OFFSET
1,1
COMMENTS
The recursion x -> (n/x + x)/2 converges to a square root of n.
These are the numerators of the first order Newton method to solve x^2-4=f(x)=0, starting at x=1 as the initial estimate: x -> x-f(x)/f'(x), where f'(x)=2x is the first derivative. - R. J. Mathar, Oct 07 2008
The equivalent sequence for n=9 starting from x=1 is 5, 17, 257,.., apparently A000215. - R. J. Mathar, Oct 14 2008
LINKS
Wikipedia, Newton's method.
EXAMPLE
(4/1+1)/2 = 5/2 = 2.5
(4/5/2+5/2)/2 = 41/20 = 2.05
(4/(41/20)+41/20)/2 = 3281/1640 = 2.000609...
PROG
(PARI) g(n, p) = x=1; for(j=1, p, x=(n/x+x)/2; print1(numerator(x)", "))
g(4, 8)
CROSSREFS
KEYWORD
frac,nonn
AUTHOR
Cino Hilliard, Sep 28 2008
EXTENSIONS
Divided the right hand side of formula in the first comment by 2. - R. J. Mathar, Oct 14 2008
STATUS
approved