A145008 Reduced numerators of the convergents to 2 = sqrt(4) using the recursion x -> (4/x + x)/2.

5, 41, 3281, 21523361, 926510094425921, 1716841910146256242328924544641, 5895092288869291585760436430706259332839105796137920554548481

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

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: A318294 A065035 A340535 * 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