A001002 -id:A001002 - cp's OEIS frontend

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

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).

A181997 G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(3n) Product_{k=1..n} (1 - 1/A(x)^k).

Original entry on oeis.org

1, 1, 2, 9, 46, 259, 1539, 9484, 59961, 386319, 2524940, 16687599, 111264335, 747080253, 5044629212, 34218868880, 232964088130, 1590660486297, 10885758313976, 74627209920879, 512254418843196, 3519150502675731, 24187028454513735, 166249089897708930

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 + 2*x^2 + 9*x^3 + 46*x^4 + 259*x^5 + 1539*x^6 +...
The g.f. satisfies:
x = (A(x)-1)/A(x)^4 + (A(x)-1)*(A(x)^2-1)/A(x)^9 + (A(x)-1)*(A(x)^2-1)*(A(x)^3-1)/A(x)^15 + (A(x)-1)*(A(x)^2-1)*(A(x)^3-1)*(A(x)^4-1)/A(x)^22 + (A(x)-1)*(A(x)^2-1)*(A(x)^3-1)*(A(x)^4-1)*(A(x)^5-1)/A(x)^30 +...

Cf. A001002, A181998, A209441, A209442, A214693 (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^3,{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 - 2*x^2 - x^3 + 4*x^4 + 4*x^5 + x^6, {x, 0, 20}], x],x] (* Vaclav Kotesovec, Dec 01 2014 *)

{a(n)=if(n<0,0,polcoeff(1 + serreverse(x - 2*x^2 - x^3 + 4*x^4 + 4*x^5 + x^6 +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)^(3*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 - 2*x^2 - x^3 + 4*x^4 + 4*x^5 + x^6.

G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(n*(n+7)/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.

A365731 G.f. satisfies A(x) = 1 + x^4A(x)^5(1 + x*A(x)).

Original entry on oeis.org

1, 0, 0, 0, 1, 1, 0, 0, 5, 11, 6, 0, 35, 120, 136, 51, 285, 1330, 2310, 1771, 3036, 14950, 35100, 40950, 47502, 175392, 503440, 791120, 927520, 2272424, 7037184, 13803405, 18643560, 33997080, 98920536, 226318196, 359255325, 578590155, 1445166360, 3584815443, 6573439928

Offset: 0

Seiichi Manyama, Sep 17 2023

nonn

Index entries for reversions of series

Cf. A365727, A365728, A365729, A365730.
Cf. A001002, A217358, A365725.
Cf. A215342.

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

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

G.f.: (1/x) * Series_Reversion( x*(1 - x^4*(1 + x)) ). - Seiichi Manyama, Sep 24 2024

A063018 Reversion of x - x^2 - x^3 - x^4.

Original entry on oeis.org

0, 1, 1, 3, 11, 44, 189, 850, 3951, 18832, 91542, 452075, 2261753, 11439372, 58394014, 300455892, 1556636807, 8113709916, 42518000652, 223868503324, 1183764310960, 6283573101960, 33470346433605, 178850415320010

Offset: 0

Olivier Gérard, Jul 05 2001

For the reversion of x - a*x^2 - b*x^3 - c*x^4 (a!=0, b!=0, c!=0) we have a(n) = Sum(k=1,n-1, (Sum(j=0..k, a^(-n+3*k-j+1) * b^(n-3*k+2*j-1) * c^(k-j) * binomial(j,n-3*k+2*j-1) * binomial(k,j) ) ) * binomial(n+k-1,n-1))/n, n>1, a(1)=1. - Vladimir Kruchinin, May 28 2011

G.f. (with offset 1) satisfies A(x) = 1 + x*A(x)^2 + x^2*A(x)^3.

Vincenzo Librandi, Table of n, a(n) for n = 0..105
Vladimir Kruchinin, The method for obtaining expressions for coefficients of reverse generating functions, arXiv:1211.3244 [math.CO], 2012.
Elżbieta Liszewska, Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.
Index entries for reversions of series

Cf. A001002 (reversion of y - y^2 - y^3).

CoefficientList[InverseSeries[Series[y - y^2 - y^3 - y^4, {y, 0, 30}], x], x]

a(n):=sum((sum(binomial(j,n-3*k+2*j-1)*binomial(k,j),j,0,k))*binomial(n+k-1,n-1),k,1,n-1)/n; \\ Vladimir Kruchinin, May 28 2011

x='x+O('x^66); /* that many terms */
Vec(serreverse(x-x^2-x^3-x^4)) /* show terms */ /* Joerg Arndt, May 28 2011 */

a(n) = Sum(k=1..n-1, (Sum(j=0..k, binomial(j,n-3*k+2*j-1) * binomial(k,j))) * binomial(n+k-1,n-1))/n, n>1, a(1)=1, a(0)=0. - Vladimir Kruchinin, May 28 2011

D-finite with recurrence 2552*n*(n-1)*(n-2)*a(n) -4*(n-1)*(n-2)*(2909*n-3951)*a(n-1) -2*(n-2)*(6839*n^2 -31331*n +36576)*a(n-2) +(-17563*n^3 +138510*n^2 -359633*n +308670)*a(n-3) -120*(4*n-15)*(2*n-7)*(4*n-17)*a(n-4)=0. - R. J. Mathar, Mar 24 2023

A218225 G.f. A(x) satisfies: (1 - xA(x)) / (1 - x^2A(x)^2)^2 = 1 - x.

Original entry on oeis.org

1, 2, 6, 23, 101, 480, 2400, 12434, 66142, 359112, 1981904, 11085198, 62696874, 357970472, 2060459256, 11943445311, 69656978837, 408466559630, 2406825745010, 14243262687023, 84618295006269, 504485687485408, 3017344000161296, 18099717207764928

Offset: 0

Paul D. Hanna, Oct 23 2012

nonn

Binomial transform of A001002. - Vladimir Kruchinin, Oct 03 2014
Conjecture: a(n) is the number of permutations of [1..n+1] that avoid one of the following sets of patterns: (2134, 42153, 24153), (3124, 42153, 24153), (2143, 42135, 24135). - Alexander Burstein, Dec 20 2017
From Alexander Burstein, Sep 29 2023: (Start)
Chern et al. (see links) proves the above conjecture for (3124, 42153, 24153) as well as the following conjecture of Hong and Li:
a(n) is the number of inversion sequences of length n+1 avoiding pattern 0021. (End)

			G.f.: A(x) = 1 + 2*x + 6*x^2 + 23*x^3 + 101*x^4 + 480*x^5 + 2400*x^6 + ...
The series reversion of x*A(x) begins:
x - 2*x^2 + 2*x^3 - 3*x^4 + 3*x^5 - 4*x^6 + 4*x^7 - 5*x^8 + 5*x^9 + ...
so A(1 - (1-x)/(1-x^2)^2) = x + 1/(1-x-x^2).
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + (1-x)*A(x))*x +
(1 + 2^2*(1-x)*A(x) + (1-x)^2*A(x)^2)*x^2/2 +
(1 + 3^2*(1-x)*A(x) + 3^2*(1-x)^2*A(x)^2 + (1-x)^3*A(x)^3)*x^3/3 +
(1 + 4^2*(1-x)*A(x) + 6^2*(1-x)^2*A(x)^2 + 4^2*(1-x)^3*A(x)^3 + (1-x)^4*A(x)^4)*x^4/4 +
(1 + 5^2*(1-x)*A(x) + 10^2*(1-x)^2*A(x)^2 + 10^2*(1-x)^3*A(x)^3 + 5^2*(1-x)^4*A(x)^4 + (1-x)^5*A(x)^5)*x^5/5 + ...

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.
Shane Chern, Shishuo Fu, and Zhicong Lin, Burstein's permutation conjecture, Hong and Li's inversion sequence conjecture, and restricted Eulerian distributions, arXiv:2209.12137 [math.CO], 2022.
Letong Hong and Rupert Li, Length-Four Pattern Avoidance in Inversion Sequences, arXiv:2112.15081 [math.CO], 2021.

Table[1/(n+1)*SeriesCoefficient[(((x-1)*(x+1)^2)/(x^2+x-1))^(n+1),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, May 22 2013 *)
Flatten[{1,Table[FullSimplify[SeriesCoefficient[(2*(1-x)^(1/3)-2^(2/3)*(-11-16*x-3*Sqrt[-15+96*x])^(1/3)+2^(2/3)*(11+16*x-3*Sqrt[-15+96*x])^(1/3))/(6*(-1+x)^(1/3)*x),{x,0,n}]],{n,1,10}]}] (* Vaclav Kotesovec, Jul 06 2013 *)
CoefficientList[Series[(-1+Cos[2/3*(ArcCot[3*Sqrt[3/5]]-ArcCot[(3*Sqrt[3])/Sqrt[5-32*x]])]+Sqrt[15]*Sin[2/3*(ArcCot[3*Sqrt[3/5]]-ArcCot[(3*Sqrt[3])/Sqrt[5-32*x]])])/(3*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Jul 06 2013 *)

{a(n)=polcoeff((1/x)*serreverse(x*(1-x-x^2)/((1-x)*(1+x)^2 +x*O(x^n))),n)}
for(n=0,30,print1(a(n),", "))

{a(n)=local(A=1);for(i=1,n,A=exp(sum(m=1,n,x^m/m*sum(k=0,m,binomial(m,k)^2*(1-x)^k*A^k)+x*O(x^n))));polcoeff(A,n)}
for(n=0,20,print1(a(n),", "))

G.f. A(x) satisfies:
(1) A(1 - (1-x)/(1-x^2)^2) = x + 1/(1-x-x^2).
(2) A(x) = (1/x) * Series_Reversion( x*(1-x-x^2)/((1-x)*(1+x)^2) ).
(3) A(x) = (1 - x*A(x)) * (1 + x*A(x))^2 / (1 - x*A(x) - x^2*A(x)^2).
(4) A(x) = exp( Sum_{n>=1} (x^n/n) * Sum_{k=0..n} binomial(n,k)^2 * (1-x)^k * A(x)^k ).
Recurrence: 5*n*(n+1)*a(n) = 21*n*(2*n-1)*a(n-1) - 3*(23*n^2-46*n+24)*a(n-2) + 16*(n-2)*(2*n-3)*a(n-3). - Vaclav Kotesovec, May 22 2013
a(n) ~ 2^(5*n+6)/(27*sqrt(Pi)*5^(n+1/2)*n^(3/2)). - Vaclav Kotesovec, May 22 2013
G.f.: (-1+cos(2/3*(arccot(3*sqrt(3/5))-arccot((3*sqrt(3))/sqrt(5-32*x))))+sqrt(15)*sin(2/3*(arccot(3*sqrt(3/5))-arccot((3*sqrt(3))/sqrt(5-32*x)))))/(3*x). - Vaclav Kotesovec, Jul 06 2013

A290646 Number of dissections of an n-gon into 3- and 4-gons counted up to rotations and reflections.

Original entry on oeis.org

1, 2, 2, 7, 14, 53, 171, 691, 2738, 11720, 50486, 224012, 1005468, 4581815, 21093190, 98093226, 459986674, 2173599817, 10340539744, 49496519950, 238240366274, 1152543685463, 5601603835982, 27341242042238, 133977037982121, 658902522544060, 3251446102879398

Offset: 3

Evgeniy Krasko, Sep 03 2017

nonn

			For a(5) = 2 the dissections of a pentagon are: a dissection into 3 triangles; a dissection into one triangle and one quadrangle.

Andrew Howroyd, Table of n, a(n) for n = 3..200
E. Krasko, A. Omelchenko, Brown's Theorem and its Application for Enumeration of Dissections and Planar Trees, The Electronic Journal of Combinatorics, 22 (2015), #P1.17.
Vladimir Shevelev, On a Luschny question, arXiv:1708.08096 [math.NT], 2017.

Cf. A001004 (counted distinctly).

Cf. A001002, A290571, A295260, A295419.

(* See A295419 for DissectionsModDihedral. *)
DissectionsModDihedral[Boole[# == 3 || # == 4]& /@ Range[1, 30]] (* Jean-François Alcover, Sep 25 2019, after Andrew Howroyd *)

\\ See A295419 for DissectionsModDihedral.
DissectionsModDihedral(apply(v->v==3||v==4, [1..25])) \\ Andrew Howroyd, Nov 22 2017

Terms a(16) and beyond from Andrew Howroyd, Nov 22 2017

A365725 G.f. satisfies A(x) = 1 + x^3A(x)^4(1 + x*A(x)).

Original entry on oeis.org

1, 0, 0, 1, 1, 0, 4, 9, 5, 22, 78, 91, 175, 680, 1224, 1938, 6270, 14630, 24794, 63756, 166980, 322920, 720720, 1900080, 4125888, 8803008, 22151360, 51778804, 111882100, 267682272, 645736432, 1442390092, 3346519020, 8094247798, 18657762006, 42890295734

Offset: 0

Seiichi Manyama, Sep 17 2023

nonn

Index entries for reversions of series

Cf. A308616, A365723, A365724, A365726.
Cf. A001002, A217358, A365731.
Cf. A054514.

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

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

G.f.: (1/x) * Series_Reversion( x*(1 - x^3*(1 + x)) ). - Seiichi Manyama, Sep 24 2024

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

A181997 G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(3*n) * Product_{k=1..n} (1 - 1/A(x)^k).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

A365731 G.f. satisfies A(x) = 1 + x^4*A(x)^5*(1 + x*A(x)).

Views

Author

Keywords

Links

Crossrefs

Programs

PARI

Formula

A063018 Reversion of x - x^2 - x^3 - x^4.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

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

A181997 G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(3n) 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.

A365731 G.f. satisfies A(x) = 1 + x^4A(x)^5(1 + x*A(x)).

A218225 G.f. A(x) satisfies: (1 - xA(x)) / (1 - x^2A(x)^2)^2 = 1 - x.

A365725 G.f. satisfies A(x) = 1 + x^3A(x)^4(1 + x*A(x)).