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A145051
Numerator of the first convergent to sqrt(n) using the recursion x = (n/x + x)/2.
9
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
1,2
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
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
EXAMPLE
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 + ...
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 *)
PROG
(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))
(Magma) [(n+1)*(3 - (-1)^(n-1))/4: n in [1..100]]; // Vincenzo Librandi, Sep 02 2015
CROSSREFS
KEYWORD
nonn,frac,easy
AUTHOR
Cino Hilliard, Sep 30 2008
STATUS
approved