%I #16 Feb 23 2018 22:03:43
%S 8,31,719,17276711,557951558165893,1713250424923433306065171045669,
%T 3960162768997467999491098138568123635738830147395528618636887,
%U 148114266323338300606167235125265318767829304330791212171374192569332869541220746054882408155611146661783688512870116687748
%N Denominators of r-Egyptian fraction expansion for Pi - 3, where r = (1,1/2,1/3,1/4,...)
%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="/A269999/b269999.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 Pi - 3 = 1/8 + 1/(2*31) + 1/(3*719) + ...
%t r[k_] := 1/k; 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 = Pi - 3; Table[n[x, k], {k, 1, z}]
%o (PARI) r(k) = 1/k;
%o x = Pi - 3;
%o f(x, k) = if(k<1, x, f(x, k - 1) - r(k)/n(x, k));
%o n(x, k) = ceil(r(k)/f(x, k - 1));
%o for(k = 1, 8, print1(n(x, k), ", ")) \\ _Indranil Ghosh_, Mar 29 2017
%Y Cf. A269993.
%K nonn,frac,easy
%O 1,1
%A _Clark Kimberling_, Mar 15 2016
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