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A144230
Numerators of the convergents to x = 1/(x^3+1).
0
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 Newton's method converge much more quickly.
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
KEYWORD
frac,nonn
AUTHOR
Cino Hilliard, Sep 15 2008
STATUS
approved