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.

A273958 G.f. A(x) satisfies: x*A(x) + x^2*A(x)^2 = C(x)^2, where C(x) = x + C(x)^2 is a g.f. of the Catalan numbers (A000108).

Original entry on oeis.org

1, 2, 4, 10, 30, 96, 313, 1038, 3512, 12100, 42286, 149440, 533076, 1917000, 6942660, 25300034, 92701746, 341319648, 1262176462, 4685724500, 17457027376, 65247128392, 244584468174, 919319871360, 3464027857900, 13082491517336, 49513139090512, 187761168576904, 713325243149880, 2714644395230080, 10347465825694361, 39500936075084958, 151005816167189208, 578038682649831252, 2215461387843704450, 8501300598574467776, 32658238309503587964
Offset: 1

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Author

Paul D. Hanna, Jun 10 2016

Keywords

Comments

It appears that a(n) = 1 (mod 2) iff n = 2*4^k - 1 for k>=0.
Radius of convergence is 1/4 with A(1/4) = 2*(sqrt(2) - 1).

Examples

			G.f.: A(x) = x + 2*x^2 + 4*x^3 + 10*x^4 + 30*x^5 + 96*x^6 + 313*x^7 + 1038*x^8 + 3512*x^9 + 12100*x^10 + 42286*x^11 + 149440*x^12 + +...
such that x*A(x) + x^2*A(x)^2 = C(x)^2, where
A(x)^2 = x^2 + 4*x^3 + 12*x^4 + 36*x^5 + 116*x^6 + 392*x^7 + 1350*x^8 + 4696*x^9 + 16500*x^10 + 58572*x^11 + 209824*x^12 +...+ A055395(n+2)*x^n +...
C(x)^2 = x^2 + 2*x^3 + 5*x^4 + 14*x^5 + 42*x^6 + 132*x^7 + 429*x^8 + 1430*x^9 + 4862*x^10 + 16796*x^11 +...+ A000108(n)*x^n +...
		

Crossrefs

Cf. A055395.

Programs

  • PARI
    /* From A(x) = (sqrt(1 + 4*C(x)^2) - 1)/(2*x) */
    {a(n) = my(A=x,C=(1 - sqrt(1 - 4*x +x*O(x^n)))/2); A = (sqrt(1 + 4*C^2) - 1)/(2*x); polcoeff(A,n)}
    for(n=1,40,print1(a(n),", "))
    
  • PARI
    /* From A(x) = (1/x)*Series_Reversion( sqrt(x^2 + x^4) - (x^2 + x^4) )^2 */
    {a(n) = my(A=x); A = ( serreverse( sqrt(x^2 + x^4 +x^3*O(x^n)) - (x^2 + x^4) ) )^2/x; polcoeff(A,n)}
    for(n=1,40,print1(a(n),", "))

Formula

G.f. A(x) satisfies:
(1) A(x) = (sqrt(1 + 4*C(x)^2) - 1)/(2*x), where C(x) = (1 - sqrt(1 - 4*x))/2.
(2) A(x) = (1/x) * Series_Reversion( sqrt(x^2 + x^4) - (x^2 + x^4) )^2.
a(n) ~ 2^(2*n - 1/2) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Jun 23 2016