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.

A192415 G.f. satisfies: A(x) = (1 + x*A(x))*(1 + x^3*A(x)^2).

Original entry on oeis.org

1, 1, 1, 2, 5, 11, 23, 51, 120, 286, 681, 1636, 3985, 9803, 24257, 60338, 150931, 379501, 958360, 2429294, 6179380, 15769380, 40361087, 103579221, 266471500, 687098810, 1775440421, 4596689688, 11922774513, 30977768907, 80615085087, 210103228155, 548352756656, 1433053608502
Offset: 0

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Author

Paul D. Hanna, Nov 02 2011

Keywords

Examples

			G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 5*x^4 + 11*x^5 + 23*x^6 + 51*x^7 +...
Related expansions:
A(x)^2 = 1 + 2*x + 3*x^2 + 6*x^3 + 15*x^4 + 36*x^5 + 82*x^6 + 190*x^7 +...
A(x)^3 = 1 + 3*x + 6*x^2 + 13*x^3 + 33*x^4 + 84*x^5 + 205*x^6 + 498*x^7 +...
where A(x) = 1 + x*A(x) + x^3*A(x)^2 + x^4*A(x)^3.
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + x^2*A(x))*x + (1 + 2^2*x^2*A(x) + x^4*A(x)^2)*x^2/2 +
(1 + 3^2*x^2*A(x) + 3^2*x^4*A(x)^2 + x^6*A(x)^3)*x^3/3 +
(1 + 4^2*x^2*A(x) + 6^2*x^4*A(x)^2 + 4^2*x^6*A(x)^3 + x^8*A(x)^4)*x^4/4 +
(1 + 5^2*x^2*A(x) + 10^2*x^4*A(x)^2 + 10^2*x^6*A(x)^3 + 5^2*x^8*A(x)^4 + x^10*A(x)^5)*x^5/5 +...
Explicitly,
log(A(x)) = x + x^2/2 + 4*x^3/3 + 13*x^4/4 + 31*x^5/5 + 70*x^6/6 + 176*x^7/7 + 469*x^8/8 + 1228*x^9/9 + 3161*x^10/10 +...
		

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)*(1 + x^3*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)=local(A=1+x); for(i=1, n, A=(1 + x*A)*(1 + x^3*(A+x*O(x^n))^2)); 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^(2*j)*(A+x*O(x^n))^j)*x^m/m))); polcoeff(A, n, x)}
    
  • PARI
    x='x; y='y; Fxy = (1+x*y) * (1 + x^3*y^2) - y;
    seq(N) = {
      my(y0 = 1 + O('x^N), y1=0);
      for (k = 1, N,
        y1 = y0 - subst(Fxy, y, y0)/subst(deriv(Fxy,y), y, y0);
        if (y1 == y0, break()); y0 = y1);
      Vec(y0);
    };
    seq(34)  \\ Gheorghe Coserea, Nov 30 2016

Formula

G.f. satisfies: A(x) = exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} C(n,k)^2 * x^(2*k) * A(x)^k ).
D-finite with recurrence: 4*(n+1)*(n+2)*(217*n^3 - 1239*n^2 + 1838*n - 336)*a(n) = 6*(n+1)*(434*n^4 - 2261*n^3 + 2339*n^2 + 1792*n - 1344)*a(n-1) - (n-1)*(2821*n^4 - 13286*n^3 + 7829*n^2 + 18464*n - 4032)*a(n-2) + 6*(868*n^5 - 6258*n^4 + 13981*n^3 - 7438*n^2 - 7769*n + 6136)*a(n-3) + 2*(n-3)*(2*n - 5)*(434*n^3 - 1393*n^2 + 211*n + 1048)*a(n-4) + 2*(n-4)*(2*n - 7)*(217*n^3 - 588*n^2 + 11*n + 480)*a(n-5). - Vaclav Kotesovec, Sep 19 2013
a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 2.730683387097269698... is the root of the equation -4 - 8*d - 24*d^2 + 13*d^3 - 12*d^4 + 4*d^5 = 0 and c = 2.078548317061344694159945441842754... is the root of the equation -1 - 67*c^2 - 19811*c^4 + 36463*c^6 - 41664*c^8 + 7936*c^10 = 0. - Vaclav Kotesovec, Sep 19 2013, updated Nov 28 2016
a(n) = Sum_{k=0..n/2+1} C(n-k+2,k-1)*C(n-k+2,2*k-1)/(n-k+2). - Vladimir Kruchinin, Feb 12 2019