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.

A143339 G.f. satisfies: A(x) = 1 + x*A(x)^2/A(-x).

Original entry on oeis.org

1, 1, 3, 7, 25, 73, 283, 911, 3697, 12561, 52467, 184471, 785929, 2829401, 12229259, 44795167, 195742177, 726541345, 3202144483, 12010174247, 53300753657, 201608659561, 899838791419, 3427434566831, 15370709035601, 58890032580913
Offset: 0

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Author

Paul D. Hanna, Aug 09 2008

Keywords

Comments

Specific values: A(2/9) = 9/5 and A(-2/9) = 9/10.
Radius of convergence: r = sqrt(2*sqrt(3)-3)/3 = 0.2270833462...
with A(r) = (2 + sqrt(1-3*r))/(1+r) = 2.0899798397...
and A(-r) = (2 - sqrt(1+3*r))/(1-r) = 3 - A(r) = r*A(r)^2/(A(r)-1) = 0.91002016...
At x=r, the equation (*) 1 - 2*y + (1+x)*y^2 - (x+x^3)*y^3 = 0, which is satisfied by y = A(x), factors out to: (y - A(r))^2 * (y - A(r)/(2*A(r)-2)) = 0; this gives the relation: (A(r)-1)*(3-A(r))/A(r)^2 = r. At x>r, the equation (*) admits complex solutions for y.
The limit a(n+1)/a(n) does not exist but oscillates between 2 attractors:
Limit a(2*n)/a(2*n-1) = sqrt(3)+3, Limit a(2*n+1)/a(2*n) = 3*(sqrt(3)+1)/2.

Examples

			A bisection of g.f. A(x) equals a bisection of A(x)^2:
A(x) = 1 + x + 3*x^2 + 7*x^3 + 25*x^4 + 73*x^5 + 283*x^6 + 911*x^7 +...
A(x)^2 = 1 + 2*x + 7*x^2 + 20*x^3 + 73*x^4 + 238*x^5 + 911*x^6 +...
that is, A(x) - x*A(x)^2 = 1 + x^2*A(x)*A(-x), where
A(x)*A(-x) = 1 + 5*x^2 + 45*x^4 + 521*x^6 + 6873*x^8 + 98061*x^10 +...
Related expressions:
A(x) = 1 + x*A(x)/A(-x) + x^2*A(x)^2/A(-x)^2 + x^3*A(x)^3/A(-x)^3 +...
log(A(x)) = A(x)/A(-x)*x + A(x)^2/A(-x)^2*x^2/2 + A(x)^3/A(-x)^3*x^3/3 +...
Illustrate the behavior of a(n+1)/a(n) as n grows:
a(301)/a(300) = 4.07522764...
a(302)/a(301) = 4.71149410...
a(303)/a(302) = 4.07537802...
a(304)/a(303) = 4.71162882...
the limits of which approach the attractors:
3*(sqrt(3)+1)/2 = 4.09807621... and sqrt(3)+3 = 4.73205080...
note that the product of the attractors equals 1/r^2, where
r = sqrt(2*sqrt(3)-3)/3  =  sqrt(2/sqrt(3))/(sqrt(3)+3)
is the radius of convergence of the g.f. A(x).
		

Crossrefs

Programs

  • Mathematica
    terms = 26; A[] = 1; Do[A[x] = 1 + x*A[x]^2/A[-x] + O[x]^terms // Normal, terms]; CoefficientList[A[x], x] (* Jean-François Alcover, Jan 15 2018 *)
  • PARI
    {a(n)=local(A=1+x+x*O(x^n));for(i=0,n,A=1+x*A^2/subst(A,x,-x));polcoeff(A,n)}
    
  • PARI
    {a(n)=local(A=1+x+x*O(x^n));for(i=1,n,A=exp(sum(m=1,n,A^m/subst(A^m,x,-x+x*O(x^n))*x^m/m)));polcoeff(A,n)}
    
  • PARI
    {a(n)=local(A=1+x+x*O(x^n));for(i=1,n,A=sum(m=0,n,x^m*A^m/subst(A^m,x,-x+x*O(x^n))));polcoeff(A,n)}

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
(1) 1 - 2*A(x) + (1+x)*A(x)^2 - (x+x^3)*A(x)^3 = 0.
(2) A(x) = exp( Sum_{n>=1} A(x)^n / A(-x)^n * x^n/n ).
(3) A(x) = Sum_{n>=0} x^n * A(x)^n / A(-x)^n.
Recurrence: (n-1)*(n+1)*(4*n^3 - 32*n^2 + 71*n - 30)*a(n) = 6*(8*n^3 - 56*n^2 + 101*n - 10)*a(n-1) + (68*n^5 - 760*n^4 + 2927*n^3 - 4202*n^2 + 683*n + 1800)*a(n-2) + 12*(4*n^3 - 40*n^2 + 136*n - 155)*a(n-3) + 3*(60*n^5 - 768*n^4 + 3509*n^3 - 6422*n^2 + 2571*n + 2950)*a(n-4) - 18*(n-4)*(8*n - 25)*a(n-5) + 27*(n-5)*(n-4)*(4*n^3 - 20*n^2 + 19*n + 13)*a(n-6). - Vaclav Kotesovec, Feb 17 2014