

A145008


Reduced numerators of the convergents to 2 = sqrt(4) using the recursion x > (4/x + x)/2.


0




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^24=f(x)=0, starting at x=1 as the initial estimate: x > xf(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

Table of n, a(n) for n=1..7.
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

Cf. A059917.
Sequence in context: A052113 A093433 A065035 * A216610 A025173 A222474
Adjacent sequences: A145005 A145006 A145007 * A145009 A145010 A145011


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



