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Numerator of the first convergent to sqrt(n) using the recursion x = (n/x + x)/2.
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%I #35 Apr 04 2024 10:14:48

%S 1,3,2,5,3,7,4,9,5,11,6,13,7,15,8,17,9,19,10,21,11,23,12,25,13,27,14,

%T 29,15,31,16,33,17,35,18,37,19,39,20,41,21,43,22,45,23,47,24,49,25,51,

%U 26,53,27,55,28,57,29,59,30,61,31,63,32,65,33,67,34,69,35,71,36,73,37,75

%N Numerator of the first convergent to sqrt(n) using the recursion x = (n/x + x)/2.

%C This is the same as A026741 without the first 2 terms in A026741. The link describes the experimental derivation of the generating function.

%C From _Jaroslav Krizek_, May 28 2010: (Start)

%C Numerators of arithmetic means of the first n positive integers for n >= 1.

%C See A040001 - denominators of arithmetic means of the first n positive integers.

%C a(n) = A026741(n+1) = A000217(n) * A040001(n) / n. (End)

%C Minimum number of line segments to draw into a circle to partition the circle into n+1 congruent circular sectors, i.e., minimum number of straight cuts required to cut a circular cake into n+1 equal slices. - _Felix Fröhlich_, Sep 01 2015

%H Cino Hilliard, <a href="http://groups.google.com/group/roots-by-recursion/web/recursion-on-polynomial">Roots by Recursion</a> [Broken link]

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,2,0,-1).

%F From _Paul Barry_, Nov 22 2009: (Start)

%F G.f.: x*(1 + 3*x - x^3)/(1 - x^2)^2.

%F a(n+1) = (n + 2)*(3 - (-1)^n)/4;

%F a(n+1) = Sum_{k=0..n, if(k=floor(n/2) or k=floor((n+1)/2),1,0)*(k+1)}. (End)

%F E.g.f.: ((x + 2)*cosh(x) + (2*x + 1)*sinh(x) - 2)/2. - _Stefano Spezia_, Apr 04 2024

%e n=1, x=1; x = (1/1+1)/2 = 1/1;

%e n=2, x=1; x = (2/1+1)/2 = 3/2;

%e n=3, x=1; x = (3/1+1)/2 = 2/1.

%e G.f.: x + 3*x^2 + 2*x^3 + 5*x^4 + 3*x^5 + 7*x^6 + 4*x^7 + 9*x^8 + 5*x^9 + ...

%t lst={};Do[a=n^2+n;b=n^2-n;c=a/b;AppendTo[lst,Denominator[c]],{n,2,5!}];lst (* _Vladimir Joseph Stephan Orlovsky_, Oct 20 2009 *)

%o (PARI) g(n, p) = x=1;for(j=1,p,x=(n/x+x)/2; if(j==1, print1(numerator(x), ",")))

%o for(k=1,100,g(k,1))

%o (Magma) [(n+1)*(3 - (-1)^(n-1))/4: n in [1..100]]; // _Vincenzo Librandi_, Sep 02 2015

%Y Cf. A000217, A026741, A040001.

%K nonn,frac,easy

%O 1,2

%A _Cino Hilliard_, Sep 30 2008