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A144230
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Numerators of the convergents to x = 1/(x^3+1).
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0
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OFFSET
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0,3
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COMMENTS
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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.
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LINKS
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PROG
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(PARI) x=0; for(j=1, 7, x=1/(x^3+1); print1((numerator(x))", "))
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CROSSREFS
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KEYWORD
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frac,nonn
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AUTHOR
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STATUS
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approved
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