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%I #10 Feb 24 2018 14:01:18
%S 2,5,19,909,866969,1391730212321,3778000141026848185401614,
%T 40721846314736913560423097666261109869051562216449,
%U 2752670766792898979726597410050364093079267724156167495561617776950485371872363605942870860465128310
%N Denominators of r-Egyptian fraction expansion for sqrt(1/3), where r(k) = 1/(2k-1).
%C Suppose that r is a sequence of rational numbers r(k) <= 1 for k >= 1, and that x is an irrational number in (0,1). Let f(0) = x, n(k) = floor(r(k)/f(k-1)), and f(k) = f(k-1) - r(k)/n(k). Then x = r(1)/n(1) + r(2)/n(2) + r(3)/n(3) + ... , the r-Egyptian fraction for x.
%C See A269993 for a guide to related sequences.
%H Clark Kimberling, <a href="/A270547/b270547.txt">Table of n, a(n) for n = 1..11</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/EgyptianFraction.html">Egyptian Fraction</a>
%H <a href="/index/Ed#Egypt">Index entries for sequences related to Egyptian fractions</a>
%e sqrt(1/3) = 1/(1*2) + 1/(3*5) + 1/(5*19) + 1/(7*909) + ...
%t r[k_] := 1/(2k-1); f[x_, 0] = x; z = 10;
%t n[x_, k_] := n[x, k] = Ceiling[r[k]/f[x, k - 1]]
%t f[x_, k_] := f[x, k] = f[x, k - 1] - r[k]/n[x, k]
%t x = Sqrt(1/3); Table[n[x, k], {k, 1, z}]
%o (PARI) r(k) = 1/(2*k-1);
%o f(k,x) = if (k==0, x, f(k-1, x) - r(k)/a(k, x););
%o a(k, x=sqrt(1/3)) = ceil(r(k)/f(k-1, x)); \\ _Michel Marcus_, Apr 03 2016
%Y Cf. A269993, A005408, A020760.
%K nonn,frac,easy
%O 1,1
%A _Clark Kimberling_, Apr 02 2016
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