A145064 Reduced numerators of the first convergent to the cube root of n using the recursion x = (2*x+n/x^2)/3.

2, 1, 4, 5, 2, 7, 8, 3, 10, 11, 4, 13, 14, 5, 16, 17, 6, 19, 20, 7, 22, 23, 8, 25, 26, 9, 28, 29, 10, 31, 32, 11, 34, 35, 12, 37, 38, 13, 40, 41, 14, 43, 44, 15, 46, 47, 16, 49, 50, 17, 52, 53, 18, 55, 56, 19, 58, 59, 20, 61, 62, 21, 64, 65, 22, 67, 68, 23, 70, 71, 24, 73, 74, 25

Offset: 0

Author

Cino Hilliard, Sep 30 2008

Keywords

• frac
• nonn
• easy

Comments

The same as A051176 without the first two terms.

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Cino Hilliard, Roots by Recursion [broken link]
• Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Crossrefs

Cf. A051176

Programs

Mathematica

CoefficientList[Series[(2+x+4*x^2+x^3-x^5)/((1-x)^2*(1+x+x^2)^2), {x, 0, 50}], x] (* G. C. Greubel, Oct 26 2017 *)

PARI

rroot3(d,p) = /* Find a root of x^3 - d */Q {
  local(x=1,x1=1,j);
  for(j=1,p,
  x=(x1+x+d/x^2)/3; /* average scheme for a cube root of d */
  x1=x; print1(numerator(x)",");
  );
}
for(k=0,100,rroot3(k,1))

PARI

Vec((2+x+4*x^2+x^3-x^5)/((1-x)^2*(1+x+x^2)^2) + O(x^100)) \\ Colin Barker, Feb 02 2016

Formula

The recursion was derived experimentally by analyzing the patterns of root recursions for polynomials

f(x) = a(n)x^n+a(n-1)x^(n-1)+...+a(1)x+a(0) and

g(x) = a(n-1)x^(n-1)+a(n-2)x^(n-2)+...+a(2)x+a(1)

where the recursion x = a(0)/g(x) may or may not converge to a root and many iterations are required to get greater accuracy. By introducing an averaging scheme, a root is found if it exists and convergence is much faster to a root of f(x) See the link for details. This cubic recursion is equivalent to Newton's Method.

From Colin Barker, Feb 02 2016: (Start)

a(n) = 2*a(n-3)-a(n-6) for n>5.

G.f.: (2+x+4*x^2+x^3-x^5) / ((1-x)^2*(1+x+x^2)^2). (End)