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A145064
Reduced numerators of the first convergent to the cube root of n using the recursion x = (2*x+n/x^2)/3.
2
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
0,1
COMMENTS
The same as A051176 without the first two terms.
LINKS
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)
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 *)
PROG
(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
CROSSREFS
Sequence in context: A182398 A214736 A051176 * A209166 A209146 A209154
KEYWORD
frac,nonn,easy
AUTHOR
Cino Hilliard, Sep 30 2008
STATUS
approved