|
%I #8 Feb 23 2018 11:00:32
%S 2,1,3,10,97,24851,510157381,695243618523592916,
%T 2521217027896573870788274798987969315,
%U 200759268273854851798439056384882383919258596635924900200845873520031055851
%N Denominators of r-Egyptian fraction expansion for (1/2)^(1/3), where r = (1,1/4,1/9,1/16,...).
%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="/A270382/b270382.txt">Table of n, a(n) for n = 1..13</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/(4*1) + 1/(9*3) + 1/(16*10) + ...
%t r[k_] := 1/k^2; 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}]
%o (PARI) r(k) = 1/k^2;
%o f(k,x) = if (k==0, x, f(k-1, x) - r(k)/a(k, x););
%o a(k, x=(1/2)^(1/3)) = ceil(r(k)/f(k-1, x)); \\ _Michel Marcus_, Mar 22 2016
%Y Cf. A269993, A270714.
%K nonn,frac,easy
%O 1,1
%A _Clark Kimberling_, Mar 22 2016
|
