login
Denominators in the approximation of sqrt(2) satisfying the recurrence: a(n)= [a(n-1)*a(n-2)+2]/[a(n-1)+a(n-2)] with a(1)=a(2)=1.
1

%I #7 Nov 03 2016 05:28:01

%S 1,1,2,5,29,408,33461,38613965,3654502875938,399133058537705128729,

%T 4125636888562548868221559797461449,

%U 4657508918199804645965719872781284840798220312648198320

%N Denominators in the approximation of sqrt(2) satisfying the recurrence: a(n)= [a(n-1)*a(n-2)+2]/[a(n-1)+a(n-2)] with a(1)=a(2)=1.

%C The recurrence is a transform of the Babylonian (Newton's) method for square root computation: a(n+1)= N/2a(n)+a(n)/2 = (a(n)^2+N)/2a(n).

%e 1/1=1.0, 1/1=1.0, 3/2=1.5, 7/5=1.4, 42/29=1.41379.., 577/408=1.4142156,... - _R. J. Mathar_, Nov 03 2016

%t a[1] = a[2] = 1; a[n_] := (a[n - 1] a[n - 2] + 2)/(a[n - 1] + a[n - 2]); Denominator@ Array[ a, 12] (* _Robert G. Wilson v_, Aug 03 2010 *)

%Y Cf. A179907, A000129, A051009.

%K nonn,frac

%O 1,3

%A _Mark Dols_, Jul 28 2010

%E a(10) - a(12) from _Robert G. Wilson v_, Aug 03 2010