A145051 Numerator of the first convergent to sqrt(n) using the recursion x = (n/x + x)/2.

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, 29, 15, 31, 16, 33, 17, 35, 18, 37, 19, 39, 20, 41, 21, 43, 22, 45, 23, 47, 24, 49, 25, 51, 26, 53, 27, 55, 28, 57, 29, 59, 30, 61, 31, 63, 32, 65, 33, 67, 34, 69, 35, 71, 36, 73, 37, 75

Offset: 1

Author

Cino Hilliard, Sep 30 2008

Keywords

• nonn
• frac
• easy

Comments

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

From Jaroslav Krizek, May 28 2010: (Start)

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

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

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

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

Continued fraction expansion of A386934. - Kelvin Voskuijl, Aug 15 2025

Examples

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

Links

• Kelvin Voskuijl, Table of n, a(n) for n = 1..10000
• Cino Hilliard, Roots by Recursion [Broken link]
• Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Crossrefs

Cf. A000217, A026741, A040001.

Programs

Magma

[(n+1)*(3 - (-1)^(n-1))/4: n in [1..100]]; // Vincenzo Librandi, Sep 02 2015

Mathematica

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 *)

PARI

g(n, p) = x=1;for(j=1,p,x=(n/x+x)/2; if(j==1, print1(numerator(x), ",")))
for(k=1,100,g(k,1))

Formula

From Paul Barry, Nov 22 2009: (Start)

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

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

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

E.g.f.: ((x + 2)*cosh(x) + (2*x + 1)*sinh(x) - 2)/2. - Stefano Spezia, Apr 04 2024