|
%I #15 Oct 22 2013 08:07:39
%S 5,41,3281,21523361,926510094425921,1716841910146256242328924544641,
%T 5895092288869291585760436430706259332839105796137920554548481
%N Reduced numerators of the convergents to 2 = sqrt(4) using the recursion x -> (4/x + x)/2.
%C The recursion x -> (n/x + x)/2 converges to a square root of n.
%C 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
%C The equivalent sequence for n=9 starting from x=1 is 5, 17, 257,.., apparently A000215. - _R. J. Mathar_, Oct 14 2008
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Newton's_method">Newton's method</a>.
%e (4/1+1)/2 = 5/2 = 2.5
%e (4/5/2+5/2)/2 = 41/20 = 2.05
%e (4/(41/20)+41/20)/2 = 3281/1640 = 2.000609...
%o (PARI) g(n,p) = x=1;for(j=1,p,x=(n/x+x)/2;print1(numerator(x)","))
%o g(4,8)
%Y Cf. A059917.
%K frac,nonn
%O 1,1
%A _Cino Hilliard_, Sep 28 2008
%E Divided the right hand side of formula in the first comment by 2. - _R. J. Mathar_, Oct 14 2008
| 