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.

Showing 1-10 of 12 results. Next

A199874 G.f. satisfies A(x) = (1 + x*A(x)^2)*(1 + x^2*A(x)^2).

Original entry on oeis.org

1, 1, 3, 10, 37, 147, 611, 2625, 11564, 51953, 237123, 1096420, 5125063, 24178427, 114974387, 550511901, 2651896733, 12843003108, 62494595022, 305400429548, 1498184696271, 7375179807191, 36421312544431, 180383163330765, 895756907248150, 4459095182031675, 22247684478181317
Offset: 0

Views

Author

Paul D. Hanna, Nov 11 2011

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 10*x^3 + 37*x^4 + 147*x^5 + 611*x^6 +...
where A( x/(1+x^2) - x^2 ) = (1+x^2)/(1-x-x^3).
Related expansions:
A(x)^2 = 1 + 2*x + 7*x^2 + 26*x^3 + 103*x^4 + 428*x^5 + 1838*x^6 +...
A(x)^4 = 1 + 4*x + 18*x^2 + 80*x^3 + 359*x^4 + 1632*x^5 + 7506*x^6 +...
where A(x) = 1 + x*(1+x)*A(x)^2 + x^3*A(x)^4.
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + x)*x*A(x) + (1 + 2^2*x + x^2)*x^2*A(x)^2/2 +
(1 + 3^2*x + 3^2*x^2 + x^3)*x^3*A(x)^3/3 +
(1 + 4^2*x + 6^2*x^2 + 4^2*x^3 + x^4)*x^4*A(x)^4/4 +
(1 + 5^2*x + 10^2*x^2 + 10^2*x^3 + 5^2*x^4 + x^5)*x^5*A(x)^5/5 +
(1 + 6^2*x + 15^2*x^2 + 20^2*x^3 + 15^2*x^4 + 6^2*x^5 + x^6)*x^6*A(x)^6/6 +...
more explicitly,
log(A(x)) = x + 5*x^2/2 + 22*x^3/3 + 101*x^4/4 + 481*x^5/5 + 2330*x^6/6 +...
		

Crossrefs

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^2)*(1+x^2*AGF^2)-AGF,x,j]==0,koef][[1]]; aa[[j]]=koef/.sol[[1]],{j,2,nmax}];Flatten[{1,aa}] (* Vaclav Kotesovec, Aug 18 2013 *)
  • PARI
    {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)*(x*A+x*O(x^n))^m/m))); polcoeff(A, n, x)}
    
  • PARI
    {a(n)=polcoeff((1/x)*serreverse(x/(1+x^2+x*O(x^n))-x^2),n)}
    
  • PARI
    {a(n)=polcoeff(((1+x^2)/(1-x-x^3+x*O(x^n)))^(n+1)/(n+1), n)}
    
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, (1-x)^(2*m+1)*sum(j=0, n, binomial(m+j, j)^2*x^j)*x^m*A^m/m))); polcoeff(A, n, x)}

Formula

G.f. A(x) satisfies:
(1) A(x) = (1/x)*Series_Reversion( x/(1+x^2) - x^2 ).
(2) A( x*(1-x-x^3)/(1+x^2) ) = (1+x^2)/(1-x-x^3).
(3) a(n) = [x^n] ((1+x^2)/(1-x-x^3))^(n+1) / (n+1).
(4) A(x) = exp( Sum_{n>=1} (Sum_{k=0..n} C(n,k)^2 * x^k) * x^n*A(x)^n/n ).
(5) A(x) = exp( Sum_{n>=1} (1-x)^(2*n+1)*(Sum_{k>=0} C(n+k,k)^2*x^k) * x^n*A(x)^n/n ).
Recurrence: 31*(n-1)*n*(n+1)*(85396*n^4 - 902916*n^3 + 3471647*n^2 - 5767203*n + 3503250)*a(n) = 2*(n-1)*n*(6319304*n^5 - 69975436*n^4 + 290875210*n^3 - 559740413*n^2 + 484175751*n - 138985722)*a(n-1) + 2*(n-1)*(2903464*n^6 - 36506072*n^5 + 179801738*n^4 - 439606930*n^3 + 553204983*n^2 - 328951215*n + 67014378)*a(n-2) + 2*(2*n - 5)*(1964108*n^6 - 24695284*n^5 + 123902749*n^4 - 317652203*n^3 + 438313617*n^2 - 307740825*n + 85471038)*a(n-3) - 32*(n-3)*(2*n - 7)*(85396*n^5 - 860218*n^4 + 3249611*n^3 - 5747414*n^2 + 4753791*n - 1471338)*a(n-4) + 8*(n-4)*(n-3)*(2*n - 9)*(85396*n^4 - 561332*n^3 + 1275275*n^2 - 1191073*n + 390174)*a(n-5). - Vaclav Kotesovec, Aug 18 2013
a(n) ~ c*d^n/n^(3/2), where d=5.28245622984... is the root of the equation -16 + 64*d - 92*d^2 - 68*d^3 - 148*d^4 + 31*d^5 = 0 and c = 0.49559010377906722118329... - Vaclav Kotesovec, Aug 18 2013
a(n) = Sum_{k=0..floor(n/2)} binomial(2*n-2*k+1,k) * binomial(2*n-2*k+1,n-2*k) / (2*n-2*k+1). - Seiichi Manyama, Jul 18 2023

A200719 G.f. satisfies A(x) = (1 + x*A(x)^2) * (1 + x^2*A(x)^5).

Original entry on oeis.org

1, 1, 3, 13, 64, 340, 1903, 11053, 65993, 402527, 2497439, 15712220, 100001459, 642719263, 4165537744, 27193644061, 178654643151, 1180282875483, 7836312619243, 52259258911091, 349902441457427, 2351240866736891, 15851508780927739, 107187240225220684, 726784821098903319
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 + 3*x^2 + 13*x^3 + 64*x^4 + 340*x^5 + 1903*x^6 +...
Related expansions:
A(x)^2 = 1 + 2*x + 7*x^2 + 32*x^3 + 163*x^4 + 886*x^5 + 5039*x^6 +...
A(x)^5 = 1 + 5*x + 25*x^2 + 135*x^3 + 765*x^4 + 4481*x^5 + 26920*x^6 +...
A(x)^7 = 1 + 7*x + 42*x^2 + 252*x^3 + 1533*x^4 + 9457*x^5 + 59101*x^6 +...
where A(x) = 1 + x*A(x)^2 + x^2*A(x)^5 + x^3*A(x)^7.
The logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (1 + x*A^3)*x*A + (1 + 2^2*x*A^3 + x^2*A^6)*x^2*A^2/2 +
(1 + 3^2*x*A^3 + 3^2*x^2*A^6 + x^3*A^9)*x^3*A^3/3 +
(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^4/4 +
(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^5/5 + ...
		

Crossrefs

Programs

  • PARI
    {a(n)=local(p=1,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=1,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=1,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)^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)^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: 4232*(n-2)*(n-1)*n*(2*n - 3)*(2*n - 1)*(2*n + 1)*(108983978975*n^7 - 1828734495225*n^6 + 13017379495661*n^5 - 50928975062019*n^4 + 118201965098732*n^3 - 162617590602876*n^2 + 122676758610192*n - 39103265134080)*a(n) = 8*(n-2)*(n-1)*(2*n - 3)*(2*n - 1)*(850837923857825*n^9 - 15127768128079400*n^8 + 116088908648008427*n^7 - 502364025369222635*n^6 + 1342887860190877280*n^5 - 2280899268898038065*n^4 + 2433907848768834828*n^3 - 1548429898790214180*n^2 + 521138603722292640*n - 68863424146977600)*a(n-1) - 30*(n-2)*(2*n - 3)*(155302170039375*n^11 - 3227155335853125*n^10 + 29807524885054600*n^9 - 161278340404759950*n^8 + 566950865855228019*n^7 - 1356848300481904461*n^6 + 2250482361655315470*n^5 - 2579665279074165840*n^4 + 1996011605601581864*n^3 - 988803599084885136*n^2 + 280851990522009984*n - 34444332223983360)*a(n-2) + 5*(7288303593953125*n^13 - 187891351713750000*n^12 + 2204843674914291875*n^11 - 15579013461781304250*n^10 + 73867718896175411475*n^9 - 247858726321141236540*n^8 + 604530296941440837821*n^7 - 1082990060568950070282*n^6 + 1421457900098213642392*n^5 - 1345695224728829837040*n^4 + 889319601933492222864*n^3 - 386196670582228097568*n^2 + 97916706472751405568*n - 10797892365692920320)*a(n-3) + 10*(n-3)*(2*n - 7)*(5*n - 18)*(5*n - 14)*(5*n - 12)*(5*n - 11)*(108983978975*n^7 - 1065846642400*n^6 + 4333636082786*n^5 - 9458655747964*n^4 + 11899609166891*n^3 - 8554104592084*n^2 + 3208950812340*n - 473478110640)*a(n-4). - Vaclav Kotesovec, Nov 17 2017
a(n) ~ s * sqrt((r*s*(r*s^3 - 1) - 3) / (7*Pi*(5*r*s*(1 + r*s^3) - 3))) / (2*n^(3/2)*r^n), where r = 0.1385102270697349252376651829944449360743895474888... and s = 1.450646440303399446510765649245639306003224666768... are real roots of the system of equations (1 + r*s^2)*(1 + r^2*s^5) = s, r*s*(2 + 5*r*s^3 + 7*r^2*s^5) = 1. - Vaclav Kotesovec, Nov 22 2017
a(n) = Sum_{k=0..floor(n/2)} binomial(2*n+k+1,k) * binomial(2*n+k+1,n-2*k) / (2*n+k+1). - Seiichi Manyama, Jul 18 2023

A200074 G.f. satisfies A(x) = (1 + x*A(x)^2)*(1 + x^2*A(x)).

Original entry on oeis.org

1, 1, 3, 9, 30, 108, 406, 1577, 6280, 25499, 105169, 439388, 1855636, 7908909, 33975250, 146954693, 639460707, 2797384235, 12295494109, 54272825103, 240480529815, 1069257987503, 4769306203838, 21334400243252, 95687482105807, 430217846136134, 1938651904470374, 8754225470415889
Offset: 0

Views

Author

Paul D. Hanna, Nov 13 2011

Keywords

Comments

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)).

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 9*x^3 + 30*x^4 + 108*x^5 + 406*x^6 + 1577*x^7 +...
Related expansions:
A(x)^2 = 1 + 2*x + 7*x^2 + 24*x^3 + 87*x^4 + 330*x^5 + 1289*x^6 +...
A(x)^3 = 1 + 3*x + 12*x^2 + 46*x^3 + 180*x^4 + 720*x^5 + 2928*x^6 +...
where A(x) = 1 + x*A(x)^2 + x^2*A(x) + x^3*A(x)^3.
The logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (A + x)*x + (A^2 + 2^2*x*A + x^2)*x^2/2 +
(A^3 + 3^2*x*A^2 + 3^2*x^2*A + x^3)*x^3/3 +
(A^4 + 4^2*x*A^3 + 6^2*x^2*A^2 + 4^2*x^3*A + x^4)*x^4/4 +
(A^5 + 5^2*x*A^4 + 10^2*x^2*A^3 + 10^2*x^3*A^2 + 5^2*x^4*A + x^5)*x^5/5 +
(A^6 + 6^2*x*A^5 + 15^2*x^2*A^4 + 20^2*x^3*A^3 + 15^2*x^4*A^2 + 6^2*x^5*A + x^6)*x^6/6 +...
more explicitly,
log(A(x)) = x + 5*x^2/2 + 19*x^3/3 + 77*x^4/4 + 331*x^5/5 + 1445*x^6/6 + 6392*x^7/7 + 28565*x^8/8 +...
		

Crossrefs

Programs

  • Maple
    a:= n-> coeff(series(RootOf(A=(1+x*A^2)*(1+x^2*A), A), x, n+1), x, n):
    seq(a(n), n=0..30);  # Alois P. Heinz, May 16 2012
  • Mathematica
    m = 28; A[_] = 0;
    Do[A[x_] = (1 + x A[x]^2)(1 + x^2 A[x]) + O[x]^m, {m}];
    CoefficientList[A[x], x] (* Jean-François Alcover, Oct 02 2019 *)
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=(1+x*A^2)*(1+x^2*A^1)+x*O(x^n));polcoeff(A,n)}
    
  • PARI
    {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^j)*(x*A+x*O(x^n))^m/m))); polcoeff(A, n, x)}
    
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, (1-x/A)^(2*m+1)*sum(j=0, n, binomial(m+j, j)^2*x^j/A^j)*x^m*A^m/m))); polcoeff(A, n, x)}

Formula

G.f. satisfies:
(1) A(x) = exp( Sum_{n>=1} (Sum_{k=0..n} C(n,k)^2 * x^k*A(x)^(n-k)) * x^n/n ).
(2) 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)^n/n ).
(3) A(x) = x / Series_Reversion( x*G(x) ) where G(x) is the g.f. of A199876.
(4) A(x) = G(x/A(x)) where G(x) = A(x*G(x)) is the g.f. of A199876.
Recurrence: (n+1)*(n+2)*(1241*n^4 - 10636*n^3 + 25417*n^2 - 7382*n - 17136)*a(n) = - 18*(n+1)*(443*n^3 - 3889*n^2 + 9734*n - 5712)*a(n-1) + 4*(6205*n^6 - 53180*n^5 + 115741*n^4 + 64762*n^3 - 370103*n^2 + 246727*n - 25704)*a(n-2) + 6*(2482*n^6 - 24995*n^5 + 76519*n^4 - 36347*n^3 - 185471*n^2 + 293092*n - 140400)*a(n-3) + 2*(4964*n^6 - 57436*n^5 + 228617*n^4 - 276802*n^3 - 361447*n^2 + 956696*n - 320496)*a(n-4) - 6*(2482*n^6 - 32441*n^5 + 140587*n^4 - 173153*n^3 - 266705*n^2 + 677518*n - 291840)*a(n-5) + 12*(n-4)*(2*n - 11)*(11*n^2 + 73*n - 748)*a(n-6) + 2*(n-5)*(2*n - 13)*(1241*n^4 - 5672*n^3 + 955*n^2 + 16508*n - 8496)*a(n-7). - Vaclav Kotesovec, Aug 18 2013
a(n) ~ c*d^n/n^(3/2), where d = 4.770539985405... is the root of the equation -4 + 12*d^2 - 8*d^3 - 12*d^4 - 20*d^5 + d^7 = 0 and c = 0.612892860188927397373456... - Vaclav Kotesovec, Aug 18 2013
a(n) = Sum_{k=0..floor(n/2)} binomial(2*n-3*k+1,k) * binomial(2*n-3*k+1,n-2*k) / (2*n-3*k+1). - Seiichi Manyama, Jul 18 2023

A200075 G.f. satisfies A(x) = (1 + x*A(x)^2)*(1 + x^2*A(x)^3).

Original entry on oeis.org

1, 1, 3, 11, 45, 198, 914, 4367, 21414, 107155, 544987, 2808978, 14640073, 77025373, 408544815, 2182206259, 11727989593, 63373962690, 344109933186, 1876562458845, 10273572074493, 56443282489240, 311097732946200, 1719707775782826, 9531914043637385, 52963938340248863, 294966593345731623
Offset: 0

Views

Author

Paul D. Hanna, Nov 13 2011

Keywords

Comments

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)).

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 11*x^3 + 45*x^4 + 198*x^5 + 914*x^6 +...
Related expansions:
A(x)^2 = 1 + 2*x + 7*x^2 + 28*x^3 + 121*x^4 + 552*x^5 + 2615*x^6 +...
A(x)^3 = 1 + 3*x + 12*x^2 + 52*x^3 + 237*x^4 + 1122*x^5 + 5463*x^6 +...
A(x)^5 = 1 + 5*x + 25*x^2 + 125*x^3 + 630*x^4 + 3211*x^5 + 16545*x^6 +...
where A(x) = 1 + x*A(x)^2 + x^2*A(x)^3 + 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 + (1 + 2^2*x*A + x^2*A^2)*x^2*A^2/2 +
(1 + 3^2*x*A + 3^2*x^2*A^2 + x^3*A^3)*x^3*A^3/3 +
(1 + 4^2*x*A + 6^2*x^2*A^2 + 4^2*x^3*A^3 + x^4*A^4)*x^4*A^4/4 +
(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^5/5 +
(1 + 6^2*x*A + 15^2*x^2*A^2 + 20^2*x^3*A^3 + 15^2*x^4*A^4 + 6^2*x^5*A^5 + x^6*A^6)*x^6*A^6/6 +...
more explicitly,
log(A(x)) = x + 5*x^2/2 + 25*x^3/3 + 129*x^4/4 + 686*x^5/5 + 3713*x^6/6 + 20350*x^7/7 +...
Given G(x) where 1+x*G(x) is the g.f. of A004148, then the coefficients in the powers of G(x) begin:
1: [(1), 1, 2, 4, 8, 17, 37, 82, 185, 423, 978, ...];
2: [1,(2), 5, 12, 28, 66, 156, 370, 882, 2112, ...];
3: [1, 3,(9), 25, 66, 171, 437, 1107, 2790, 7009, ...];
4: [1, 4, 14,(44), 129, 364, 1000, 2696, 7172, 18892, ...];
5: [1, 5, 20, 70,(225), 686, 2015, 5760, 16135, 44500, ...];
6: [1, 6, 27, 104, 363,(1188), 3713, 11214, 32994, 95106, ...];
7: [1, 7, 35, 147, 553, 1932,(6398), 20350, 62734, 188650, ...];
8: [1, 8, 44, 200, 806, 2992, 10460,(34936), 112585, 352560, ...];
9: [1, 9, 54, 264, 1134, 4455, 16389, 57330,(192726), 627406, ...]; ...;
the coefficients in parenthesis form the initial terms of this sequence:
[1/1, 2/2, 9/3, 44/4, 225/5, 1188/6, 6398/7, 34936/8, 192726/9, ...].
The coefficients in the logarithm of the g.f. is also a diagonal in the above table.
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[1/x*InverseSeries[Series[x*(1-x-x^2 + Sqrt[(1+x+x^2)*(1-3*x+x^2)])/2,{x,0,20}],x],x] (* Vaclav Kotesovec, Sep 19 2013 *)
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=(1+x*A^2)*(1+x^2*A^3)+x*O(x^n));polcoeff(A,n)}
    
  • PARI
    {a(n)=polcoeff(1/x*serreverse(x*(1-x-x^2 + sqrt((1+x+x^2)*(1-3*x+x^2)+x*O(x^n)))/2),n)}
    
  • PARI
    {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^j)*(x*A+x*O(x^n))^m/m))); polcoeff(A, n, x)}
    
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, (1-x*A)^(2*m+1)*sum(j=0, n, binomial(m+j, j)^2*x^j*A^j)*x^m*A^m/m))); polcoeff(A, n, x)}

Formula

G.f.: (1/x)*Series_Reversion( x*(1-x-x^2 + sqrt((1+x+x^2)*(1-3*x+x^2)))/2 ).
a(n) = [x^n] G(x)^(n+1)/(n+1), where 1+x*G(x) is the g.f. of A004148.
G.f. A(x) satisfies:
(1) A(x) = (1/x)*Series_Reversion( x/G(x) ) where 1+x*G(x) is the g.f. of A004148.
(2) A(x) = G(x*A(x)) where G(x) = A(x/G(x)) 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)^k) * x^n*A(x)^n/n ).
(4) 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)^n/n ).
Recurrence: 8*n*(2*n+1)*(4*n+1)*(4*n+3)*(1557671*n^7 - 18939961*n^6 + 94817789*n^5 - 252067387*n^4 + 381880748*n^3 - 327052012*n^2 + 145198992*n - 25583040)*a(n) = (2026529971*n^11 - 24640889261*n^10 + 122927623620*n^9 - 322351865586*n^8 + 467303512311*n^7 - 343677276405*n^6 + 61590777290*n^5 + 76066203476*n^4 - 45605627832*n^3 + 4625651136*n^2 + 1916801280*n - 338688000)*a(n-1) + 2*(800642894*n^11 - 10936104295*n^10 + 62803409541*n^9 - 196202081616*n^8 + 357730085364*n^7 - 370711524567*n^6 + 174415015309*n^5 + 25877389846*n^4 - 63266190708*n^3 + 19055552472*n^2 + 1313789760*n - 861840000)*a(n-2) + 6*(308418858*n^11 - 4675368852*n^10 + 30103912361*n^9 - 106665982366*n^8 + 223860428776*n^7 - 274000455628*n^6 + 166116940489*n^5 - 2432493994*n^4 - 54297743044*n^3 + 22033617000*n^2 + 936446400*n - 1315440000)*a(n-3) + 6*(n-2)*(2*n-7)*(3*n-10)*(3*n-8)*(1557671*n^7 - 8036264*n^6 + 13889114*n^5 - 7559372*n^4 - 2491645*n^3 + 2975476*n^2 - 179460*n - 187200)*a(n-4). - Vaclav Kotesovec, Sep 19 2013
a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 1301/1024 + 1/(1024*sqrt(3/(7183147 - (2002819072*2^(2/3))/(3725055779 + 42057117*sqrt(16305))^(1/3) + 1024*(7450111558 + 84114234*sqrt(16305))^(1/3)))) + (1/2)*sqrt(7183147/393216 - (3725055779 + 42057117*sqrt(16305))^(1/3)/(384*2^(2/3)) + 977939/(192*(7450111558 + 84114234*sqrt(16305))^(1/3)) + (1/131072)*(4194454317*sqrt(3/(7183147 - (2002819072*2^(2/3))/(3725055779 + 42057117*sqrt(16305))^(1/3) + 1024*(7450111558 + 84114234*sqrt(16305))^(1/3))))) = 5.89828930084513611... is the root of the equation -108 - 1188*d - 1028*d^2 - 1301*d^3 + 256*d^4 = 0 and c = 0.656947859044624009263362998790812821830934... - Vaclav Kotesovec, Sep 19 2013
a(n) = Sum_{k=0..floor(n/2)} binomial(2*n-k+1,k) * binomial(2*n-k+1,n-2*k) / (2*n-k+1). - Seiichi Manyama, Jul 18 2023

A186241 G.f. satisfies A(x) = 1 + x*A(x)^2 + x^2*A(x)^4 + x^3*A(x)^6.

Original entry on oeis.org

1, 1, 3, 12, 54, 262, 1337, 7072, 38426, 213197, 1202795, 6879160, 39794416, 232429030, 1368806610, 8118934656, 48458809586, 290832756606, 1754059333738, 10625545472716, 64620970743082, 394409682103262, 2415084675723048, 14832185219521152, 91339478577683664
Offset: 0

Views

Author

Vladimir Kruchinin, Feb 15 2011

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 12*x^3 + 54*x^4 + 262*x^5 + 1337*x^6 +...
where A(x) = (1 + x*A(x)^2)*(1 + x^2*A(x)^4).
Related expansions:
A(x)^2 = 1 + 2*x + 7*x^2 + 30*x^3 + 141*x^4 + 704*x^5 + 3666*x^6 +...
A(x)^4 = 1 + 4*x + 18*x^2 + 88*x^3 + 451*x^4 + 2392*x^5 + 13022*x^6 +...
A(x)^6 = 1 + 6*x + 33*x^2 + 182*x^3 + 1014*x^4 + 5718*x^5 + 32623*x^6 +...
where A(x) = 1 + x*A(x)^2 + x^2*A(x)^4 + x^3*A(x)^6.
From _Paul D. Hanna_, Nov 11 2011: (Start)
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 + ...
which involves squares of binomial coefficients. (End)
		

Crossrefs

Programs

  • Maple
    F:= proc(n) if n::even then
      simplify((1/2)*hypergeom([-(1/2)*n, -2*n-1, -(1/2)*n+1/2], [(1/2)*n+1, 3/2+(1/2)*n], -1)*(2*n+2)!/((2*n+1)*((n+1)!)^2))
      else
      simplify((1/2)*hypergeom([-(1/2)*n, -2*n-1, -(1/2)*n+1/2], [(1/2)*n+1, 3/2+(1/2)*n], -1)*(2*n+2)!/((2*n+1)*((n+1)!)^2))
      fi
    end proc:
    map(F, [$0..30]); # Robert Israel, Jun 22 2015
  • Mathematica
    a[n_] := 1/(2n + 1) Sum[Binomial[2n + 1, k] Binomial[2n + 1, n - 2k], {k, 0, n/2}];
    (* or: *)
    a[n_] := (Binomial[2n + 1, n] HypergeometricPFQ[{-2n - 1, 1/2 - n/2, -n/2}, {n/2 + 1, n/2 + 3/2}, -1])/(2n + 1);
    Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Nov 17 2017 *)
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=(1+x*A^2)*(1+x^2*A^4)+x*O(x^n));polcoeff(A,n)} /* Paul D. Hanna */
    
  • PARI
    {a(n)=polcoeff(sqrt((1/x)*serreverse(x/(1 + x + x^2 + x^3 +x*O(x^n))^2)), n)} /* Paul D. Hanna */
    
  • PARI
    {a(n)=polcoeff( (1 + x + x^2 + x^3+x*O(x^n))^(2*n+1)/(2*n+1), n)} /* Paul D. Hanna */
    
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, (x*A+x*O(x^n))^m/m*sum(j=0, m, binomial(m, j)^2*x^j*A^(2*j))))); polcoeff(A, n, x)} /* Paul D. Hanna */
    
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n,x^m*A^m/m*(1-x*A^2)^(2*m+1)*sum(j=0, n, binomial(m+j, j)^2*x^j*A^(2*j))))); polcoeff(A, n, x)} /* Paul D. Hanna */

Formula

a(n) = (1/(2*n-1))*Sum_{j=0..2*n-1} binomial(2*n-1,j)*Sum_{i=j..n+j-1} binomial(j,i-j)*binomial(2*n-j-1,3*j-3*n-i+1), n>0.
From Paul D. Hanna, Nov 11 2011: (Start)
G.f. A(x) satisfies:
(1) A(x) = sqrt( (1/x)*Series_Reversion( x/(1 + x + x^2 + x^3)^2 ) ).
(2) A( x/(1 + x + x^2 + x^3)^2 ) = 1 + x + x^2 + x^3.
(3) A(x) = G(x*A(x)) where G(x) = A(x/G(x)) = g.f. of A036765 (number of rooted trees with a degree constraint).
(4) a(n) = [x^n] (1 + x + x^2 + x^3)^(2*n+1) / (2*n+1).
(5) 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)] ).
(6) 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)] ).
(End)
From Peter Bala, Jun 21 2015: (Start)
a(n) = 1/(2*n + 1)*Sum_{k = 0..floor(n/2)} binomial(2*n + 1,k)*binomial(2*n + 1,n - 2*k).
More generally, the coefficient of x^n in A(x)^r equals r/(2*n + r)*Sum_{k = 0..floor(n/2)} binomial(2*n + r,k)*binomial(2*n + r,n - 2*k) by the Lagrange-Bürmann formula.
O.g.f. A(x) = exp(Sum_{n >= 1} 1/2*b(n)*x^n/n), where b(n) = Sum_{k = 0..floor(n/2)} binomial(2*n,k)*binomial(2*n,n - 2*k). Cf. A036765, A198951, A200731. (End)
Recurrence: 5*n*(5*n - 1)*(5*n + 1)*(5*n + 2)*(5*n + 3)*(13144*n^4 - 57784*n^3 + 90149*n^2 - 59354*n + 13980)*a(n) = 8*(2*n - 1)*(16259128*n^8 - 71478808*n^7 + 108653137*n^6 - 60530902*n^5 - 2811173*n^4 + 12694433*n^3 - 2398482*n^2 - 352503*n + 78570)*a(n-1) + 128*(n-1)*(2*n - 3)*(2*n - 1)*(52576*n^6 - 178560*n^5 + 136156*n^4 + 22938*n^3 - 16067*n^2 - 3138*n - 405)*a(n-2) + 2048*(n-2)*(n-1)*(2*n - 5)*(2*n - 3)*(2*n - 1)*(13144*n^4 - 5208*n^3 - 4339*n^2 + 168*n + 135)*a(n-3). - Vaclav Kotesovec, Nov 17 2017
A(x^2) = (1/x) * series reversion of x/(1 + x^2 + x^4 + x^6). - Peter Bala, Jul 27 2023

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

Views

Author

Paul D. Hanna, Aug 16 2012

Keywords

Comments

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)).

Examples

			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 +...
		

Crossrefs

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^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 *)
  • PARI
    {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),", "))
    
  • PARI
    {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)}
    
  • PARI
    {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)}
    
  • PARI
    {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)}

Formula

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 + x*A(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

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 + 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 + ...
		

Crossrefs

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^2) - 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)=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),", "))
    
  • 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) = (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

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 + ...
		

Crossrefs

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 + ...
		

Crossrefs

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.
		

Crossrefs

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
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