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Denominators of r-Egyptian fraction expansion for (1/2)^(1/3), where r = (1,1/2,1/3,1/4,...)
2

%I #10 Feb 23 2018 22:04:10

%S 2,2,8,123,149367,19877572990,3398650153657920854371,

%T 38501744904404393452660892011327652171148221,

%U 1751742507912624184333715455628345093210972368514121272905550101268413741408122585972087

%N Denominators of r-Egyptian fraction expansion for (1/2)^(1/3), 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="/A270316/b270316.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 (1/2)^(1/3) = 1/2 + 1/(2*2) + 1/(3*8) + ...

%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 = (1/2)^(1/3); Table[n[x, k], {k, 1, z}]

%Y Cf. A269993.

%K nonn,frac,easy

%O 1,1

%A _Clark Kimberling_, Mar 17 2016