m=1,n,-x^m*prod(j=1,m,(1-(1+y)^j)/(1-x*(1+y)^j)+x*O(x^n)));polcoeff(polcoeff(Axy,n,x),k,y)}
{for(n=1,10,for(k=1,n*(n+1)/2,print1(T(n,k),", "));print(""))}

{a(n,p)=local(A=[1, 1]); for(i=1, n, A=concat(A, 0); A[#A]=-polcoeff(sum(m=1, #A, 1/Ser(A)^(p*m)*prod(k=1, m, 1-1/Ser(A)^k)), #A-1)); A[n+1]}
{for(n=1,8,Tn=Vec(serreverse(sum(m=1,n*(n+1)/2,a(m,n)*x^m)+x*O(x^(n*(n+1)/2))));for(k=1,n*(n+1)/2,print1(Tn[k],", "));print(""))}

G.f.: A(x,y) = Sum_{n>=1} -x^n * Product_{k=1..n} (1 - (1+y)^k) / (1 - x*(1+y)^k).
G.f. for row n is R(y,n) = Sum_{k=1..n*(n+1)/2} y^k*T(n,k) defined by:
A(x,y) = Sum_{n>=1} x^n * R(y,n) such that:
R(y,n) = Series_Reversion( G(y,n) - 1 ) where G(y,n) satisfies:
y = Sum_{m>=1} 1/G(y,n)^(n*m) * Product_{k=1..m} (1 - 1/G(y,n)^k), for n>=1.
Row polynomials R(y,n) satisfy:
(1) R(1,n) = (-1)^(n-1) * A005014(n) for n>=1.
(2) R(-1,n) = 1 for n>=1.
(3) R'(-1,n) = 0 for n>1.
(4) R'(1,n) = A214669(n) for n>=1.

cp's OEIS Frontend

A001002 Number of dissections of a convex (n+2)-gon into triangles and quadrilaterals by nonintersecting diagonals.

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A214693 G.f. A(x) satisfies: x = Sum_{n>=1} 1/A(x)^(6*n) * Product_{k=1..n} (1 - 1/A(x)^(2*k-1)).

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A209441 G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(5*n) * Product_{k=1..n} (1 - 1/A(x)^k).

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A181998 G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(4*n) * Product_{k=1..n} (1 - 1/A(x)^k).

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A209442 G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(6*n) * Product_{k=1..n} (1 - 1/A(x)^k).

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A214670 Triangle, read by rows of n*(n+1)/2 terms, where row n equals the coefficients in the series reversion of the function G(x,n)-1 such that: x = Sum_{m>=1} 1/G(x,n)^(n*m) * Product_{k=1..m} (1 - 1/G(x,n)^k), for n>=1.

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A214693 G.f. A(x) satisfies: x = Sum_{n>=1} 1/A(x)^(6n) Product_{k=1..n} (1 - 1/A(x)^(2*k-1)).

A209441 G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(5n) Product_{k=1..n} (1 - 1/A(x)^k).

A181998 G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(4n) Product_{k=1..n} (1 - 1/A(x)^k).

A209442 G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(6n) Product_{k=1..n} (1 - 1/A(x)^k).

A214670 Triangle, read by rows of n(n+1)/2 terms, where row n equals the coefficients in the series reversion of the function G(x,n)-1 such that: x = Sum_{m>=1} 1/G(x,n)^(nm) * Product_{k=1..m} (1 - 1/G(x,n)^k), for n>=1.