A036765 -id:A036765 - cp's OEIS frontend

A181734 G.f. A(x) satisfies: A(x) = (1 + xA(x)) (1 + x^2*A(x)^3).

1, 1, 2, 6, 19, 64, 227, 832, 3126, 11980, 46646, 184003, 733783, 2953434, 11982265, 48949631, 201182110, 831292029, 3451336467, 14390479996, 60232976244, 252992172572, 1066000599632, 4504710385216, 19086728370308, 81069926894797

Offset: 0

Paul D. Hanna, Feb 14 2011

nonn

			G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 19*x^4 + 64*x^5 + 227*x^6 +...
The logarithm of the g.f. may be expressed as the series:
log(A(x)) = (1 + x*A(x)^2)*x +
(1 + 4*x*A(x)^2 + x^2*A(x)^4)*x^2/2 +
(1 + 9*x*A(x)^2 + 9*x^2*A(x)^4 + x^3*A(x)^6)*x^3/3 +
(1 + 16*x*A(x)^2 + 36*x^2*A(x)^4 + 16*x^3*A(x)^6 + x^4*A(x)^8)*x^4/4 +...
which involves the squared binomial coefficients.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A036765.

Table[Sum[Binomial[n+k,k]*Binomial[n+k+1, n-2k]/(n+1),{k,0,Floor[n/2]}],{n,0,20}] (* Vaclav Kotesovec, Sep 18 2013 *)

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

{a(n)=local(A=1+x+x*O(x^n));for(i=1,n,A=(1+x*A)*(1+x^2*A^3));polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

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

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

a(n) = Sum_{k=0..[n/2]} C(n+k, k)*C(n+k+1, n-2k)/(n+1).
G.f.: A(x) = (1/x)*Series_Reversion( x/(1+x) - x^3 ).
G.f. satisfies:
(1) A(x) = G(x*A(x)) where A(x/G(x)) = G(x) = (1 + x)/(1 - x^2 - x^3).
(2) A(x) = exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^2*x^k*A(x)^(2k)] * x^n/n ).
(3) A(x) = exp( Sum_{n>=1} [Sum_{k>=0} C(n+k,k)^2*x^k*A(x)^(2k)]*(1-x*A(x)^2)^(2*n+1)* x^n/n ).
Recurrence: 23*(n-1)*n*(n+1)*(429*n^2 - 1903*n + 2004)*a(n) = 3*(n-1)*n*(12870*n^3 - 63525*n^2 + 87653*n - 28312)*a(n-1) + 3*(n-1)*(3861*n^4 - 24849*n^3 + 59286*n^2 - 61456*n + 22956)*a(n-2) + 3*(23166*n^5 - 207009*n^4 + 712578*n^3 - 1173947*n^2 + 916800*n - 266772)*a(n-3) - 3*(n-3)*(3*n - 10)*(3*n - 5)*(429*n^2 - 1045*n + 530)*a(n-4). - Vaclav Kotesovec, Sep 18 2013
a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 1/46*(45 + 36*sqrt(3) + 23*sqrt(3429/529 + (3792*sqrt(3))/529)) = 4.50735893936524052... is the root of the equation 27 - 162*d - 27*d^2 - 90*d^3 + 23*d^4 = 0 and c = 1/(2*sqrt(-29 - 17*sqrt(3) + 5*sqrt(69 + 40*sqrt(3)))) = 0.842957337580479516110291505734... - Vaclav Kotesovec, Sep 18 2013, updated Mar 18 2024

Name changed slightly by Paul D. Hanna, Nov 14 2012

A215576 G.f. satisfies A(x) = (1 + x^2)(1 + xA(x)^2).

Original entry on oeis.org

1, 1, 3, 8, 24, 80, 278, 997, 3670, 13782, 52588, 203314, 794726, 3135540, 12470444, 49942305, 201233170, 815205699, 3318291966, 13565162636, 55669063762, 229257178198, 947142023262, 3924380904498, 16303716754884, 67899954924360, 283425070356740, 1185551594834910

Offset: 0

Paul D. Hanna, Aug 16 2012

nonn

More generally, for fixed parameters p, q, r, and s, if F(x) satisfies:
F(x) = exp( Sum_{n>=1} x^(n*r)*F(x)^(n*p)/n * [Sum_{k=0..n} C(n,k)^2 * x^(k*s)*F(x)^(k*q)] ),
then F(x) = (1 + x^r*F(x)^(p+1))*(1 + x^(r+s)*F(x)^(p+q+1)).

			G.f.: A(x) = 1 + x + 3*x^2 + 8*x^3 + 24*x^4 + 80*x^5 + 278*x^6 + 997*x^7 +...
Related expansions:
A(x)^2 = 1 + 2*x + 7*x^2 + 22*x^3 + 73*x^4 + 256*x^5 + 924*x^6 + 3414*x^7 +...
where A(x) = 1+x^2 + x*(1+x^2)*A(x)^2.
The logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (1 + x/A^2)*A*x + (1 + 2^2*x/A^2 + x^2/A^4)*A^2*x^2/2 +
 (1 + 3^2*x/A^2 + 3^2*x^2/A^4 + x^3/A^6)*A^3*x^3/3 +
 (1 + 4^2*x/A^2 + 6^2*x^2/A^4 + 4^2*x^3/A^6 + x^4/A^8)*A^4*x^4/4 +
 (1 + 5^2*x/A^2 + 10^2*x^2/A^4 + 10^2*x^3/A^6 + 5^2*x^4/A^8 + x^5/A^10)*A^5*x^5/5 +
 (1 + 6^2*x/A^2 + 15^2*x^2/A^4 + 20^2*x^3/A^6 + 15^2*x^4/A^8 + 6^2*x^5/A^10 + x^6/A^12)*A^6*x^6/6 +...
more explicitly,
log(A(x)) = x + 5*x^2/2 + 16*x^3/3 + 57*x^4/4 + 231*x^5/5 + 938*x^6/6 + 3830*x^7/7 + 15833*x^8/8 +...

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A199876, A199877, A198951, A198953, A198957, A192415, A198888, A200717, A004148, A036765.

Cf. A186241, A199874, A200074, A200075, A200718, A200719.

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[Coefficient[(1+x^2)*(1+x*AGF^2)-AGF,x,j]==0,koef][[1]];aa[[j]]=koef/.sol[[1]],{j,2,nmax}];Flatten[{1,aa}] (* Vaclav Kotesovec, Aug 19 2013 *)
CoefficientList[Series[(1 - Sqrt[1 - 4*x*(1 + x^2)^2]) / (2*x*(1 + x^2)), {x, 0, 30}], x] (* Vaclav Kotesovec, Oct 11 2018 *)
Table[Sum[Binomial[2*n - 4*i + 1, i] * Binomial[2*n - 4*i + 1, n - 2*i]/(2*n - 4*i + 1), {i, 0, Floor[n/2]}], {n, 0, 30}] (* Vaclav Kotesovec, Oct 11 2018, after Vladimir Kruchinin *)

{a(n)=polcoeff((1 - sqrt(1 - 4*x*(1+x^2 +x*O(x^n))^2)) / (2*x*(1+x^2 +x*O(x^n))),n)}
for(n=0,31,print1(a(n),", "))

{a(n)=local(A=1+x); for(i=1, n, A=(1+x*A^2)*(1+x^2)+x*O(x^n)); polcoeff(A, n)}

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

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

G.f. satisfies:
(1) A(x) = (1 - sqrt(1 - 4*x*(1+x^2)^2)) / (2*x*(1+x^2)).
(2) A(x) = exp( Sum_{n>=1} (x^n/n) * A(x)^n * (Sum_{k=0..n} C(n,k)^2 * x^k / A(x)^(2*k)) ).
(3) A(x) = exp( Sum_{n>=1} (1-x/A(x)^2)^(2*n+1) * (Sum_{k>=0} C(n+k,k)^2*x^k/A(x)^(2*k)) * x^n*A(x)^n/n ).
(4) A(x) = x / Series_Reversion( x*G(x) ) where G(x) is the g.f. of A200717.
(5) A(x) = G(x/A(x)) where G(x) = A(x*G(x)) is the g.f. of A200717.
Recurrence: (n+1)*a(n) = 2*(2*n-1)*a(n-1) - (n+1)*a(n-2) + 6*(2*n-5)*a(n-3) + 6*(2*n-9)*a(n-5) + 2*(2*n-13)*a(n-7). - Vaclav Kotesovec, Aug 19 2013
a(n) ~ c*d^n/n^(3/2), where d = 4.41997678... is the root of the equation -4-8*d^2-4*d^4+d^5=0 and c = sqrt(d*(8 + 16*d^2 + 8*d^4 + 3*d^5 + d^7) / (Pi*(1 + d^2)^3))/4 = 0.648259186485429075561822659694489853... - Vaclav Kotesovec, Aug 19 2013, updated Oct 11 2018
a(n) = Sum_{i=0..floor(n/2)} C(2*n-4*i+1,i)*C(2*n-4*i+1,n-2*i)/(2*n-4*i+1). - Vladimir Kruchinin, Oct 11 2018

A200717 G.f. satisfies: A(x) = (1 + xA(x)^3) (1 + x^2*A(x)^2).

Original entry on oeis.org

1, 1, 4, 18, 93, 521, 3073, 18806, 118297, 760162, 4968480, 32928392, 220766739, 1494635330, 10203884795, 70167751762, 485574854049, 3379064343829, 23631314301088, 165998001901786, 1170706810318259, 8286253163771045, 58842370488310336, 419102145275264242, 2993221125640617827

Offset: 0

Paul D. Hanna, Nov 21 2011

nonn

More generally, for fixed parameters p and q, if F(x) satisfies:

F(x) = exp( Sum_{n>=1} x^n * F(x)^(n*p)/n * [Sum_{k=0..n} C(n,k)^2 * x^k * F(x)^(k*q)] ), then F(x) = (1 + x*F(x)^(p+1))*(1 + x^2*F(x)^(p+q+1)).

			G.f.: A(x) = 1 + x + 4*x^2 + 18*x^3 + 93*x^4 + 521*x^5 + 3073*x^6 +...
Related expansions:
A(x)^2 = 1 + 2*x + 9*x^2 + 44*x^3 + 238*x^4 + 1372*x^5 + 8256*x^6 +...
A(x)^3 = 1 + 3*x + 15*x^2 + 79*x^3 + 447*x^4 + 2655*x^5 + 16324*x^6 +...
A(x)^5 = 1 + 5*x + 30*x^2 + 180*x^3 + 1110*x^4 + 7006*x^5 + 45075*x^6 +...
where A(x) = 1 + x*A(x)^3 + x^2*A(x)^2 + x^3*A(x)^5.
The logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (1 + x/A)*x*A^2 + (1 + 2^2*x/A + x^2/A^2)*x^2*A^4/2 +
(1 + 3^2*x/A + 3^2*x^2/A^2 + x^3/A^3)*x^3*A^6/3 +
(1 + 4^2*x/A + 6^2*x^2/A^2 + 4^2*x^3/A^3 + x^4/A^4)*x^4*A^8/4 +
(1 + 5^2*x/A + 10^2*x^2/A^2 + 10^2*x^3/A^3 + 5^2*x^4/A^4 + x^5/A^5)*x^5*A^10/5 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Vaclav Kotesovec, Recurrence (of order 9)

Cf. A200716, A200718, A200719, A200074, A200075, A215576, A199876, A199874, A199877, A198951, A198953, A198957, A192415, A198888, A036765.

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[Coefficient[(1 + x*AGF^3) * (1 + x^2*AGF^2) - AGF,x,j]==0,koef][[1]];aa[[j]]=koef/.sol[[1]],{j,2,nmax}]; Flatten[{1,aa}] (* Vaclav Kotesovec, Sep 19 2013 *)

{a(n)=polcoeff((1/x)*serreverse( x*(1 + sqrt(1 - 4*x*(1+x^2)^2 +x*O(x^n))) / (2*(1+x^2)) ),n)}
for(n=0,30,print1(a(n),", "))

{a(n)=local(p=2,q=-1,A=1+x);for(i=1,n,A=(1+x*A^(p+1))*(1+x^2*A^(p+q+1))+x*O(x^n));polcoeff(A,n)}

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

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

G.f. A(x) satisfies:
(1) A(x) = (1/x)*Series_Reversion( x*(1 + sqrt(1 - 4*x*(1+x^2)^2)) / (2*(1+x^2)) ).
(2) A(x) = exp( Sum_{n>=1} x^n * A(x)^(2*n)/n * [Sum_{k=0..n} C(n,k)^2 * x^k / A(x)^k] ).
(3) A(x) = exp( Sum_{n>=1} x^n * A(x)^(2*n)/n * [(1-x/A(x)^2)^(2*n+1) * Sum_{k>=0} C(n+k,k)^2*x^k / A(x)^k] ).
a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 7.60435909657327146... is the root of the equation -108 + 27*d^2 + 1620*d^3 - 216*d^4 - 1456*d^5 - 2556*d^6 - 716*d^7 + 20*d^8 + 16*d^9 = 0 and c = 0.45780648099092640511434469483084555191269495951... - Vaclav Kotesovec, Sep 19 2013

A215623 G.f. satisfies A(x) = (1 + xA(x)) (1 + x*A(x)^4).

Original entry on oeis.org

1, 2, 11, 89, 836, 8551, 92445, 1039030, 12019135, 142151324, 1711116646, 20894534324, 258195565959, 3222677162409, 40569811695707, 514520507077695, 6567611974106756, 84310605465652750, 1087798325715407703, 14098475168420865396, 183465816241394787196

Offset: 0

Paul D. Hanna, Aug 17 2012

The radius of convergence of g.f. A(x) is r = 0.0712256396327314729661274986100... with A(r) = 1.4248895273944523042559975726479124492235978714420... where y=A(r) satisfies 3*y^7 - 4*y^6 + 16*y^5 - 28*y^4 + 8*y^3 - 4 = 0.

			G.f.: A(x) = 1 + 2*x + 11*x^2 + 89*x^3 + 836*x^4 + 8551*x^5 + 92445*x^6 + ...
Related expansions.
A(x)^4 = 1 + 8*x + 68*x^2 + 652*x^3 + 6750*x^4 + 73544*x^5 + 831078*x^6 + ...
A(x)^5 = 1 + 10*x + 95*x^2 + 965*x^3 + 10350*x^4 + 115507*x^5 + ...
where A(x) = 1 + x*(A(x) + A(x)^4) + x^2*A(x)^5.
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + A(x)^3)*x + (1 + 2^2*A(x)^3 + A(x)^6)*x^2/2 +
  (1 + 3^2*A(x)^3 + 3^2*A(x)^6 + A(x)^9)*x^3/3 +
  (1 + 4^2*A(x)^3 + 6^2*A(x)^6 + 4^2*A(x)^9 + A(x)^12)*x^4/4 +
  (1 + 5^2*A(x)^3 + 10^2*A(x)^6 + 10^2*A(x)^9 + 5^2*A(x)^12 + A(x)^15)*x^5/5 + ...
more explicitly,
log(A(x)) = 2*x + 18*x^2/2 + 209*x^3/3 + 2550*x^4/4 + 32082*x^5/5 + 411705*x^6/6 + 5356416*x^7/7 + ....

Seiichi Manyama, Table of n, a(n) for n = 0..876

Cf. A198953, A215624, A036765, A198951, A181734, A364331, A364336, A364376.

{a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2*(A+x*O(x^n))^(3*j))*x^m/m))); polcoeff(A, n)}

{a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=(1+x*A)*(1+x*A^4)+x*O(x^n)); polcoeff(A, n)}
for(n=0,21,print1(a(n),", "))

G.f. satisfies A(x) = exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} C(n,k)^2 * A(x)^(3*k)).
The formal inverse of g.f. A(x) is (sqrt((1-x^3)^2 + 4*x^4) - (1+x^3))/(2*x^4).
a(n) = Sum_{k=0..n} binomial(n+3*k+1,k) * binomial(n+3*k+1,n-k) / (n+3*k+1). - Seiichi Manyama, Jul 19 2023
From  Peter Bala, Sep 10 2024: (Start)
x/series_reversion(x*A(x)) = 1 + 2*x + 7*x^2 + 39*x^3 + 242*x^4 + 1634*x^5 + ..., the g.f. of A364336.
(1/x) * series_reversion(x/A(x)) = 1 + 2*x + 15*x^2 + 163*x^3 + 2070*x^4 + 28698*x^5 + ..., the g.f. of A364331. (End)

A215624 G.f. satisfies A(x) = (1 + xA(x)) (1 + x*A(x)^5).

Original entry on oeis.org

1, 2, 13, 130, 1518, 19358, 261323, 3670828, 53100530, 785657529, 11834135909, 180863294507, 2797643204500, 43715591710804, 689030031494554, 10941710269299893, 174889301792724294, 2811464199460768704, 45426696813655278251

Offset: 0

Paul D. Hanna, Aug 17 2012

nonn

The radius of convergence of g.f. A(x) is r = 0.05685644444171304880925020950930... with A(r) = 1.3208055627586104770123863310077013110788003146438630... where y=A(r) satisfies 4*y^9 - 5*y^8 + 25*y^6 - 40*y^5 + 10*y^4 - 5 = 0.

			G.f.: A(x) = 1 + 2*x + 13*x^2 + 130*x^3 + 1518*x^4 + 19358*x^5 +...
Related expansions.
A(x)^5 = 1 + 10*x + 105*x^2 + 1250*x^3 + 16120*x^4 + 219162*x^5 +...
A(x)^6 = 1 + 12*x + 138*x^2 + 1720*x^3 + 22803*x^4 + 315840*x^5 +...
where A(x) = 1 + x*(A(x) + A(x)^5) + x^2*A(x)^6.
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + A(x)^4)*x + (1 + 2^2*A(x)^4 + A(x)^8)*x^2/2 +
  (1 + 3^2*A(x)^4 + 3^2*A(x)^8 + A(x)^12)*x^3/3 +
  (1 + 4^2*A(x)^4 + 6^2*A(x)^8 + 4^2*A(x)^12 + A(x)^16)*x^4/4 +
  (1 + 5^2*A(x)^4 + 10^2*A(x)^8 + 10^2*A(x)^12 + 5^2*A(x)^16 + A(x)^20)*x^5/5 +...
more explicitly,
log(A(x)) = 2*x + 22*x^2/2 + 320*x^3/3 + 4886*x^4/4 + 76962*x^5/5 + 1236784*x^6/6 + 20152260*x^7/7 +...

Seiichi Manyama, Table of n, a(n) for n = 0..807

Cf. A198953, A215623, A036765, A198951, A181734.

{a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2*(A+x*O(x^n))^(4*j))*x^m/m))); polcoeff(A, n)}

{a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=(1+x*A)*(1+x*A^5)+x*O(x^n)); polcoeff(A, n)}
for(n=0,21,print1(a(n),", "))

G.f. satisfies A(x) = exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} C(n,k)^2 * A(x)^(4*k)).
The formal inverse of g.f. A(x) is (sqrt((1-x^4)^2 + 4*x^5) - (1+x^4))/(2*x^5).
a(n) = Sum_{k=0..n} binomial(n+4*k+1,k) * binomial(n+4*k+1,n-k) / (n+4*k+1). - Seiichi Manyama, Jul 19 2023

A200716 G.f. satisfies: A(x) = (1 + xA(x)^3) (1 + x^2*A(x)).

Original entry on oeis.org

1, 1, 4, 17, 84, 453, 2574, 15185, 92119, 571022, 3600981, 23029021, 149000790, 973581692, 6415198045, 42580369370, 284427460919, 1910594331920, 12898153658337, 87461992473577, 595455441375978, 4068652368270955, 27891991988552554, 191783482751813061, 1322319472577803761

Offset: 0

Paul D. Hanna, Nov 21 2011

nonn

More generally, for fixed parameters p and q, if F(x) satisfies:
F(x) = exp( Sum_{n>=1} x^n * F(x)^(n*p)/n * [Sum_{k=0..n} C(n,k)^2 * x^k * F(x)^(k*q)] ),
then F(x) = (1 + x*F(x)^(p+1))*(1 + x^2*F(x)^(p+q+1)).

			G.f.: A(x) = 1 + x + 4*x^2 + 17*x^3 + 84*x^4 + 453*x^5 + 2574*x^6 +...
Related expansions:
A(x)^3 = 1 + 3*x + 15*x^2 + 76*x^3 + 414*x^4 + 2370*x^5 + 14047*x^6 +...
A(x)^4 = 1 + 4*x + 22*x^2 + 120*x^3 + 685*x^4 + 4048*x^5 + 24558*x^6 +...
where A(x) = 1 + x*A(x)^3 + x^2*A(x) + x^3*A(x)^4.
The logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (1 + x/A^2)*x*A^2 + (1 + 2^2*x/A^2 + x^2/A^4)*x^2*A^4/2 +
(1 + 3^2*x/A^2 + 3^2*x^2/A^4 + x^3/A^6)*x^3*A^6/3 +
(1 + 4^2*x/A^2 + 6^2*x^2/A^4 + 4^2*x^3/A^6 + x^4/A^8)*x^4*A^8/4 +
(1 + 5^2*x/A^2 + 10^2*x^2/A^4 + 10^2*x^3/A^6 + 5^2*x^4/A^8 + x^5/A^10)*x^5*A^10/5 + ...

Vaclav Kotesovec, Table of n, a(n) for n = 0..500
Vaclav Kotesovec, Recurrence (of order 11)

Cf. A200717, A200718, A200719, A200074, A200075, A199874, A199876, A199877, A198951, A198953, A198957, A192415, A198888, A036765.

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[Coefficient[(1 + x*AGF^3) * (1 + x^2*AGF) - AGF,x,j]==0,koef][[1]];aa[[j]]=koef/.sol[[1]],{j,2,nmax}]; Flatten[{1,aa}] (* Vaclav Kotesovec, Sep 19 2013 *)

{a(n)=local(p=2,q=-2,A=1+x);for(i=1,n,A=(1+x*A^(p+1))*(1+x^2*A^(p+q+1))+x*O(x^n));polcoeff(A,n)}

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

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

G.f. A(x) satisfies:
(1) A(x) = exp( Sum_{n>=1} x^n * A(x)^(2*n)/n * [Sum_{k=0..n} C(n,k)^2 * x^k / A(x)^(2*k)] ).
(2) A(x) = exp( Sum_{n>=1} x^n * A(x)^(2*n)/n * [(1-x/A(x)^2)^(2*n+1) * Sum_{k>=0} C(n+k,k)^2 * x^k/A(x)^(2*k)] ).
a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 7.342019160707096169... is the root of the equation -27 + 108*d^2 - 162*d^4 + 54*d^5 + 108*d^6 + 216*d^7 - 27*d^8 - 18*d^9 - 27*d^10 + 4*d^11 = 0 and c = 0.468554406193087607276981923311829947714908080994... - Vaclav Kotesovec, Sep 19 2013

A200731 G.f. satisfies: A(x) = (1 + xA(x)^3) (1 + x^2*A(x)^6).

Original entry on oeis.org

1, 1, 4, 22, 139, 953, 6894, 51796, 400269, 3161262, 25403536, 207043048, 1707345547, 14219399626, 119431172630, 1010495472960, 8604568715969, 73683710894255, 634142349130800, 5482062214763436, 47582484748270453, 414503778412715065, 3622792181209018168, 31758958747482608912

Offset: 0

Paul D. Hanna, Nov 21 2011

More generally, for fixed parameters p and q, if F(x) satisfies:
F(x) = exp( Sum_{n>=1} x^n * F(x)^(n*p)/n * [Sum_{k=0..n} C(n,k)^2 * x^k * F(x)^(k*q)] ),
then F(x) = (1 + x*F(x)^(p+1))*(1 + x^2*F(x)^(p+q+1)); here p=2, q=3.

			G.f.: A(x) = 1 + x + 4*x^2 + 22*x^3 + 139*x^4 + 953*x^5 + 6894*x^6 +...
where A(x) = (1 + x*A(x)^3)*(1 + x^2*A(x)^6).
Related expansions:
A(x)^3 = 1 + 3*x + 15*x^2 + 91*x^3 + 609*x^4 + 4335*x^5 + 32197*x^6 +...
A(x)^6 = 1 + 6*x + 39*x^2 + 272*x^3 + 1989*x^4 + 15054*x^5 + 116955*x^6 +...
A(x)^9 = 1 + 9*x + 72*x^2 + 570*x^3 + 4545*x^4 + 36639*x^5 + 298662*x^6 +...
where A(x) = 1 + x*A(x)^3 + x^2*A(x)^6 + x^3*A(x)^9.
The logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (1 + x*A^3)*x*A^2 + (1 + 2^2*x*A^3 + x^2*A^6)*x^2*A^4/2 +
(1 + 3^2*x*A^3 + 3^2*x^2*A^6 + x^3*A^9)*x^3*A^6/3 +
(1 + 4^2*x*A^3 + 6^2*x^2*A^6 + 4^2*x^3*A^9 + x^4*A^12)*x^4*A^8/4 +
(1 + 5^2*x*A^3 + 10^2*x^2*A^6 + 10^2*x^3*A^9 + 5^2*x^4*A^12 + x^5*A^15)*x^5*A^10/5 + ...
which involves squares of binomial coefficients.

Vaclav Kotesovec, Table of n, a(n) for n = 0..1000

Cf. A036765, A200716, A200717, A200718, A200719, A200725, A200074, A200075, A199874, A199876, A199877, A198951, A198953, A198957, A192415, A198888. Cf. A186241.

nmax = 23; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x] - (1 + x A[x]^3)*(1 + x^2 A[x]^6) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
sol /. Rule -> Set;
a /@ Range[0, nmax] (* Jean-François Alcover, Nov 02 2019 *)

{a(n)=polcoeff( ((1/x)*serreverse(x/(1 + x + x^2 + x^3 +x*O(x^n))^3))^(1/3), n)}

{a(n)=polcoeff( (1 + x + x^2 + x^3 +x*O(x^n))^(3*n+1)/(3*n+1), n)}

{a(n)=local(p=2,q=3,A=1+x);for(i=1,n,A=(1+x*A^(p+1))*(1+x^2*A^(p+q+1))+x*O(x^n));polcoeff(A,n)}

{a(n)=local(p=2,q=3,A=1+x);for(i=1,n,A=exp(sum(m=1,n,x^m*(A+x*O(x^n))^(p*m)/m*sum(j=0,m,binomial(m, j)^2*x^j*(A+x*O(x^n))^(q*j))))); polcoeff(A, n, x)}

{a(n)=local(p=2,q=3,A=1+x);for(i=1,n,A=exp(sum(m=1,n,x^m*(A+x*O(x^n))^(p*m)/m*(1-x*A^q)^(2*m+1)*sum(j=0, n, binomial(m+j, j)^2*x^j*(A+x*O(x^n))^(q*j))))); polcoeff(A, n, x)}

G.f. A(x) satisfies:
(1) a(n) = [x^n] (1 + x + x^2 + x^3)^(3*n+1) / (3*n+1).
(2) A(x) = ( (1/x)*Series_Reversion( x/(1 + x + x^2 + x^3)^3 ) )^(1/3).
(3) A( x/(1 + x + x^2 + x^3)^3 ) = 1 + x + x^2 + x^3.
(4) A(x) = G(x*A(x)^2) where G(x) = A(x/G(x)^2) = g.f. of A036765 (number of rooted trees with a degree constraint).
(5) A(x) = exp( Sum_{n>=1} x^n*A(x)^(2*n)/n * [Sum_{k=0..n} C(n,k)^2 * x^k*A(x)^(3*k)] ).
(6) A(x) = exp( Sum_{n>=1} x^n*A(x)^(2*n)/n * [(1-x*A(x)^2)^(2*n+1)*Sum_{k>=0} C(n+k,k)^2*x^k*A(x)^(3*k) )] ).
From Peter Bala, Jun 21 2015: (Start)
a(n) = 1/(3*n + 1)*Sum_{k = 0..floor(n/2)} binomial(3*n + 1,k)*binomial(3*n + 1,n - 2*k).
More generally, the coefficient of x^n in A(x)^r equals r/(3*n + r)*Sum_{k = 0..floor(n/2)} binomial(3*n + r,k)*binomial(3*n + r,n - 2*k) by the Lagrange-Bürmann formula.
O.g.f. A(x) = exp(Sum_{n >= 1} 1/3*b(n)x^n/n), where b(n) = Sum_{k = 0..floor(n/2)} binomial(3*n,k)*binomial(3*n,n - 2*k). Cf. A036765, A186241, A198951. (End)
Recurrence: 128*n*(2*n - 1)*(4*n - 1)*(4*n + 1)*(8*n - 3)*(8*n - 1)*(8*n + 1)*(8*n + 3)*(511073753*n^7 - 4871850365*n^6 + 19478089219*n^5 - 42349790393*n^4 + 54094962928*n^3 - 40605677522*n^2 + 16589611340*n - 2846611200)*a(n) = 3*(3*n - 2)*(3*n - 1)*(3047149994898003*n^13 - 32094344705469618*n^12 + 145743661212727337*n^11 - 373710048777443810*n^10 + 593788894662012231*n^9 - 600683242386376410*n^8 + 377600776651518819*n^7 - 130595257353511374*n^6 + 11334217618972546*n^5 + 8004135084547148*n^4 - 2618300200112616*n^3 + 152383960257264*n^2 + 33025238671680*n - 3264156403200)*a(n-1) - 576*(n-1)*(3*n - 5)*(3*n - 4)*(3*n - 2)*(3*n - 1)*(495741540410*n^10 - 3982082543435*n^9 + 12891395244590*n^8 - 21360691645174*n^7 + 18695904340190*n^6 - 7495052530111*n^5 + 212344193250*n^4 + 656210670544*n^3 - 106487698440*n^2 - 7969373424*n + 1477828800)*a(n-2) + 110592*(n-2)*(n-1)*(3*n - 8)*(3*n - 7)*(3*n - 5)*(3*n - 4)*(3*n - 2)*(3*n - 1)*(511073753*n^7 - 1294334094*n^6 + 979535842*n^5 - 149518418*n^4 - 72732399*n^3 + 16154432*n^2 + 843684*n - 192240)*a(n-3). - Vaclav Kotesovec, Nov 17 2017
a(n) ~ s/(2*sqrt(3*Pi*(4 - 9*r*s^2*(1 + r*s^3)))*n^(3/2)*r^n), where r = 0.1068159753611743655799981945670627355827110854720... and s = 1.345561337338583233012136458010090420775336284226... are real roots of the system of equations (1 + r*s^3)*(1 + r^2*s^6) = s, 3*r*s^2*(1 + 2*r*s^3 + 3*r^2*s^6) = 1. - Vaclav Kotesovec, Nov 22 2017

A203717 A Catalan triangle by rows.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 8, 4, 1, 1, 20, 15, 5, 1, 1, 50, 53, 21, 6, 1, 1, 126, 182, 84, 28, 7, 1, 1, 322, 616, 326, 120, 36, 8, 1, 1, 834, 2070, 1242, 495, 165, 45, 9, 1, 1, 2187, 6930, 4680, 1997, 715, 220, 55, 10, 1, 1, 5797, 23166, 17512, 7942, 3003, 1001, 286, 66, 11, 1

Offset: 1

Gary W. Adamson, Jan 04 2012

Row sums = the Catalan sequence starting with offset 1: (1, 2, 5, 14, 42,...).
T(n,k) is the number of Dyck n-paths whose maximum ascent length is k. - David Scambler, Aug 22 2012
T(n,k) is the number of ordered rooted trees with n non-root nodes and maximal outdegree k. T(4,3) = 4:
.    o      o      o      o
.    |     /|\    /|\    /|\
.    o    o o o  o o o  o o o
.   /|\   |        |        |
.  o o o  o        o        o   - Alois P. Heinz, Jun 29 2014
T(n,k) also is the number of permutations p of [n] such that in 0p the largest up-jump equals k and no down-jump is larger than 1. An up-jump j occurs at position i in p if p_{i} > p_{i-1} and j is the index of p_i in the increasingly sorted list of those elements in {p_{i}, ..., p_{n}} that are larger than p_{i-1}. A down-jump j occurs at position i in p if p_{i} < p_{i-1} and j is the index of p_i in the decreasingly sorted list of those elements in {p_{i}, ..., p_{n}} that are smaller than p_{i-1}. First index in the lists is 1 here. T(4,3) = 4: 1432, 3214, 3241, 3421. - Alois P. Heinz, Aug 29 2017

			First few rows of the array begin:
1,...1,...1,...1,...1,...;
1,...2,...4,...9,..21,...; = A001006
1,...2,...5,..13,..36,...; = A036765
1,...2,...5,..14,..41,...; = A036766
1,...2,...5,..14,..42,...; = A036767
... Taking finite differences of array terms starting from the top by columns, we obtain row terms of the triangle. First few rows of the triangle are:
  1;
  1,    1;
  1,    3,    1;
  1,    8,    4,    1;
  1,   20,   15,    5,    1;
  1,   50,   53,   21,    6,   1;
  1,  126,  182,   84,   28,   7,   1;
  1,  322,  616,  326,  120,  36,   8,  1;
  1,  834, 2070, 1242,  495, 165,  45,  9,  1;
  1, 2187, 6930, 4680, 1997, 715, 220, 55, 10, 1;
  ...
Example: Row 4 of the triangle = (1, 8, 4, 1) = the finite differences of (1, 9, 13, 14), column 4 of the array. Term (3,4) = 13 of the array is the upper left term of M^4, where M is an infinite square production matrix with four diagonals of 1's starting at (1,2), (1,1), (2,1), and (3,1); with the rest zeros.

Alois P. Heinz, Rows n = 1..141, flattened

Cf. A000108, A001006, A036765, A036766, A036767, A291680, A288942.
Columns k=1-3 give: A057427, A140662(n-1) for n>1, A303271.
T(2n,n) gives A291662.
T(2n+1,n+1) gives A005809.
T(n,ceiling(n/2)) gives A303259.

b:= proc(n, t, k) option remember; `if`(n=0, 1, `if`(t>0,
      add(b(j-1, k$2)*b(n-j, t-1, k), j=1..n), b(n-1, k$2)))
    end:
T:= (n, k)-> b(n, k-1$2) -`if`(k=1, 0, b(n, k-2$2)):
seq(seq(T(n, k), k=1..n), n=1..14);  # Alois P. Heinz, Jun 29 2014
# second Maple program:
b:= proc(u, o, k) option remember; `if`(u+o=0, 1,
      add(b(u-j, o+j-1, k), j=1..min(1, u))+
      add(b(u+j-1, o-j, k), j=1..min(k, o)))
    end:
T:= (n, k)-> b(0, n, k)-`if`(k=0, 0, b(0, n, k-1)):
seq(seq(T(n, k), k=1..n), n=1..14);  # Alois P. Heinz, Aug 28 2017

b[n_, t_, k_] := b[n, t, k] = If[n == 0, 1, If[t > 0, Sum[b[j-1, k, k]*b[n - j, t-1, k], {j, 1, n}], b[n-1, k, k]]]; T[n_, k_] := b[n, k-1, k-1] - If[k == 1, 0, b[n, k-2, k-2]]; Table[T[n, k], {n, 1, 14}, {k, 1, n}] // Flatten (* Jean-François Alcover, May 27 2016, after Alois P. Heinz *)

from sympy.core.cache import cacheit
@cacheit
def b(u, o, k): return 1 if u + o==0 else sum([b(u - j, o + j - 1, k) for j in range(1, min(1, u) + 1)]) + sum([b(u + j - 1, o - j, k) for j in range(1, min(k, o) + 1)])
def T(n, k): return b(0, n, k) - (0 if k==0 else b(0, n, k - 1))
for n in range(1, 16): print([T(n, k) for k in range(1, n + 1)]) # Indranil Ghosh, Aug 30 2017

Finite differences of antidiagonals of an array in which n-th array row is generated from powers of M, extracting successive upper left terms. M for n-th row of the array is an infinite square production matrix composed of (n+1) diagonals of 1's and the rest zeros. Given the upper left term of the array is (1,1), the diagonals begin at (1,2), (1,1), (2,1), (3,1), (4,1),...

T(n,k) = A288942(n,k) - A288942(n,k-1). - Alois P. Heinz, Sep 01 2017

A219977 Expansion of 1/(1+x+x^2+x^3).

Original entry on oeis.org

1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0

Offset: 0

Harvey P. Dale, Dec 02 2012

			G.f. = 1 - x + x^4 - x^5 + x^8 - x^9 + x^12 - x^13 + x^16 - x^17 + x^20 - x^21 + ...

G. C. Greubel, Table of n, a(n) for n = 0..5000
Elena Barcucci, Antonio Bernini, Stefano Bilotta, Renzo Pinzani, Non-overlapping matrices, arXiv:1601.07723 [cs.DM], 2016.
Stefano Bilotta, Variable-length Non-overlapping Codes, arXiv preprint arXiv:1605.03785, 2016
Kyu-Hwan Lee, Se-jin Oh, Catalan triangle numbers and binomial coefficients, arXiv:1601.06685 [math.CO], 2016.
Index entries for linear recurrences with constant coefficients, signature (-1,-1,-1).

Cf. A036765, A038505, A049347, A077962, A133872.

m:=100; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(1/(1+x+x^2+x^3))); // Vincenzo Librandi, Apr 22 2015

CoefficientList[Series[1/(1+x+x^2+x^3),{x,0,80}],x] (* or *) PadRight[{},120,{1,-1,0,0}]
LinearRecurrence[{-1,-1,-1},{1,-1,0},80] (* Harvey P. Dale, May 22 2021 *)

{a(n) = [1, -1, 0, 0][n%4 + 1]} /* Michael Somos, Dec 12 2012 */

Vec(1/(1+x+x^2+x^3) + O(x^100)) \\ Michel Marcus, Jan 28 2016

G.f.: 1/(1 +x +x^2 +x^3).
Euler transform of length 4 sequence [ -1, 0, 0, 1]. - Michael Somos, Dec 12 2012
a(n) = a(n+4) = -a(1-n). |a(n)| = A133872(n). REVERT transform is A036765. INVERT transform is A077962. - Michael Somos, Dec 12 2012
A038505(n+2) = p(-1)  where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 0, 1, ..., n. - Michael Somos, Dec 12 2012
From Wesley Ivan Hurt, Apr 22 2015: (Start)
a(n) +a(n-1) +a(n-2) +a(n-3) = 0.
a(n) = (-1)^n/2 +(-1)^(n/2 +1/4 -(-1)^n/4)/2. (End)

A200725 G.f. A(x) satisfies A(x) = (1+x^2)(1 + xA(x)^3).

Original entry on oeis.org

1, 1, 4, 16, 76, 399, 2206, 12664, 74790, 451420, 2772313, 17267652, 108821293, 692609446, 4445642625, 28744599748, 187047449289, 1224027357216, 8050074481917, 53179900898596, 352726704965748, 2348036826102013, 15682048658695168, 105052549830928908, 705678173069959645

Offset: 0

Paul D. Hanna, Nov 21 2011

nonn

More generally, for fixed parameters p and q, if F(x) satisfies:
F(x) = exp( Sum_{n>=1} x^n * F(x)^(n*p)/n * [Sum_{k=0..n} C(n,k)^2 * x^k * F(x)^(k*q)] ),
then F(x) = (1 + x*F(x)^(p+1))*(1 + x^2*F(x)^(p+q+1)); here p=2, q=-3.

			G.f.: A(x) = 1 + x + 4*x^2 + 16*x^3 + 76*x^4 + 399*x^5 + 2206*x^6 +...
Related expansion:
A(x)^3 = 1 + 3*x + 15*x^2 + 73*x^3 + 384*x^4 + 2133*x^5 + 12280*x^6 +...
where a(3) = 1 + 15; a(4) = 3 + 73; a(5) = 15 + 384; a(6) = 73 + 2133; ...
The logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (1 + x/A^3)*x*A^2 + (1 + 2^2*x/A^3 + x^2/A^6)*x^2*A^4/2 +
(1 + 3^2*x/A^3 + 3^2*x^2/A^6 + x^3/A^9)*x^3*A^6/3 +
(1 + 4^2*x/A^3 + 6^2*x^2/A^6 + 4^2*x^3/A^9 + x^4/A^12)*x^4*A^8/4 +
(1 + 5^2*x/A^3 + 10^2*x^2/A^6 + 10^2*x^3/A^9 + 5^2*x^4/A^12 + x^5/A^15)*x^5*A^10/5 + ...
which involves the squares of the binomial coefficients C(n,k).

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A200716, A200717, A200718, A200719, A200074, A200075, A199874, A199876, A199877, A198951, A198953, A198957, A192415, A198888, A036765.

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[Coefficient[(1+x^2)*(1+x*AGF^3)-AGF,x,j]==0,koef][[1]];aa[[j]]=koef/.sol[[1]],{j,2,nmax}];Flatten[{1,aa}] (* Vaclav Kotesovec, Aug 19 2013 *)

{a(n)=local(p=2,q=-3,A=1+x);for(i=1,n,A=(1+x*A^(p+1))*(1+x^2*A^(p+q+1))+x*O(x^n));polcoeff(A,n)}

{a(n)=local(p=2,q=-3,A=1+x);for(i=1,n,A=exp(sum(m=1,n,x^m*(A+x*O(x^n))^(p*m)/m*sum(j=0,m,binomial(m, j)^2*x^j*(A+x*O(x^n))^(q*j))))); polcoeff(A, n, x)}

{a(n)=local(p=2,q=-3,A=1+x);for(i=1,n,A=exp(sum(m=1,n,x^m*(A+x*O(x^n))^(p*m)/m*(1-x*A^q)^(2*m+1)*sum(j=0, n, binomial(m+j, j)^2*x^j*(A+x*O(x^n))^(q*j))))); polcoeff(A, n, x)}

G.f. A(x) satisfies:
(1) A(x) = exp( Sum_{n>=1} x^n * A(x)^(2*n)/n * (Sum_{k=0..n} C(n,k)^2 * x^k / A(x)^(3*k)) ).
(2) A(x) = exp( Sum_{n>=1} x^n * A(x)^(2*n)/n * (1 - x/A(x)^3)^(2*n+1) * Sum_{k>=0} C(n+k,k)^2*x^k / A(x)^(3*k) ).
Recurrence: 2*(n-4)*(n-2)*n*(2*n+1)*a(n) = 3*(n-4)*(n-2)*(3*n-2)*(3*n-1)*a(n-1) - 2*(n-4)*(n-2)*n*(2*n-1)*a(n-2) + 6*(n-4)*(3*n-8)*(6*n^2 - 17*n + 2)*a(n-3) + 6*(3*n-14)*(9*n^3 - 66*n^2 + 114*n - 4)*a(n-5) + 6*n*(3*n-20)*(6*n^2 - 47*n + 78)*a(n-7) + 3*(n-2)*n*(3*n-26)*(3*n-19)*a(n-9). - Vaclav Kotesovec, Aug 19 2013
a(n) ~ c*d^n/n^(3/2), where d = 7.1535029565... is the root of the equation -27 - 81*d^2 - 81*d^4 - 27*d^6 + 4*d^7 = 0 and c = 0.26300783791885411389369671... - Vaclav Kotesovec, Aug 19 2013
a(n) = Sum_{k=0..floor(n/2)} binomial(3*n-6*k+1,k) * binomial(3*n-6*k+1,n-2*k)/(3*n-6*k+1). - Seiichi Manyama, Dec 17 2024

cp's OEIS Frontend

A181734 G.f. A(x) satisfies: A(x) = (1 + x*A(x)) * (1 + x^2*A(x)^3).

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A215576 G.f. satisfies A(x) = (1 + x^2)*(1 + x*A(x)^2).

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A200717 G.f. satisfies: A(x) = (1 + x*A(x)^3) * (1 + x^2*A(x)^2).

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A215623 G.f. satisfies A(x) = (1 + x*A(x)) * (1 + x*A(x)^4).

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A215624 G.f. satisfies A(x) = (1 + x*A(x)) * (1 + x*A(x)^5).

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A200716 G.f. satisfies: A(x) = (1 + x*A(x)^3) * (1 + x^2*A(x)).

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A181734 G.f. A(x) satisfies: A(x) = (1 + xA(x)) (1 + x^2*A(x)^3).

A215576 G.f. satisfies A(x) = (1 + x^2)(1 + xA(x)^2).

A200717 G.f. satisfies: A(x) = (1 + xA(x)^3) (1 + x^2*A(x)^2).

A215623 G.f. satisfies A(x) = (1 + xA(x)) (1 + x*A(x)^4).

A215624 G.f. satisfies A(x) = (1 + xA(x)) (1 + x*A(x)^5).

A200716 G.f. satisfies: A(x) = (1 + xA(x)^3) (1 + x^2*A(x)).

A200731 G.f. satisfies: A(x) = (1 + xA(x)^3) (1 + x^2*A(x)^6).

A200725 G.f. A(x) satisfies A(x) = (1+x^2)(1 + xA(x)^3).