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%I #12 Apr 09 2017 12:42:57
%S 1,3,2,2,9,5,11,3,3,19,10,7,11,23,4,4,33,17,35,9,37,19,39,5,5,51,26,
%T 53,27,11,28,57,29,59,6,6,73,37,25,19,77,13,79,20,27,41,83,7,7,99,50,
%U 101,51,103,52,15,53,107,54,109
%N Numerators of a rational guess r(n) for the input for Newton's algorithm to find sqrt(n).
%C The corresponding denominators are given in A256098.
%C This educated guess for the rational input R(n) = x(n;k=0) for the so-called Babylonian (also called Heron's) iteration to find sqrt(n) (Newton's method for sqrt(n)), x(n; k+1) = (x(n; k) + n/x(n; k))/2, k >= 0, was used in Vedic Mathematics (see the H.-W. Alhen et al. reference, pp. 145-146, and the MacTutor link on Sulbasutras). In the Wikipedia link on Shulba Sutras another suggestion is given how the approximation 1 + 1/3 + 1/(3*4) - 1/(3*4*34) for sqrt(2) was obtained in Sulbasutras. The explanation given in the H.-W. Alten et al. reference seems to me more convincing.
%C This R(n) is obtained by n = s(n)^2 + r(n) with s(n)^2 = A048760(n) (largest square not exceeding n) and the remainder r(n). Then the approximation of the square root is used sqrt(n) = sqrt(s(n)^2 + r(n)) approximately s(n)*(1 + r(n)/(2*s(n)^2)) = s(n) + r(n)/(2*s(n)). Note that A048760(n) = A000196(n)^2, that is, s(n) = floor(sqrt(n)).
%D H.-W. Alten et al., 4000 Jahre Algebra, 2. Auflage, Springer, 2014, p. 145-146.
%H The MacTutor History of Mathematics archive, <a href="http://www-history.mcs.st-and.ac.uk/HistTopics/Indian_sulbasutras.html"> Sulbasutras</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Shulba_Sutras#Square_roots"> Shulba Sutras</a>.
%F a(n) = numerator(R(n)) with the rational (in lowest terms) R(n) = f(n) + (n - f(n)^2)/(2*f(n)) = (f(n) + n/f(n))/2 with f(n) := floor(sqrt(n)) = A000196(n), for n >= 1. See the comment above for this formula.
%e n = 2: s(n) = floor(sqrt(2)) = sqrt(A048760(2)) = 1, r(n) = 2 - 1^2 = 1. R(2) = s(2) + r(2)/(2*s(2)) = 1 + 1/(2*1) = 3/2. That is a(2) = 3 and A256098(2) = 2.
%e n = 17: s(n) = floor(sqrt(17)) = sqrt(A048760(17)) = 4 , r(n) = 17 - 4^2 = 1. R(17) = s(17) + r(17)/(2*s(17)) = 4 + 1/(2*4) = 33/8. That is, a(n) = 33 and A256098(17) = 8.
%e The rationals R(n) for n = 1..60 are: [1, 3/2, 2, 2, 9/4, 5/2, 11/4, 3, 3, 19/6, 10/3, 7/2, 11/3, 23/6, 4, 4, 33/8, 17/4, 35/8, 9/2, 37/8, 19/4, 39/8, 5, 5, 51/10, 26/5, 53/10, 27/5, 11/2, 28/5, 57/10, 29/5, 59/10, 6, 6, 73/12, 37/6, 25/4, 19/3, 77/12, 13/2, 79/12, 20/3, 27/4, 41/6, 83/12, 7, 7, 99/14, 50/7, 101/14, 51/7, 103/14, 52/7, 15/2, 53/7, 107/14, 54/7, 109/14,...]
%e For n=2 the Newton (Babylonian also called Heron) iteration produces. with x(2; k=0) = R(2) = 3/2: x(2; 1) = (3/2 + 4/3)/2 = 17/12 = 1 + 5/2 = 1 + 1/3 + 1/(3*4).
%e x(2; 2) = (17/12 + 24/17)/2 = 577/408 = 17/12 + (577/408 - 17*34/408) = 17/12 - 1/408 = 1 + 1/3 + 1/(3*4) - 1/(3*4*34) = 1.4142156... versus sqrt(2) = 1.4142135... (see A002193).
%Y Cf. A256098, A048760, A000196, A002193.
%K nonn,frac,easy
%O 1,2
%A _Wolfdieter Lang_, Mar 24 2015
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