cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 11-16 of 16 results.

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

Views

Author

Paul D. Hanna, Nov 21 2011

Keywords

Comments

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)).

Examples

			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 + ...

Links

Crossrefs

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

Programs

  • Mathematica
    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 *)
  • PARI
    {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)}
  • PARI
    {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)}
  • PARI
    {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)}

Formula

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

A104545 Number of Motzkin paths of length n having no consecutive (1,0) steps.

Original entry on oeis.org

1, 1, 1, 3, 5, 11, 25, 55, 129, 303, 721, 1743, 4241, 10415, 25761, 64095, 160385, 403263, 1018369, 2581887, 6569089, 16767871, 42927105, 110194175, 283574017, 731427583, 1890600193, 4896499455, 12704869633, 33021750015, 85966113281
Offset: 0

Views

Author

Emeric Deutsch, Mar 14 2005

Keywords

Comments

a(n) = A104544(n,0) (n > 0).

Examples

			a(3)=3 because we have UDH, HUD and UHD, where U=(1,1), D=(1,-1) and H=(1,0) (HHH does not qualify).
The logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (1 + x*A^2)*x/A + (1 + 2^2*x*A^2 + x^2*A^4)*x^2/A^2/2 +
(1 + 3^2*x*A^2 + 3^2*x^2*A^4 + x^3*A^6)*x^3/A^3/3 +
(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^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)*x^5/A^5/5 + ...

Links

Crossrefs

Cf. A001006, A104544, A198957, A200718.

Programs

  • Maple
    G:=(1-sqrt(1-4*z^2*(1+z)^2))/2/z^2/(1+z): Gser:=series(G,z=0,35): 1,seq(coeff(Gser,z^n),n=1..31);
  • Mathematica
    Array[Sum[Binomial[2 k, k]/(k + 1) (Binomial[2 k, # - 2 k + 1] + Binomial[2 k, # - 2 k]), {k, Ceiling[#/4], (# + 1)/2}] &[# - 1] &, 31, 0] (* Michael De Vlieger, Feb 18 2020 *)
CoefficientList[Series[(1-Sqrt[1-4x^2 (1+x)^2])/(2x^2 (1+x)),{x,0,30}],x] (* Harvey P. Dale, Mar 02 2020 *)
  • Maxima
    b(n):=sum(binomial(2*k,k)/(k+1)*(binomial(2*k,n-2*k+1)+binomial(2*k,n-2*k)),k,ceiling(n/4),(n+1)/2); a(n):=if n=0 then 1 else b(n-1); /* Vladimir Kruchinin, Mar 14 2012 */
  • PARI
    {a(n)=local(p=-1,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)}
  • PARI
    {a(n)=local(p=-1,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)}
  • PARI
    {a(n)=local(p=-1,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)}

Formula

G.f.: (1-sqrt(1-4z^2*(1+z)^2))/(2z^2*(1+z)).
G.f. A(x) satisfies:
(1) A(x) = (1+x)*(1 + x^2*A(x)^2).
(2) A(x) = exp( Sum_{n>=1} x^n * A(x)^(-n)/n * [Sum_{k=0..n} C(n,k)^2 * x^k * A(x)^(2*k)] ).
(3) A(x) = exp( Sum_{n>=1} x^n * A(x)^(-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+1) = sum(binomial(2*k,k)/(k+1)*(binomial(2*k,n-2*k+1)+binomial(2*k,n-2*k)),k,ceiling(n/4),(n+1)/2), a(0)=1. - Vladimir Kruchinin, Mar 14 2012
(n+2)*a(n) + (n+2)*a(n-1) + 4*(-n+1)*a(n-2) + 12*(-n+2)*a(n-3) + 12*(-n+3)*a(n-4) + 4*(-n+4)*a(n-5) = 0. - R. J. Mathar, Jul 23 2017
a(n) ~ 3^(1/4) * (1+sqrt(3))^(n + 1/2) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Nov 17 2017

A200731 G.f. satisfies: A(x) = (1 + x*A(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

Views

Author

Paul D. Hanna, Nov 21 2011

Keywords

Comments

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.

Examples

			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.

Links

Crossrefs

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

Programs

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

Formula

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

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

Views

Author

Paul D. Hanna, Nov 21 2011

Keywords

Comments

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.

Examples

			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).

Links

Crossrefs

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

Programs

  • Mathematica
    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 *)
  • PARI
    {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)}
  • PARI
    {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)}
  • PARI
    {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)}

Formula

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

A211248 G.f. satisfies: A(x) = (1 + x*A(x)^3) * (1 + x^2*A(x)^4).

Original entry on oeis.org

1, 1, 4, 20, 114, 703, 4565, 30752, 212921, 1505916, 10833164, 79018804, 583062388, 4344431508, 32641910199, 247033970128, 1881402836376, 14408753414558, 110897147057354, 857307054338476, 6653979156676983, 51831065993122915, 405060413133136902, 3175019470333290488
Offset: 0

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Author

Paul D. Hanna, Apr 05 2012

Keywords

Comments

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 and q=1.

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 20*x^3 + 114*x^4 + 703*x^5 + 4565*x^6 +...
where A( x*(1-x-x^3)^2/(1+x^2)^2 ) = (1+x^2)/(1-x-x^3).
Related expansions:
A(x)^3 = 1 + 3*x + 15*x^2 + 85*x^3 + 522*x^4 + 3381*x^5 + 22735*x^6 +...
A(x)^4 = 1 + 4*x + 22*x^2 + 132*x^3 + 841*x^4 + 5588*x^5 + 38288*x^6 +...
A(x)^7 = 1 + 7*x + 49*x^2 + 343*x^3 + 2429*x^4 + 17430*x^5 +...
where A(x) = 1 + x*A(x)^3 + x^2*A(x)^4 + x^3*A(x)^7.
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + x*A(x))*x*A(x)^2 + (1 + 2^2*x*A(x) + x^2*A(x)^2)*x^2*A(x)^4/2 +
(1 + 3^2*x*A(x) + 3^2*x^2*A(x)^2 + x^3*A(x)^3)*x^3*A(x)^6/3 +
(1 + 4^2*x*A(x) + 6^2*x^2*A(x)^2 + 4^2*x^3*A(x)^3 + x^4*A(x)^4)*x^4*A(x)^8/4 +
(1 + 5^2*x*A(x) + 10^2*x^2*A(x)^2 + 10^2*x^3*A(x)^3 + 5^2*x^4*A(x)^4 + x^5*A(x)^5)*x^5*A(x)^10/5 +
(1 + 6^2*x*A(x) + 15^2*x^2*A(x)^2 + 20^2*x^3*A(x)^3 + 15^2*x^4*A(x)^4 + 6^2*x^5*A(x)^5 + x^6*A(x)^6)*x^6*A(x)^12/6 +...
more explicitly,
log(A(x)) = x + 7*x^2/2 + 49*x^3/3 + 359*x^4/4 + 2706*x^5/5 + 20767*x^6/6 +...

Crossrefs

Cf. A200716, A200717, A199876, A211249, A200731, A199874, A199877.
Cf. A198951, A198953, A198957, A192415, A198888, A036765.

Programs

  • Mathematica
    CoefficientList[Sqrt[1/x * InverseSeries[Series[x*(1 - x - x^3)^2/(1 + x^2)^2, {x, 0, 20}], x]], x] (* Vaclav Kotesovec, Nov 22 2017 *)
  • PARI
    {a(n)=polcoeff(sqrt( (1/x)*serreverse( x*(1-x-x^3)^2/(1+x^2+x*O(x^n))^2 ) ), n)}
for(n=0,30,print1(a(n),", "))
  • PARI
    {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)}
  • PARI
    {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)}
  • PARI
    {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)}

Formula

G.f. A(x) satisfies:
(1) A(x) = sqrt( (1/x)*Series_Reversion( x*(1-x-x^3)^2/(1+x^2)^2 ) ).
(2) A( x*(1-x-x^3)^2/(1+x^2)^2 ) = (1+x^2)/(1-x-x^3).
(3) a(n) = [x^n] ((1+x^2)/(1-x-x^3))^(2*n+2) / (n+1).
(4) A(x) = exp( Sum_{n>=1} (Sum_{k=0..n} C(n,k)^2 * x^k*A(x)^k) * x^n*A(x)^(2*n)/n ).
(5) A(x) = exp( Sum_{n>=1} (1-x*A(x))^(2*n+1) * (Sum_{k>=0} C(n+k,k)^2*x^k*A(x)^k) * x^n*A(x)^(2*n)/n ).
(6) A(x) = (1/x)*Series_Reversion(x/G(x)) where A(x) = G(x*A(x)) and A(x/G(x)) = G(x) = (1 + x*G(x)^2)*(1 + x^2*G(x)^2) is the g.f. of A199874.
a(n) ~ s * sqrt((1 + 2*r*s + 3*r^2*s^4) / (3*Pi*(1 + 2*r*s + 7*r^2*s^4))) / (2*n^(3/2)*r^n), where r = 0.1194948955213353102456218138370139612914667337222... and s = 1.428770161302757679335810379290625953730830139744... are real roots of the system of equations (1 + r*s^3)*(1 + r^2*s^4) = s, r*s^2*(3 + 4*r*s + 7*r^2*s^4) = 1. - Vaclav Kotesovec, Nov 22 2017

A211249 G.f. satisfies: A(x) = (1 + x*A(x)^3) * (1 + x^2*A(x)^5).

Original entry on oeis.org

1, 1, 4, 21, 126, 819, 5611, 39900, 291719, 2179181, 16560175, 127617168, 994951887, 7833555324, 62196300997, 497425570173, 4003607595960, 32404662671330, 263586896132154, 2153631763231319, 17666722629907960, 145449082369322208, 1201414340736684702
Offset: 0

Views

Author

Paul D. Hanna, Apr 05 2012

Keywords

Comments

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 and q=2.

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 21*x^3 + 126*x^4 + 819*x^5 + 5611*x^6 +...
where A( x*(1-x-x^3)^2/(1+x^2)^2 ) = (1+x^2)/(1-x-x^3).
Related expansions:
A(x)^3 = 1 + 3*x + 15*x^2 + 88*x^3 + 564*x^4 + 3828*x^5 + 27040*x^6 +...
A(x)^5 = 1 + 5*x + 30*x^2 + 195*x^3 + 1335*x^4 + 9486*x^5 + 69305*x^6 +...
A(x)^8 = 1 + 8*x + 60*x^2 + 448*x^3 + 3374*x^4 + 25704*x^5 +...
where A(x) = 1 + x*A(x)^3 + x^2*A(x)^5 + x^3*A(x)^8.
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + x*A(x)^2)*x*A(x)^2 +
(1 + 2^2*x*A(x)^2 + x^2*A(x)^4)*x^2*A(x)^4/2 +
(1 + 3^2*x*A(x)^2 + 3^2*x^2*A(x)^4 + x^3*A(x)^6)*x^3*A(x)^6/3 +
(1 + 4^2*x*A(x)^2 + 6^2*x^2*A(x)^4 + 4^2*x^3*A(x)^6 + x^4*A(x)^8)*x^4*A(x)^8/4 +
(1 + 5^2*x*A(x)^2 + 10^2*x^2*A(x)^4 + 10^2*x^3*A(x)^6 + 5^2*x^4*A(x)^8 + x^5*A(x)^10)*x^5*A(x)^10/5 +
(1 + 6^2*x*A(x)^2 + 15^2*x^2*A(x)^4 + 20^2*x^3*A(x)^6 + 15^2*x^4*A(x)^8 + 6^2*x^5*A(x)^10 + x^6*A(x)^12)*x^6*A(x)^12/6 +...
more explicitly,
log(A(x)) = x + 7*x^2/2 + 52*x^3/3 + 403*x^4/4 + 3211*x^5/5 + 26050*x^6/6 +...

Crossrefs

Cf. A200716, A200717, A199876, A211248, A200731, A200075, A199877.
Cf. A199874, A198951, A198953, A198957, A192415, A198888, A036765.

Programs

  • Mathematica
    CoefficientList[Sqrt[1/x * InverseSeries[Series[x*(1-2*x-x^2+x^4 + (1-x-x^2) * Sqrt[(1+x+x^2)*(1-3*x+x^2)])/2, {x, 0, 20}], x]], x] (* Vaclav Kotesovec, Nov 22 2017 *)
  • PARI
    {a(n)=polcoeff(sqrt( (1/x)*serreverse( x*(1-2*x-x^2+x^4 + (1-x-x^2)*sqrt( (1+x+x^2)*(1-3*x+x^2) +x*O(x^n)))/2 ) ), n)}
for(n=0,30,print1(a(n),", "))
  • PARI
    {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)}
  • PARI
    {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)}
  • PARI
    {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)}

Formula

G.f.: sqrt( (1/x)*Series_Reversion( x*(1-2*x-x^2+x^4 + (1-x-x^2)*sqrt( (1+x+x^2)*(1-3*x+x^2) ))/2 ) ).
G.f. A(x) satisfies:
(1) A(x) = (1/x)*Series_Reversion(x/G(x)) where A(x) = G(x*A(x)) and A(x/G(x)) = G(x) is the g.f. of A200075.
(2) A(x) = sqrt( (1/x)*Series_Reversion( x/G(x)^2 ) ) where A(x) = G(x*A(x)^2) and G(x) = A(x/G(x)^2) and 1+x*G(x) is the g.f. of A004148.
(3) A(x) = exp( Sum_{n>=1} (Sum_{k=0..n} C(n,k)^2 * x^k*A(x)^(2*k)) * x^n*A(x)^(2*n)/n ).
(4) 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)^(2*n)/n ).
a(n) ~ s * sqrt((1 + 2*r*s^2 + 3*r^2*s^5) / (Pi*(3 + 10*r*s^2 + 28*r^2*s^5))) / (2*n^(3/2)*r^n), where r = 0.1130413665724951344267888513870607581680912144315... and s = 1.385648922830296011590145919380626723251960276539... are real roots of the system of equations (1 + r*s^3)*(1 + r^2*s^5) = s, r*s^2*(3 + 5*r*s^2 + 8*r^2*s^5) = 1. - Vaclav Kotesovec, Nov 22 2017
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