A144230 Numerators of the convergents to x = 1/(x^3+1).

1, 1, 8, 729, 1911240521, 12145762597269451258228301000

Offset

0,3

Comments

These numbers are cubes. The recursion provides a method of solving the quartic x^4 + x - 1. In general, extending this notion, we can use the recursion x = 1/(x^(n-1)+1) to find a solution for n-th degree equations of the form x^n+x-1=0. However the bisection method and Newtons method converges much more quickly.

Links

Table of n, a(n) for n=0..5.

Prog

(PARI) x=0; for(j=1, 7, x=1/(x^3+1); print1((numerator(x))", "))

Crossrefs

Sequence in context: A068895 A115590 A134923 * A309377 A110039 A271741

Adjacent sequences: A144227 A144228 A144229 * A144231 A144232 A144233

Keyword

frac,nonn

Author

Cino Hilliard, Sep 15 2008

Status

approved